System Dynamics 2nd Edition Solution manual

Solutions Manual c©

to accompany

System Dynamics, Second Edition

by

William J. Palm III

University of Rhode Island

Solutions to Problems in Chapter One

c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.

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1.1 W = mg = 3(32.2) = 96.6 lb.

1.2 m = W/g = 100/9.81 = 10.19 kg. W = 100(0.2248) = 22.48 lb. m = 10.19(0.06852) =0.698 slug.

1.3 d = (50 + 5/12)(0.3048) = 15.37 m.

1.4 n = 1/[60(1.341× 10−3)] = 12.43, or approximately 12 bulbs.

1.5 5(70− 32)/9 = 21.1 C

1.6 ω = 3000(2π)/60 = 314.16 rad/sec. Period P = 2π/ω = 60/3000− 1/50 sec.

1.7 ω = 5 rad/sec. Period P = 2π/ω = 2π/5 = 1.257 sec. Frequency f = 1/P = 5/2π =0.796 Hz.

c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which

the textbook has been adopted. Any other use without publisher’s consent is unlawful.

1.8 Physical considerations require the model to pass through the origin, so we seek a modelof the form f = kx. A plot of the data shows that a good line drawn by eye is given byf = 0.2x. So we estimate k to be 0.2 lb/in.

Skipping ahead to Section 1.5, we can solve this problem using the least squares method,based on equation (1.5.3). The script file is

x = [4.7,7.2,10.6,12.9]-4.7;f = [0,0.47,1.15,1.64];num = sum(x.*f);den = sum(x.^2);k = num/den

The result is k = 0.1977.



1.9 The script file is

x = [0:0.01:1];subplot(2,2,1)plot(x,sin(x),x,x),xlabel(′x (radians)′),ylabel(′x and sin(x)′),...gtext(′x′),gtext(′sin(x)′)subplot(2,2,2)plot(x,sin(x)-x),xlabel(′x (radians)′),ylabel(′Error: sin(x) - x′)subplot(2,2,3)plot(x,100*(sin(x)-x)./sin(x)),xlabel(′x (radians)′),...ylabel(′Percent Error′),grid

The plots are shown in the figure.

0 0.5 10

0.2

0.4

0.6

0.8

1

x (radians)

x an

d si

n(x)

x

sin(x)

0 0.5 1−0.2

−0.15

−0.1

−0.05

0

x (radians)

Err

or: s

in(x

) − x

0 0.5 1−20

−15

−10

−5

0

x (radians)

Per

cent

Err

or

Figure : for Problem 1.9.

From the third plot we can see that the approximation sin x ≈ x is accurate to within5% if |x| ≤ 0.5 radians.



1.10 For θ near π/4,

f(θ) ≈ sinπ

4+

(cos

π

4

) (θ − π

4

)

For θ near 3π/4,

f(θ) ≈ sin3π

4+

(cos

3π

4

) (θ − 3π

4

)



1.11 For θ near π/3,

f(θ) ≈ cosπ

3−

(sin

π

3

) (θ − π

3

)

For θ near 2π/3,

f(θ) ≈ cos2π

3−

(sin

2π

3

) (θ − 2π

3

)



1.12 For h near 25,

f(h) ≈√

25 +1

2√

25(h − 25) = 5 +

110

(h − 25)



1.13 For r near 5,f(r) ≈ 52 + 2(5)(r− 5) = 25 + 10(r − 5)

For r near 10,f(r) ≈ 102 + 2(10)(r− 10) = 100 + 20(r − 10)



1.14 For h near 16,

f(h) ≈√

16 +1

2√

16(h − 16) = 4 +

18(h − 16)

f(h) ≥ 0 if h > −16.



1.15 Construct a straight line the passes through the two endpoints at p = 0 and p = 900.At p = 0, f(0) = 0. At p = 900, f(900) = 0.002

√900 = 0.06. This straight line is

f(p) =0.06900

p =1

15, 000p



1.16 (a) The data is described approximately by the linear function y = 54x − 1360. Theprecise values given by the least squares method are y = 53.5x−1354.5 (see Problem 1.34a).

(b) Only the loglog plot of the data gives something close to a straight line, so the datais best described by a power function y = bxm where the approximate values are m = −0.98and b = 3600. The precise values given by the least squares method are y = 3582.1x−0.9764

(see Problem 1.34b).(c) Both the loglog and semilog plot (with the y axis logarithmic) give something close

to a straight line, but the semilog plot gives the straightest line, so the data is best describedby a exponential function y = b(10)mx where the approximate values are m = −0.007 andb = 2.1 × 105. The precise values given by the least squares method are y = 2.0622 ×105(10)−0.0067x (see Problem 1.34c).



1.17 With this problem, it is best to scale the data by letting x = year − 1990, to avoidraising large numbers like 1990 to a power. Both the loglog and semilog plot (with they axis logarithmic) give something close to a straight line, but the semilog plot gives thestraightest line, so the data is best described by a exponential function y = b(10)mx. Theapproximate values are m = 0.035 and b = 9.98.

Set y = 20 to determine how long it will take for the population to increase from 10 to20 million. This gives 20 = 9.98(10)0.03x. Solve it for x: x = (log(20) − log(9.98))/0.035.The answer is 8.63 years, which corresponds to 8.63 years after 1990.

More precise values are given by the least squares method (see Problem 1.35).



1.18 (a) If C(t)/C(0) = 0.5 when t = 500 years, then 0.5 = e−5500b, which gives b =− ln(0.5)/5500 = 1.2603× 10−4.

(b) Solve for t to obtain t = − ln[C(t)/C(0)]/b using C(t)/C(0) = 0.9 and b = 1.2603×10−4. The answer is 836 years. Thus the organism died 836 years ago.

(c) Using b = 1.1(1.2603 × 10−4) in t = − ln(0.9)/b gives 760 years. Using b =0.9(1.2603× 10−4) in t = − ln(0.9)/b gives 928 years.



1.19 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx where y is the temperaturein degrees C and x is the time in seconds. The approximate values are m = −3.67 andb = 356. The alternate exponential form is y = be(m ln 10)x = 356e−8.451x. The timeconstant is 1/8.451 = 0.1183 s.

The precise values given by the least squares method are y = 356.0199(10)−3.6709x (seeProblem 1.37).



1.20 Only the semilog plot of the data gives something close to a straight line, so the data isbest described by an exponential function y = b(10)mx where y is the bearing life thousandsof hours and x is the temperature in degrees F. The approximate values are m = −0.007and b = 142. The bearing life at 150 F is estimated to be y = 142(10)−0.007(150) = 12.66,or 12,600 hours. The alternate exponential form is y = be(m ln 10)x = 142e−0.0161x. Thetime constant is 1/0.0161 = 62.1 or 6.21× 104 hr.




1.21 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx where y is the voltage andx is the time in seconds. The first data point does not lie close to the straight line onthe semilog plot, but a measurement error of ±1 volt would account for the discrepancy.The approximate values are m = −0.43 and b = 96. The alternate exponential form isy = be(m ln 10)x = 96e−0.99x. The time constant is 1/0.99 = 1.01 s.




1.22 A semilog plot generated by the following script file shows that the exponential functionT − 70 = bemt fits the data well.

t = [0:300:3000];temp = [207,182,167,155,143,135,128,123,118,114,109];DT = temp-70;semilogy(t,DT,t,DT,’o’)

Fitting a line by eye gives the approximate values m = −4 × 10−4 and b = 125. Thecorresponding function is T (t) = 70 + 125e−4×10−4t.

The precise values given by the least squares method are m = −4.0317 × 10−4 andb = 125.1276 (see Problem 1.40).



1.23 Plots of the data on a log-log plot and rectilinear scales both give something close toa straight line, so we try both functions. (Note that the flow should be 0 when the heightis 0, so we do not consider the exponential function and we must force the linear functionto pass through the origin by setting b = 0.) The three lowest heights give the same time,so we discard the heights of 1 and 2 cm.

The power function fitted by eye in terms of the height h is approximately f = 4h0.9.Note that the exponent is not close to 0.5, as it is for orifice flow. This is because theflow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye isapproximately f = 3.2h.

Using the least squares method gives more precise results: f = 4.1595h0.8745 and f =3.2028h (see Problem 1.41)



1.24 Plots of the data on a log-log plot and rectilinear scales both give something close toa straight line, so we try both functions. (Note that the flow should be 0 when the heightis 0, so we do not consider the exponential function and we must force the linear functionto pass through the origin by setting b = 0.) The variable x is the height and the variabley is the flow rate. The three lowest heights give the same time, so we discard the heightsof 1 and 2 cm.

The power function fitted by eye in terms of the height h is approximately f = 4h0.9.Note that the exponent is not close to 0.5, as it is for orifice flow. This is because theflow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye isapproximately f = 3.7h.

Using the least squares method gives more precise results: f = 4.1796h0.9381 and f =3.6735h (see Problem 1.42).



1.25 Fitting a straight line by eye gives the approximate values m = 15 and b = 7. Theprecise values given by the least squares method are m = 15.0750, b = 7.1500, J = 43.5750,S = 4.5451× 103, r2 = 0.9904 (see Problem 1.43).



1.26 Plot the data on a loglog plot. We must delete the first data point to avoid takingthe logarithm of 0. The power function fitted by eye is approximately y = 7x3. The precisevalues given by the least squares method are m = 2.9448, b = 7.4053, J = 2.7494 × 103,S = 1.1218× 105, and r2 = 0.9755 (see Problem 1.44).



1.27 Plot the data on a semilog plot. The exponential function fitted by eye is approximatelyy = 6e3x. The precise values given by the least squares method are y = 6.4224e2.8984x,J = 77.4488, S = 2.5496× 104, and r2 = 0.9970 (see Problem 1.45).



1.28 Fitting a straight line by eye through the origin gives the approximate values m = 17and b = 0. The precise values given by the least squares method are m = 16.6071, J =116.6071, S = 5.5420× 103, and r2 = 0.9790 (see Problem 1.46).



1.29 Plot the data on a loglog plot. We must delete the first data point to avoid takingthe logarithm of 0. The power function fitted by eye is approximately y = 7x3. The precisevalues given by the least squares method are b = 7.4793, J = 799.4139, S = 1.6176× 105,and r2 = 0.9951 (see Problem 1.47).



1.30 a) The equation for b is obtained as follows.

J =n∑

i=1

(bemxi − yi)2

∂J

∂b= 2b

n∑

i=1

e2mxi − 2n∑

i=1

yiemxi = 0

b =∑

yemx

∑e2mx

b) Plot the data on a semilog plot. The exponential function fitted by eye is approxi-mately y = 6e3x. The precise values given by the least squares method are y = 5.8449e3x

(see Problem 1.48).



1.31 The integral form of the sum of squares to fit the function y = 5x2 is

J =∫ 4

0

(5x2 − mx − b

)2dx =

∫ 4

0

[25x4 − 10mx3 + (m2 − 10b)x2 + 2mbx + b2

]dx

This evaluates to

J = 5120− 640m + (m2 − 10b)643

+ 16mb + 4b2

Thus∂J

∂m= −640 +

1283

m + 16b = 0

∂J

∂b= −640

3+ 16m + 8b = 0

These give m = 20 and b = −13.333. Thus the line is y = 20x− 13.333.



1.32 a) The integral form of the sum of squares to fit the function y = Ax2 + Bx is

J =∫ L

0

(Ax2 + Bx − mx − b

)2dx

or

J =∫ L

0

[A2x4 + (2AB − 2mA)x3 + (−2bA + B2 + m2 − 2mB)x2 + (2mb− 2bB)x + b2

]dx

This evaluates to

J = A2 L5

5+ 2A(B − m)

L4

4+ (−2bA + B2 + m2 − 2mB)

L3

3+ 2b(m− B)

L2

2+ b2L

Thus∂J

∂m= −AL4

2+

23mL3 − 2

3BL3 + bL2 = 0

∂J

∂b= −2

3AL3 + (m − B)L2 + 2bL = 0

These give4Lm + 6b = 4BL + 3AL2

3Lm + 6b = 2AL2 + 3BL

which can be solved for m and b, given values of A, B, and L.b) With L = 2, A = 3, and B = 5, we obtain m = 11 and b = −2. The fitted straight

line is y = 11x − 2.



1.33 a) The integral form of the sum of squares to fit the function y = BeMx is

J =∫ L

0

(BeMx − mx − b

)2dx

or

J =∫ L

0

[B2e2Mx − 2mBxeMx − 2bBeMx + m2x2 + 2bmx + b2

]dx =

This evaluates to

J =B2

2M

(e2ML − 1

)− 2mBL

MeML +

2mB

M2

(eML − 1

)

+2bB

M

(1 − eML

)− m2 L3

3+ bmL2 + b2L

Thus∂J

∂m= −2BL

MeML +

2B

M2

(eML − 1

)+

2mL3

3+ bL2 = 0

∂J

∂b= −2B

MeML +

2B

M+ mL2 + 2bL = 0

These give2L3

3m + L2b =

2BL

MeML − 2B

M2

(eML − 1

)

L2m + 2Lb =2B

MeML − 2B

M

which can be solved for m and b, given values of M , B, and L.b) With L = 1, M = −5, and B = 15, we obtain m = −10.973 and b = 8.466. The

fitted straight line is y = −10.973x + 8.466.



1.34 (a) The script file is

x = [25:5:45];y = [5, 260, 480, 745, 1100];p = polyfit(x,y,1)p =

1.0e+003 *0.0535 -1.3545

The function is y = 53.5x− 1354.5.



1.34 (b) Only the loglog plot of the data gives something close to a straight line, so thedata is best described by a power function y = bxm. The script file to find the coefficientsm and b is

x = [2.5:0.5:6,7:10];y = [1500,1220,1050,915,810,745,690,620,520,480,410,390];p = polyfit(log10(x),log10(y),1);m = p(1)b = 10^p(2)

The results are m = −0.9764 and b = 3582.1. Thus the power function is y = 3582.1x−0.9764.



1.34 (c) Both the loglog and semilog plot (with the y axis logarithmic) give somethingclose to a straight line, but the semilog plot gives the straightest line, so the data is bestdescribed by a exponential function y = b(10)mx. The script file to find the coefficients mand b is

x = [550:50:750];y =[41.2,18.62,8.62,3.92,1.86];p = polyfit(x,log10(y),1);m = p(1)b = 10^p(2)m =

-0.0067b =

2.0622e+005

This gives the results m = −0.0067 and b = 2.0622× 105. Thus the exponential function isy = 2.0622× 105(10)−0.0067x.

The results for the power function are obtained from

p = polyfit(log10(x),log10(y),1);m = p(1)b = 10^p(2)

This gives the results m = −9.9949 and b = 1.0601 × 1029. Thus the power function isy = 1.0601× 1029x−9.9949.

A plot of the data and the two functions shows that both functions describe the datawell, but the exponential curve passes closer to the first data point than the power curve.In addition, we should be careful about using this power function because its coefficient bhas a very large value. This large coefficient means that any predictions of y values madewith this function will be very prone to error unless we use very precise values of x. Thusthe exponential function is the best choice to describe this data.



1.35 With this problem, it is best to scale the data by letting x = year − 1990, to avoidraising large numbers like 1990 to a power. Both the loglog and semilog plot (with they axis logarithmic) give something close to a straight line, but the semilog plot gives thestraightest line, so the data is best described by a exponential function y = b(10)mx. Thescript file to find the coefficients m and b is

year = [1990:1995];x = x = year-1990;pop=[10,10.8,11.7,12.7,13.8,14.9];p=polyfit(x,log10(pop),1)p =

0.0349 0.9992m=p(1);b=10^p(2)

This gives the results m = 0.0349 and b = 9.9817. Thus the exponential function isy = 9.9817(10)0.0349x. Set y = 20 to determine how long it will take for the populationto increase from 10 to 20 million. This gives 20 = 9.9817(10)0.0349x. Solve it for x: x =(log(20)− log(9.9817))/0.0349 = 8.6483 years, which corresponds to 8.6483 years after 1990.



1.36 (a) If C(t)/C(0) = 0.5 when t = 500 years, then 0.5 = e−5500b, which gives b =− ln(0.5)/5500. In Matlab this calculation is

b = -log(0.5)/5500

The answer is b = 1.2603× 10−4.(b) Solve for t to obtain t = − ln[C(t)/C(0)]/b using C(t)/C(0) = 0.9 and b = 1.2603×

10−4. In MATLAB this calculation is

t = -log(0.9)/b

The answer is 836.0170 years. Thus the organism died 836 years ago.(c) Using b = 1.1(1.2603 × 10−4) in t = − ln(0.9)/b gives 760 years. Using b =

0.9(1.2603× 10−4) in t = − ln(0.9)/b gives 928 years.



1.37 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx. The script file to find thecoefficients m and b is

time = [0:0.1:0.6];temp = [300,150,75,35,12,5,2];p=polyfit(time,log10(temp),1)m=p(1)b=10^p(2)

This gives the results: m = −3.6709 and b = 356.0199. Thus the exponential function isy = 356.0199(10)−3.6709x, where y is the temperature in degrees C and x is the time inseconds. The alternate exponential form is y = be(m ln 10)x = 356.0199e−8.4526x. The timeconstant is 1/8.4526 = 0.1183 s.



1.38 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx. The script file to find thecoefficients m and b is

temp = [100:20:220];life = [28,21,15,11,8,6,4];p = polyfit(temp,log10(life),1);m = p(1)b = 10^p(2)

This gives the results: m = −0.0070 and b = 141.8603. Thus the exponential function is y =141.8603(10)−0.0070x, where y is the bearing life thousands of hours and x is the temperaturein degrees F. The bearing life at 150 F is estimated to be y = 141.8603(10)−0.0070(150) =12.6433, or 12,643 hours.

The alternate exponential form is y = be(m ln 10)x = 141.8603e−0.0161x. The time con-stant is 1/0.0161 = 62.0421 or 6.20421× 104 hr.



1.39 Only the semilog plot of the data gives something close to a straight line, so the datais best described by an exponential function y = b(10)mx. The first data point does not lieclose to the straight line on the semilog plot, but a measurement error of ±1 volt wouldaccount for the discrepancy. The script file to find the coefficients m and b is

time = [0:0.5:4];voltage = [100,62,38,21,13,7,4,2,3];p = polyfit(time,log10(voltage),1);m=p(1)b = 10^p(2)

This gives the results: m = −0.4333 and b = 95.8063. Thus the exponential functionis y = 95.8063(10)−0.4333x, where y is the voltage and x is the time in seconds. Thealternate exponential form is y = be(m ln 10)x = 95.8063e−0.9977x. The time constant is1/0.9977 = 1.002 s.



1.40 The semilog plot generated by the following script file shows that the exponentialfunction T − 70 = bemt fits the data well.

t = [0:300:3000];temp = [207,182,167,155,143,135,128,123,118,114,109];DT = temp-70;semilogy(t,DT,t,DT,’o’)p = polyfit(t,log(DT),1)m = p(1)b = exp(p(2))

The results are m = −4.0317 × 10−4 and b = 125.1276. The time constant is 1/4.0317 ×10−4 = 2.4803× 103 s, or 0.689 hr.



1.41 The calculations are shown in the following script file. Plots of the data on a log-logplot and rectilinear scales both give something close to a straight line, so we try all bothfunctions. (Note that the flow should be 0 when the height is 0, so we do not consider theexponential function and we must force the linear function to pass through the origin bysetting b = 0.) The variable x is the height and the variable y is the flow rate. The threelowest heights give the same time, so we discard the heights of 1 and 2 cm.

For the power function, we use (1.5.1) and (1.5.2) in terms of the variables x andY = log y.

t = [7,8,9,10,11,13,15,17,23]; x = [11:-1:3];y = 250./t; X = log10(x); Y = log10(y);a1 = sum(X.^2); a2 = sum(X);a3 = sum(Y.*X); a4 = sum(Y); n = length(x);A = [a1, a2; a2, n]; C = [a3; a4];solution = A\C;m = solution(1)b = 10^solution(2)J = sum((b*x.^m-y).^2)S = sum((b*x.^m-mean(y)).^2)r2 = 1 - J/S

The results are m = 0.8745, b = 4.1595, J = 5.3644, S = 493.5634, and r2 = 0.9891. So thefitted function in terms of the height h is f = 4.1595h0.8745. Note that the exponent is notclose to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow.

For the linear function f = mx, we use (1.5.3).

m = sum(x.*y)/sum(x.^2)J = sum((m*x-y).^2)S = sum((m*x-mean(y)).^2)r2 = 1 - J/S

The results are m = 3.2028, J = 8.0247, S = 615.9936, and r2 = 0.9870. The fitted functionin terms of the height h is f = 3.2028h. Using only r2 as the criterion, it is impossible todecide whether the linear or the power function is the best model.



1.42 The calculations are shown in the following script file. Plots of the data on a log-logplot and rectilinear scales both give something close to a straight line, so we try all bothfunctions. (Note that the flow should be 0 when the height is 0, so we do not consider theexponential function and we must force the linear function to pass through the origin bysetting b = 0.) The variable x is the height and the variable y is the flow rate. The threelowest heights give the same time, so we discard the heights of 1 and 2 cm.

For the power function, we use (1.5.1) and (1.5.2)in terms of the variables x and Y =log y.

t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:-1:3];y = 250./t; X = log10(x); Y = log10(y);a1 = sum(X.^2); a2 = sum(X);a3 = sum(Y.*X); a4 = sum(Y); n = length(x);A = [a1, a2; a2, n]; C = [a3; a4];solution = A\C;m = solution(1)b = 10^solution(2)J = sum((b*x.^m-y).^2)S = sum((b*x.^m-mean(y)).^2)r2 = 1 - J/S

The results are m = 0.9381, b = 4.1796, J = 13.5531, S = 729.3505, and r2 = 0.9814. Sothe fitted function in terms of the height h is f = 4.1796h0.9381.

For the linear function f = mx, we use (1.5.3).

t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:-1:3];y = 250./t;m = sum(x.*y)/sum(x.^2)J = sum((m*x-y).^2)S = sum((m*x-mean(y)).^2)r2 = 1 - J/S

The results are m = 3.6735, J = 14.7546, S = 809.8919, and r2 = 0.9818. The fittedfunction in terms of the height h is f = 3.6735h. Using only r2 as the criterion, it isimpossible to decide whether the linear or the power function is the best model.




x = [0:2:6];y = [4.5, 39, 72, 94];a1 = sum(x.^2);a2 = sum(x);a3 = sum(y.*x);a4 = sum(y);n = length(x);A = [a1, a2; a2, n];B = [a3; a4];solution = A\B;m = solution(1)b = solution(2)J = sum((m*x+b-y).^2)S = sum((m*x+b-mean(y)).^2)r2 = 1 - J/S

The results are m = 15.0750, b = 7.1500, J = 43.5750, S = 4.5451× 103, r2 = 0.9904.



1.44 Following the procedure shown in Example 1.5.1, we fit X = log x and Y = log y toa linear function. We must delete the first data point to avoid taking the logarithm of 0.The script file is

x = [1:4];y = [8, 50, 178, 490];X = log10(x);Y = log10(y);a1 = sum(X.^2);a2 = sum(X);a3 = sum(Y.*X);a4 = sum(Y);n = length(x);A = [a1, a2; a2, n];C = [a3; a4];solution = A\C;m = solution(1)b = 10^solution(2)J = sum((b*x.^m-y).^2)S = sum((b*x.^m-mean(y)).^2)r2 = 1 - J/S

The results are m = 2.9448, b = 7.4053, J = 2.7494×103, S = 1.1218×105, and r2 = 0.9755.



1.45 We fit x and Y = log y to the linear function Y = mx + B, where B = log b. Thescript file is

x = [0:0.4:1.2];y = [6.3, 22, 60, 215];Y = log10(y);a1 = sum(x.^2);a2 = sum(x);a3 = sum(Y.*x);a4 = sum(Y);n = length(x);A = [a1, a2; a2, n];C = [a3; a4];solution = A\C;m = solution(1)/log10(exp(1))b = 10^solution(2)J = sum((b*exp(m*x)-y).^2)S = sum((b*exp(m*x)-mean(y)).^2)r2 = 1 - J/S

The results are m = 2.8984, b = 6.4244, J = 77.4488, S = 2.5496× 104, and r2 = 0.9970.The fitted function is y = 6.4244e2.8984x.



1.46 The equation for m, obtained from (1.5.3), is

m =∑

xiyi∑x2

i

The script file is

x = [0:2:6];y = [4.5, 39, 72, 94];m = sum(x.*y)/sum(x.^2)J = sum((m*x-y).^2)S = sum((m*x-mean(y)).^2)r2 = 1 - J/S

The results are m = 16.6071, J = 116.6071, S = 5.5420× 103, and r2 = 0.9790.



1.47 The equation for b is obtained from equation (1) in Example 1.5.3 with m = 3.

b =∑

x3y∑x6

x = [0:4];y = [1, 8, 50, 178, 490];b = sum((x.^3).*y)/sum(x.^6)J = sum((b*x.^3-y).^2)S = sum((b*x.^3-mean(y)).^2)r2 = 1 - J/S

The results are b = 7.4793, J = 799.4139, S = 1.6176× 105, and r2 = 0.9951.



1.48 a) The equation for b is obtained as follows.

J =n∑

i=1

(bemxi − yi)2

∂J

∂b= 2b

n∑

i=1

e2mxi − 2n∑

i=1

yiemxi = 0

b =∑

yemx

∑e2mx

b) The script file is

x = [0:0.4:1.2];y = [6.3, 22, 60, 215];b = sum(y.*exp(3*x))/sum(exp(6*x))J = sum((b*exp(3*x)-y).^2)S = sum((b*exp(3*x)-mean(y)).^2)r2 = 1 - J/S

The results are b = 5.8449, J = 27.7380, S = 2.7279× 104, and r2 = 0.9990.




to accompany


by

William J. Palm III


Solutions to Problems in Chapter Two


CD1

Highlight

2.1 mv = mg. Thus

v(t) = gt = 32.2t x(t) =12gt2 = 16.1t2 h(t) = 20 − x(t) = 20 − 16.1t2

Thus

t =

√20− h(t)

16.1

For h = 10, t = 0.788 sec. For h = 0, t = 1.115 sec.



2.2

t =60/5280

903600 = 0.455 sec

x =12gt2 = 16.1(0.455)2 = 3.326 ft



2.3 Summing forces in the direction parallel to the plane gives

mv = f1 − mg sin φ − µmg cos φ

Substituting the given values,

10v = f1 − 98.1 sin 25 − 0.3(98.1) cos 25 = f1 − 68.132

Thus v > 0 if f1 > 68.132. If f1 = 100 the block will continue to move up the plane. Iff1 = 50, v = −1.813 and the speed is given by

v(t) = −1.813t + v(0) = −1.813t + 2

Thus v(0) = 0 at t = 1.103 s.



2.4 Let d be the vertical distance dropped by the time the mass leaves the surface. See thefigure on the following page. Then

d =L

tan θ

The speed v0 of the mass when it leaves the surface is found from conservation of energy:

KE =12mv2

0 = PE =mgL

tan θ

Thus

v0 =

√2gL

tan θ(1)

The horizontal and vertical velocity components are

vx = v0x sin θ

v0y = v0 cos θ (2)

Establish a coordinate system at the point where the mass leaves the surface, with xpositive to the right and y positive down. In the vertical direction : y = g and

y =12gt2 + v0yt

The mass hits the ground when y = H , and time-to-hit tH is found from

12gt2H + 2v0ytH = H

or

tH =−2v0y ±

√4v2

0y + 8gH

2g(3)

There will be one positive solution and one negative solution. Take the positive solution.(Continued on the next page)



Problem 2.4 continued:In the horizontal direction: mx = 0 and

x = (v0 sin θ) t

andD = (v0 sin θ) tH (4)

The solution is given by (1) through (4).

Figure : for Problem 4



2.5 Let the origin of the launch point be at x = y = 0. Assuming the projectile is launchedwith a speed v0 at an angle θ from the horizontal, Newton’s law in the x and y directionsgives

x = 0 y = −g

vx = v0 cos θ vy = v0 sin θ − gt

x = (v0 cos θ)t y = −g

2t2 + (v0 sin θ)t

These can be manipulated to show that

v2y = (v0 − sin θ)2 − 2gy

which givesv0 sin θ =

√v2y − 2gy

Using this relation, the above can also be manipulated to show that

vy = −gx

vx+√

v2y − 2gy

Solve for x.x =

vx

g

√v2y − 2gy − vxvy

g(1)

where y is computed fromy = R sin φ (2)

The desired distance D can be computed from

D = x + R cos φ

where x is computed from (1).



2.6 IG = 2mr2/5. Apply the parallel axis theorem.

IO = IG + mR2 =25mr2 + mR2



2.7 a) Let point O be the pivot point and G be the center of mass. Let L be the distancefrom O to G. Treat the pendulum as being composed of three masses:

1) m1, the rod mass above point O, whose center of mass is 1 ft above point O;2) m2, the rod mass below point O, whose center of mass is 1.5 ft below point O, and3) m3, the mass of the 10 lb block.

Then, summing moments about G gives

m1g(L + 1) − m2g(1.5− L)− m3g(3.5− L) = 0 (1)

wherem1g =

253 =

65

lb

m2g =353 =

95

lb

m3g = 10 lb

From equation (1):

65(L + 1) − 9

5(1.5− L) − 10(3.5− L) = 0

which gives L = 2.808 ft.b) Summing moments about the pivot point O gives

IOθ = −mgL sin θ

where m is the total mass. From the parallel-axis theorem, treating the rod as a slenderrod, we obtain

IO =112

(3g

)(5)2 +

(3g

)(0.5)2 +

(10g

)(3.5)2 = 4.022 slug − ft2

and mgL = 13(2.808) = 36.504 ft-lb. Thus the equation of motion is

4.022θ = −36.504 sin θ

orθ + 9.076 sin θ = 0



2.8 a)IA = mL2

2 + mCL21

IAθ = −mCL1g sin θ + mgL2 cos(β − θ)

b) Set θ = 0 and solve for mg.

mg =mCL1g sin θ

L2 cos(β − θ)

c) Substitute the given values to obtain

mg =5(0.2)g sin 20

0.15 cos 10= 2.315g = 22.713 N



2.9 The easiest way is to use the energy-equivalence method. Let ωB the speed of pulley B(positive counterclockwise) and ωC the speed of pulley C (positive clockwise). The kineticenergy of the entire system is

KE =12IBω2

B +12ICω2

C +12mLv2

C +12mCv2

C =12mev

2A

where me is the mass of the equivalent translational system. The potential energy is

PE = (mL + mC)g(−sC) = FesA

where Fe is the equivalent gravity force. Thus

Fe = −mL + mC

2g

Using the kinematic relations:

sC = −12sA vC = −1

2vA ωB =

1RB

vA ωC =1

RCvC

the kinetic energy expression becomes

KE =12

(IB

R2B

+IC

4R2C

+mL + mC

4

)v2A =

12mev

2A

Thusme =

IB

R2B

+IC

4R2C

+mL + mC

4

The model ismevA = FA − mL + mC

2g



2.10 a) Let T be the tension in the cable attached to mass m2. See the figure on thefollowing page. Then the cable force pulling up on m1 is T/2 because of the pulleys. Notealso that because of the pulleys, x = 2y. Summing forces acting on m2 parallel to the plane,we obtain

m2y = T − m2g sin θ (1)

Summing the vertical forces acting on m1, we obtain

m1x = m1g − 12T (2)

Since x = 2y, this becomes

2m1y = m1g − 12T (3)

Solve for T :T = 2m1g − 4m1y (4)

Substitute this into (1) and collect the y terms to obtain

(4m1 + m2)y = 2m1g − m2g sin θ (5)

The mass m1 will lift m2 if y > 0; that is, if 2m1 − m2 sin θ > 0.




Problem 2.10 continued:b) Follow the same procedure as in part (a) but include the friction force. Equation (1)

becomesm2y = T − m2g sin θ − µdm2g cos θ (6)

Equations (2) through (4) remain the same, but (5) becomes

(4m1 + m2)y = 2m1g − m2g(µd cos θ + sin θ) (7)

The mass m1 will lift m2 if y > 0; that is, if

2m1 − m2(µd cos θ + sin θ) > 0

For the case m1 = m2/2, this becomes

1 − (µd cos θ + sin θ) > 0



2.11 The assumption in Example 2.3.2 is that the only inertia on each shaft is the concen-trated inertia I1 and I2 at each end. This implies that the inertias of the shafts and gearsare negligible. If, on the other hand, the inertias of the shafts are not negligible, then thekinetic energy expression becomes

KE =12

(I1 + Is1)ω21 +

12

(I2 + Is2)ω22

or

KE =12

(I1 + Is1)ω21 +

12

(I2 + Is2)(

ω1

N

)2

Therefore the equivalent inertia felt on the input shaft is

Ie = I1 + Is1 +I2 + Is2

N2



2.12 Let F be the contact force between the two gears. For gear 1,

IG1ω1 = T1 − r1F

For gear 2,IG2ω1 = T2 − r2F

If ω1 = ˙ ω2 = 0, or if IG1 = IG2 = 0,

T1 = r1F T2 = r2F

which give

T1 =r1

r2T2 =

1N

T2



2.13 The equivalent inertia felt on shaft 1 is

Ie = I1 +1

N2I2 +

1N2

m2R2 +

1N2

m3R2

With N = 2,

Ie = I1 +14

(I2 + m2R

2 + m3R2)

The equation of motion is

Ieω1 = T1 −m2gR

N+

m3gR

N= T1 −

gR

2(m3 − m2)



2.14 The total kinetic energy is

KE =12

(Is + I) θ2 +12mx2

Substituting x = Rθ we obtain

KE =12

(Is + I + mR2

)θ2

Thus the equivalent inertia isIe = Is + I + mR2




KE =12

(I1 + IS1)ω21 +

12

(I2 + IS2)ω22 +

12mv2

Substituting ω2 = r1ω1/r2 and v = r1ω1 we obtain

KE =12

(I1 + IS1) ω21 +

12

(I2 + IS2)(

r1ω1

r2

)2

+12m (r1ω1)

2

or

KE =12

[I1 + IS1 + (I2 + IS2)

(r1

r2

)2

+ mr21

]ω2

1

Thus the equivalent inertia is

Ie = I1 + IS1 + (I2 + IS2)(

r1

r2

)2

+ mr21



2.16 a) With everything reflected to the load shaft (shaft 2), Newton’s law gives

ω2 =T2 + NT1

I2 + N2I1

b) To maximize ω2, differentiate the above expression with respect to N , set the deriva-tive equal to 0. This gives

∂ω2

∂N=

(I2 + N2I1)T1 − 2I1N(T2 + NT1)(I2 + N2I1)2

= 0

This is true if the numerator is 0. Thus

I1T1N2 + 2I1T2N − I2T1 = 0

The positive solution for N is

N = −T2

T1+

√(T2

T1

)2

+I2

I1

This is the ratio that maximizes ω2. (This can be confirmed to give a maximum ratherthan a minimum by showing that ∂2ω2/∂N2 < 0).

If the load torque T2 is 0, the optimal ratio for this case is denoted No and is: No =√I2/I1, or I1 = I2/N

2o . This says that the ratio that maximizes the load acceleration is the

ratio that makes the reflected load inertia (felt by the motor) equal to the motor’s inertia.This is the principle of inertia matching.

If T2 = 0 and the actual ratio N differs from the optimal value No such that N = γNo,the efficiency E is

E =actual ω2

max ω2=

2γ

1 + γ2

Because E(γ) = E(1/γ), the efficiency when “overgearing” is the same as when “under-gearing”. For example, if γ = 2 or γ = 1/2, E = 0.8, so the acceleration is 80% of themaximum possible.



2.17 With I1 = I2 = I3 = 0, the total kinetic energy is

KE =12I4ω

21 +

12I5ω

23

Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2ω3 = 1.96ω3, and I4 = 0.02, I5 = 0.1, weobtain

KE =12(0.177)ω2

3

and the equivalent inertia is Ie = 0.177 kg·m2.The equation of motion is Ieω3 = (1.4)2T , or 0.177ω3 = 1.96T




KE =12

(I4 + I1)ω21 +

12I2ω

22 +

12

(I3 + I5)ω23

Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2ω3 = 1.96ω3 and the given values of theinertias, we obtain

KE =12(0.203)ω2

3

and the equivalent inertia is Ie = 0.203 kg·m2.The equation of motion is Ieω3 = (1.4)2T , or 0.203ω3 = 1.96T



2.19 a)ω4

ω1=

ω4

ω3

ω3

ω2

ω2

ω1=

76

353 = 2.1

b) The torque T1 felt on shaft 4 is T1/2.1 and the equation of motion is

Iω4 =T1

2.1



2.20 Using kinetic energy equivalence,

KE =12mv2 +

12Isω

2 =12Ieω

2

The mass translates a distance x when the screw rotates by θ radians. When θ = 2π, x = L.Thus x = Lθ/2π and x = v = Lθ/2π. Because ω = θ, we have

KE =12m

(Lω

2π

)2

+12Isω

2 =12Ieω

2

Solve for Ie to obtain

Ie =mL2

4π2+ Is



2.21 The expression for the kinetic energy is

KE = KE of 2 rear wheels + KE of front wheel + KE of body

= 2(

12Irω

2r

)+

12Ifω2

f +12mbv

2

But for the rear wheels, v = Rrωr = 4ωr, and for the front wheel, v = Rfωf = 2ωf . Theinertias are calculated as follows. For the rear wheels:

Ir =12mrR

2r =

12

500g

42 =4000

g

For the front wheel:If =

12mfR2

f =12

800g

22 =1600

g

Also, mb = 9000/g. Thus

KE =4950

gv2

and the equivalent mass is me = 2(4950)/g = 9000/g.The equation of motion is mev = mg sin 10, where

m = 2mr + mf + mb = 2500g

+800g

+9000

g=

10, 800g

Thus9900

gv = 10, 800 sin 10

or v = 0.1894. Thus v = 0.1894t ft/sec.



2.22 Using the equivalent mass approach, the equivalent mass referenced to the coordinatex is

me = m1 + m2 +I

R2

where I is the inertia of the cylinder about its center. The force acting on me due to theweight of the cylinder is m1g sin β. The force acting on me due to the weight of m2 ism2g sin φ. See the following figure. The equation of motion of the equivalent system is

mex = m1g sin β − m2g sin φ




2.23 Let y be the translational displacement of the cylinder to the right. Using the equiv-alent mass approach, the kinetic energy of the system is

KE =12m2y

2 +12Iω2 +

12m1x

2

where I = m2R2/2 is the inertia of the cylinder about its center. Because y = 2x and

ω = ˙ y/R, we have

KE =12m2

(4x2

)+

12

(m2R

2

2

)(2x

R

)2

+12m1x

2 =12

(6m2 + m1) x2

Thus the effective mass referenced to the coordinate x is me = 6m2 + m1.The equation of motion is

mex = m1g

or(6m2 + m1)x = m1g



2.24 Summing moments about the pivot gives

IOθ = T − mgL sin θ

where the effect of the motor torque is T = [2(1.5)]Tm = 3Tm and the inertia is

IO = mL2 + Im[2(1.5)]2 + IG1[2(1.5)]2 + IG2(1.5)2 + IG3(1.5)2 + IG4 = 1.936 kg · m2

Thus1.936θ = 3Tm − 29.43 sin θ



2.25 a) The speed ratio of the sprocket drive at the driving shaft is the ratio of the sprocketdiameters, which is 0.05/0.15 = 1/3. The equivalent inertia felt at the motor shaft is

Ie = I1 + I2 +[IS1 + mc1(0.05)2

] ( 110

)2

+[IS2 + Id + 4Iw + 2mc2r

2d + mLr2

d

](13

110

)2

where IS1, IS2, Id, and Iw are the inertias of the shafts 1 and 2, the drive shaft, and thedrive wheels. The masses mc1, mc2, and mL are the masses of the sprocket chain, the drivechain, and the load. This expression evaluates to Ie = 0.0143 kg·m2. The magnitude of thefriction torque felt at the motor shaft is 54/[3(10)] = 1.8 N·m. The equation of motion is

0.0143ω1 = T1 − 1.8sgn(ω1) (1)

b) If T1 = 10, ω1 = 573.427 and ω1 = 573.427t rad/s.c) Solve (1) for T1 assuming ω1 > 0:

T1 = 0.0143ω1 + 1.8

From the trapezoidal profile,

ω1 =

300 0 ≤ t ≤ 0.50 0.5 < t < 2.5−300 2.5 ≤ t ≤ 3

Thus

T1 =

300Ie + 1.8 0 ≤ t ≤ 0.51.8 0.5 < t < 2.51.8− 300Ie 2.5 ≤ t ≤ 3

=

6.09 0 ≤ t ≤ 0.51.8 0.5 < t < 2.5−249 2.5 ≤ t ≤ 3



2.26 a) The equation of motion for rotation is

Iω = Rft

where I = mR2/2 = 1.631 kg·m2 is the inertia of the cylinder about its center, and ft

is the tangential force between the cylinder and the ground. The equation of motion fortranslation is

mx = f cos φ − ft = f cos φ − Iω

R

Substituting x = Rω we obtain(mR2 + I

)ω = Rf cos φ

Noting that m = 800/9.81 = 81.549 kg, and substituting the given values, we obtain

16.636ω = f cos φ

b) Using the equivalent mass approach, the equivalent mass referenced to the coordinatex is

me = m +I

R2

where I = mR2/2 = 1.631 kg·m2 is the inertia of the cylinder about its center. The equationof motion is

mex = f cos φ

Substituting the given values, we obtain

83.18x = f cos φ



2.27 See the following figure for the coordinate definitions and the definition of the reactionforce R. Let P be the point on the axle. Note that yP = 0. The coordinates of the masscenter of the rod are

xG = xP − L

2sin θ yG = −L

2cos θ

ThusxG = xP − L

2θ cos θ +

L

2θ2 sin θ

yG =L

2θ sin θ +

L

2θ2 cos θ

Let m be the mass of the rod. Summing forces in the x direction:

mxG = f or m

(xP − L

2θ cos θ +

L

2θ2 sin θ

)= f (1)

Summing forces in the y direction:

myG = R − mg or m

(L

2θ sin θ +

L

2θ2 cos θ

)= R − mg (2)




Problem 2.27 continued:Summing moments about the mass center:

IGθ = fL

2cos θ − R

L

2sin θ

Substituting for R from (2) and using the fact that IG = mL2/12, we obtain

112

mL2θ =fL

2cos θ − mgL

2sin θ − mL2

4sin2 θ θ − mL2

4θ2 sin θ cos θ (3)

The model consists of (1) and (3) with m = 20 kg and L = 1.4 m.



2.28 See the following figure for the coordinate definitions and the definition of the reactionforces Rx, Ry, N , and ft. Let P be the point on the axle. Note that yP = 0. The coordinatesof the mass center of the rod are

xG = xP − L

2sin θ yG = −L

2cos θ

ThusxG = xP − L

2θ cos θ +

L

2θ2 sin θ

yG =L

2θ sin θ +

L

2θ2 cos θ




Problem 2.28 continued:Let m1 be the mass of the rod, m2 the mass of the wheel, and I the moment of inertia

of the wheel about its center. Summing forces on the wheel in the x direction:

m2xP = f − Rx − ft (1)

Summing forces on the wheel in the y direction:

m2yP = N − Ry − m2g (2)

Summing moments about the mass center of the wheel gives :Iω = Rft. But xP = Rω andxP = Rω. Thus

I

R2xP = ft (3)

Substitute this into (1) to obtain(m2 +

I

R2

)xP = f − Rx (4)

Summing forces on the rod in the x direction:

m1xG = Rx or m1

(xP − L

2θ cos θ +

L

2θ2 sin θ

)= Rx (5)

Summing forces on the rod in the y direction:

m1yG = Ry − m1g or m1

(L

2θ sin θ +

L

2θ2 cos θ

)= Ry − m1g (6)

(Continued on the next page)



Problem 2.28 continued:Summing moments about the mass center of the rod:

IGθ = RxL

2cos θ − Ry

L

2sin θ

Substituting for Rx and Ry from (5) and (6), canceling terms using the identity sin2 θ +cos2 θ = 1, and using the fact that IG = m1L

2/12, we obtain

112

m1L2θ =

m1L

2xP cos θ − m1gL

2sin θ − m1L

2

4θ

which can be rearranged as follows:

13m1L

2θ =m1L

2xP cos θ − m1gL

2sin θ (7)

Substituting for Rx from (4) into (5) gives(

m1 + m2 +I

R2

)xP = F − m1L

2

(θ2 sin θ − θ cos θ

)(8)

The model consists of (7) and (8) with m1 = 20 kg, m2 = 3 kg, L = 1.4 m, R = 0.05 m,and I = m2R

2/2.



2.29 Summing forces in the x direction:

maGx = mg sin θ − F (1)


maGy = N − mg cos θ − F (2)

For no bounce, aGy = 0 andN = mg cos θ (3)

Summing moments about G:IGα = Fr (4)

From (1) and (4):

maGx = mg sin θ − IGα

r(5)

With slipping, aGx 6= rα, butF = µdN (6)

From (4) and (6):IGα = rµdN = rµdmg cos θ (7)

Thus,

α =rµdmg cos θ

IG

From (1),aGx = g sin θ − µdg cos θ



2.30 Follow the procedure of Example 2.4.2. Summing forces in the x direction:

maGx = mg sin θ − F (1)


maGy = N − mg cos θ − F (2)

For no bounce, aGy = 0 andN = mg cos θ (3)

a) Summing moments about G:

IGα = Fr (4)

From (1) and (4):

maGx = mg sin θ − IGα

r(5)

For no slip,aGx = rα (6)

From (5) and (6):

aGx =mgr2 sin θ

mr2 + IG(7)

Since IG = mr2,

aGx =g sin θ

2(8)

andα =

aGx

r=

g sin θ

2r

b) Summing moments about P :

MP = (mg sin θ)r = IGα + maGxr = IGaGx

r+ maGxr

Thus

aGx =mgr2 sin θ

mr2 + IG=

g sin θ

2

andα =

aGx

r=

g sin θ

2r




Problem 2.30 continued:c) The maximum friction force is Fmax = µsN = µsmg cos θ. From (4), (6), and (7),

F =IGaGx

r2=

IGmg sin θ

mr2 + IG

If Fmax > F there will be no slip; that is, if

µs cos θ >IG sin θ

mr2 + IG=

sin θ

2

there will be no slip.



2.31 See the following figure. Summing forces in the x and y directions gives:

maGx = µsNB − mg sin θ (1)

NA + NB = mg cos θ (2)

Summing moments about G:

NALA − NBLB + µsNBH = 0 (3)

Solve (2) and (3):

NA =mg cos θ (µsH − LB)

µsH − LA − LB=

16, 677 cos θ (µs − 2.1)µs − 4.6

NB = − mgLA cos θ

µsH − LA − LB=

40, 025µs cos θ

4.6− µs

The maximum acceleration is, from (1),

aGx =23.544µs cos θ

4.6− µs− 9.81 sin θ




2.32 Use (2.4.6) with MP = −mgL sin θ, IP = mL2, aPx = −a(t), and aPy = 0. Also,referring to Figure 2.4.1, rx = L sin θ and ry = −L cos θ. Thus

mL2θ − mLa(t) cos θ = −mgL sin θ

Rearrange to cancel mL and to put the input a(t) on the right-hand side.

Lθ + g sin θ = a(t) cos θ



2.33 Use (2.4.6) with MP = mgL sin θ, IP = mL2, aPx = 0, and aPy = a(t). Also, referringto Figure 2.4.1, rx = −L sin θ and ry = L cos θ. Thus

mL2θ − mLa(t) sin θ = mgL sin θ

Rearrange to cancel mL and to put the input a(t) on the right-hand side.

Lθ − g sin θ = a(t) sin θ




to accompany


by

William J. Palm III


Solutions to Problems in Chapter Three


CD1

Highlight

3.1 a) Nonlinear because of the yy term. b) Nonlinear because of the sin y term. c)Nonlinear because of the

√y term.



3.2 a)

4∫ x

2dx = 3

∫ t

0t dt

x(t) = 2 +38t2

b)

5∫ x

3dx = 2

∫ t

0e−4t dt

x(t) = 3.1− 0.1e−4t

c) Let v = ˙ x.

3∫ v

7dv = 5

∫ t

0t dt

v(t) =dx

dt= 7 +

56t2

∫ x

2dx =

∫ t

0

(7 +

56t2)

dt

x(t) = 2 + 7t +518

t3

d) Let v = x.

4∫ v

2dv = 7

∫ t

0e−2t dt

v(t) =238

− 78e−2t

∫ x

4dx =

∫ t

0

(238

− 78e−2t

)dt

x(t) =5716

+238

t +716

e−2t



3.3 a) ∫ x

3

dx

25 − 5x2=∫ t

0dt = t

110

√5

ln

(√5 + x√5 − x

)∣∣∣∣∣

x

3

Let

A = ln√

5 + 3√5− 3

Thus

ln√

5 + x√5 − x

= 10√

5t + A

x(t) =√

5eAe10

√5t − 1

eAe10√

5t + 1b) ∫ x

10

dx

36 + 4x2=∫ t

0dt = y

112

tan−1 x

3

∣∣∣∣x

10= t

x(t) = 3 tan(12t + C) C = tan−1 103

c) ∫ x

4

dx

5x + 25=∫ t

0dt

15

ln(5x + 25)∣∣∣∣x

4= t

ln(5x + 25) = 5t + ln 45

x(t) = 9e5t − 5

(continued on the next page)



Problem 3.3 continued:

d) ∫ x

5

dx

x= 2

∫ t

0e−4t dt

ln x|x5 = − 12e−4t

∣∣∣∣t

0

ln x = ln 5− 12

(e−4t − 1

)

x(t) = 5e−(e−4t−1)/2



3.4 a) The roots are −3 and −5. The form of the free response is

x(t) = A1e−3t + A2e

−5t

Evaluating this with the given initial conditions gives

x(t) = 27e−3t − 17e−5t

The steady-state solution is xss = 30/15 = 2. Thus the form of the forced response is

x(t) = 2 + B1e−3t + B2e

−5t

Evaluating this with zero initial conditions gives

x(t) = 2 − 5e−3t + 3e−5t

The total response is the sum of the free and the forced response. It is

x(t) = 2 + 22e−3t − 14e−5t

The transient response consists of the two exponential terms.





b) The roots are −5 and −5. The form of the free response is

x(t) = A1e−5t + A2te

−5t


x(t) = e−5t + 9te−5t

The steady-state solution is xss = 75/25 = 3. Thus the form of the forced response is

x(t) = 3 + B1e−5t + B2te

−5t

Evaluating this with zero initial conditions gives

x(t) = 3 − 3e−5t − 15te−5t


x(t) = 3 − 2e−5t − 6te−5t






c) The roots are ±5j. The form of the free response is

x(t) = A1 sin 5t + A2 cos 5t


x(t) =45

sin 5t + 10 cos 5t

The form of the forced response is

x(t) = B1 + B2 sin 5t + B3 cos 5t

Thus the entire forced response is the steady-state forced response. There is no transientforced response. Evaluating this function with zero initial conditions shows that B2 = 0and B3 = −B1. Thus

x(t) = B1 − B1 cos 5t

Substituting this into the differential equation shows that B1 = 4 and the forced responseis

x(t) = 4− 4 cos 5t


x(t) = 4 + 6 cos 5t +45

sin 5t

The entire response is the steady-state response. There is no transient response.





d) The roots are −4 ± 7j. The form of the free response is

x(t) = A1e−4t sin 7t + A2e

−4t cos 5t


x(t) =447

e−4t sin 7t + 10e−4t cos 7t

The form of the forced response is

x(t) = B1 + B2e−4t sin 7t + B3e

−4t cos 7t

The steady-state solution is xss = 130/65 = 2. Thus B1 = 2. Evaluating this function withzero initial conditions shows that B2 = −8/7 and B3 = −2. Thus the forced response is

x(t) = 2 − 87e−4t sin 7t − 2e−4t cos 7t


x(t) = 2 +367





3.5 a) The root is s = 5/3, which is positive. So the model is unstable.b) The roots are s = 5 and −2, one of which is positive. So the model is unstable.c) The roots are s = 3 ± 5j, whose real part is positive. So the model is unstable.d) The root is s = 0, so the model is neutrally stable.e) The roots are s = ±2j, whose real part is zero. So the model is neutrally stable.f) The roots are s = 0 and −5, one of which is zero and the other is negative. So the

model is neutrally stable.



3.6 a) The system is stable if both of its roots are real and negative or if the roots arecomplex with negative real parts. Assuming that m 6= 0, we can divide the characteristicequation by m to obtain

s2 +c

ms +

k

m= s2 + as + b = 0

where a = c/m and b = k/m. The roots are given by the quadratic formula:

s =−a ±

√a2 − 4b

2

Thus the condition that m, c, and k have the same sign is equivalent to a > 0 and b > 0.There are three cases to be considered:

1. Complex roots (a2 − 4b < 0). In this case the real part of both roots is −a/2 and isnegative if a > 0.

2. Repeated, real roots (a2 − 4b = 0). In this case both roots are −a/2 and are negativeif a > 0.

3. Distinct, real roots (a2 − 4b > 0). Let the two roots be denoted r1 and r2. We canfactor the characteristic equation as s2 + as + b = (s − r1)(s − r2) = 0. Expandingthis gives

(s − r1)(s− r2) = s2 − (r1 + r2)s + r1r2 = 0

Comparing the two forms shows that

r1r2 = b (1) and r1 + r1 = −a (2)

If b > 0, condition (1) shows that both roots have the same sign. If a < 0, condition(2) shows that the roots must be negative. Therefore, if the roots are distinct andreal, the roots will be negative if a > 0 and b > 0.

b) Neutral stability occurs if either 1) both roots are imaginary or 2) one root is zerowhile the other root is negative. Imaginary roots occur when a = 0 (the roots are s = ±

√b)

In this case the free response is a constant-amplitude oscillation. Case 2 occurs when b = 0and a > 0 (the roots are s = 0 and s = −a). In this case the free response decays to anon-zero constant.



3.7 From the transform definition, we have

L[mt] = limT→∞

[∫ T

0mte−stdt

]= m lim

T→∞

[∫ T

0te−stdt

]

The method of integration by parts states that∫ T

0u dv = uv|T0 −

∫ T

0v du

Choosing u = t and dv = e−stdt, we have du = dt, v = −e−st/s, and

L[mt] = m limT→∞

[∫ T

0te−stdt

]= m lim

T→∞

t

e−st

−s

∣∣∣∣∣

T

0

−∫ T

0

e−st

−sdt

= m limT→∞

t

e−st

−s

∣∣∣∣∣

T

0

− e−st

(−s)2

∣∣∣∣∣

T

0

= m lim

T→∞

[Te−sT

−s− 0 − e−sT

(−s)2+

e0

(−s)2

]

=m

s2

because, if we choose the real part of s to be positive, then from Table 3.3-2,

limT→∞

Te−sT = 0



3.8 From the transform definition, we have

L[t2] = limT→∞

[∫ T

0t2e−stdt

]

The method of integration by parts states that∫ T

0u dv = uv|T0 −

∫ T

0v du

Choosing u = t2 and dv = e−stdt, we have du = 2t dt, v = −e−st/s, and

L[t2] = limT→∞

[∫ T

0t2e−stdt

]= lim

T→∞

t2

e−st

−s

∣∣∣∣∣

T

0

−∫ T

0

e−st

−s2t dt

= limT→∞

[−T 2e−st

s+

2s

∫ T

0te−stdt

]= lim

T→∞

[−T 2 e−st

s

]+

2s

(1s2

)

=2s3

because, if we choose the real part of s to be positive, then,

limT→∞

T 2e−sT = 0



3.9 a)

X(s) =10s

+2s3

b)

X(s) =6

(s + 5)2+

1s + 3

c) From Property 8,

X(s) = −dY (s)ds

where y(t) = e−3t sin 5t. Thus

Y (s) =5

(s + 3)2 + 52=

5s2 + 6s + 34

dY (s)ds

= − 10s + 30(s2 + 6s + 34)2

ThusX(s) =

10s + 30(s2 + 6s + 34)2

d) X(s) = e−5sG(s), where g(t) = t. Thus G(s) = 1/s2 and

X(s) =e−5s

s2



3.10f(t) = 5us(t) − 7us(t − 6) + 2us(t − 14)

Thus

F (s) =5s− 7

e−6s

s+ 2

e−14s

s



3.11 a)

x(0+) = lims→∞

s5

3s + 7=

53

x(∞) = lims→0

s5

3s + 7= 0

b)

x(0+) = lims→∞

s10

3s2 + 7s + 4= 0

x(∞) = lims→0

s10

3s2 + 7s + 4= 0



3.12 a)

X(s) =32

(1s− 1

s + 4

)

x(t) =32

(1− e−4t

)

b)

X(s) =53

1s

+313

1s + 3

x(t) =53

+313

e−3t

c)

X(s) = −13

1s + 2

+133

1s + 5

x(t) = −13e−2t +

133

e−5t

d)

X(s) =5/2

s2(s + 4)=

58

1s2

− 532

1s

+532

1s + 4

x(t) =58t − 5

32+

532

e−4t

e)

X(s) =25

1s2

+1325

1s− 13

251

s + 5

x(t) =25t +

1325

− 1325

e−5t

f)

X(s) = −314

1(s + 3)2

+7916

1s + 3

− 7916

1s + 7

x(t) = −314

te−3t +7916

e−3t − 7916

e−7t



3.13 a)

X(s) =7s + 2

(s + 3)2 + 52= C1

5(s + 3)2 + 52

+ C2s + 3

(s + 3)2 + 52

orX(s) = −19

55

(s + 3)2 + 52+ 7

s + 3(s + 3)2 + 52

x(t) = −195


b)

X(s) =4s + 3

s[(s + 3)2 + 52]=

C1

s+ C2

5(s + 3)2 + 52

+ C3s + 3

(s + 3)2 + 52

orX(s) =

334

1s

+127170

5(s + 3)2 + 52

− 334

s + 3(s + 3)2 + 52

x(t) =334

+127170

e−3t sin 5t − 334

e−3t cos 5t

c)

X(s) =4s + 9

[(s + 3)2 + 52][(s + 2)2 + 42]

= C15

(s + 3)2 + 52+ C2

s + 3(s + 3)2 + 52

+ C34

(s + 2)2 + 42+ C4

s + 2(s + 2)2 + 42

or

X(s) = − 44205

5(s + 3)2 + 52

− 1982

s + 3(s + 3)2 + 52

+69328

4(s + 2)2 + 42

+1982

s + 2(s + 2)2 + 42

x(t) = − 44205

e−3t sin 5t − 1982

e−3t cos 5t +69328

e−2t sin 4t +1982

e−2t cos 4t

d)

X(s) = 2.6251

s + 2− 18.75

1s + 4

+ 21.1251

s + 6

x(t) = 2.625e−2t − 18.75e−4t + 21.125e−6t



3.14 a) x = 7t/5 ∫ x

3dx =

75

∫ t

0t dt

x(t) =710

t2 + 3

b) x = 3e−5t/4 ∫ x

0dx =

34

∫ t

0e−5t dt

x(t) =320

(1 − e−5t

)

c) x = 4t/7

x(t) − x(0) =47

∫ t

0t dt

x(t) =414

t2 + 5

∫ x

3dx =

∫ t

0

(414

t2 + 5)

dt

x(t) =442

t3 + 5t + 3

d) x = 8e−4t/3

x(t)− x(0) =83

∫ t

0e−4t dt

x(t) =173

− 812

e−4t

∫ x

3dx =

∫ t

0

(173

− 812

e−4t)

dt

x(t) =173

t +16e−4t +

176



3.15 a) The root is −7/5 and the form is x(t) = Ce−7t/5. With x(0) = 4, C = 4 andx(t) = 4e−7t/5

b) The root is −7/5 and the form is x(t) = C1e−7t/5 + C2. At steady state, x = 15/7 =

C2. With x(0) = 0, C1 = −15/7. Thus

x(t) =157

(1 − e−7t/5

)

c) The root is −7/5 and the form is x(t) = C1e−7t/5 + C2. At steady state, x = 15/7 =

C2. With x(0) = 4, C1 = 13/7. Thus

x(t) =137

(1 + e−7t/5

)

d)

sX(s)− x(0) + 7X(s) =4s2

X(s) =5s2 + 4

s2(s + 7)=

47s2

− 449

+24949

e−7t

x(t) =47t − 4

49+

24949

e−7t



3.16 a) The roots are −7 and −3. The form is

x(t) = C1e−7t + C2e

−3t

Evaluating C1 and C2 for the initial conditions gives

x(t) = −94e−7t +

254

e−3t

b) The roots are −7 and −7. The form is

x(t) = C1e−7t + C2te

−7t


x(t) = e−7t + 10te−7t

c) The roots are −7 ± 3j. The form is

x(t) = C1e−7t sin 3t + C2e

−7t cos 3t


x(t) =203




3.17 a) The roots are −3 and −7. The form is

x(t) = C1e−3t + C2e

−7t + C3

At steady state, x = 5/63 so C3 = 5/63. Evaluating C1 and C2 for the initial conditionsgives

x(t) = − 35252

e−3t +15252

e−7t +563

b) The roots are −7 and −7. The form is


−7t + C3

At steady state, x = 98/49 = 2 so C3 = 2. Evaluating C1 and C2 for the initial conditionsgives

x(t) = −2e−7t − 16te−7t + 2

c) The roots are −7 ± 3j. The form is


−7t cos 3t + C3

At steady state, x = 174/58 = 3 so C3 = 3. Evaluating C1 and C2 for the initial conditionsgives

x(t) = −7e−7t sin 3t − 3e−7t cos 3t + 3



3.18 a)X(s)F (s)

=15

5s + 7

The root is s = −7/5.b)

X(s)F (s)

=5

3s2 + 30s + 63

The roots are s = −7 and s = −3.c)

X(s)F (s)

=4

s2 + 10s + 21

The roots are s = −7 and s = −3.d)

X(s)F (s)

=7

s2 + 14s + 49

The roots are s = −7 and s = −7.e)

X(s)F (s)

=6s + 4

s2 + 14s + 58

The roots are s = −7 ± 3j.f)

X(s)F (s)

=4s + 155s + 7

The root is s = −7/5.



3.19 Transform each equation using zero initial conditions.

3sX(s) = Y (s)

sY (s) = F (s) − 3Y (s) − 15X(s)

Solve for X(s)/F (s) and Y (s)/F (s).

X(s)F (s)

=1

3s2 + 9s + 15

Y (s)F (s)

=3s

3s2 + 9s + 15



3.20 Transform each equation using zero initial conditions.

sX(s) = −2X(s) + 5Y (s)

sY (s) = F (s) − 6Y (s)− 4X(s)

Solve for X(s)/F (s) and Y (s)/F (s).

X(s)F (s)

=5

s2 + 8s + 32

Y (s)F (s)

=s + 2

s2 + 8s + 32



3.21 a)6

s(s + 5)=

65s

− 65

1s + 5

x(t) =65

(1− e−5t

)

b)4

s + 3)(s + 8)=

45

1s + 3

− 45

1s + 8

x(t) =45

(e−3t − e−8t

)

c)8s + 5

2s2 + 20s + 48=

12

8s + 5(s + 4)(s + 6)

= −274

1s + 4

+434

1s + 6

x(t) = −274

e−4t +434

e−6t

d) The roots are s = −4 ± 10j.

4s + 13s2 + 8s + 116

+4s + 13

(s + 4)2 + 102= C1

10(s + 4)2 + 102

+ C2s + 4

(s + 4)2 + 102

= − 310

10(s + 4)2 + 102

+ 4s + 4

(s + 4)2 + 102

x(t) = − 310




3.22 a)3s + 2

s2(s + 10)=

15

1s2

− 150

1s

+150

1s + 10

x(t) =15t − 1

50

(1− e−10t

)

b)5

(s + 4)2(s + 1)= 1.25

1(s + 4)2

− 0.31251

s + 4+ 0.3125

1s + 1

x(t) = 1.25te−4t − 0.3125e−4t + 0.3125e−t

c)s2 + 3s + 5s3(s + 2)

=54

1s3

+716

1s2

+964

1s− 9

641

s + 4

x(t) =58t2 +

716

t +964

− 964

e−4t

d)s3 + s + 6s4(s + 2)

= 31s4

− 1s3

+12

1s2

+14

1s− 1

41

s + 2

x(t) =12t3 − 1

2t2 +

12t +

14− 1

4e−2t



3.23 a)

5[sX(s)− 2] + 3X(s) =10s

+2s3

X(s) =10s3 + 10s2 + 2

5s3(s + 3)=

2s3 + 2s2 + 2/5s3(s + 3/5)

=23

1s3

− 109

1s2

+1409

1s− 86

271

s + 3/5

x(t) =13t2 − 10

9t +

14027

− 8627

e−3t/5

b)

4[sX(s)− 5] + 7X(s) =6

(s + 5)2+

1s + 3

X(s) =14

20s3 + 261s2 + 1116s + 1525(s + 5)2(s + 7/4)(s + 3)

=14

[6013

1(s + 5)2

− 474169

1s + 5

+16904845

1s + 7/4

+145

1s + 3

]

x(t) = 1.1538te−5t − 0.7012e−5t + 5.0012e−7t/4 + 0.7e−3t

c) This simple-looking problem actually requires quite a lot of algebra to find the solu-tion, and thus it serves as a good motivating example of the convenience of using MATLAB.The algebraic complexity is due to a pair of repeated complex roots.

First obtain the transform of the forcing function. Let f(t) = te−3t sin 5t. From Prop-erty 8,

F (s) = −dY (s)ds

where y(t) = e−3t sin 5t. Thus

Y (s) =5

(s + 3)2 + 52=

5s2 + 6s + 34

dY (s)ds

= − 10s + 30(s2 + 6s + 34)2

ThusF (s) =

10s + 30(s2 + 6s + 34)2

(1)





Using the same technique, we find that the transform of te−3t cos 5t is

2s2 + 12s + 18(s2 + 6s + 34)2

− 1s2 + 6s + 34

(2)

This fact will be useful in finding the forced response.From the differential equation,

4[s2X(s)− 10s + 2] + 3X(s) = F (s) =10s + 30

(s2 + 6s + 34)2

Solve for X(s).

X(s) =40s − 84s2 + 3

+10s + 30

[(s + 3)2 + 25]2(4s2 + 3)

The free response is given by the first fraction, and is

xfree(t) = − 4√3

sin√

32

t + 10 cos√

32

t (3)

The forced response is given by the second fraction, which can be expressed as

2.5s + 7.5[(s + 3)2 + 25]2(s2 + 3/4)

(4)

The roots of this are s = ±j√

3/2 and the repeated pair s = −3 ± 5j. Thus, referring to(1), (2), and (3), we see that the form of the forced response will be

xforced(t) = C1te−3t sin 5t + C2te

−3t cos 5t

+ C3e−3t sin 5t + C4e

−3t cos 5t

+ C5 sin√

32

t + C6 cos√

32

t (5)





The forced response can be obtained several ways. 1) You can substitute the form (5)into the differential equation and use the initial conditions to obtain equations for the Ci

coefficients. 2) You can use (1) and (2) to create a partial fraction expansion of (4) in termsof the complex factors. 3) You can perform an expansion in terms of the six roots, of theform

A1

(s + 3 + 5j)2+

A2

s + 3 + 5j+

A3

(s + 3 − 5j)2+

A4

s + 3 − 5j

+√

3A5/2s2 + 3/4

+A6s

s2 + 3/4

4) You can use the MATLAB residue function.The solution for the forced response is

xforced(t) = −0.0034te−3t sin 5t + 0.0066te−3t cos 5t

− 0.0026e−3t sin 5t + 2.308× 10−4e−3t cos 5t

+ 0.00796 sin 0.866t− 2.308× 10−4 cos 0.866t

The initial condition x(0) = 0 is not exactly satisfied by this expression because of thelimited number of digits used to display it.



3.24 The denominator roots are s = −3 and s = −5, which are distinct. Factor thedenominator so that the highest coefficients of s in each factor are unity:

X(s) =7s + 4

2s2 + 16s + 30=

12

[7s + 4

(s + 3)(s + 5)

]

The partial-fraction expansion has the form

X(s) =12

[7s + 4

(s + 3)(s + 5)

]=

C1

s + 3+

C2

s + 5

Using the coefficient formula (3.5-4), we obtain

C1 = lims→−3

[(s + 3)

7s + 42(s + 3)(s + 5)

]= lim

s→−3

[7s + 4

2(s + 5)

]= −17

4

C2 = lims→−5

[(s + 5)

7s + 42(s + 3)(s + 5)

]= lim

s→−5

[7s + 4

2(s + 3)

]=

314

Using the LCD method we have

12

7s + 4(s + 3)(s + 5)

=C1

s + 3+

C2

s + 5=

C1(s + 5) + C2(s + 3)(s + 3)(s + 5)

=(C1 + C2)s + 5C1 + 3C2

(s + 3)(s + 5)

Comparing numerators, we see that C1 + C2 = 7/2 and 5C1 + 3C2 = 4/2 = 2, which giveC1 = −17/4 and C2 = 31/4.

The inverse transform is

x(t) = C1e−3t + C2e

−5t = −174

e−3t +314

e−5t

In this example the LCD method requires more algebra, including the solution of twoequations for the two unknowns C1 and C2.



3.25 a)7[sX(s)− 3] + 5X(s) = 4

X(s) =25

7s + 5=

25/7s + 5/7

x(t) =257

e−5t/7

Note that this gives x(0+) = 25/7. From the initial value theorem

x(0+) = lims→∞

s25/7

s + 5/7)=

257

which is not the same as x(0).b)

(3s2 + 30s + 63)X(s) = 5

X(s) =5

3s2 + 30s + 63=

5/3s2 + 10s + 21

=512

1s + 3

− 512

1s + 7

x(t) =512

(e−3t − e−7t

)

From the initial value theorem

x(0+) = lims→∞

s5/3

s2 + 10s + 21= 0

which is the same as x(0). Also

x(0+) = lims→∞

s2 5/3s2 + 10s + 21

=53

which is not the same as x(0).





c)s2X(s)− 2s − 3 + 14[sX(s)− 2] + 49X(s) = 3

X(s) =2s + 34

s2 + 14s + 49= 20

1(s + 7)2

+ 21

s + 7

x(t) = 20te−7t + 2e−7t


x(0+) = lims→∞

s2s + 35

s2 + 14s + 49= 2

which is the same as x(0). However, the initial value theorem is invalid for computingx(0+) and gives an undefined result because the orders of the numerator and denominatorof sX(s) are equal.

d)s2X(s)− 4s − 7 + 14[sX(s)− 4] + 58X(s) = 4

X(s) =4s + 67

s2 + 14s + 58=

4s + 67(s + 7)2 + 32

= 133

(s + 7)2 + 32+ 4

s + 7(s + 7)2 + 32

x(t) = 13e−7t sin 3t + 4e−7t cos 3t


x(0+) = lims→∞

s4s + 67

s2 + 14s + 58= 4

which is the same as x(0). However, the initial value theorem is invalid for computing x(0+)and gives an undefined result because the order of the numerator of sX(s) is greater thanthe denominator.



3.26 a)

7[sX(s)− 3] + 5X(s) = 4s1s

= 4

X(s) =25

7s + 5=

25/7s + 5/7

x(t) =257

e−5t/7


x(0+) = lims→∞

s25/7

s + 5/7=

257

b)

7[sX(s)− 3] + 5X(s) = 4s1s

+6s

X(s) =25s + 6

s(7s + 5)=

17

25s + 6s(s + 5/7)

=65

1s

+8335

1s + 5/7

x(t) =65

+8335

e−5t/7

which gives x(0+) = 25/7. However, the initial value theorem is invalid for computingx(0+) and gives an undefined result because the orders of the numerator and denominatorof X(s) are equal (see the bottom of page 146).

c)

3[s2X(s)− 2s − 3] + 30[sX(s)− 2] + 63X(s) = 4s1s

= 4

X(s) =13

6s + 73(s + 3)(s + 7)

=5512

1s + 3

− 3112

1s + 7

x(t) =5512

e−3t − 3112

e−7t


x(0+) = lims→∞

s13

6s + 73(s + 3)(s + 7)

= 2

which is the same as x(0). However, the initial value theorem is invalid for computing x(0+)and gives an undefined result because the order of the numerator of sX(s) is greater thanthe denominator (see the bottom of page 146).





d)

3[s2X(s)− 4s − 7] + 30[sX(s)− 4] + 63X(s) = 4s1s

+6s

X(s) =13

12s2 + 145s + 6s(s2 + 10s + 21)

= 0.09521s

+ 8.91671

s + 3− 5.0119

1s + 7

x(t) = 0.0952 + 8.9167e−3t − 5.0119e−7t

The initial value theorem gives x(0+) = 4 but is invalid for computing x(0+) because theorders of the numerator and denominator of sX(s) are equal (see the bottom of page 146).



3.27 Transform each equation.

3[sX(s)− 5] = Y (s)

sY (s) − 10 = 4− 3Y (s) − 15X(s)

Solve for X(s) and Y (s).

X(s) =15s + 59

3s2 + 9s + 15=

13

15s + 59s2 + 3s + 5

Y (s) =42s − 225

3s2 + 9s + 15=

14s − 75s2 + 3s + 5

The denominator roots are s = −1.5± 1.658j. Thus

X(s) =13

[C1

1.658(s + 1.5)2 + 2.75

+ C2s + 1.5

(s + 1.5)2 + 2.75

]

= 7.33681.658

(s + 1.5)2 + 2.75+ 5

s + 1.5(s + 1.5)2 + 2.75

andx(t) = 7.3368e−1.5t sin 1.658t + 5e−1.5t cos 1.658t

Also,

Y (s) = C11.658

(s + 1.5)2 + 2.75+ C2

s + 1.5(s + 1.5)2 + 2.75

= −57.9011.658

(s + 1.5)2 + 2.75+ 14

s + 1.5(s + 1.5)2 + 2.75

andy(t) = −57.901e−1.5t sin 1.658t + 14e−1.5t cos 1.658t



3.28 Transform each equation.

sX(s) − 5 = −2X(s) + 5Y (s)

sY (s) − 2 = −6Y (s) − 4X(s) +10s

Solve for X(s) and Y (s).

X(s) =5s2 + 40s + 50s3 + 8s2 + 32s

Y (s) =2s2 − 6s + 20s3 + 8s2 + 32s

The denominator roots are s = 0 and s = −4 ± 4j. Thus

X(s) =C1

s+ C2

4(s + 4)2 + 42

+ C3s + 4

(s + 4)2 + 42

=2516s

+5516

4(s + 4)2 + 42

+5516

s + 4(s + 4)2 + 42

x(t) =2516

+5516

e−4t sin 4t +5516

e−4t cos 4t

Also,

Y (s) =C1

s+ C2

4(s + 4)2 + 42

+ C3s + 4

(s + 4)2 + 42

=58s

− 338

4(s + 4)2 + 42

+118

s + 4(s + 4)2 + 42

y(t) =58− 33

8e−4t sin 4t +

118

e−4t cos 4t



3.29 Transforming both sides of the equation we obtain

s2Y (s) − sy(0)− y(0) + Y (s) =1

s + 1

which gives

Y (s) =(s + 1) [sy(0) + y(0)] + 1

(s + 1)(s2 + 1)=

s2y(0) + [y(0) + y(0)] + y(0) + 1(s + 1)(s2 + 1)

This can be expanded as follows.

Y (s) = C11

s + 1+ C2

1s2 + 1

+ C3s

s2 + 1

We find the coefficients following the usual procedure and obtain C1 = 1/2, C2 = y(0)+1/2,and C3 = y(0)− 1/2. Thus the solution is

y(t) =12e−t +

[y(0) +

12

]sin t +

[y(0)− 1

2

]cos t

Because the initial values can be arbitrary, the general form of the solution is

y(t) =12e−t + A1 sin t + A2 cos t (1)

This form can be used to obtain a solution for cases where y(t) or y(t) are specified at pointsother than t = 0. For example, suppose we are given that y(0) = 5/2 and y(π/2) = 3. Thenevaluation of equation (1) at t = 0 and at t = π/2 gives

y(0) =12

+ A2 =52

y

(π

2

)=

12e−π/2 + A1 = 3

The solution of these two equations is A1 = 3 − e−π/2/2 = 2.896 and A2 = 2, and thesolution of the differential equation is

y(t) =12e−t + 2.896 sin t + 2 cos t



3.30 For zero initial conditions, the transform gives

(s2 + 4)X(s) =3s2

orX(s) =

3s2(s2 + 4)

=C1

s2+

C2

s+ C3

2s2 + 4

+ C4s

s2 + 4

The solution form is thus

x(t) = C1t + C2 + C3 sin 2t + C4 cos 2t

This form satisfies the differential equation if C1 = 3/4 and C2 = 0. From x(0) = 10, weobtain C4 = 10. From x(5) = 30, we obtain C3 = −63.675. Thus

x(t) =34t − 63.675 sin 2t + 10 cos 2t



3.31 The denominator roots are s = −3 ± 5j and s = ±6j. Thus we can express X(s) asfollows.

X(s) =30

[(s + 3)2 + 52] (s2 + 62)

which can be expressed as the sum of terms that are proportional to entries 8 through 11in Table 3.3-1.

X(s) = C15

(s + 3)2 + 52+ C2

s + 3(s + 3)2 + 52

+ C36

s2 + 62+ C4

s

s2 + 62(1)

We can obtain the coefficients by noting that X(s) can be written as

X(s) =5C1(s2 + 62) + C2(s + 3)(s2 + 62) + 6C3

[(s + 3)2 + 52

]+ C4s

[(s + 3)2 + 52

]

[(s + 3)2 + 52] (s2 + 62)(2)

Comparing the numerators of equations (1) and (2), and collecting powers of s, we see that

(C2 + C4)s3 + (5C1 + 3C2 + 6C3 + 6C4)s2 + (36C2 + 36C3 + 34C4)s

+180C1 + 108C2 + 204C3 = 30

orC2 + C4 = 0 5C1 + 3C2 + 6C3 + 6C4 = 0

36C2 + 36C3 + 34C4 = 0 180C1 + 108C2 + 204C3 = 30

These are four equations in four unknowns. Note that the first equation gives C4 = −C2.Thus we can easily eliminate C4 from the equations and obtain a set of three equations inthree unknowns. The solution is C1 = 6/65, C2 = 9/65, and C3 = −1/130, and C4 = −9/65.

The inverse transform is


−3t cos 5t + C3 sin 6t + C2 cos 6t

=665

e−3t sin 5t +965

e−3t cos 5t − 1130

sin 6t − 965

cos 6t



3.32 Transform the equation.

(s2 + 12s + 40)X(s) = 35

s2 + 25

The characteristic roots are s = −6 ± 2j. Thus

X(s) =15

(s2 + 25)(s2 + 12s + 40)

= C15

s2 + 25+ C2

s

s2 + 25+ C3

2(s + 6)2 + 4

+ C4s + 6

(s + 6)2 + 4

orX(s) =

185

5s2 + 25

− 485

s

s2 + 25+

19170

2(s + 6)2 + 4

+485

s + 6(s + 6)2 + 4

Thusx(t) =

185

sin 5t − 485

cos 5t +19170

e−6t sin 2t +485

e−6t cos 2t



3.33 From Example 3.7.1, the form A sin(ωt + φ) has the transform

As sin φ + ω cos φ

s2 + ω2

For this problem, ω = 5. Comparing numerators gives

s sin φ + 5 cos φ = 4s + 9

ThusA sin φ = 4 5A cosφ = 9

With A > 0, φ is seen to be in the first quadrant.

φ = tan−1 sin φ

cos φ= tan−1 4/A

9/5A= tan−1 20

9= 1.148 rad

Because sin2 φ + cos2 φ = 1, (4A

)2

+(

95A

)2

= 1

which gives A = 4.386. Thus

x(t) = 4.386 sin(5t + 1.148)



3.34 Taking the transform of both sides of the equation and noting that both initial con-ditions are zero, we obtain

s2X(s) + 6sX(s) + 34X(s) = 56

s2 + 62

Solve for X(s).

X(s) =30

(s2 + 6s + 34)(s2 + 62)

The inverse transform was obtained in Problem 3.31. It is

x(t) =665

e−3t sin 5t +965

e−3t cos 5t − 1130

sin 6t − 965

cos 6t




(s2 + 12s + 40)X(s) =10s

or, since the characteristic roots are s = −6 ± 2j,

X(s) =10

s[(s + 6)2 + 22](1)

From Example 3.7.1(b), the form Ae−at sin(ωt + φ) has the transform

As sin φ + a sin φ + ω cos φ

(s + a)2 + ω2

For this problem, a = 6 and ω = 2. Thus

X(s) =10

s[(s + 6)2 + 22]=

C1

s+ C2

s sin φ + 6 sin φ + 2 cos φ

(s + 6)2 + 22

or

X(s) =C1(s2 + 12s + 40) + C2s

2 sin φ + 6C2s sin φ + 2C2s cos φ

s[(s + 6)2 + 22](2)

Collecting terms and comparing the numerators of equations (1) and (2), we have

(C1 + C2 sinφ)s2 + (12C1 + 6C2 sin φ + 2C2 cos φ)s + 40C1 = 10

Thus comparing terms, we see that C1 = 1/4 and

14

+ C2 sin φ = 0

3 + 6C2 sin φ + 2C2 cos φ = 0

SoC2 sinφ = −1

4C2 cos φ = −3

4





Thus φ is in the third quadrant and

φ = tan−1 −1/4−3/4

= 0.322 + π = 3.463 rad

Because sin2 φ + cos2 φ = 1, (1

4C2

)2

+(

34C2

)2

= 1

which gives C2 = 0.791. Thus

x(t) =14

+ 0.791e−6t sin(2t + 3.463)




X(s) =F (s)

s2 + 8s + 1Thus

F (s) − X(s) = F (s) − F (s)s2 + 8s + 1

=s2 + 8s

s2 + 8s + 1F (s)

Because F (s) = 6/s2,

F (s) − X(s) =s2 + 8s

s2 + 8s + 16s2

=s + 8

s2 + 8s + 16s

From the final value theorem,

fss − xss = lims→0

s[F (s) − X(s)] = lims→0

ss + 8

s2 + 8s + 16s

= 8



3.37 The roots are s = −2 and −4. Thus

X(s) =1 − e−3s

(s + 2)(s + 4)

LetF (s) =

1(s + 2)(s + 4)

=12

(1

s + 2− 1

s + 4

)

sof(t) =

12

(e−2t − e−4t

)

From Property 6 of the Laplace transform,

x(t) =12

(e−2t − e−4t

)− 1

2

[e−2(t−3) − e−4(t−3)

]us(t − 3)



3.38f(t) =

C

Dtus(t) −

2C

D(t − D)us(t − D) +

C

D(t − 2D)us(t − 2D)


F (s) =C

Ds2− 2C

Ds2e−Ds +

C

Ds2e−2Ds =

C

Ds2

(1− 2e−Ds + e−2Ds

)



3.39f(t) =

C

Dtus(t) −

C

D(t − D)us(t − D)− Cus(t − D)


F (s) =C

Ds2− C

Ds2e−Ds − C

se−Ds



3.40f(t) = Mus(t) − 2Mus(t − T ) + Mus(t − 2T )

From Property 6,

F (s) =M

s− 2M

se−Ts +

M

se−2Ts



3.41P (t) = 3us(t) − 3us(t − 5)

From Property 6,

P (s) =3s− 3

se−5s

X(s) =P (s)4s + 1

=3(1 − e−5s

)

s(4s + 1)=

34

1 − e−5s

s(s + 1/4)

LetF (s) =

34

1s(s + 1/4)

= 3(

1s− 1

s + 1/4

)

Thenf(t) = 3

(1 − e−t/4

)

SinceX(s) = F (s)

(1 − e−5s

)

we have

x(t) = f(t) − f(t − 5)us(t − 5) = 3(1 − e−t/4

)− 3

[1 − e−(t−5)/4

]us(t − 5)



3.42 Let

f(t) = t +t3

3+

2t5

15Then

F (s) =1s2

+2s4

+16s6

=s4 + 2s2 + 16

s6

From the differential equation,

X(s) =F (s)s + 1

=s4 + 2s2 + 16

s6(s + 1)

=16s6

− 16s5

+18s4

− 18s3

+19s2

− 19s

+19

s + 1

Thusx(t) =

215

t5 − 23t4 + 3t3 − 9t2 + 19t − 19 + 19e−t

On a plot of this and the solution obtained from the lower-order approximation, the twosolutions are practically indistinguishable.



3.43 From the derivative property of the Laplace transform, we know that

L[x(t)] =∫ ∞

0x(t)e−st dt = sX(s)− x(0)

Thereforelims→∞

[sX(s)] = lims→∞

[x(0) +

∫ ∞

0x(t)e−st dt

]

= lims→∞

x(0) + lims→∞

lim

ε→0+

[∫ ε

0x(t)e−st dt

]+ lim

ε→0+

∫ ε

0lim

s→∞

[x(t)e−st dt

]

The limits on ε and s can be interchanged because s is independent of t. Within the interval[0, 0+], e−st = 1, and so

lims→∞

[sX(s)] = x(0) + lims→∞

lim

ε→0+

[∫ ε

0x(t) dt

]+ lim

ε→0+

∫ ε

0lims→∞

[x(t)e−st dt

]

= x(0) + x(t)|t=0+t=0 + 0 = x(0+)

This proves the theorem.



3.44 From the derivative property of the Laplace transform, we know that

L[x(t)] =∫ ∞

0x(t)e−st dt = sX(s)− x(0)

Therefore,

lims→0

[sX(s)] = lims→0

x(0) + lims→0

[∫ ∞

0x(t)e−st dt

]

= x(0) +∫ ∞

0lims→0

[x(t)e−st dt

]= x(0) +

∫ ∞

0x(t) dt

because s is independent of t and lims→0 e−st = 1. Thus

lims→0

[sX(s)] = x(0) + limT→∞

[∫ T

0x(t) dt

]= x(0) + lim

T→∞

[x(t)|t=T

t=0

]

= x(0) + limT→∞

x(T )− x(0) = limT→∞

x(T ) = limt→∞

x(t)

This proves the theorem.



3.45 Letg(t) =

∫ t

0x(t) dt

ThenL[∫ t

0x(t) dt

]= L[g(t)] =

∫ t

0g(t)e−st dt

To use integration by parts we define u = g and dv = e−stdt, which give du = dg = x(t) dtand v = −e−st/s. Thus

∫ t

0g(t)e−st dt =

g(t)e−st

−s

∣∣∣∣∣

t=∞

t=0

−∫ ∞

0

e−st

−sx(t) dt

= 0 +g(0)s

+1s

∫ ∞

0x(t)e−st dt =

g(0)s

+X(s)

s

=1s

∫x(t) dt

∣∣∣∣t=0

+X(s)

s

This proves the property.If there is an impulse in x(t) at t = 0, then g(0) equals the strength of the impulse. If

there is no impulse at t = 0, then g(0) = 0.



3.46 a)

[r,p,k] = residue([8,5],[2,20,48])

The result is r = [10.7500, -6.7500], p = [-6.0000, -4.0000], and k = [ ]. Thesolution is

x(t) = 10.75e−6t − 6.75e−4t

b)

[r,p,k] = residue([4,13],[2,8,116])

The result is r = [1.0000 - 0.1701i, 1.0000 + 0.1701i], p = [-2.0000 + 7.3485i,-2.0000 - 7.3485i], and k = [ ]. The solution is

x(t) = (1 − 0.1701j)e(−2+7.3485j)t + (1 + 0.1701j)e(−2−7.3485j)t

From (3.8.4) and (3.8.5), the solution is

x(t) = 2e−2t (cos 7.3485t + 0.1701 sin 7.3485t)

c)

[r,p,k] = residue([3,2],[1,10,0,0])

The result is r = [ -0.2800, 0.2800, 0.2000], p = [-10, 0, 0], and k = [ ]. Thesolution is

x(t) = −0.28e−10t + 0.28 + 0.2t

d)

[r,p,k] = residue([1,0,1,6],[1,2,0,0,0,0])

The result is r = [-0.2500, 0.2500, 0.5000, -1.0000, 3.0000], p =[ -2, 0, 0, 0,0], and k = [ ]. The solution is

x(t) = −0.25e−2t + 0.25 + 0.5t − 12t2 +

12t3





e)

[r,p,k] = residue([4,3],[1,6,34,0])

The result is r = [-0.0441 - 0.3735i, -0.0441 + 0.3735i, 0.0882], p = [-3.0000+ 5.0000i, -3.0000 - 5.0000i, 0], and k = [ ].The solution is

x(t) = (−0.0441− 0.3735j)e(−3+5j)t + (−0.0441 + 0.3735j)e(−3−5j)t + 0.0882


x(t) = 2e−3t (−0.0441 cos 5t + 0.3735 sin 5t) + 0.0882

f)

[r,p,k] = residue([5,3,7],[1,12,144,48])

The result is r = [2.4759 + 1.3655i, 2.4759 - 1.3655i, 0.0482], p = -5.8286 +10.2971i,-5.8286 -10.2971i, -0.3428], and k = [ ]. The solution is

x(t) = (2.4759+1.3655j)e(−5.8286+10.2971j)t+(2.4759−1.3655j)e(−5.8286−10.2971j)t+0.0482e−0.3428t


x(t) = 2e−5.8286t (2.4759 cos 10.29t− 1.3655 sin 10.29t) + 0.0482e−0.3428t



3.47 a)

[r,p,k] = residue(5,conv([1,8,16],[1,1]))

The result is r = [-0.5556, -1.6667, 0.5556], p = [-4.0000, -4.0000, -1.0000], k= [ ]. The solution is

x(t) = −0.5556e−4t − 1.6667te−4t + 0.5556e−t

b)

[r,p,k] = residue([4,9],conv([1,6,34],[1,4,20]))

The result is r = [-0.1159 + 0.1073i, -0.1159 - 0.1073i, 0.1159 - 0.1052i, 0.1159+ 0.1052i], p = -3.0000 + 5.0000i, -3.0000 - 5.0000i, -2.0000 + 4.0000i, -2.0000- 4.0000i], and k = [ ]. The solution is

x(t) = (−0.1159 + 0.1073j)e(−3+5j)t + (−0.1159− 0.1073j)e(−3−5j)t

+ (0.1159− 0.1052j)e(−2+4j)t + (0.1159 + 0.1052j)e(−2−4j)t


x(t) = 2e−3t (−0.1159 cos 5t − 0.1073 sin 5t) + 2e−2t (0.1159 cos 4t + 0.1052 sin 4t)



3.48 a)

sys = tf(1,[3,21,30]);step(sys)

b)

sys = tf(1,[5,20, 65]);step(sys)

c)

sys = tf([3,2],[4,32,60]);step(sys)



3.49 a)

sys = tf(1,[3,21,30]);impulse(sys)

b)

sys = tf(1,[5,20, 65]);impulse(sys)



3.50

sys = tf(5,[3,21,30]);impulse(sys)



3.51

sys = tf(5,[3,21,30]);step(sys)



3.52 a)

sys = tf(1,[3,21,30]);t = [0:0.001:1.5];f = 5*t;[x,t] = lsim(sys,f,t);plot(t,x)

b)

sys = tf(1,[5,20,65]);t = [0:0.001:1.5];f = 5*t;[x,t] = lsim(sys,f,t);plot(t,x)

c)

sys = tf([3,2],[4,32,60]);t = [0:0.001:1.5];f = 5*t;[x,t] = lsim(sys,f,t);plot(t,x)



3.53 a)

sys = tf(1,[3,21,30]);t = [0:0.001:6];f = 6*cos(3*t);[x,t] = lsim(sys,f,t);plot(t,x)

b)

sys = tf(1,[5,20,65]);t = [0:0.001:6];f = 6*cos(3*t);[x,t] = lsim(sys,f,t);plot(t,x)

c)

sys = tf([3,2],[4,32,60]);t = [0:0.001:6];f = 6*cos(3*t);[x,t] = lsim(sys,f,t);plot(t,x)




to accompany


by

William J. Palm III


Solutions to Problems in Chapter Four


CD1

Highlight

4.1 For a helical coil spring

k =Gd4

64nR3= 1.7 × 109 (1/24)4

64(6)(2/12)3= 2.88× 103 lb/ft



4.2 Let D be the length of the dead space to the left of spring k2. The plot follows.

Figure : for Problem 4.2



4.3 a) Series.b) Let kb be the stiffness of the beam and kc be the stiffness of the cable.

ke =kbkc

kb + kc



4.4 Sum moments about the pivot.

0 = fL1 − k1(L2θ)L2 − k2(L3θ)L3

Thus

f =k1L

22 + k2L

23

L1θ

But x = L1θ, so

f =k1L

22 + k2L

23

L21

x = kex



4.5 Series combination.ke =

k1k2

k1 + k2

where

ki =EAi

Li=

E

Liπ

(Di

2

)2

=EπD2

i

4Li



4.6

ke = 4(

EA

L

)=

4Eπ(d/2)2

L=

4(2× 1011)π(0.03)2

4(1)= 1.8π × 108 N/m



4.7 The beam stiffness is

kb =4Ewh3

L3=

4(3× 107)(144)(1)(1/12)3

(3)3= 3.7037× 105 lb/ft

a) The springs are in parallel, so ke = kb + k. We want ke = 2kb, so we must requirethat k = kb.

b)

ωn =

√ke

m=

√2(3.7037× 105)

40= 136.1 rad/sec



4.8 First reduce the system to the equivalent one shown in part (a) of the figure, where

1k1

=12k

+1k

=32k

Thus k1 = 2k/3.From part (b) of the figure,

1ke

=1k

+1

k1 + k=

k1 + 2k

k(k1 + k)

Thus, solving for ke and substituting for k1, we obtain

ke =k(k1 + k)k1 + 2k

=5k

8




4.9 The stiffnesses are in series. Thus

kTe =kT1kT2

kT1 + kT2

where

kT1 =πG(D4 − d4)

32L=

π(8× 1010)(0.44 − 0.34)32(2)

= 6.874× 107

kT2 =πG(D4 − d4)

32L=

π(8× 1010)(0.354 − 0.254)32(3)

= 2.906× 107

ThuskTe = 2.043× 107 N ·m/rad



4.10 The plot is the following.




4.11 Since x1 and x2 are measured from the equilibrium positions, the gravity forces arecanceled by the static spring forces in parts (a) and (b). Thus the equations of motion arethe same for all three cases. They are

m1x1 = −k1x1 + k2(x2 − x1)

m2x2 = f − k2(x2 − x1)



4.12 Sum moments about the pivot to obtain

mL23θ = −(k1L1θ)L1 + k2(x − L2θ)L2 − mgL3θ

Collecting terms we obtain

mL23θ + (k1L

21 + k2L

22 + mgL3)θ = k2L2x



4.13 Let T be the tension in the cable. See the figure. Since the pulley is consideredmassless,

2T =kx

2For the mass m,

mx = f − T + mg = f − kx

4+ mg

Thusmx +

kx

4= f + mg

Note that the static spring force does not cancel the weight mg because x is not measuredfrom the equilibrium position.




4.14 a) The buoyancy force is

B = (ρV )g = ρgπ

(D

2

)2

x

or

B =πρgD2

4x

b) Summing forces in the vertical direction,

mx = −B

or

mx = −πρgD2

4x

Thus, the natural frequency is

ωn =

√πρgD2

4m=

D

2

√πρg

m

c) The period is

P =2π

ωn=

4π

D

√m

πρg=

4π

gD

√W

πρ

or

P =4π

32.2(2)

√10001.94π

= 2.5 sec



4.15 a) Sum moments about the mass center G to obtain

Iθ = −(B sin θ)h

Because B = mg = W , the ship’s weight,

Iθ + Wh sin θ = 0

where I is the ship’s moment of inertia about G.b) For small angles, sin θ ≈ θ, and the equation becomes

Iθ + Whθ = 0

The characteristic roots are s = ±j√

Wh/I , and the roll period is given by 2π√

I/Wh.



4.16 Sum moments about the rotation axis.

Iθ = k1(φ − θ) − k2θ

Collecting terms we obtainIθ + (k1 + k2)θ = k1φ

The transfer function isΘ(s)Φ(s)

=k1

Is2 + k1 + k2



4.17 a) Refer to the following figure. Summing forces in the x direction gives

mx = k(y − x)− ft (1)

Summing moments about the mass center of the wheel gives Iθ = Rft. But x = Rθ, andthus θ = x/R. Therefore

ft =I

Rθ =

I

R2x (2)

Combine (1) and (2):

mx + kx = ky − I

R2x

or (m +

I

R2

)x + kx = ky

where I = mR2/2. Thus1.5mx + kx = ky (3)






b) Substituting the values into (3) gives

15x + 1000x = 1000y

or3x + 200x = 200y

The roots are s = ±j10√

2/3. The response to a unit-step input is

x(t) = 1 − cos 10√

23t



4.18 Summing forces in the horizontal direction on each mass gives

m1x1 = −k1x1 + k2(x2 − x1)

m2x2 = −k3x2 − k2(x2 − x1)



4.19 Summing moments about the axis of rotation of each inertia gives

I1θ1 = −k1θ1 − k2(θ1 − θ2)

I2θ2 = T2 + k2(θ1 − θ2) − k3θ2



4.20 Summing moments about each pivot point and assuming small angles, we obtain

m1L22θ1 = −m1gL2θ1 − kL1(θ1 − θ2)L1

m2L22θ2 = −m2gL2θ2 + kL1(θ1 − θ2)L1

Collecting terms givesm1L

22θ1 + (m1gL2 + kL2

1)θ1 = kL21θ2

m2L22θ2 + (m2gL2 + kL2

1)θ2 = kL21θ1



4.21 The torsional stiffness of the torsion bar is

kT =πGD4

32L=

1.7 × 109π(1.5/12)4

32(4)= 104 lb − ft/rad

The characteristic equation isIs2 + kT = 0

where I = mL2 = (40/32.2)(2)2 = 4.9689. Thus the natural frequency is

ωn =

√kT

I= 44.86 rad/sec



4.22 The kinetic energy of the system is

KE =12mx2 +

12Iω2 =

12mx2 +

12I

(x

R

)2

=12(1.5m)x2

since x = Rω and I = mR2/2. The potential energy is

PE =12(k1 + k2)x2

From conservation of mechanical energy,

KE + PE =12(1.5m)x2 +

12(k1 + k2)x2 = constant

Differentiating with respect to time gives

1.5mxx + (k1 + k2)xx = 0

Canceling x gives the equation of motion.

1.5mx + (k1 + k2)x = 0



4.23 Because x = 0 at equilibrium, the static spring force cancels the constant weight m2g.Taking the potential energy to be zero at x = 0, we obtain

PE =12kx2

The kinetic energy of the system is

KE =12m1x

2 +12m2y

2 =12(m1 + 4m2)x2

since y = 2x.From conservation of energy,

KE + PE = constant

or12(m1 + 4m2)x2 +

12kx2 = constant

Differentiating with respect to time and setting the derivative to 0, we obtain

(m1 + 4m2)xx + kxx = 0


(m1 + 4m2)x + kx = 0



4.24 Because x = 0 at equilibrium, the static spring force cancels the constant weight m1g.Taking the potential energy to be zero at x = 0, and noting that the spring extension y

from its equilibrium position is y = 2x, we obtain

PE =12k(2x)2 = 2kx2

Note that y = Rω = 2x and the inertia of the cylinder is I = m2R2/2. The kinetic energy

of the system is

KE =12m2y

2 +12Iω2 +

12m1x

2

=12

(4m2 + 4

I

R2+ m1

)x2

=12

(4m2 + 2m2 + m1) x2

=12

(6m2 + m1) x2

From conservation of energy,

KE + PE = constant

or12

(6m2 + m1) x2 + 2kx2 = constant

Differentiating with respect to time and setting the derivative to 0, we obtain

(6m2 + m1)xx + 4kxx = 0


(6m2 + m1)x + 4kx = 0



4.25 Because x = 0 at equilibrium, the static spring force cancels the constant weight mg.Taking the potential energy to be zero at x = 0, and noting that the spring extension y isy = x/2, we obtain

PE =12ky2 =

18kx2

Note that x = Rω, where ω is the angular velocity of the pulley, and that the inertia of thecylinder is I = mR2/2. The kinetic energy of the system is

KE =12mx2 +

12Iω2

=12mx2 +

12

mR2

2x2

R2

=12

(m +

m

2

)x2

=12(1.5m)x2

Let A = |xmax|. Then |x| = ωnA for simple harmonic motion, and Rayleigh’s principlegives

12(1.5m)ω2

nA2 =18kA2

This gives

ωn =

√k

6m




T =12mx2 +

12Iω2 =

12mx2 +

12I

(x

R

)2

=12(1.5m)x2

since x = Rω and I = mR2/2. The potential energy is

V =12kx2

From the Rayleigh principle, Tmax = Vmax − Vmin. For simple harmonic motion, x = ωnA,and this implies that

12(1.5m)ω2

nA2 =12kA2

Thus

ωn =

√k

1.5m




T =12mv2 =

12m

L21

L22

x2

since v = L1θ and x = L2θ.The potential energy is

V =12kx2

From the Rayleigh principle, Tmax = Vmax − Vmin. For simple harmonic motion, x = ωnA,and this implies that

12m

L21

L22

ω2nA2 =

12kA2

Thus

ωn =L2

L1

√k

m



4.28 See the figure. Assuming that θ is small, summing moments about the pivot gives

IOθ = fcL1 − k2L2x = fcL1 − k2L2(L2θ)

The equivalent mass at x due to the valve and spring is

me = mv +13ms

Since IO = Ir + meL22, we have

(Ir + meL22)θ = fcL1 − k2L

22θ

Thus the force on the cam is given by

fc =(Ir + meL

22)θ + k2L

22θ

L1

Knowing the cam profile and the cam rotation speed, we can compute θ(t) and θ(t) to usein the expression for fc.




4.29 The equivalent mass is me = 12 + 0.23(3) = 12.69 kg. From statics,

k(0.02) = 12(9.81)

or k = 5886 N/m.



4.30 The equivalent mass is me = 3 + 0.38(1) = 3.38 slug. From statics,

k(0.01) = 3(32.2)

or k = 9660 lb/ft.



4.31 The equation given in the problem statement is

xy =P

6EIAy2(3L− y) (1)

The deflection xL at the end of the beam, where y = L, is found from (1) to be:

xL =PL3

3EIA(2)

Comparing (1) and (2) shows that

xy =y2(3L− y)

2L3xL (3)

Differentiate this equation with respect to time for a fixed value of y to obtain

xy =y2(3L− y)

2L3xL (4)

Because (1) describes the static deflection, it does not account for inertia effects, and (4) isnot exactly true. However, lacking another reasonable expression for xy , we will use (4).

The kinetic energy of a beam mass element dm at position y is x2ydm/2. Let ν be the

beam’s mass per unit length. Then dm = ν dy, and the total kinetic energy KE in thebeam is

KE =12

∫ L

0x2

y dm

=ν

8L6x2

L

∫ L

0y4(3L − y)2 dy

=ν

8L6x2

L

33L7

35=

33νL

2(140)x2

L

Because the beam mass mb = νL,

KE =33mb

2(140)x2

L

If a mass me is located at the end of the beam, and moves with a velocity xL, its kineticenergy is mex

2L/2. Comparing this with the beam’s energy, we see that me = (33/140)mb ≈

0.23mb. Thus the equivalent mass of the cantilever spring is 23% of the beam mass.



4.32 a) See part (a) of the figure. The equivalent mass of the beam and winch is me =0.23mb + mw. Assuming that x1 and x2 are measured from the equilibrium positions, theequation of motion for me is

mex1 = T − kx1 (1)

where T is the tension in the rope and the beam stiffness is k = Ewh3/4L3.For the hoisted mass,

mhx2 = −T (2)

Solve this for T and substitute into (1) to obtain

mex1 + mhx2 = −kx1 (3)

If the rope does not stretch, then x1 = x2 and (3) becomes

(me + mh)x1 = −kx1






b) If the rope acts like a spring, then T = kr(x2 − x1), and we obtain the free-bodydiagrams shown in part (b) of the figure. For me,

mex1 = kr(x2 − x1) − kx1

For mh,mhx2 = −kr(x2 − x1)

These are the equations of motion.



4.33 a) Summing forces parallel to the plane, we obtain

mv = mg sin 30 − cv

or6v + v = 29.43

b)v(t) = 29.43− 25.43e−t/6

c) The steady-state speed is 29.43 m/s, and it takes approximately 4τ = 4(6) = 24 s toreach that speed.

d) Because the steady-state speed can be found directly from the equation of motion,by setting v = 0, it is independent of the initial speed.



4.34 a) The characteristic equation is 40s2 + 680s + 1200 = 0, which has the roots s = −2and s = −15. The form of the solution is

x(t) = A1e−2t + A2e

−15t + 6000/1200

For zero initial conditions, A1 = −75/13 and A2 = 10/13, so

x(t) = −7513

e−2t +1013

e−15t + 5

b) The characteristic equation is 40s2 + 400s + 1200 = 0, which has the roots s =−5 ± j

√5. The form of the solution is

x(t) = A1e−5t sin

√5t + A2e

−5t cos√

5t + 6000/1200

For zero initial conditions, A1 = −5√

5 and A2 = −5, so

x(t) = −5√

5e−5t sin√

5t − 5e−5t cos√

5t + 5



4.35 a) Summing the horizontal forces on each mass gives

m1x1 = −cx1 + k(x2 − x1)

m2x2 = f − k(x2 − x1)

b) Summing the horizontal forces on each mass gives

m1x1 = −kx1 + c(x2 − x1)

m2x2 = f − c(x2 − x1)

c) Summing the horizontal forces on each mass gives

m1x1 = −cx1 − k1x1 + k2(x2 − x1)

m2x2 = f − k2(x2 − x1)

d) Summing the horizontal forces on each mass gives

m1x1 = −cx1 − k1x1 + k2(x2 − x1) + c2(x2 − x1)

m2x1 = f − k2(x2 − x1) − c2(x2 − x1)



4.36 a) See the diagram. Let F be the tension in the cable. Summing moments about thedrum center gives

Iω = T − cTω − FR (1)

Summing vertical forces on the mass m gives

mv = F − mg (2)

Solve (2) for F , substitute for F in (1), and use the fact that v = Rω to obtain

(I + mR2)v + cTv = TR − mgR2






b) Using the given values, we obtain

24v + v = 243.3

The solution has the formv(t) = 243.3 + Ae−t/24

For v(0) = 0, A = −243.3, and

v(t) = 243.3(1 − e−t/24

)



4.37 Summing moments about the pivot, and assuming small, angles, we obtain

Iθ = fL1 − c(L2θ

)L2 − k (L3θ) L3

orIθ + cL2

2θ + kL23θ = fL1



4.38 Summing horizontal forces gives

mx = −k2x + k1(y − x) + c(y − x)

Collect terms to obtainmx + cx + (k1 + k2)x = k1y + cy

The transfer function isX(s)Y (s)

=cs + k1

ms2 + cs + k1 + k2



4.39 Summing moments about the rotation axis of the pulley gives

Ipθp = kT (φ − θp) − cT

(θp − θd

)

Summing moments about the center of mass of the damper gives

Idθd = cT

(θp − θd

)

These two equations form the system model.



4.40 Summing moments about the pivot, and assuming small, angles, we obtain

IOθ = −mgL3θ − c(L2θ

)L2 + k (z − L1θ) L1

where IO = mL23 + I . Collecting terms gives

(mL23 + I)θ + cL2

2θ + (kL21 + mgL3)θ = kL1z



4.41 Summing horizontal forces on m gives

mx = f + k(xA − x)

From statics, at point A,0 = −k(xA − x) − cxA

These can be combined into a single equation by using the Laplace transform.

ms2X(s) = F (s) + k [XA(s) − X(s)]

0 = −k [XA(s) − X(s)]) − csXA(s)

These can be solved for the transfer function relating the given output xA to the given inputf .

XA(s)F (s)

=k

s(mcs2 + mks + ck)

This is the required model.



4.42 Summing horizontal forces on m gives

mx = c(xA − x)− k2x

From statics, at point A,0 = k1(y − xA) − c(xA − x)

These can be combined into a single equation by using the Laplace transform.

ms2X(s) = csXA(s) − csX(s)− k2X(s)

0 = k1 [Y (s) − XA(s)]) − csXA(s) + csX(s)

These can be solved for the transfer function relating the given output x to the given inputy.

X(s)Y (s)

=ck1s

mcs3 + mk1s2 + c(k1 + k2)s + k1k2

This is the required model.



4.43 See the diagram. Let F be the force on the pinion due to the rack. Summing momentson the pinion gives

(Im + Ip)θ = T − RF (1)

Summing horizontal forces on the rack gives

mrx = F − cx− kx (2)

Solving (2) for F , substituting in (1), and using the fact that x = Rθ, gives

(Im + Ip + mrR2)θ + cR2θ + kR2θ = T




4.44 We can represent the system as shown in the diagram, where

I1e = N2I1 T1e = NT1

Summing moments on I1e gives

I1eθ2 = T1e + kT (θ3 − θ2) (1)

Summing moments on I2 gives

I2θ3 = −kT (θ3 − θ2) − cT θ3 (2)

To obtain the model in terms of θ1, substitute θ2 = θ1/N into (1) and (2) to obtain

I1e

Nθ1 = T1e + kT

(θ3 −

θ1

N

)(3)

andI2θ3 = −kT

(θ3 −

θ1

N

)− cT θ3 (4)

The system model consists of (3) and (4).




4.45 a) Assume that the equilibrium position corresponds to θ = 0. Let φ = 45 + θ andδ be the static spring deflection at equilibrium. Summing moments about the pivot andassuming small angles, we obtain

mL2θ = (mg sin φ)L − 2k(δ + L1θ)L1

But

sin φ = sin(45 + θ) =√

22

cos θ +√

22

sin θ ≈√

22

+√

22

θ

for small angles. Thus

mL2θ = mgL

(√2

2+

√2

2θ

)− 2kL1δ − 2k1L

21θ

At equilibrium,

mgL

√2

2− 2kL1δ = 0

and thus the equation of motion becomes

mL2θ +

(mgL

√2

2− 2k1L

21

)θ = 0

The characteristic equation is

mL2s2 + mgL

√2

2− 2k1L

21 = 0

This will be neutrally stable if

mgL

√2

2− 2k1L

21 ≥ 0

and unstable if

mgL

√2

2− 2k1L

21 < 0





b) Assume that the equilibrium position corresponds to θ = 0. Let φ = 135 + θ andδ be the static spring deflection at equilibrium. Summing moments about the pivot andassuming small angles, we obtain

mL2θ = (mg sin φ)L − 2k(δ + L1θ)L1

But

sin φ = sin(135 + θ) =√

22

cos θ −√

22

sin θ ≈√

22

−√

22

θ


mL2θ = mgL

(√2

2−

√2

2θ

)− 2kL1δ − 2k1L

21θ

At equilibrium,

mgL

√2

2− 2kL1δ = 0


mL2θ +

(mgL

√2

2+ 2k1L

21

)θ = 0


mL2s2 + mgL

√2

2+ 2k1L

21 = 0

This will always be neutrally stable since

mgL

√2

2+ 2k1L

21 > 0



4.46 Assume that the equilibrium position corresponds to θ = 0. Let φ = 135 + θ and δbe the static spring deflection at equilibrium. Let L be the distance from the pivot to themass m. Let L1 be the distance from the pivot to the connection point of the spring anddamper.

Summing moments about the pivot and assuming small angles, we obtain

mL2θ = (mg sin φ)L− k(δ + L1θ)L1 − c(L1θ)L1

But

sin φ = sin(135 + θ) =√

22

cos θ −√

22

sin θ ≈√

22

−√

22

θ


mL2θ = mgL

(√2

2−

√2

2θ

)− kL1δ − k1L

21θ − cL2

1θ

At equilibrium,

mgL

√2

2− kL1δ = 0


mL2θ + cL21θ +

(mgL

√2

2+ k1L

21

)θ = 0


mL2s2 + cL21s + mgL

√2

2+ k1L

21 = 0

This will always be stable sincemL2 > 0

cL21 > 0

and

mgL

√2

2+ k1L

21 > 0



4.47 Let I be the inertia of the pendulum about the pivot point. Then summing momentsabout the pivot point, and assuming small angles, we give

Iθ = −k1(L1θ)L1 + k2(y − L2θ)L2 + c(y − L2θ)L2

Collecting terms we obtain

Iθ + cL22θ + (k1L

21 + k2L

22)θ = k2L2y + cL2y



4.48 Assuming that x1, x2, and x3 are measured from equilibrium, summing vertical forceson m1 gives

m1x1 = k1(x2 − x1) + c1(x2 − x1)− k3(x1 − x3) − c3(x1 − x3)

Summing vertical forces on m2 gives

m2x2 = k2(y − x2) − k1(x2 − x1) − c1(x2 − x1)

Summing vertical forces on m3 gives

m3x3 = k3(x1 − x3) + c3(x1 − x3)

The system model consists of these three equations.



4.49 From impulse-momentum,

(m + 5m)v(0+)− [mv1 + 5m(0)] = 0

Thus v(0+) = v1/6.The equation of motion is 6mx + kx = 0, and its solution with x(0+) = 0, x(0+) =

v(0+) = v1/6 is

x(t) =v(0+)

ωnsin

√k

6mt =

v1

6

√6m

ksin

√k

6mt



4.50 With m1 = m and m2 = 5m, (4.6.7) becomes

v3 =m − 5m

m + 5mv1 = −2

3v1

The change in momentum is

m

(−2

3v1

)− mv1 =

∫ 0+

0f(t) dt

Thus ∫ 0+

0f(t) dt = −5

3mv1

The linear impulse applied to m2 is (5/3)mv1, so

5mx + kx =53mv1δ(t)

orx + ω2

nx =13v1δ(t)

where ωn =√

k/5m. The response is given by

x(t) =v1

3ωnsin ωnt =

v1

3

√5m

ksin

√k

5mt



4.51 a) Summing horizontal forces on each mass gives

m1x1 = f − k1x1 − k2(x1 − x2)

m2x2 = k2(x1 − x2)

Substituting the parameter values and collecting terms, we obtain

20x1 = f − 9 × 104x1 + 6 × 104x2

60x2 = 6 × 104(x1 − x2)

b) Apply the Laplace transform to both sides of each equation, using zero initial condi-tions: (

20s2 + 9× 104)

X1(s) − 6 × 104X2(s) = F (s)

−6 × 104X1(s) +(60s2 + 6 × 104

)X2(s) = 0

The solutions areX1(s)F (s)

=s2 + 103

20s4 + 1.1× 105s2 + 3 × 107

X2(s)F (s)

=103

20s4 + 1.1× 105s2 + 3 × 107

c) The MATLAB session is

sys = tf([1, 0, 1e+3],[20, 0, 1.1e+5, 0, 3e+7]);step(sys)



4.52 a) Summing moments about the axis of rotation of each inertia gives

I1θ1 = −k1θ1 − k2(θ1 − θ2)

I2θ2 = T2 + k2(θ1 − θ2) − k3θ2

b) Substitute the given parameter values and collect terms to obtain

Iθ1 = −2kθ1 + kθ2

2Iθ2 = T2 − 2kθ2 + kθ1

Apply the Laplace transform to both sides of each equation, using zero initial conditions:

(Is2 + 2k)Θ1(s) − kΘ2(s) = 0

−kΘ1(s) + (2Is2 + 2k)Θ2(s) = T2(s)

The solutions areΘ1(s)T2(s)

=k

2I2s4 + 6kIs2 + 3k2

Θ2(s)T2(s)

=Is2 + 2k

2I2s4 + 6kIs2 + 3k2

c) With I = 10 and k = 60,

Θ1(s)T2(s)

=60

200s4 + 3600s2 + 10, 800

The MATLAB session is

sys = tf(60,[200, 0, 3600, 0, 10800]);impulse(sys)



4.53 Summing horizontal forces on each mass gives

m1x1 = f − k1x1 − k2(x1 − x2)

m2x2 = k2(x1 − x2)

Substituting the parameter values and collecting terms, we obtain

20x1 = f − 9 × 104x1 + 6 × 104x2

60x2 = 6 × 104(x1 − x2)


(20s2 + 9 × 104)X1(s) − 6 × 104X2(s) = F (s)

−6 × 104X1(s) + (60s2 + 6 × 104)X2(s) = 0

The transfer functions can be found by using Cramer’s method. The session is

D = conv([20, 0, 9e+4], [60, 0, 6e+4])-[0, 0, 0, 0, 3.6e+9];D1 = [60, 0, 6e+4];D2 = 6e+4;sys1 = tf(D1, D)sys2 = tf(D2, D)

The displayed transfer functions agree with the answers found by hand in Problem 4.51.



4.54 Summing moments about the axis of rotation of each inertia gives

I1θ1 = −k1θ1 − k2(θ1 − θ2)

I2θ2 = T2 + k2(θ1 − θ2) − k3θ2

Substitute the given parameter values and collect terms to obtain

10θ1 = −120θ1 + 60θ2

20θ2 = T2 − 120θ2 + 60θ1


(10s2 + 120)Θ1(s) − 60Θ2(s) = 0

−60Θ1(s) + (20s2 + 120)Θ2(s) = T2(s)

The transfer functions can be found by using Cramer’s method. The session is

D = conv([10, 0, 120], [20, 0, 120])-[0, 0, 0, 0, 60*60];D1 = 60;D2 = [10, 0, 120];sys1 = tf(D1, D)sys2 = tf(D2, D)

The displayed transfer functions agree with the answers found by hand in Problem 4.52.



4.55 a) Summing moments about each pivot point and assuming small angles, we obtain

m1L22θ1 = −m1gL2θ1 − kL1(θ1 − θ2)L1

m2L22θ2 = −m2gL2θ2 + kL1(θ1 − θ2)L1

Collecting terms givesm1L

22θ1 + (m1gL2 + kL2

1)θ1 = kL21θ2

m2L22θ2 + (m2gL2 + kL2

1)θ2 = kL21θ1

b) Substitute the given values: m1 = 1, m2 = 4, L1 = 2, L2 = 5, and k = 10, we obtain

25θ1 + 89.05θ1 − 40θ2 = 0

400θ2 + 118.48θ2 − 40θ1 = 0

Apply the Laplace transform to both sides of each equation, using the given initial condi-tions: θ1(0) = 0.1, θ2(0) = 0, θ1(0) = 0, and θ2(0) = 0.

(25s2 + 89.05)Θ1(s) − 40Θ2(s) = 2.5s

−40Θ1(s) + (400s2 + 118.48)Θ2(s) = 0

The transfer functions can be found by using Cramer’s method. Note that we must appenda 0 to the numerator polynomial D1, so that we can use the step function to obtain thefree response. See Example 4.7.1. The session is

D = conv([25, 0, 89.05], [400, 0, 118.48]) - [0, 0, 0, 0, 40*40];D1 = conv([2.5, 0, 0], [400, 0, 118.48]);sys1 = tf(D1, D);step(sys1)



4.56 a) Summing horizontal forces on each mass gives

m1x1 = −k1x1 − k2(x1 − x2)

m2x2 = −k3x2 − k2(x1 − x2)

b) Substitute the parameter values and collect terms to obtain

x1 = −3.2 × 104x1 + 1.6× 104x2

2x2 = −3.2x2 + 1.6× 104x2

Apply the Laplace transform to both sides of each equation, using the initial conditions:x1(0) = 0.1, x2(0) = 0, x1(0) = 0, and x2(0) = 0.

(s2 + 3.2× 104)X1(s)− 1.6× 104X2(s) = 0.1s

−1.6× 104X1(s) + (2s2 + 3.2 × 104)X2(s) = 0

The transfer functions can be found by using Cramer’s method. Note that we must appenda 0 to the numerator polynomial D1, so that we can use the step function to obtain thefree response. See Example 4.7.1. The session is

D = conv([1, 0, 3.2e+4], [2, 0, 3.2e+4]) - [0, 0, 0, 0, 2.56e+8];D1 = conv([0.1, 0, 0], [2, 0, 3.2e+4]);sys1 = tf(D1, D);step(sys1)




to accompany


by

William J. Palm III


Solutions to Problems in Chapter Five


CD1

Highlight

5.1 Define the following state variables: x1 = x and x2 = ˙ x. Then the state equations are

x1 = x2 x2 =15[−4x1 − 7x2 + f(t)]



5.2 Define the following state variables: x1 = y, x2 = y, and x3 = y. Then the stateequations are

x1 = x2 x2 = x3 x3 =12[f(t) − 7x1 − 4x2 − 5x3]




x1 = x2 x2 =12(4y − 4x1 − 5x2)



5.4 From the transfer function we obtain:

3y + 6y + 10y = 6f(t)

Define the following state variables: x1 = x and x2 = ˙ x. Then the state equations are

x1 = x2 x2 =13[6f(t) − 10x1 − 6x2]



5.5 Define the following state variables: z1 = x1, z2 = ˙ x1, z3 = x2, and z4 = x2. Then thestate equations are

z1 = z2 z2 =1

m1[f(t)− k1z1 + k1z3]

z3 = z4 z4 =1

m2[k1z1 − (k1 + k2)z3]



5.6 Define the following state variables: z1 = x1, z2 = ˙ x1, z3 = x2, and z4 = x2. Then thestate equations are

z1 = z2 z2 =110

(−40z1 − 8z2 + 25z3 + 5z4)

z3 = z4 z4 =15[f(t) + 25z1 + 5z2 − 25z3 − 5z4]




x1 = x2 x2 =12[4y(t) − 4x1 − 5x2]

Then

A =

[0 1−2 −5/2

]B =

[02

]

C =[

1 0]

D = [0]



5.8

A =

[−5 30 −4

]B =

[2 00 6

]

C =

[1 30 1

]D =

[2 00 0

]



5.9 a)

A =

[−5 31 −4

]B =

[05

]

C =

[1 00 1

]D =

[00

]

b)

A =

[−5 30 −4

]B =

[4 00 5

]

C =

[1 00 0

]D =

[0 00 0

]



5.10 Define the following state variables: z1 = x1, z2 = x1, z3 = x2, and z4 = ˙ x2. Then thestate equations are

z1 = z2 z2 =110

(−40z1 − 8z2 + 25z3 + 5z4)

z3 = z4 z4 =15[f(t) + 25z1 + 5z2 − 25z3 − 5z4]

The matrices are

A =

0 1 0 0−4 −4/5 5/2 1/20 0 0 15 1 −5 −1

B =

000

1/5

C =

[1 00 1

]D =

[00

]



5.11 First way: Isolate a 1 in the denominator:

Y (s)F (s)

=6 + 7/s

1 + 3/s

ThusY (s) = −3

sY (s) + 6F (s) +

7sF (s) =

1s

[7F (s) − 3Y (s)] + 6F (s) (1)

LetX(s) =

1s

[7F (s) − 3Y (s)] (2)

Thus, from (1),Y (s) = X(s) + 6F (s) (3)

Substitute this into (2):

X(s) =1s7F (s) − 3 [X(s) + 6F (s)] =

1s

[−3X(s)− 11F (s)]

Thusx = −3x − 11f(t)

wherey = x + 6f

Thus x(0) = y(0)− 6f(0) = y(0) if we take f(0) = 0.(Continued on the next page)




Second way: Write the equation as

Y (s) = (6s + 7)F (s)s + 3

(1)

LetX(s) =

F (s)s + 3

(2)

Then(s + 3)X(s) = F (s)

which givessX(s) = −3X(s) + F (s) (3)

andx = −3x + f(t)

From (1), (2), and (3),

Y (s) = 6sX(s) + 7X(s) = 6[−3X(s) + F (s)] + 7X(s) = −11X(s) + 6F (s)

Thusy(t) = −11x(t) + 6f(t)

Sox(0) = − 1

11y(0) +

611

f(0) = − 111

y(0)

if we take f(0) = 0.Thus the state model is

x = −3x + f(t)

wherex(0) = − 1

11y(0)



5.12 First way: Divide by s2 to obtain a 1 in the denominator.

Y (s)F (s)

=2s−2 + s−1

1 + 4s−1 + 3s−2

Use the 1 in the denominator to solve for Y (s).

Y (s) =(2s−2 + s−1

)F (s) −

(4s−1 + 3s−2

)Y (s)

=1s

−4Y (s) + F (s) +

1s

[2F (s) − 3Y (s)]

(2)

This equation shows that Y (s) is the output of an integration. Thus y can be chosen as astate-variable.

X1(s) = Y (s)

The term within square brackets in (2) is the input to an integration, and can be chosen asthe second state-variable:

X2(s) =1s

[2F (s) − 3Y (s)] =1s

[2F (s) − 3X1(s)]

Then from equation (2)

X1(s) =1s

[−4X1(s) + F (s) + X2(s)]

The state equations arex1 = −4x1 + x2 + f (3)

x2 = 2f − 3x1 (4)

and the output equation is y = x1.(Continued on the next page)



Problem 5.12 continued.

The matrices of the standard form are

A =

−4 1

−3 0

B =

[12

]

C =[

1 0]

D = [0]

Because x1 = y, we have x1(0) = y(0). From (3), x2 = x1 + 4x1 − f = y + 4y − f . Thus,taking f(0) = 0, we obtain x2(0) = y(0) + 4y(0).

Second way: Express the model as

Y (s) =s + 2s + 1

F (s)s + 3

Let the first state variable beX1(s) =

F (s)s + 3

(1)

Thenx1 = f − 3x1 (2)

ThenY (s) =

s + 2s + 1

X1(s) (3)





Let the second state variable be

X2(s) =X1(s)s + 1

(4)

This givesx2 = x1 − x2 (5)

From (3)Y (s) = (s + 2)X2(s)

andy = ˙ x2 + 2x2

From (5),y = x1 − x2 + 2x2 = x1 + x2 (6)

Thus the model isx1 = f − 3x1

x2 = x1 − x2

y = x1 + x2

For the initial conditions, note that (1) and (3) give

y + y = ˙ x1 + 2x1 = f − x1

orx1 = f − y − y





Taking f(0) = 0, this gives

x1(0) = −y(0)− y(0) (7)

From (6),x2 = y − x1

which givesx2(0) = y(0)− x1(0) = 2y(0) + y(0) (8)

This initial conditions are given by (7) and (8).The state model matrices are

A =

−3 0

1 −1

B =

[10

]

C =[

1 1]

D = [0]



5.13 a) The matrices are

A =

[−5 30 −4

]B =

[05

]

C =

[1 00 1

]D =

[00

]


A = [-5, 3; 1, -4]; B = [0; 5];C = [1, 0; 0, 1]; D = [0;0];sys = ss(A,B,C,D);systf = tf(sys)

The resulting transfer functions displayed on the screen are for x1 and x2 in that order.

X1(s)U(s)

=15

s2 + 9s + 17

X2(s)U(s)

=5s + 25

s2 + 9s + 17





b) The matrices are

A =

[−5 31 −4

]B =

[4 00 5

]

C =[

1 0]

D =[

0 0]


A = [-5, 3; 1, -4]; B = [4, 0;0, 5];C = [1, 0]; D = [0, 0];sys = ss(A,B,C,D);systf = tf(sys)

The resulting transfer functions displayed on the screen are for u1 and u2 in that order.They are

X1(s)U1(s)

=4s + 16

s2 + 9s + 17

X1(s)U2(s)

=15

s2 + 9s + 17



5.14 a) The session is

sys1 = tf(1, [2, 5, 4, 7]);sys2 = ss(sys1)

The matrices displayed on the screen are

A =

−2.5 −0.5 −0.21884 0 00 4 0

B =

0.12500

C =[

0 0 0.25]

D = [0]

b) The session is

sys1 = tf(6, [3, 6, 10]);sys2 = ss(sys1)


A =

[−2 −0.41678 0

]B =

[0.50

]

C =[

0 0.5]

D = [0]



5.15 a) The session is

sys1 = tf([6, 7], [1, 3]);sys2 = ss(sys1)

The matrices, which are scalars in this case, displayed on the screen are

A = −3 B = 4

C = −2.75 D = 6

So the model isx = −3x + 4f

y = −2.75x + 6f

b) The session is

sys1 = tf([1, 2], [1, 4, 3]);sys2 = ss(sys1)


A =

[−4 −0.3758 0

]B =

[10

]

C =[

1 0.25]

D = [0]

So the model isx1 = −4x1 − 0.375x2 + f x2 = 8x1

y = x1 + 0.25x2



5.16 The matrices are

A =

[−5 31 −4

]B =

[05

]

C =[

1 0]

D = [0]

The session is

A = [-5 ,3; 1, -4]; B = [0; 5]; C = [1, 0]; D = 0;sys = ss(A,B,C,D);initial(sys, [3, 5])

b) The session is

A = [-5 ,3; 1, -4]; B = [0; 5]; C = [1, 0]; D = 0;sys = ss(A,B,C,D);step(sys)

c) The session is

A = [-5 ,3; 1, -4]; B = [0; 5]; C = [1, 0]; D = 0;sys = ss(A,B,C,D);t = [0:0.01:2];f = 3*sin(10*pi*t);lsim(sys,f,t)

This plots both the input f(t) and the output x1(t) on the same plot. To plot only theoutput, replace the last line with

y = lsim(sys, f, t); plot(t, y)



5.17 To obtain the roots, all we need is the A matrix, which is

A =

[−5 30 −4

]

The session is

A = [-5, 3; 0, -4];poly(A)ans =

1 9 20eig(A)ans =-5 -4

Thus the characteristic equation is s2 + 9s + 20 = 0 and the roots are s = −5 and s = −4.



5.18 a) Let z1 = x1, z2 = ˙ x1, z3 = x2, and z4 = ˙ x2. Then the state equations are

z1 = z2

z2 =1

m1[−k1z1 − c1z2 + k1z3 + c1z4]

z3 = z4

z4 =1

m2[k1z1 + c1z2 − k1z3 − c1z4 + k2y − k2z3]

The following script file creates the state model.

m1 = 36;m2 = 240;k1 = 1.6e+5;k2 = 1.6e+4;c1 = 98;A = [0, 1, 0, 0; -k1/m1, -c1/m1, k1/m1, c1/m1; ....

0, 0, 0, 1; k1/m2, c1/m2, -(k1+k2)/m2, -c1/m2];B = [0; 0; 0; k2/m2];C = [1, 0, 0, 0; 0, 0, 1, 0]; D = [0; 0];sys = ss(A,B,C,D);

(continued of the next page)




b), c) To compute the impulse response, add the following lines to the script file.

impulse(sys)format longpoly(A)eig(A)

We must use the format long command to see the results adequately. The characteristicpolynomial given by the poly(A) function is

s4 + 3.13106s3 + 517777s2 + 181.481s + 2.96296× 105 = 0

The roots given by the eig(A) function are

s = −1.56527± 71.5364j s = −3.07931× 10−5 ± 7.60733j

The period of the first root pair is 2π/71.5364 = 0.0878 s, whereas the dominant timeconstant of the system is 1/(3.07931×10−5) = 3.247×104 s. So there will be approximately4(3.247×104)/0.0878 = 1.479×105 oscillations before the impulse response disappears. Thusthe impulse(sys) will produce a plot on which the individual oscillations are impossibleto discern. To remedy this, you can use the syntax [y, t] = impulse(sys); plot(t,y),axis([0 1000 -10 10]). This will show several oscillations.

d) Add the following line to the script file.

tfsys = tf(sys)

The following transfer functions are displayed on the screen.

X1(s)Y (s)

=181.5s + 2.963× 105

s4 + 3.131s3 + 5178s2 + 181.5s + 2.963× 105

X2(s)Y (s)

=66.67s2 + 181.5s + 2.963× 105

s4 + 3.131s3 + 5178s2 + 181.5s + 2.963× 105



5.19 (a) The small displacement assumption implies that the suspension forces are nearlyperpendicular to the centerline of the mass m, and thus are nearly vertical. To obtain theequations of motion, assume arbitrarily that

y1 > x − L1θ

y1 > x − L1θ

y2 > x + L2θ

y2 > x + L2θ

We obtain the following moment equation about the mass center G.

IGθ = − c1(y1 − x + L1θ)L1 − k1(y1 − x + L1θ)L1

+ c2(y2 − x − L2θ)L2 + k2(y2 − x − L2θ)L2

Rearranging gives

IGθ + (c2L22 + c1L

21)θ + (k1L

21 + k2L

22)θ

− (c1L1 − c2L2)x − (k1L1 − k2L2)x= −c1L1y1 + c2L2y2 − k1L1y1 + k2L2y2 (1)

Summing forces in the vertical direction gives

mx = c1(y1 − x + L1θ) + k1(y1 − x + L1θ) + c2(y2 − x − L2θ) + k2(y2 − x − L2θ)

Rearranging gives

mx + (c1 + c2)x + (k1 + k2)x− (c1L1 − c2L2)θ − (k1L1 − k2L2)θ= c1y1 + c2y2 + k1y1 + k2y2 (2)

Equations (1) and (2) are the desired model.(Continued on the next page)




b) The transformed equations of motion have the following forms:

AΘ(s) + BX(s) = CY1(s) + DY2(s)

EΘ(s) + FX(s) = GY1(s) + HY2(s)

whereA = IGs2 + (c2L

22 + c1L

21)s + k1L

21 + k2L

22

B = (c2L2 − c1L1)s + k2L2 − k1L1

C = −c1L1s + k1L1

D = c2L2s + k2L2

E = −B

F = ms2 + (c1 + c2)s + k1 + k2

G = c1s + k1

H = c2s + k2

Because these models have input derivatives, we must be careful in deriving a statevariable model. First obtain the transfer functions using Cramer’s method, and then usethe MATLAB functions tf and ss to obtain the state model.(continued on the next page)




For Cramer’s method, let

M =

∣∣∣∣∣A B

E F

∣∣∣∣∣ = AF − BE

N1 =

∣∣∣∣∣C BG F

∣∣∣∣∣ = CF − BG N2 =

∣∣∣∣∣D BH F

∣∣∣∣∣ = DF − BH

N3 =

∣∣∣∣∣A C

E G

∣∣∣∣∣ = AG − CE N4 =

∣∣∣∣∣A D

E H

∣∣∣∣∣ = AH − DE

Then the transfer functions are found as follows

Θ(s)Y1(s)

=N1

M

Θ(s)Y2(s)

=N2

M

X(s)Y1(s)

=N3

M

X(s)Y2(s)

=N4

M

We can implement Cramer’s method in MATLAB by using the conv function. To dothis, note that A and F are second-order polynomials, and that B, C, D, E, G, and Hare first-order polynomials. This distinction is important because will need to add leadingzeros to subtract polynomials of different orders. Note that with a multi-input, multi-output(MIMO) model whose transfer functions all have the same denominator, we must store thenumerators in a cell array. The MATLAB script file is shown on the following page.(continued on the next page)




% Store the parameter values.k1 = 1100; k2 = 1525; c1 = 4; c2 = c1;L1 = 4.8; L2 = 3.6; m = 50; IG = 1000;% Create the coefficients of the transformed equations.A = [IG, c2*L2^2 + c1*L1^2, k1*L1^2 + k2*L2^2];B = [c2*L2 - c1*L1, k2*L2 - k1*L1];C = -[c1*L1, k1*L1];D = [c2*L2, k2*L2];E = -B;F = [m, c1 + c2, k1 + k2];G = [c1, k1];H = [c2, k2];% Create the Cramer determinants.M = conv(A, F) - [0, 0, conv(E, B)];N1 = conv(C, F) - [0, conv(G, B)];N2 = conv(D, F) - [0, conv(B, H)];N3 = conv(A, G) - [0, conv(C, E)];N4 = conv(A, H) - [0, conv(D, E)];% Create the transfer functions.% Store the numerators in a cell array.numerators = N1, N2; N3, N4 ;tfsys = tf(numerators, M);% Create the state variable models.statesys = ss(tfsys)





The results are (note that MATLAB returns the state model matrices as lowercasesymbols, a, b, c, d. We use these symbols to avoid confusion with the equation coefficientsA, B,C, and D defined above.)

a =

[a1 a2

a3 a4

]

where

a1 =

−0.304 −6.102 −0.1151 −2.31416 0 0 00 8 0 00 0 8 0

a2 =

0 0 0 00 0 0 00 0 0 00 0 0 0

a3 = a2 a4 = a1

b =

2 00 00 00 00 20 00 00 0

c =

[−0.0096 −0.1651 −0.006891 −0.1376 0.0072 0.1716 0.006891 0.1376

0.04 0.6879 0.02814 0.4737 0.04 0.9534 0.02943 0.683

]

d =

[0 00 0

]





c) With y1 = 0, we must redefine b as

b =

0 00 00 00 00 20 00 00 0

To do this, and to compute and plot the impulse response, add the following lines to thescript file.

[a, b, c, d] = ssdata(statesys)% Obtain unit-impulse response for y_1 = 0.b(:,1) = zeros(size(b(:,1)));statesys2 = ss(a,b,c,d);impulse(statesys2)

Note that we do not need the state model to compute the impulse response, which canalso be computed from the transfer functions N2/M and N4/M . The problem statement,however, specifically asks for the state space model.



5.20 a) Create the following function file.

function ydot = problem20(t,y)ydot = cos(t);

Then type

[ta, ya] = ode45(′problem20′,[0, 12], 6);

b) The closed-form solution is found as follows.∫ y

6dy =

∫ t

0cos t dt

Thusy(t) − 6 = sin t

To compare the two solutions, continue the session as follows:

tb = [0:0.01:12];yb = 6 + sin(tb);plot(ta,ya,tb,yb)

The plots are identical.




function ydot = problem21(t,y)ydot = 5*exp(-4*t);

Then type

[ta, ya] = ode45(′problem21′,[0, 1], 2);

b) The closed-form solution is found as follows.∫ y

2dy = 5

∫ t

0e−4t dt = −5

4e−4t

∣∣∣t

0

Thusy(t) =

134

− 54e−4t


tb = [0:0.01:1];yb = 5*exp(-4*tb);plot(ta,ya,tb,yb)





function ydot = problem22(t,y)ydot = 5*exp(4*t)-3*y;

Then type

[ta, ya] = ode45(′problem22′,[0, 1], 10);

b) The closed-form solution is found with the Laplace transform.

sY (s) − y(0) + 3Y (s) =5

s − 4

With y(0) = 10, this gives

Y (s) =10s − 35

(s + 3)(s− 4)=

65/7s + 3

+5/7

s − 4

andy(t) =

657

e−3t +57e4t


tb = [0:0.01:1];yb = (65*exp(-3*tb)+5*exp(4*tb))/7;plot(ta,ya,tb,yb)





function ydot = problem23(t,y)ydot = -sin(y);

Then type

[ta, ya] = ode45(′problem23′,[0, 4], 0.1);

b) The closed-form solution to the approximate equation y = −y is found as follows.The root is s = −1, so the solution form is

y(t) = Ae−t

Because y(0) = 0.1, we obtain A = 0.1. Thus the solution is

y(t) = 0.1e−t


tb = [0:0.01:4];yb = 0.1*exp(-tb);plot(ta,ya,tb,yb)





function ydot = problem24(t,y)if t <= 2

f = 3*t;elseif t <= 5

f = 6;else

f = -3*(t - 5) + 6;endydot = f - 2*y;

Then type

[ta, ya] = ode45(′problem24′,[0, 7], 2);



5.25 A MATLAB solution is as follows. First create the following function file.

function vdot = problem25(t,v)vdot = (8000-20*v-0.05*v.^2)/50;

Then type

[t,v] = ode45(′problem25′,[0, 5], 0);plot(t,v),xlabel(′t (sec)′),ylabel(′v (ft/sec)′)

The final time used (5 sec) was an estimate. The solution actually reaches steady-state inabout 4.6 sec.



5.26 Put the model into state variable form. Let x1 = y and x2 = y. Then x1 = x2 andx2 = 9.81 − 180x1 − 340x3

1. A MATLAB solution is as follows. First create the followingfunction file.

function xdot = problem26(t,x)xdot = [x(2);9.81-180*x(1)-340*x(1).^3;

Then type

[ta,xa] = ode45(′problem26′,[0, 2], [0.06, 0]);[tb,xb] = ode45(′eqn′,[0, 2], [0.1, 0];plot(ta,xa(:,1),tb,xb(:,1)),xlabel(′t (s)′),ylabel(′y(m)′)



5.27 The state variable form, with x1 = y and x2 = ˙ y, is

x1 = x2

x2 = −x1 + b(1− x21)x2

Create the following function file:

function xdot = vander(t,x)global bxdot(1) = x(2);xdot(2) = b*(1-x(1)^2)*x(2) - x(1);xdot = [xdot(1);xdot(2)];

Then use the following script file to solve the equation and plot x1 = y. For Case 1:

global bb = 0.1;[t, x] = ode45(′vander′, [0, 25], [1, 1]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y(t)′)

For Case 2:

global bb = 0.1;[t, x] = ode45(′vander′, [0, 25], [3, 3]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y(t)′)

For Case 3:

global bb = 3;[t, x] = ode45(′vander′, [0, 25], [1, 1]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y(t)′)





Note on Passing Additional Parameters to an ODE Function. Rather than using theglobal statement, the extended syntax of the ODE solvers using the options argument letsyou pass any input parameters that follow the options argument to the ODE function andany function you specify in options. This syntax uses function handles. For example, youcan pass the parameter b directly to the van der Pol function, as follows:

function xdot = vander(t,x,b)xdot(1) = x(2);xdot(2) = b*(1-x(1)^2)*x(2) - x(1);xdot = [xdot(1);xdot(2)];

Then pass the parameter b to the function vander by specifying it after the options argumentin the call to the solver, using options = [ ] as a placeholder, as follows:

[t,x] = ode45(@vander, [0, 25], [1, 1], [ ],b)

The function handle @vander calls vander(t,x,b).



5.28 Put the model in state variable form as follows. Let x1 = y and x2 = ˙ y.

x1 = x2

x2 = −x1 + b(1− x21)x2

Create the following function file.

function xdot = vander(t,x);b = 1000;xdot = [x(2); -x(1) + b*(1-x(1)^2)*x(2)];

The following script file uses the ode23s function.

[t, x] = ode23s(′vander′, [0, 3000], [2, 0]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y′)

The ode23 function successfully solves the equation, but the functions ode45 and ode23fail to converge to a solution for this problem. They can, however, obtain a solution for“nonstiff” values of b, such as b = 1.



5.29 The state variable form, with x1 = θ and x2 = θ, is

x1 = x2

x2 =a(t) cos θ − g sin θ

L

Express the acceleration a(t) as the linear function a(t) = mt + b. Thus for cases (a)and (b), m = 0 and b = 5. For case (c), m = 0.5 and b = 0.

Create the following function file:

function xdot = accbase(t,x)global m bL = 1; g = 9.81;xdot(1) = x(2);xdot(2) = ((m*t + b)*cos(x(1)) -g*sin(x(1)))/L;xdot = [xdot(1);xdot(2)];

To solve the equation and plot the solution, create the following script file.

global m bm = 0; b = 5;[t, x] = ode45(′accbase′, [0, 10], [0.5, 0]);plot(t,x(:,1)),xlabel(′t (seconds)′),ylabel(′theta(t) (radians)′)

For case (b), change the third line to

[t, x] = ode45(′accbase′, [0, 10], [3, 0]);

For case (c), change the second line to

m = 0.5; b = 0;





Note on Passing Additional Parameters to an ODE Function. Rather than using theglobal statement, the extended syntax of the ODE solvers using the options argument letsyou pass any input parameters that follow the options argument to the ODE function andany function you specify in options. This syntax uses function handles. For example, youcan pass the parameters L, g, m, and b directly to the accbase function, as follows:

function xdot = accbase(t,x,L,g,m,b)xdot(1) = x(2);xdot(2) = ((m*t + b)*cos(x(1)) -g*sin(x(1)))/L;xdot = [xdot(1);xdot(2)];

Then pass the parameters L, g, m, and b to the function accbase by specifying them afterthe options argument in the call to the solver, using options = [ ] as a placeholder, asfollows:

[t,x] = ode45(@accbase, [0, 10], [3, 0], [ ], L, g, m, b)

The function handle @accbase calls accbase(t,x,L,g,m,b).



5.30 The cosine function is obtained with the Sine Wave Function block by using a phaseshift of π/4 (pi/2). Make sure the Sine Wave blocks are set to use simulation time. Oth-erwise a Clock block will be required. Set the initial condition of the Integrator block to 1and the initial condition of the Integrator1 block to 4.




5.31 The time variable tout is automatically placed in the workspace by selecting DataImport/Export under the Configuration Parameters menu. Thus a Clock block is notneeded. You can enter the initial condition x(0) as 100*cos(30*pi/180), with a similarexpression for y(0).

To plot the trajectory y versus x, type

plot(simout(:,1),simout(:,2)),xlabel(′x′),ylabel(′y′)




5.32 The model is shown in the following figure. To plot the percent relative error in theseries solution, type

t = tout;x = (t.^3)/3-t.^2+3*t-3+3*exp(-t);plot(t,100*(simout-x)/simout),xlabel(′t′),ylabel(′Percent Error′),grid

The plot shows that the percent relative error in the series solution is less than 1% fort < 0.75. It is about 6% at t = 1.




5.33 The model is shown in the following figure. Set the Step time and Final value for theStep block to 0 and 4, respectively. Set the Step time and Final value for the Step1 blockto 2 and 4, respectively. Set the Initial condition of the Integrator to 2. You can plot theresults by typing

plot(tout,simout),xlabel(′t′),ylabel(′x′)




5.34 The model is shown in the following figure. Set the Step time and Final value for theStep block to 0 and 5, respectively. Set the Step time and Final value for the Step1 blockto 2 and 5, respectively. Set the numerator of the Transfer Function block to [1] and thedenominator to [2, 12, 10]. You can plot the results by typing





5.35 The Simulink model is shown in the figure. Set the Amplitude of the Sine Wave blockto 5 and the Frequency to 0.8. Make sure the Initial conditions of the Integrator blocks areset to 0. The Save format of the To Workspace block has been selected as Array, and thesimulation time tout has been saved to the workspace (Use the Workspace I/O tab on theSimulation Parameters menu under the Simulation menu to do this). Then the output canbe plotted in MATLAB by typing

plot(tout,simout)




5.36 The model is shown in the following figure. Set the Sine Wave block to use simulationtime. Set the Amplitude to 10 and the Frequency to 3. In the Saturation block set thelimits to ±8. Set the Initial condition of the Integrator to 2. You can plot the results bytyping





5.37 The model is shown in the following figure. Set the Sine Wave block to use simulationtime. Set the Amplitude to 10 and the Frequency to 4. In the Saturation block set thelimits to ±5. Set the Initial condition of the Integrator to 0. You can plot the results bytyping





5.38 The model is shown in the following figure. Set the Sine Wave block to use simulationtime. Set the Amplitude to 2 and the Frequency to 4. In the Math Function block selectthe square function. Set the Initial condition of the Integrator to 1. You can plot the resultsby typing


Using a stop time of 3 results in an error message that indicates that Simulink is havingtrouble finding a small enough step size to handle the rapidly decreasing solution. Byexperimenting with the stop time, we find that, to two decimal places, a stop time of 1.38will not generate an error.

This model is unstable, and the output x → −∞ if the time span is long enough. Wecan see this by writing the equation as

x = 2 sin 4t − 10x2

Once x drops below√

2/10, x remains negative and thus x continues to decrease throughnegative values.




5.39 The model is shown in the following figure. Set the Start and End of the Dead Zoneto −0.5 and 0.5 respectively. Set the Sine Wave block to use simulation time. Set theAmplitude to 2 and the Frequency to 4. In the Math Function block select the squarefunction. Set the Initial condition of the Integrator to 1. You can plot the results by typing


Using a stop time of 3 results in an error message that indicates that Simulink is havingtrouble finding a small enough step size to handle the rapidly decreasing solution. Byexperimenting with the stop time, we find that, to two decimal places, a stop time of 1.49will not generate an error.

This model is unstable, and the output x → −∞ if the time span is long enough. Wecan see this by writing the equation as (not including the effect of the dead zone)

x = 2 sin 4t − 10x2

Once x drops below√

2/10, x remains negative and thus x continues to decrease throughnegative values.




5.40 The model is shown in the following figure. Set the Initial condition of Integrator to 0.5and the Initial condition of Integrator1 to 0. In the Fcn block type 900*u(1)+1700*u(1)^3for the expression. You can plot the results by typing





5.41 The model is shown in the following figure. Set the Initial condition of Integra-tor to 30*pi/180 and the Initial condition of Integrator1 to 0. In the Fcn block type-17500*cos(u(1))+626000*sin(1.33+u(1))/sqrt(2020+1650*cos(1.33+u(1)) for the ex-pression. You can plot the results by typing





5.42 The model is shown in the following figure. Set the Step time and Final value for theStep block to 0 and f1 − mg cos(30π/180) = 5 − 3(9.8) cos(30π/180), respectively. Set thegain in the top gain block to µmg sin(30π/180) = 0.5(2)(9.8) sin(30π/180). Set the Initialcondition of the Integrator to 3. The block comes to rest in about 0.35 s. You can plot theresults by typing





5.43 a) The model is shown in the following figure. Set the Step time and Final value forthe Step block to 0 and 4, respectively. Set the Step time and Final value for the Step1block to 2 and 4, respectively. Set the Initial condition of the Integrator to 2. You can plotthe results by typing


b) Delete the Step block and replace it with a Sine Wave block. Set the Amplitude to10 and the frequency to 2.5.





to accompany


by

William J. Palm III


Solutions to Problems in Chapter Six


CD1

Highlight

6.1 See the following figure. From part (a) of the figure,

R1 = R +1

1R + 1

R

=3R

2

From part (b) of the figure,1

R2=

1R

+1

R1

which gives R2 = 3R/5.From part (c) of the figure,

Re = R + R2 =8R

5Thus

vs =8R

5i




6.2 From the figure on the following page,

1R1

=15

+12

=710

So R1 = 10/7. Also, R2 = 8 + 2 = 10. From conservation of charge,

i1 + i2 + i3 = i (1)

Kirchhoff’s voltage law applied to the three loops gives

vs = 1i + 4i3 (2)

4i3 = R1(i1 + i2) + 3i2 (3)

3i2 = R2i1 (4)

Also,v1 = 2i1 (5)

Use (1) and (2) to eliminate i:

i1 + i2 + 5i3 = vs (6)

Collect terms, substitute the values of R1 and R2, and rearrange (3) and (4):

107

i1 +317

i2 − 4i3 = 0 (7)

10i1 − 3i2 + 0i3 = 0 (8)

Equations (6), (7), and (8) have the solution i1 = 0.1193vs. (These three equations can besolved with MATLAB by setting vs = 1.) From (5) we have the answer: v1 = 0.2386vs.



6.3 Let i1 be the current through R1 and R4, and i2 be the current through R2 and R3.Then vs = (R1 + R4)i1 = (R2 + R3)i2 and

v1 = R4i1 − R3i2 =(

R4

R1 + R4− R3

R2 + R3

)vs

This gives

v1 =R2R4 − R1R3

(R1 + R4)(R2 + R3)vs

If the changes in R3 and R4 are small enough so that (R1 + R4)(R2 + R3) ≈ constant = K,then

v1 =R2vs

KR4 −

R1vs

KR3 = αR4 − βR3

where α and β are approximately constant.



6.4 The two resistors in series are equivalent to one 70 kΩ resistor, which is in parallel withthe 80 kΩ resistor. Thus the total equivalent resistance R is found from

1R

=180

+170

which gives R = 112/3 kΩ.Thus

i =9

112/3× 103= 2.411× 10−4 A

The power P is computed from

P = iv = (2.411× 10−4)9 = 2.17× 10−3 W



6.5 The model isv1 =

1C

∫ (is −

1R

v1

)dt

Differentiate both sides with respect to t to obtain a differential equation.

RCdv1

dt+ v1 = Ris



6.6 a) Kirchhoff’s voltage law gives

vs = R1i + vo + R2i = (R1 + R2)i + vo (1)

For the capacitor:

vo =1C

∫i dt

which gives

Cdvo

dt= i (2)

Solve equation (1) for i and substitute into equation (2) to obtain the answer:

Cdvo

dt=

vs − vo

R1 + R2

which, in standard form, is

(R1 + R2)Cdvo

dt+ vo = vs



6.7 a) From Kirchhoff’s voltage law,

vs = R2i2 = R1i1 + vo

which gives

i1 =vs − vo

R1(1)

For the capacitor,

vo =1C

∫i1 dt (2)

Differentiate (2) with respect to time, and substitute for i1 from (1):

Cdvo

dt= i1 =

vs − vo

R1

Thus the model isR1C

dvo

dt+ vo = vs (3)

Note the we did not need to find i2 or i3, and that the model is independent of R2

because there is no element between vs and R2, so the voltage input to the right-most loopis vs.

b) From (3) with vs = 0, we obtain the free response:

vo(t) = vo(0)e−t/R1C

If vs(t) = V us(t) and if vo(0) = 0, the response is

vo(t) = V(1− e−t/R1C

)

This is the forced response. So the total response is

vo(t) = vo(0)e−t/R1C + V(1 − e−t/R1C

)



6.8 a) Kirchhoff’s voltage law gives

vs = R1i1 + vo (1)

For the capacitor:

vo = R2i2 =1C

∫i3 dt (2)

From conservation of charge,i1 = i2 + i3

Substitute for the currents to obtain:

vs − vo

R1=

vo

R2+ C

dvo

dt

This gives

R1R2Cdvo

dt+ (R1 + R2)vo = R2vs

b) The free response isvo(t) = vo(0)e−t/τ

whereτ =

R1R2C

R1 + R2

The forced response is

vo(t) =R2V

R1 + R2

(1 − e−t/τ

)



6.9 a) The model of the circuit is

Ri + Ldi

dt= vs

b) If vs(t) = V us(t) and i(0) = 0, then

I(s) =Vs(s)

Ls + R=

V

s(Ls + R)=

V

R− V

R

1s + R/L

Thusi(t) =

V

R

(1 − e−Rt/L

)



6.10 For the circuit,

Ldi

dt+ Ri = vi(t)

or5di

dt+ 10i = vi(t)

Since the applied voltage is 12 volts, the impulse strength is 12(0.3) = 3.6. So we modelvi(t) as 3.6δ(t), and thus

(5s + 10)I(s) = 3.6

which gives

I(s) =3.6

5s + 10=

0.72s + 2

andi(t) = 0.72e−2t



6.11 Define i to be the current passing through R and L. From Kirchhoff’s voltage law,

vs = Ri + v0

Solve for i:i =

vs − v0

R

For the inductor:v0 = L

di

dt

Substitute for i to obtain

v0 = Ld

dt

(vs − v0

R

)=

L

R

(dvs

dt− dv0

dt

)

Move the output terms (the v0 terms) to the left side, putting the highest-order derivativefirst, and move the input terms to the right side, to obtain the answer in a somewhatstandard form:

L

R

dv0

dt+ v0 =

L

R

dvs

dt



6.12 Define i to be the current passing through R1, C, and R2. Then from Kirchhoff’svoltage law:

vs = R1i + v0 (1)

From the definition of v0,

v0 =1C

∫i dt + R2i (2)

Differentiate this equation to obtain:

dv0

dt=

1C

i + R2di

dt(3)

Solve equation (1) for i:

i =vs − v0

R1(4)

and differentiate to obtain:

di

dt=

1R1

(dvs

dt− dv0

dt

)(5)

Substitute equations (4) and (5) into equation (3) to obtain

dv0

dt=

1R1C

(vs − v0) +R2

R1

(dvs

dt− dv0

dt

)

Putting this into standard form gives the answer:

(R1 + R2)Cdv0

dt+ v0 = R2C

dvs

dt+ vs



6.13 From the voltage law,

vs = Ri + Ldi

dt+

1C

∫i dt

Differentiate this with respect to time:

LCd2i

dt2+ RC

di

dt+ i = C

dvs

dt



6.14is = i1 + i2

vo = Ri1

Thusv1 =

1C

∫i2 dt =

1C

∫(is − i1) dt =

1C

∫(is −

vo

R) dt

Also,

Ldi1dt

= v1 − vo =1C

∫(is −

vo

R) dt − vo

orL

R

dvo

dt=

1C

∫(is −

vo

R) dt− vo

ThusL

R

d2vo

dt2=

1C

(is −

vo

R

)− dvo

dt

Rearranging gives

LCd2vo

dt2+ RC

dvo

dt+ vo = Ris



6.15v1 = L

di1dt

+ Ri3 (1)

Ri3 =1C

∫i2 dt + v2

ThusRC

di3dt

= i2 + Cdv2

dt(2)

i1 = i2 + i3 (3)

Transform (1) through (3) to obtain

LsI1(s) + RI3(s) = V1(s)

I2(s)− RCsI3(s) = −CsV2(s)

I1(s) − I2(s) − I3(s) = 0

These have the following solution

I1(s) =(RCs + 1)V1(s) − RCsV2(s)

D(s)

I2(s) =RCsV1(s) − (RCs + LCs2)V2(s)

D(s)

I3(s) =V1(s) + LCs2V2(s)

D(s)

whereD(s) = LRCs2 + Ls + R

The differential equation model is

LRCd2i1dt2

+ Ldi1dt

+ Ri1 = RCdv1

dt+ v1 − RC

dv2

dt

LRCd2i2dt2

+ Ldi2dt

+ Ri2 = RCdv1

dt− RC

dv2

dt− LC

d2v2

dt2

LRCd2i3dt2

+ Ldi3dt

+ Ri3 = v1 + LCd2v2

dt2



6.16v1 = R1i1 + L1

di1dt

+ v3 (1)

v3 =1C

∫i3 dt

which gives

i3 = Cdv3

dt

v3 = L2di2dt

+ v2 (2)

i1 = i2 + i3 = i2 + Cdv3

dt(3)


(R + L1s)I1(s) + V3(s) = V1(s)

L2sI2(s) − V3(s) = −V2(s)

I1(s) − I2(s) − CsV3(s) = 0

These have the following solution

I1(s) =(L2Cs2 + 1)V1(s)− V2(s)

D(s)

I2(s) =V1(s) − (L1Cs2 + RCs + 1)V2(s)

D(s)

V3(s) =L2sV1(s) + (L1s + R)V2(s)

D(s)

whereD(s) = L1L2Cs3 + RL2Cs2 + (L1 + L2)s + R





The differential equation model is

L1L2Cd3i1dt3

+ RL2Cd2i1dt2

+ (L1 + L2)di1dt

+ Ri1 = L2Cd2v1

dt2+ v1 − v2

L1L2Cd3i2dt3

+ RL2Cd2i2dt2

+ (L1 + L2)di2dt

+ Ri2 = v1 − L1Cd2v2

dt2− RC

dv2

dt− v2

L1L2Cd3v3

dt3+ RL2C

d2v3

dt2+ (L1 + L2)

dv3

dt+ Rv3 = L2

dv1

dt+ L1

dv2

dt+ Rv2



6.17 a) Choose v1 and i1 as the state variables because they describe the energy storage inthe circuit. The basic circuit equations are

v1 =1C

∫(is − i1) dt

v1 = Ldi1dt

+ vo vo = Ri1

These give the state equations

Cdv1

dt= is − i1 L

di1dt

= v1 − Ri1



6.18 Choose i1 and vc as the state variables because they describe the energy storage inthe circuit (vc is the voltage drop across the capacitor). The basic circuit equations are

v1 = L1di1dt

+ Ri3 Ri3 = vc + v2

vc =1C

∫i2 dt i1 = i2 + i3


Ldi1dt

= v1 − v2 − vc Cdvc

dt= i1 −

1R

vc −1R

v2



6.19 a) Choose i1, i2, and v3 as the state variables because they describe the energy storagein the circuit. The basic circuit equations are

v1 = R1i1 + L1di1dt

+ v3 v3 =1C

∫i3 dt

v3 = L2di2dt

+ v2 i1 = i2 + i3


L1di1dt

= v1 − v3 − Ri1 L2di2dt

= v3 − v2

Cdv3

dt= i1 − i2



6.20 It can be seen that R2 and C are in parallel. Thus, their equivalent impedance Z(s)is found from

1Z(s)

=1

1/Cs+

1R2

orZ(s) =

R2

R2Cs + 1

It can be seen that Z and R1 are in series. Thus,

Vs(s)I(s)

= Z(s) + R1

andVo(s) = R1I(s)

Eliminating I(s) from the last two relations yields the desired transfer function.

Vs(s) =Vo(s)R1

[Z(s) + R1]

or

T (s) =Vo(s)Vs(s)

=R1

Z(s) + R1

=R1R2Cs + R1

R1R2Cs + R2 + R1

The network is seen to be a first-order system with numerator dynamics.



6.21 The impedance of R1 and C is their parallel combination:

11

R1+ 1

1/Cs

=R1

R1Cs + 1

The series combination of this impedance and R2 is:

R2 +R1

R1Cs + 1=

R1R2Cs + R1 + R2

R1Cs + 1=

Vs(s)I1(s)

and the required transfer function is

I1(s)Vs(s)

=R1Cs + 1

R1R2Cs + R1 + R2



6.22 Kirchhoff’s voltage law givesvs = Ri + vo

Solve this for i:i =

vs − vo

R(1)

For the capacitor:

vo =1C

∫i2 dt

which gives

Cdvo

dt= i2 (2)

For the inductor:vo = L

di1dt

Solve this for i1:

i1 =1L

∫vo dt (3)

From conservation of charge:i = i1 + i2 (4)

Substitute i, i1, and i2 from equations (1), (2), and (3) into equation (4) to obtain:

vs − vo

R=

1L

∫vo dt + C

dvo

dt

Differentiate both sides to get

1R

(vs − vo) =vo

L+ Cvo

Rearrange in standard form to get the answer:

RLCvo + Lvo + Rvo = Lvs



6.23 From conservation of charge:

is = i1 + i2 + i3 (1)

R, L, and C have the same voltage drop vo across them, so

vo = Ldi2dt

= Ri1 =1C

∫i3 dt

Integrate the first equation for i2 to obtain:

i2 =1L

∫vo dt (2)

Solve the second equation for i1:i1 =

vo

R(3)

Solve the third equation for i3:

i3 = Cdvo

dt(4)

Substitute equations (2), (3), and (4) into (1) to obtain:

is = Cdvo

dt+

vo

R+

1L

∫vo dt

Differentiate both sides to obtain

disdt

= Cd2vo

dt2+

1R

dvo

dt+

1L

vo

If desired, this can be “cleaned up” somewhat by multiplying both sides by RL to obtain

RLCd2vo

dt2+ L

dvo

dt+ Rvo = RL

disdt



6.24 Let Z1(s) = Vs(s)/I(s) be the series impedance of L, C1, C2, and R.

Z1(s) = Ls +1

C1s+

1C2s

+ R =LC1C2s

2 + RC1C2s + C1 + C2

C1C2s

Also,

Vo(s) =1

C2sI(s) =

1C2s

Vs(s)Z1(s)

=C1s

LC1C2s3 + RC1C2s2 + (C1 + C2)sVs(s)

and

LC1C2d3vo

dt3+ RC1C2

d2vo

dt2+ (C1 + C2)

dvo

dt= C1

dvs

dt



6.25 Let i1 be the current through R1; i2 be the current through the left-side C; i3 be thecurrent through R2; i4 be the current through the right-side C; and i5 be the current to theoutput vo. Then from Kirchhoff’s voltage law,

Vs(s) =1

CsI2(s) + R2I2(s) R2I3(s) =

1Cs

I4(s) + Vo(s)

R1I1(s) =1

CsI2(s) +

1Cs

I4(s)

From conservation of charge,

I2(s) = I3(s) + I4(s) I5(s) = I1(s) + I4(s) = 0

Thus I4(s) = −I1(s). Use these equations to eliminate the current variables, and find theratio Vo(s)/Vs(s). The answer is

Vo(s)Vs(s)

=R1R2C

2s2 + 2R2Cs + 1R1R2C2s2 + (R1 + 2R2)Cs + 1



6.26 The impedance of the R1C1 combination is

Z1(s) =R1

R1C1s + 1

For the R2C2 combination:

Z2(s) = R2 +1

C2s=

R2C2s + 1C2s

ThusVs(s) = Z1(s)I(s) + Z2(s)I(s) = [Z1(s) + Z2(s)] I(s)

andVo(s) = Z2(s)I(s) =

Z2(s)Z1(s) + Z2(s)

Vs(s)

ThusVo(s)Vs(s)

=Z2(s)

Z1(s) + Z2(s)=

(R1C1s + 1)(R2C2s + 1)R1C2s + (R1C1s + 1)(R2C2s + 1)



6.27 The transfer function isVo(s)Vi(s)

= −Tf(s)Ti(s)

where Ti(s) = R1 and

Tf (s) = R2 +1

Cs=

R2Cs + 1Cs

ThusVo(s)Vi(s)

= −R2Cs + 1R1Cs

andR1C

dvo

dt= −R2C

dvi

dt− vi




= −Tf(s)Ti(s)

where Tf (s) = R3 andTi(s) = Z(s) + R2

1Z(s)

=1

R1+

11

Cs

ThusZ(s) =

R1

R1Cs + 1

andTi(s) =

R1

R1Cs + 1+ R2 =

R1R2Cs + R1 + R2

R1Cs + 1Thus

Vo(s)Vi(s)

= − R3(R1Cs + 1)R1R2Cs + R1 + R2

andR1R2C

dvo

dt+ (R1 + R2)vo = −R1R3C

dvi

dt− vi




= −Tf(s)Ti(s)

whereTf(s) = R2 +

1C2s

=R2C2s + 1

C2s

andTi(s) = Z(s) + R3

1Z(s)

=1R1

+11

C1s

ThusZ(s) =

R1

R1C1s + 1

andTi(s) =

R1

R1C1s + 1+ R3 =

R3R1C1s + R3 + R1

R1C1s + 1Thus

Vo(s)Vi(s)

= − (R2C2s + 1)(R1C1s + 1)C2s(R3R1C1s + R3 + R1)

and

R3R1C1C2d2vo

dt2+ (R3 + R1)C2

dvo

dt= −R2R1C1C2

d2vi

dt2− (R1C1 + R2C2)

dvi

dt− vi



6.30 Let Z1(s) be the forward impedance, and Z2(s) be the feedback impedance. Then

1Z1(s)

=1

R1+

11

C1s

Z1(s) =R1

R1C1s + 1

Similarly,

Z2(s) =R2

R2C2s + 1and

Vo(s)Vi(s)

= −Z2(s)Z1(s)

= −R2C2s + 1R1C1s + 1

R1

R2



6.31 Let Z1(s) be the impedance of the series R1C1 connection, Z2(s) the forward impedance,Z3(s) the impedance of the series R3C2 connection, and Z4(s) the feedback impedance.Then

Z1(s) = R1 +1

C1s=

R1C1s + 1C1s

1Z2(s)

=1

R2+

1Z1(s)

=(R1C1 + R2C1)s + 1

R1C1s + 1

Z3(s) =R3C2s + 1

C2s

Z4(s) =R3C2s + 1

(R3C2 + R4C2)s + 1

ThusVo(s)Vi(s)

= −Z4(s)Z2(s)

= − R3C2s + 1(R3C2 + R4C2)s + 1

(R1C1 + R2C1)s + 1R1C1s + 1



6.32 a) From equations (3) and (4) of Example 6.5.1,

(Is2 + cs + KT )Θ(s) = nBLrI(s)

(Ls + R)I(s) + nBLrsΘ(s) = Vi(s)

Let A = nBLr. The solution for Θ(s) gives

Θ(s)Vi(s)

=A

(Is2 + cs + KT )(Ls + R) + A2=

A

LIs3 + (RI + Lc)s2 + (LKT + Rc)s + RKT + A2

b) Assuming all the parameters are positive, the system is stable if and only if

(RI + Lc)(LKT + Rc) > LI(RKT + A2)

If the system is stable, the final value theorem can be applied. It gives

θss = lims→0

sA

LIs3 + (RI + Lc)s2 + (LKT + Rc)s + RKT + A2

Vi

s=

A

RKT + A2Vi



6.33 a) With vf = 0, if = 0, and

IsΩ(s) = −cΩ(s)− TL(s)

ThusΩ(s)TL(s)

= − 1Is + c

b)

Id2θ

dt2= KT if − c

dθ

dt− TL

Thus (Is2 + cs

)Θ(s) = KT If (s) − TL(s)

Also,Vf(s) = (Lfs + Rf)If (s)

and therefore (Is2 + cs

)Θ(s) = KT

Vf(s)(Lfs + Rf)

− TL(s)

This gives

Θ(s)Vf(s)

=KT

(Lfs + Rf)(Is2 + cs)=

KT

s[ILfs2 + (RfI + Lfc)s + Rfc]

andΘ(s)TL(s)

= − 1s(Is + c)



6.34 The equivalent inertia and damping at the motor shaft are

Ie = Im +IL

N2ce =

cL

N2+ cm

The transfer functions are

Ωf(s)Vf(s)

=KT /N

(Lfs + Rf)(Ies + ce)

Ωf(s)TL(s)

=1/N2

Ies + ce



6.35 Summing moments on I1 and I2 gives

I1d2θ1

dt2= T − kT (θ1 − θ2) = KT if − kT (θ1 − θ2)

I2d2θ2

dt2= kT (θ1 − θ2)− cθ2

For the field circuit,

vf = Lfdifdt

+ Rif

Transforming these equations and solving for Θ2(s) gives

Θ2(s) =kTKT

s2(Lfs + Rf)[I1I2s2 + cI1s + kT (I1 + I2)]

The differential equation is

I1I2Lfd5θ2

dt5+(cI1I2Lf+Rf )

d4θ2

dt4+[LfkT (I1+I2)+RfcI1]

d3θ2

dt3+RfkT (I1+I2)

d2θ2

dt2= kT KTvf



6.36 There are two circuits and one inertia in this system, and we must write an equationfor each. For the generator circuit, Kirchhoff’s voltage law gives

vf = Rf if + Lfdifdt

For the motor circuit, Kirchhoff’s voltage law gives

va = Raia + Ladiadt

+ Kbω

where va = Kf if . For the inertia I , Newton’s law gives:

Idω

dt= T − cω − TL

where T = KT ia. Substitute for va and T , and rearrange to obtain:

Lfdifdt

+ Rf if = vf

Idω

dt+ cω = KT ia − TL

Ladiadt

+ Raia = Kf if − Kbω



6.37 Equations (6.6.5) and (6.6.6) give

ia =5 × 10−4Va + 0.2TL

(0.8)5× 10−4 + (0.2)2=

7.5 + 200TL

40.4(1)

ω =0.2Va − 0.8TL

4.04× 10−2=

300− 80TL

4.04(2)

Setting TL = 0 in (2) gives the no-load speed ω = 300/4.04 = 74.26 rad/s or 709 rpm. From(1) we obtain the no-load current: ia = 7.5/40.4 = 0.186 A.

Setting ω = 0 in (2) gives the stall torque TL = 0.2(15)/0.8 = 3.75 N·m. From (1) weobtain the stall current: ia = [7.5 + 200(15/4)]/4.04 = 187.5 A.



6.38 (a) The characteristic equation is

LaIs2 + (Lac + RaI)s + Rac + KbKT = 0

The roots are

s =−(Lac + RaI)±

√R2

aI2 − 2RaILac + L2

ac2 − 4LaIKbKT

2LaI

b) For c = 0 the roots are −219 and −47.5. The speed and the current will not oscillate.It will take about 4(1/47.5) = 0.08 second for the speed and the current to become constant.

For c = 0.01 the roots are −196 ± 73.5i. The speed and current will oscillate with afrequency of 73.5 rad/s and a period of 0.085 s. It will take about 4(1/196) = 0.02 secondfor the speed and current to become constant. Because this time is less than the period, wewill not see any oscillations in the plots.

For c = 0.1 the roots are −1240 and −277. The speed and current will not oscillate. Itwill take about 4(1/277) = 0.014 second for them to become constant.



6.39 Use the transfer functions given by (6.6.1) through (6.6.4). All four transfer functionshave the same denominator, which is

D(s) = 20× 10−7s2 + 4.02× 10−4s + 4× 10−4

Current Response to Command Input Assuming va is a unit step input, we have from(6.6.1):

Ia(s) =5 × 10−4(s + 1)

sD(s)=

250(s + 1)s(s2 + 201s + 2020)

(1)

The characteristic roots are s = −100.5± 100.5j. Thus the denominator can be expressedas (s + 100.5)2 + (100.5)2. Expanding Ia(s) in a partial fraction expansion gives

Ia(s) =C1

s+

100.5C2

(s + 100.5)2 + (100.5)2+

(s + 100.5)C3

(s + 100.5)2 + (100.5)2(2)

This gives a solution of the form:

ia(t) = C1 + C2e−100.5t sin 100.5t + C3e

−100.5t cos 100.5t

Reducing (2) to a single fraction using the least common denominator s[(s + 100.5)2 +(100.5)2], we obtain

Ia(s) =C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s

s[(s + 100.5)2 + (100.5)2](3)

Comparing the numerators of (1) and (3), we obtain

C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s = 250(s + 1)

or(C1 + C3)s2 + (201C1 + 100.5C2 + 100.5C3)s + 20, 200C1 = 250s + 250

Comparing like powers of s gives

C1 = −C3 = 0.01238 C2 = 2.47518

Thus, after multiplying by 10, the magnitude of the step input, we have

ia(t) = 0.1238 + 24.7518e−100.5t sin 100.5t− 0.1238e−100.5t cos 100.5t





Speed Response to Command Input Assuming va is a unit step input, we have from(6.6.3):

Ω(s) =0.2

sD(s)=

105

s(s2 + 201s + 2020)(4)

Expanding Ω(s) in a partial fraction expansion gives

Ω(s) =C1

s+

100.5C2

(s + 100.5)2 + (100.5)2+

(s + 100.5)C3

(s + 100.5)2 + (100.5)2(5)


ω(t) = C1 + C2e−100.5t sin 100.5t + C3e

−100.5t cos 100.5t


Ω(s) =C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s

s[(s + 100.5)2 + (100.5)2](6)


C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s = 105

or(C1 + C3)s2 + (201C1 + 100.5C2 + 100.5C3)s + 20, 200C1 = 105


C1 = −C3 = 4.9505 C2 = −C1 = −4.9505 (7)

Thus, after multiplying by 10, the magnitude of the step input, we have

ω(t) = 49.505(1 − e−100.5t sin 100.5t− e−100.5t cos 100.5t

)





Current Response to Disturbance Input Comparing (6.6.2) with (6.6.3), and notingthat Kb = KT , we see that

Ia(s)TL(s)

=Kb

KT

Ω(s)Va(s)

=Ω(s)Va(s)

Thus the response of ia to a unit-step disturbance TL will be the same as the response of ωto a unit-step voltage va. Thus we can use the coefficients given in (7) and multiply themby 0.2, the magnitude of TL, to obtain

ıa(t) = 0.9901(1 − e−100.5t sin 100.5t− e−100.5t cos 100.5t

)





Speed Response to Disturbance Input To obtain the response of ω to TL, use (6.6.4).Assuming TL is a unit step input, we have from (6.6.4):

Ω(s) =4 × 10−3s + 0.8

sD(s)=

5 × 10−4s + 0.1s(s2 + 201s + 2020)

(8)

Expanding Ω(s) in a partial fraction expansion gives

Ω(s) =C1

s+

100.5C2

(s + 100.5)2 + (100.5)2+

(s + 100.5)C3

(s + 100.5)2 + (100.5)2(9)


ω(t) = C1 + C2e−100.5t sin 100.5t + C3e

−100.5t cos 100.5t


Ω(s) =C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s

s[(s + 100.5)2 + (100.5)2](10)


C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s = 5× 10−4s + 0.1

or(C1 + C3)s2 + (201C1 + 100.5C2 + 100.5C3)s + 20, 200C1 = 5 × 10−4s + 0.1


C1 = −C3 = 4.9505× 10−6 C2 = −4.187× 10−7

Thus, after multiplying by 0.2, the magnitude of the step input, we have

ω(t) = 9.901× 10−7 − 8.37× 10−8e−100.5t sin 100.5t− 9.901× 10−7e−100.5t cos 100.5t



6.40 Given Va = 20 and

Istall = 25 ino load = 0.6 ωno load =240060

2π = 80π rad/s

ThusRa =

Va

istall=

2025

= 0.8 Ω

From (6.6.6) and (6.6.6) with TL = 0 and Kb = KT ,

ino load = 0.6 =20c

0.8c + K2T

ωno load = 80π =20KT

0.8c + K2T

These two equations have the solution KT = Kb = 0.078 N·m/A and c = 1.85 × 10−4

N·m·s/rad.



6.41 Refer to Table 6.6.1. θf = 3π/4 rad. Also,

I = 0.215 Td = 4.2 tf = 2

t1 = 0.3 t2 = 1.7 R = 4

KT = 0.3 = Kb

Note that L is not needed.The calculated quantities are

E = 35.7 J per cycle

ωmax = 1.386 rad/s = 13.2 rpm

Tmax = 5.19 N · m

Trms = 4.24 N ·m

imax = 17.3 A

irms = 14.1 A

vmax = 69.6 V



6.42 Refer to Table 6.6.1. θf = 11(2π) = 22π rad. Also,

I = 0.05 Td = 3.6 tf = 3

t1 = 0.5 t2 = 2.5 R = 3

KT = 0.4 = Kb

Note that L is not needed.The calculated quantities are

E = 872 J per cycle

ωmax = 27.65 rad/s = 264 rpm

Tmax = 6.36 N · m

Trms = 3.94 N ·m

imax = 15.9 A

irms = 9.85 A

vmax = 58.8 V



6.43 a) The transfer function is

Y (s)Z(s)

= T (s) =s2

s2 + 18s + 100

With s = 120j we obtain

M = |T (120j)| =∣∣∣∣∣

−(120)2

100− (120)2 + 18(120)j

∣∣∣∣∣ = 0.9957

φ = 6 T (120j) = − tan−1 18(120)100− (120)2

= −2.992

Thus the steady state response is

y(t) = 9.957 sin(120t− 2.992)

b) The transfer function is

Y (s)Z(s)

= T (s) =s2

s2 + 1800s + 106

With s = 120j we obtain

M = |T (120j)| =∣∣∣∣∣

−(120)2

106 − (120)2 + 1800(120)j

∣∣∣∣∣ = 0.0144

φ = 6 T (120j) = − tan−1 1800(120)106 − (120)2

= −0.2157


y(t) = 0.144 sin(120t− 0.2157)

For case (a), the amplitude of the response (9.957) is almost identical to the amplitudeof the displacement input (10). Thus the instrument functions as a vibrometer.

For case (b), the amplitude of the response (0.144) is equal to 10−6 times the amplitudeof the acceleration of the input, which is 10(120)2 = 0.144 × 106. Thus the instrumentfunctions as an accelerometer with a gain of 10−6.



6.44 Summing forces on the mass m gives

md2x

dt2+ c

dx

dt+ kx = fs + Kf i

or(ms2 + cs + k)X(s) = Fs(s) + KfI(s) (1)

Since the applied voltage is zero, the circuit equation is

Ldi

dt+ Ri + Kb

dx

dt= 0

or(Ls + R)I(s) + KbsX(s) = 0 (2)

Solving (1) for X(s) and substituting into (2), and rearranging, gives the transfer func-tion.

I(s)Fs(s)

=−Kbs

mLs3 + (cL + mR)s2 + (kL + cR + KbKf )s + kR



6.45 a) The roots of the second-order model are s = −4333± 13 462j.b) The roots of the third-order model are s = −1643 ± 16 513j, which is the dominant

pair, and s = −8715.




KT = 0.2; Kb = 0.2; c = 5e-4; Ra = 0.8;La = 4e-3; I = 5e-4;den = [La*I, Ra*I + c*La, c*Ra + Kb*KT];% Speed transfer functionsys1 = tf(KT, den);% Current transfer functionsys2 = tf([I, c], den);[om, t1] = step(sys1);[ia, t2] = step(sys2);subplot(2,1,1)plot(t1, 10*om),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)subplot(2,1,2)plot(t1, 10*ia),xlabel(′t (s)′),ylabel(′i_a (A)′)

The plot is shown in the following figure. The peak current is approximately 8 A. The moreaccurate value of 8.06 A can be found with the step(sys2) function by right-clicking onthe plot, selecting characteristics, then peak response, and multiplying the answer by 10.

0 0.01 0.02 0.03 0.04 0.05 0.060

10

20

30

40

50

60

t (s)

ω (ra

d/s)

0 0.01 0.02 0.03 0.04 0.05 0.06−2

0

2

4

6

8

10

t (s)

i a (A)





KT = 0.2; Kb = 0.2; c = 5e-4; Ra = 0.8; La = 4e-3; I = 5e-4;den = [La*I, Ra*I + c*La, c*Ra + Kb*KT];% Speed transfer function: sys1 = tf(KT, den);% Current transfer function: sys2 = tf([I, c], den);% Applied voltaget = [0:0.001:0.05];va = 10*ones(size(t));om = lsim(sys1, va, t);ia = lsim(sys2, va, t);subplot(2,1,1)plot(t, om),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)subplot(2,1,2)plot(t, ia),xlabel(′t (s)′),ylabel(′i_a (A)′)

The plot is shown in the following figure. The peak current is approximately 8 A. Themore accurate value of 8.06 A can be found with the lsim(sys2, va, t) function byright-clicking on the plot, selecting characteristics, then peak response.

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050

10

20

30

40

50

60

t (s)

ω (ra

d/s)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−2

0

2

4

6

8

10

t (s)

i a (A)




6.48 See Example 6.3.3 for the transfer functions. The script file is

R = 1e+3; C = 2e-6; L = 2e-3;den = [L*R*C, L, R];% Transfer function for v1:sys1 = tf(1, den);% Transfer function for v2:sys2 = tf([R*C, 0], den);% Response to unit-step voltage v1:t = [0:0.00001:0.025];i31 = step(sys1, t);% Voltage v2:v2 = 4*sin(2*pi*60*t);% Response to voltage v2:i32 = lsim(sys2, v2, t);% Total response:i3 = 5*i31 + i32;plot(t, i3),xlabel(′t (s)′),ylabel(′i_3 (A)′)

The plot is shown in the figure on the following page. The steady-state response is a sinewave.



0 0.005 0.01 0.015 0.02 0.0250

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

t (s)

i 3 (A)




6.49 See equations (6.6.1) through (6.6.4) for the transfer functions, where I and c mustbe replaced with the equivalent inertia and equivalent damping: Ie = I + 4× 10−4/N2 andce = c + 1.8× 10−3/N2. The script file is

KT = 0.2; Kb = 0.2; c = 3e-4; Ra = 0.8;La = 4e-3; I = 4e-4; N = 3;% Equivalent inertia:Ie = I + 1e-3/N^2;% Equivalent damping:ce = c + 1.8e-3/N^2;den = [La*Ie, Ra*Ie + ce*La, ce*Ra + Kb*KT];% Speed command transfer functionsys1c = tf(KT, den);% Current command transfer functionsys2c = tf([I, c], den);% Speed disturbance transfer functionsys1d = tf([La, Ra], den);% Current disturbance transfer functionsys2d = tf(Kb, den);% Unit Command response:[omc, t1c] = step(sys1c);[iac, t2c] = step(sys2c);% Unit Disturbance response:omd = step(sys1d, t1c);iad = step(sys2d, t2c);% Total response:om = 20*omc + 0.04*omd;ia = 20*iac + 0.04*iad;subplot(2,1,1)plot(t1c, om),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)subplot(2,1,2)plot(t2c, ia),xlabel(′t (s)′),ylabel(′i_a (A)′)

The plot is shown in the figure on the following page. The peak current is 12.73 A, whichcan be found by displaying the values in the array ia.





0 0.01 0.02 0.03 0.04 0.05 0.060

20

40

60

80

100

120

t (s)

ω (ra

d/s)

0 0.01 0.02 0.03 0.04 0.05 0.06−5

0

5

10

15

t (s)

i a (A)




6.50 The main script file is Problem6p50.m.

% Program Problem6p50.mKT = 0.05; Kb = KT; c = 0;Ra = 0.8; La = 3e-3; I = 8e-5;trapezoidmotortfcurrent = lsim(currenttf, v, t);speed = lsim(speedtf, v, t);subplot(2,1,1)plot(t,current),xlabel(′t (s)′),ylabel(′Current (A)′)subplot(2,1,2)plot(t,speed),xlabel(′t (s)′),ylabel(′Speed (rad/s)′)performance

This program calls the following files.

% trapezoid.m Trapezoidal voltage profilet1 = 0.5; t2 = 2; tfinal = 2.5; t3 = 4;max_v = 30;dt = t3/1000;t = [0:dt:t3];for k = 1:1001

if t(k) <= t1v(k) = (max_v/t1)*t(k);

elseif t(k) <= t2v(k) = max_v;

elseif t(k) <= tfinalv(k) = (max_v/t1)*(tfinal - t(k));

elsev(k) = 0;

endend





% motortf.m Motor transfer functions for voltage input.% current:currenttf = tf([I,c],[La*I,Ra*I+c*La,c*Ra+Kb*KT]);% speed:speedtf = tf(KT,[La*I,Ra*I+c*La,c*Ra+Kb*KT]);

% performance.m Computes motor performance measures.ia = current;dt = t(2) - t(1);E = trapz(t,Ra*ia.^2) + trapz(t,c*speed.^2)i_max = max(ia)i_rms = sqrt(trapz(t,ia.^2)/t3)T_max = KT*i_maxT_rms = KT*i_rmsspeed_max = max(speed)v_max = Ra*i_max+Kb*speed_max

The plots are shown in the figure on the following page. The performance results are

E = 2.8203 J/cycle imax = 1.9200 A irms = 0.9388 A

Tmax = 0.0960 N · m Trms = 0.0469 N · m

speedmax = 600 rad/s vmax = 31.536 V





0 0.5 1 1.5 2 2.5 3 3.5 4−2

−1

0

1

2

t (s)

Cur

rent

(A)

0 0.5 1 1.5 2 2.5 3 3.5 40

100

200

300

400

500

600

700

t (s)

Spe

ed (r

ad/s

)




6.51 The equations are

Ladiadt

= va − Raia − Kbω (1)

dθ

dt= ω (2)

Iedω

dt= KT ia − ceω − 4.2 sin θ (3)

Choose the three state variables to be x1 = ia, x2 = θ, and x3 = ω.The slew speed required for a rotation angle of 3π/4 rad is ωslew = 3π/4/t2 = 1.386

rad/s. Thus, letting A represent ωslew, the specified trapezoidal speed profile is

ω(t) =

Att1

0 ≤ t ≤ t1A t1 ≤ t ≤ t2−A

t1(t − t2) + A t2 ≤ t ≤ tfinal

The specified angle θ(t) resulting from the trapezoidal speed profile can be found by inte-grating that profile. The result is

θ(t) =

At2

2t10 ≤ t ≤ t1

A(t − t1) + At12 t1 ≤ t ≤ t2

−A(t−t2)2

2t1+ A(t − t2) + At2 − At1

2 t2 ≤ t ≤ tfinal





The voltage required to move the arm at the desired angle θ(t) is found from (1):

va = Ladiadt

+ Raia − Kbω (4)

We obtain ia from (3):

ia(t) =1

KT

(Ie

dω

dt+ ceω + 4.2 sin θ

)(5)

Note that dω/dt = α, the angular acceleration, where

α(t) =

At1

0 ≤ t ≤ t10 t1 ≤ t ≤ t2−A

t1t2 ≤ t ≤ tfinal

Thus (5) can be expressed as

ia(t) =1

KT(Ieα + ceω + 4.2 sin θ) (6)

Differentiate this to obtain dia/dt:

diadt

=1

KT

(Ie

dα

dt+ ceα + 4.2θ cos θ

)(7)

Note that θ = ω and that dα/dt = 0. Also, for this problem, ce = 0. Thus (7) becomes

diadt

=1

KT4.2ω cos θ (8)

Thus, with ce = 0, (4) becomes

va =4.2La

KTω cos θ +

Ra

KT(Ieα + 4.2 sin θ) − Kbω (9)

In summary, va is computed from (9) using the expressions for α(t), ω(t) and θ(t). Thecomputed values of va are then used with (2) and (3) to compute the actual values of ω(t)and θ(t). The m-files required for this are shown on the next page.





% Program problem6p51.mglobal KT Kb La Ra Ieglobal t1 t2 tfinal AKT = 0.3; Kb = 0.3; Ra = 4;La = 3e-3; Ie = 0.215;t1 = 0.3;t2 = 1.7; tfinal = 2;% A is the slew speedA = (3*pi/4)/t2;[t, x] = ode45(′problem6p52a′, [0, tfinal], [0, 0, 0]);subplot(3,1,1)plot(t, x(:,1)),ylabel(′i_a (A)′)subplot(3,1,2)plot(t, x(:,2)),ylabel(′\theta (rad)′)subplot(3,1,3)plot(t, x(:,3)),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)





function xdot = problem6p51a(t,x)% Differential equation file for program problem6p52.mglobal KT Kb La Ra Ieglobal t1 t2 tfinal A% Evaluate trapezoidal speed profile% om is the speedif t <= t1

om = (A/t1)*t;elseif t <= t2

om = A;else om = -(A/t1)*(t-t2) + A;end% Evaluate angle profile% theta is the angleif t <= t1

theta = (A/(2*t1))*t.^2;elseif t <= t2

theta = A*(t-t1)+A*t1/2;else theta = -(A/(2*t1))*(t-t2).^2 + A*(t-t2) + A*t2-A*t1/2;end% Evaluate angular acceleration profile% alpha is the accelerationif t <= t1

alpha = A/t1;elseif t <= t2

alpha = 0;else alpha = -A/t1;end% Evaluate the voltage profile% va is the voltageva = 4.2*(La/KT)*om*cos(theta) + (Ra/KT)*(Ie*alpha + 4.2*sin(theta)) - Kb*om;% x(1) is the current; x(2) is the angle; x(3) is the speedxdot = [(-Ra*x(1)-Kb*x(3)+va)/La;x(3);(KT*x(1)-4.2*sin(x(2)))/Ie];





The simulation results are shown in the following figure. The angle θ does not quitereach the desired value of 3π/4. From the velocity plot we see that the velocity starts tofall below that specified by the trapezoidal profile. This is due to the approximation madeby neglecting dα/dt at the corners of the velocity profile, at the time t1 and t2. Because α,the slope of the profile, changes suddenly at these times, dα/dt is undefined, and thus ourapproximation does not represent a true solution of the differential equations.

This problem is an example of open-loop control in which the voltage is precalculatedfrom the differential equations. Thus the effectiveness of this method depends on the ac-curacy of the differential equation model and on any approximation made to solve for thevoltage. The method is also tedious, as this problem illustrates. A better way to controlthe arm angle is to use feedback control, which is discussed in Chapter 10.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

5

10

15

i a (A)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1

2

3

θ (r

ad)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−2

−1

0

1

2

t (s)

ω (r

ad/s

)




6.52 From Example 6.3.1,

I3(s) =V1(s) + RCsV2(s)LRCs2 + Ls + R

To avoid using a Transfer Function block containing the illegal transfer function RCs (whichis a differentiator), we can write the equation as follows:

I3(s) =RCsV1(s)

RCs(LRCs2 + Ls + R)+

RCsV2(s)LRCs2 + Ls + R

The corresponding Simulink model is shown in the following figure. In the Sine block, setthe Amplitude to 4 and the Frequency to 120*pi. In the Pulse Generator block, set theAmplitude to 5, the Period to 1, and the Pulse Width to 5. Set the Stop Time to whatevervalue you want, greater than 0.05 to see the full pulse but less than 1 so that the pulse willnot repeat; say 0.075. In the To Workspace block, set the Save Format to array. Beforerunning the model, type the following in the MATLAB Command window.

R = 1e4; C = 2e-6; L = 2e-3;

You can plot the response by typing plot(tout, simout(:,1)). Note that if you use alarger Stop time, the buffer for tout may overflow. The number of points to be retained intout can be increased if necessary. This is done under the Configuration Parameters menu.




6.53 From Figure 6.3.2b we can create the corresponding Simulink model, which is shownin the following figure. In the Sine block, set the Amplitude to 3 and the Frequency to120*pi. In the Step block, set the Step time to 0, the Initial value to 0, and the Final valueto 12. Set the Stop Time to an appropriate value, say 0.5. In the To Workspace block, setthe Save Format to Array. Before running the model, type the following in the MATLABCommand window.

R = 2e4; C = 3e-6;

You can plot the response v1(t) by typing plot(tout, simout(:,1)). You can plot theresponse v2(t) by typing plot(tout, simout(:,2)).




6.54 The Simulink model is like that shown in Figure 6.9.5, with the disturbance torque Tdset to 0. In the Step block, set the Step time to 0, the Initial value to 0, and the Final valueto 10. Set the Stop Time to an appropriate value, say 0.1. In the To Workspace block, setthe Save Format to Array. Before running the model, in the MATLAB Command window,type

KT = 0.2; Kb = KT; c = 5e-4;Ra = 0.8; La = 4e-3; I = 5e-4;

For part (a), set the limits on the saturation block very high, say ±100. You can plot thetorque response by typing plot(tout, simout(:,1)). You can plot the speed response bytyping plot(tout, simout(:,2)). The following figure shows the two plots created withthe subplot function. The maximum torque is 1.6 N·m.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−0.5

0

0.5

1

1.5

2

t (s)

Tor

que

(N m

)

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

10

20

30

40

50

60

t (s)

Spe

ed (

rad/

s)

Figure : for Problem 6.54a





For part (b), set the limits on the saturation block to ±0.8. The resulting responses areshown in the following figure.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−0.2

0

0.2

0.4

0.6

0.8

t (s)

Tor

que

(N m

)

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

10

20

30

40

50

60

t (s)

Spe

ed (

rad/

s)

Figure : for Problem 6.54b



6.55 The model is shown in the following figure. Set the Initial output to 0 in all fourRamp blocks. In the topmost Ramp block, set the Slope to 60 and the Start time to 0. Inthe Ramp1 block, set the Slope to −60 and the Start time to 0.5. In the Ramp2 block, setthe Slope to −60 and the Start time to 2. In the Ramp3 block, set the Slope to 60 and theStart time to 2.5. Set the Stop Time to 4. In the To Workspace block, set the Save Formatto array. Before running the model, type the following in the MATLAB Command window.

KT = 0.05;Kb = 0.05;c = 0;Ra = 0.8;La = 3e-3;I = 8e-5;

You can plot the current response by typing plot(tout, simout(:,1)) and the speedresponse by typing plot(tout, simout(:,2)).





to accompany


by

William J. Palm III


Solutions to Problems in Chapter Eight


CD1

Highlight

8.1 a)

yss =25√

(14ω)2 + 18215 =

25√[14(1.5)]2 + 182

15 = 13.5582

b)

yss =15ω√

[3ω]2 + 425 =

15(2)√(3(2))2 + 42

5 = 20.8013

c)

yss =√

(ω)2 + 502

√(ω)2 + 1502

3 =√

104 + 502

√104 + 1502

3 = 1.8605

d)

yss =33200

√ω2 + 1002

√ω2 + 332

8 =33200

√502 + 1002

√502 + 332

8 = 2.46



8.2 a) In standard form of (8.1.2), the transfer function is

T (s) =15

6s + 2=

152

13s + 1

Here τ = 3 and the multiplicative factor 15/2 shifts the m curve up by 20 log 152 = 17.501

dB. The m plot looks like that in Figure 8.1.6, but shifted up by 17.501 dB. Because6 (15/2) = 0, the phase curve is identical to that shown in Figure 8.1.6.

b) In standard form of τs/(τs + 1), the transfer function is

T (s) =9s

8s + 4=

92

2s

2s + 1

Here τ = 2 and the multiplicative factor 9/2 shifts the m curve up by 20 log 92 = 13.064 dB.

The m plot looks like that in Figure 8.1.9, but shifted up by 13.064 dB. Because 6 (9/2) = 0,the phase curve is identical to that shown in Figure 8.1.9.

c) In standard form of (8.1.13), the transfer function is

T (s) = 614s + 710s + 2

=72

2s + 15s + 1

Here τ1 = 2 and τ2 = 5. The multiplicative factor 7/2 shifts the m curve up by 20 log 72 =

10.881 dB. The m plot looks like that in part (a) of Figure 8.1.12 (because τ1 < τ2), butshifted up by 10.881 dB. Note that Figure 8.1.12 applies to the case where K = 1 in (8.1.13).

Because 6 (7/2) = 0, the phase curve is given by

φ(ω) = 6 (1 + 2ωj)− 6 (1 + 5ωj)

At low frequencies, φ ≈ 0 but negative. At ω = 1/τ2 = 1/5 and at ω = 1/τ1 = 1/2,φ = −23 (this identical value is a coincidence because here τ1 = 1/τ2). As ω → ∞, φ → 0

through negative values.



8.3 The input is

hi(t) = 10 + 3 sin2π

12t

We are given that the steady-state response is

h(t) = 10 + 2 sin(

2π

12t + φ

)

Thus the amplitude ratio is M = 2/3.The model is

Ah = − g

R(h − hi)

so that the time constant is τ = RA/g. Thus the model can be written as

τ h + h = hi

The amplitude ratio for this model is

M =1√

1 + ω2τ2=

23

where ω = 2π/12 = π/6. Solving for τ gives

τ =2π

√5 = 1.423 hr

The phase shift isφ = − tan−1(ωτ) = −0.641 rad

and the time shift is |φ|/ω = 1.22 hr.



8.4 The total wall surface area is A = 4(5×3) = 60 m2, and the air volume is V = 3(5×5) =75 m3. The thermal capacitance of the room air is

C = mcp = ρV cp = 1.289(75)(1004) = 9.704× 104

The outside temperature is To = 15 + 5 sin ωt where

ω =2π

24(3600)= 7.272× 10−5 rad/sec

The model isC

dT

dt=

1RT

(To − T )

orRTC

dT

dt+ T = To

where RT is the total wall resistance, which is

RT = AR = 60(4.5× 10−3) = 0.27

The time constant isτ = RTC = 2.62× 104 s

The amplitude ratio is

M =1√

1 + ω2τ2= 0.4647

Thus the amplitude of oscillation of the indoor temperature is 5(0.4647) = 2.32C. Thephase shift is

φ = − tan−1 ωτ = −1.087 rad

ThusT (t) = 15 + 2.32 sin(7.272× 10−5t − 1.087)



8.5 The model is Iω + cω = T , or 2ω + 4ω = T . The transfer function is

Ω(s)T (s)

=1

2s + 4=

0.250.5s + 1

and the amplitude ratio is

M(ω) =0.25√

1 + 0.25ω2

The phase shift is φ(ω) = − tan−1(ωτ) = − tan−1(0.5ω).Because there are three terms in the forcing function, there are three corresponding

terms in the steady-state response. These are

ωss(t) = A1M1 + A2M2 sin(3t + φ2) + A3M3 cos(5t + φ3)

From the M(ω) and φ(ω) equations we obtain

M1 = M(0) = 0.25 M2 = M(3) = 0.1387 M3 = M(5) = 0.0928

φ2 = φ(3) = − tan−1(1.5) = −0.983 φ3 = φ(5) = − tan−1(2.5) = −1.19

Thus, noting from the input that A1 = 30, A2 = 5, and A3 = 2, we have

ωss(t) = 7.5 + 0.6935 sin(3t − 0.983) + 0.1856 cos(5t − 1.19)



8.6 The model isAh = − g

Rh + qvi

orRA

gh + h =

R

gqvi

With the given values,585.39h + h = 46.58qvi

The time constant is τ = 585.39 sec.The amplitude ratio is

M(ω) =46.58√

1 + ω2τ2

where ω = 0.002.The phase shift is

φ(ω) = − tan−1(ωτ)

Since the input consists of the sum of two terms, the steady-state response has the form

hss(t) = A1M1 + A2M2 sin(0.002t + φ2)

where A1 = 0.2, A2 = 0.1, and

M1 = M(0) = 46.58 M2 = M(0.002) = 30.25

φ2 = φ(0.002) = − tan−1[0.002(585.39)] = −0.8639

Thushss(t) = 9.316 + 3.025 sin(0.002t− 0.8639)

The lag is given by |φ2|/ω = 432 sec.



8.7 The transfer function given in the text is

T (s) =R1R2Cs + R2

R1R2Cs + R1 + R2=

R2

R1 + R2

τ1s + 1τ2s + 1

whereτ1 = R1C τ2 =

R1R2C

R1 + R2

For the circuit to be a low-pass filter, the corner frequency of the numerator must begreater than the corner frequency of the denominator; that is,

1τ1

>1τ2

This implies that τ2 > τ1, which says

R1R2C

R1 + R2> R1C

Canceling R1C givesR2

R1 + R2> 1

which is impossible to satisfy. Thus the circuit cannot be a low-pass filter.



8.8 a)

yss =10√

(10ω)2 + 1√

(4ω)2 + 110 =

10√[10(0.2)]2 + 1

√[4(0.2]2 + 1

10 = 34.92

b)

yss =1√

(200− 4ω2)2 + (20ω)216 =

1√(200− 100)2 + [20(5)]2

16 = 0.1131



8.9 a)

T (jω) =8

jω(100− ω2 + 10ωj)

M(ω) =8

ω√

(100− ω2)2 + 100ω2

Thus M(9) = 0.0096.

φ(ω) = −6 jω − 6 [(100− ω2) + 10ωj]

So φ(9) = π/2− tan−1(90/19) = 0.2081. Thus

yss(t) = 6M(9) sin[9t + φ(9)] = 0.0576 sin(9t − 0.2081)

b)

T (jω) =10

−ω2(1 + ωj)

M(ω) =10

ω2√

1 + ω2

Thus M(2) = 1.118.

φ(ω) = −6 (−10/ω2) − 6 [(1 + ωj)

So φ(2) = π − tan−1(2) = 2.0344. Thus

yss(t) = 9M(2) sin[2t + φ(2)] = 10.062 sin(2t + 2.0344)





c)

T (jω) =ωj

(1 + 2ωj)(1 + 5ωj)

M(ω) =ω√

1 + 4ω2√

1 + 25ω2

Thus M(0.7) = 0.1118.

φ(ω) = 6 ωj − 6 (1 + 2ωj)− 6 (1 + 5ωj)

So φ(0.7) = π/2− tan−1(1.4)− tan−1(3.5) = −0.6722. Thus

yss(t) = 9M(0.7) sin[0.7t + φ(0.7)] = 1.0062 sin(0.7t− 0.6722)

d)

T (jω) =−ω2

(1 + 2ωj)(1 + 5ωj)

M(ω) =ω2

√1 + 4ω2

√1 + 25ω2

Thus M(0.7) = 0.0782.

φ(ω) = 6 (−ω2) − 6 (1 + 2ωj)− 6 (1 + 5ωj)

So φ(0.7) = −π/2 − tan−1(1.4)− tan−1(3.5) = −3.8138. Thus

yss(t) = 9M(0.7) sin[0.7t + φ(0.7)] = 0.7038 sin(0.7t− 3.8138)



8.10 The resonant frequency and peak magnitude are given by

ωr = ωn

√1 − 2ζ2 Mr =

12ζ

√1 − ζ2

Note that ωn =√

2. Thusa) For ζ = 0.1, ωr = 1.4 and Mr = 5.025.b) For ζ = 0.3, ωr = 1.28 and Mr = 1.747.



8.11 The maximum amplitude occurs at ω = ωr and is 22Mr, where

Mr =1

2ζ√

1 − ζ2

The damping ratio isζ =

c

2√

50=

c

10√

2Thus

22Mr =22

2ζ√

1 − ζ2= 3

Solve for ζ by squaring each side to obtain

4ζ4 − 4ζ2 − 13.44 = 0

which has the positive root ζ2 = 4.2, or ζ = 2.05. Thus c = 10√

2(2.05) = 28.99.



8.12 The transfer function isT (s) =

113s2 + 2s + k

From (8.2.20)

ωr = ωn

√1− 2ζ2

where

ωn =

√k

13

ζ =2

2√

13k=

1√13k

We want ωr = 4. Thus we must solve the following for k.

4 =

√k

13

√1 − 2

13k=

113

√13k − 2

This gives k = 208 lb/ft. This gives ζ = 0.0192.From (8.2.21),

Mr =1

2ζ√

1− ζ2= 26.295

Thus the amplitude of the steady-state response is 26.295(10) = 262.95The phase shift of the response is (with ω = 4)

φ = − 6 [(k − 13ω2) + 2ωj] = −6 [−8 + 8j] = −2.356

Thus the steady-state response is

xss(t) = 262.95 sin(4t − 2.356)



8.13 a)

T (jω) =7

ωj(58− ω2 + 6ωj)

M(ω) =7

ω√

(58− ω2)2 + 36ω2

Since M(0) = ∞, the resonant frequency is at ω = 0.b)

T (jω) =7

(174− 3ω2 + 18ωj)(58− 2ω2 + 8ωj)

M(ω) =7√

(174− 3ω2)2 + 324ω2√

(58− 2ω2)2 + 64ω2

A plot of M versus ω shows that M has a single peak near ω = 4.97, which is the resonantfrequency. We might have expected two peaks since there are two quadratic factors in thedenominator. These quadratic factors have resonant frequencies of 4.58 and 6.32. Becausethese frequencies are so close in value, their two peaks merge into a single peak at ω = 4.97.



8.14 Applying the voltage law to each loop gives

v1 − Ri1 − Ldi3dt

= 0

v2 −1C

∫i2 dt − L

di3dt

= 0

From conservation of charge, i3 = i1 + i2.Take the Laplace transform of each equation using zero initial conditions. Solve the first

two for I1(s) and I2(s):

I1(s) =V1(s) − LsI3(s)

R

I2(s) = CsV2(s) − LCs2I3(s)

Substitute these into the third equation to obtain

(RLCs2 + Ls + R)I3(s) = V1(s) + RCsV2(s)

The two transfer functions are

I3(s)V1(s)

=1

RLCs2 + Ls + R=

110−5s2 + 0.1s + 100

I3(s)V2(s)

=RCs

RLCs2 + Ls + R=

10−4s

10−5s2 + 0.1s + 100

The roots are s = −1127 and s = −8873. Thus the corner frequencies are ω1 = 1127 rad/sand ω2 = 8873 rad/s.(continued on the next page)




First consider the transfer function I3(s)/V1(s). At low frequencies the m curve ap-proaches 20 log(105/[(1127)(8873)] = −40 dB. The m curve breaks down at ω1 = 1127 andagain at ω2 = 8873 rad/s. So the system acts like a low pass filter for the input voltage v1,and it tends to filter out frequency components in v1 that are higher than ω1 = 1127 rad/s.The low frequency gain is M = 10m/20 = 10−40/20 = 0.01.

Now consider the transfer function I3(s)/V2(s). At low frequencies the m curve ap-proaches 20 log(0) = −∞. At ω = 10,

m = 20 log(

10(10)√11272 + 102

√88732 + 102

)= −100 dB

The m curve has a slope of 20 dB/decade until ω1 = 1127, when the slope becomes approx-imately zero. The curve breaks downward at ω2 = 8873 rad/s, and the slope becomes −20dB/decade for higher frequencies. So the system acts like a band pass filter for the inputvoltage v2, and it passes frequency components in v2 that are between ω1 = 1127 rad/s andω2 = 8873 rad/s.



8.15 a) The equation of motion is

mx = k(y − x) − cx

orx + cx + 600x = 600y

The transfer function isT (s) =

X(s)Y (s)

=600

s2 + cs + 600

This has the same form as (8.2.14) with ωn =√

600 = 10√

6 and ζ = c/(20√

6). So,provided that c is such that ζ ≤ 0.707, we can use (8.2.20) and (8.2.21) to obtain

ωr = ωn

√1 − 2ζ2 = 10

√6√

1 − 2ζ2 rad/s

Mr =1

2ζ√

1 − ζ2

Since c = 20ζ√

6, ζ will be no greater than 0.707 if 0 ≤ c ≤ 34.64 N·s/m.For 0 ≤ c ≤ 34.64, a plot of ωr versus c shows that the resonant frequency varies from

3.6 (for c = 34.64) to 24.5 rad/s (for c = 0). For 0 ≤ c ≤ 34.64, a plot of Mr versus c showsthat Mr varies from ∞ (for c = 0) to 1 rad/s (for c = 34.64).

For values of ζ > 0.707 (for c > 34.64), there is no resonant peak and thus no resonantfrequency.

b) The transfer function is

T (s) =X(s)Y (s)

=k

ms2 + cs + k=

k/m

s2 + (c/m)s + k/m=

ω2n

s2 + 2ζωns + ω2n

This has the same form as (8.2.14), so provided ζ ≤ 0.707, we can use (8.2.20) and (8.2.21)to compute ωr and Mr. For values of ζ > 0.707, there is no resonant peak and thus noresonant frequency.



8.16 Applying the voltage law gives

vs −1C

∫i dt − L

di

dt− vo = 0

where v0 = Ri. Use the latter equation to eliminate i:

LCd2vo

dt2+ RC

dvo

dt+ vo = RC

dvs

dt

For the given values,

5 × 10−8 d2vo

dt2+ 10−5R

dvo

dt+ vo = 10−5R

dvs

dt

The transfer function isVo(s)Vs(s)

=103Rs

5s2 + 103Rs + 108

For R = 10,Vo(s)Vs(s)

=104s

5s2 + 104s + 108

The roots are s = −1000± 4359j.For R = 100,

Vo(s)Vs(s)

=105s

5s2 + 105s + 108

The roots are s = −100.05 and s = −199, 900.The plots are shown on the next page. The peak is approximately the same for each. It

is −0.000185 dB, or M = 10−0.000185/20 = 1 at 5170 rad/s.(continued on the next page)



100

101

102

103

104

105

106

107

−90

−80

−70

−60

−50

−40

−30

−20

−10

0

Mag

nitu

de (

dB)

Frequency (rad/sec)

R = 1000

R = 10

Figure : Log magnitude plots for Problem 8.16

Thus an increase in the value of R by a factor of 100 does not significantly change thepeak value or the peak frequency, but it does change the spread of the plot about the peak.



8.17 The equations of motion are

I1ω1 = T1 − c1(ω1 − ω2)

I2ω2 = c1(ω1 − ω2) − c2ω2

Applying the Laplace transform for zero initial conditions, and eliminating Ω1(s), we obtainthe transfer function:

Ω2(s)T1(s)

=100

s2 + 5s + 2

From this we obtain the magnitude ratio and the phase angle.

M(ω) =100√

(2 − ω2)2 + (5ω)2

φ(ω) = − tan−1 5ω

2 − ω2

Evaluation of M and φ for ω = 0, 1.5, and 2 gives the following results.

M(0) = 50 M(1.5) = 13.3259 M(2) = 9.8058

φ(0) = 0 φ(1.5) = −1.604 rad φ(2) = −1.763 rad


ω2ss = 50(4) + 13.3259(2) sin(1.5t− 1.604) + 9.8058(0.9) sin(2t − 1.763)

orω2ss = 200 + 26.6518 sin(1.5t− 1.604) + 8.8252 sin(2t− 1.763)



8.18 The forcing frequency is ω = 5.2 and the natural frequency is ωn =√

75/3 = 5. Thebeat period is

2π

|ω − ωn|=

2π

|5.2− 5|= 31.4159

The vibration period is4π

ω + ωn=

4π

10.2= 1.232



8.19 From (8.3.4) with fo = 0.2, k = 64, and ωn = 8,

x(t) = 0.0125(

sin 8t

8− t cos 8t

)

The plot is shown below. It takes approximately 8.2 sec for |x(t)| to become greater than0.1 ft.

0 1 2 3 4 5 6 7 8 9 10−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

t (sec)

x(t)

(ft)

Figure : Plot for Problem 8.19



8.20 From the given data

ω =3400(2π)

60= 356 rad/s

c = 2ζ√

mk = 548 N · s/m

AlsoFt(s)Fr(s)

=cs + k

ms2 + cs + k

Thus

|Ft(ω)| = |Fr||k + cωj|

|k − mω2 + cωj| = |Fr|√

k2 + c2ω2

√(k − mω2)2 + c2ω2

For rotating unbalance

|Fr| = muεω2 = 0.1(0.02)(356)2 = 253.5 N

Substituting the values of k, c, m, and ω, we obtain

|Ft| = 253.5(0.1027) = 26 N



8.21 For base motion,X

Y=

√k2 + c2ω2

√(k − mω2)2 + c2ω2

Plotting this expression versus ω for the given values of m, c, and k, we find that the ratioX/Y is never less than 1. Therefore, X will never be greater than 2 mm as long as Y isnever greater than 2 mm.



8.22 The vibration frequency of the base motion as a function the vehicle velocity v is

ω =(

528020

) (1

3600

)(2π)v = 0.4608v

From the results of Example 4.5.9,

D3 = (m1s2+c1s+k1)(m2s

2+c1s+k1+k2)−(c1s+k1)2 = m1m2s4+(m1k1+m1k2+m2k1)s2+k1k2

where we have taken c1 = 0. Using the given values, we obtain

114.472s4 + 4.0838× 105s2 + 8 × 107 = 0

The roots are s = ±14.4j and s = ±57.96j. Thus the resonant frequencies are 14.4 and57.96 rad/s, and the resonant speeds are

v =14.4

0.4608= 31.25 ft/sec and v =

57.960.4608

= 125.8 ft/sec

or v = 21.3 mph and 85.8 mph.Note that the amplitude of the road surface variation, although given in the problem

statement, was not needed.



8.23 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be

mb = 2388(2)(0.3)(0.03) = 42.98 kg

The equivalent system mass is

me = 50 + 0.23mb = 59.886 kg

The stiffness of the beam is

k =Ewh3

4L3=

2 × 1011(0.3)(0.03)3

4(2)3= 1.6875× 106 N/m

The forcing frequency is

ω =3400(2π)

60= 356 rad/s

and the amplitude of the unbalance force is

|Fr| = muεω2 = 0.02(356)2 = 2535 N

Thus the displacement amplitude is, since c = 0,

X =|Fr|√

(k − meω2)2=

|Fr||k − meω2| = 4.29× 10−4 m

or 0.428 mm.



8.24 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be

mb = 2388(2)wh = 4776wh

where w is the beam width and h is the beam thickness.The equivalent system mass is

me = 50 + 0.5mb = 50 + 2388wh

The stiffness of the beam is

k =16Ewh3

L3= 4 × 1011wh3

The forcing frequency is

ω =3400(2π)

60= 356 rad/s

and the amplitude of the unbalance force is

|Fr| = muεω2 = 0.02(356)2 = 2535 N

Thus the displacement amplitude is, since c = 0,

X = 0.01 =|Fr|√

(k − meω2)2=

|Fr||k − meω2| (1)

This expression is a complicated function of w and h since both k and me are functionsof w and h. So let us temporarily neglect the beam mass. Then me ≈ 50 kg and theprevious expression becomes

X = 0.01 =2535

|k − 6.3368× 106|

This gives k = 6.5903× 106 N/m. Thus

wh3 =6.5903× 106

4 × 1011= 1.6× 10−5





There is an infinite number of solutions to this equation, but we should choose one thatresults in an equivalent beam mass that is small compared to 50 kg, because we neglectedthe beam mass in obtaining this solution.

One solution is h = 0.04 m, which gives w = 0.25 m, both reasonable dimensions.These give an equivalent beam mass of 0.5(47.76) = 23.88 kg and a total equivalent massof me = 73.88 kg. Using this value in equation (1) we obtain

X =2535

|6.59× 106 − 73.88(356)2| = 9.14× 10−4 m

which is much less than the maximum allowable value of 0.01 m. So our solution is valid.



8.25 We are given m = 1500 kg, k = 20 000 N/m, ζ = 0.04, and Y = 0.01 m. Thusc = 2ζ

√mk = 438.18. From (8.3.9), with s = jω,

|Ft| = |Y | mω2√

k2 + (cω)2√(k − mω2)2 + (cω)2

At resonance, from (8.2.20),

ω = ωr = ωn

√1− 2ζ2 = 0.998

√k

m= 3.6442

Thus|Ft| = 2500 N



8.26 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also,

ω = 40 Hz = 251 rad/s

ωn =

√k

m=

√66 667

(200/9.81)= 57.2 rad/s

From (8.3.7) with c = 0,

|X ||Y | =

k√(k − mω2)2

= 0.055

Thus 5.5% of the airframe motion is transmitted to the module.



8.27 a) Neglect damping in the isolator, and determine its required stiffness k. From (8.3.7)with c = 0,

|X ||Y | =

k√(k − mω2)2

=k/m√

(k/m− ω2)2=

ω2n√

(ω2n − ω2)2

= 0.1

which gives (ω/ωn)2 = 11. Thus

ω2n =

ω2

11=

[3000(2π)/60]2

11=

(314)2

11

and

k = mω2n =

232.2

(314)2

11= 556.7 lb/ft

b) Let r = ω/ωn. From part (a), ωn = 314/√

11, we have

r1 =2500(2π)/60

314/√

11= 2.77

andr2 =

3500(2π)/60314/

√11

= 3.87

From (8.3.7) with c = 0,|X ||Y | =

1|1− r2|

Thus the highest percentage of motion will be transmitted at the lowest r value, whichis r1, the value corresponding to 2500 rpm. For 2500 rpm,

|X ||Y | =

1|1 − r2

1|= 0.15

For 3500 rpm,|X ||Y | =

1|1 − r2

2|= 0.07

Thus at most, 15% of the crane motion will be transmitted to the module.



8.28 The transmitted force is given by (8.3.14) and (8.3.17):

|Ft| = muεω2

√k2 + (cω)2√

(k − mω2)2 + (cω)2

If we letr = ω/ωn

then we can express the previous equation as

|Ft| = muεω2

√1 + 4ζ2r2

(1 − r2)2 + 4ζ2r2

We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. ε =0.1.12 = 0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus

r =ω

ωn=

104.7√500/1.55

= 5.83

Thus

|Ft| = (0.00155)(0.0083)(104.7)2√

1 + 136ζ2

1088 + 136ζ2= 0.141

√1 + 136ζ2

1088 + 136ζ2

a) For ζ = 0.05, Ft = 0.0049 lb.b) For ζ = 0.7, Ft = 0.034 lb.



8.29 The transmitted force is given by (8.3.14) and (8.3.17):

|Ft| = muεω2

√k2 + (cω)2√

(k − mω2)2 + (cω)2

If we letr = ω/ωn

then if c = 0, we can express the previous equation as

|Ft| = muεω2 1|1− r2|

Here Ft = 15 and ωR = 200(2π)/60 = 20.9 rad/s. Thus

r =ω

ωn=

20.9√k/m

=20.9√

2500/75= 3.62

Thus|Ft| = muε(20.9)2(0.0826) = 15

Solve for muε to obtain muε = 0.416 kg·m.



8.30 For a vibrometer we require that the natural frequency ωn be much less than theforcing frequency ω. Here

ω = 2π(200) = 400π rad/s

and

ωn =

√k

0.1

So we require that √k

0.1 400π

ork 1.579× 105 N/m

Suppose that k = 1.6×104. Then the static deflection will be δst = mg/k = 6.1×10−5 m.The designer must verify that there will be enough space in the instrument to accommodatethis deflection.



8.31 The equation of motion is

mx = k(y − x) − cx

or200x + 2000x + 2 × 104x = 2 × 104y

The transfer function is

T (s) =X(s)Y (s)

=2 × 104

200s2 + 2000s + 2 × 104=

100s2 + 10s + 100

This has the same form as (8.2.14) with ωn = 10 and ζ = 0.5. So we can use (8.2.20) and(8.2.21) to obtain

ωr = ωn

√1 − 2ζ2 = 7.07 rad/s

Mr =1

2ζ√

1 − ζ2= 1.155

For the bandwidth, equation (2) of Example 8.4.2 gives

r =

√1 − 2ζ2 ± 2ζ

√1 − ζ2 = 1.167, 0.605j

Because there is only one positive, real solution, the bandwidth extend from ω = 0 toω = 1.167ωn = 11.69 rad/s.



8.32 The transfer function becomes

T (s) =10−6s

3 × 10−4s2 + 10−2s + 1

From the frequency response plot shown below we can tell that the bandwidth is between45 and 76 rad/s.

You can obtain this plot in MATLAB by typing:

sys = tf([1e-6,0],[3e-4,1e-2,1]);bodemag(sys)

100

101

102

103

−125

−120

−115

−110

−105

−100

−95

−90

−85

−80

Mag

nitu

de (

dB)

Bode Diagram

Frequency (rad/sec)




8.33 a) The transfer function for the circuit in part (a) of the figure was derived in Chapter6. It is

Vo(s)Vs(s)

=G

R2C2s2 + 2RCs + 1

Its natural frequency is

ωn =√

1R2C2

=1

RC

Its damping ratio is

ζ =2RC

2√

R2C2= 1

From Equation (1) in Example 8.4.2, with r = ω/ωn = RCω,

kM =1√

(1− r2)2 + 4r2

which has a maximum of 1 at r = 0. Setting kM = 1/√

2 and solving for r gives r = 0.644.Thus the bandwidth corresponds to 0 ≤ r ≤ 0.644 or

0 ≤ ω ≤ 0.644RC





b) The transfer function for the circuit in part (b) of the figure was derived in Chapter6. It is

Vo(s)Vs(s)

=G

R2C2s2 + 3RCs + 1

Its natural frequency is

ωn =√

1R2C2

=1

RC

Its damping ratio is

ζ =3RC

2√

R2C2= 1.5

From Equation (1) in Example 8.4.2, with r = ω/ωn = RCω,

kM =1√

(1− r2)2 + 9r2

which has a maximum of 1 at r = 0. Setting kM = 1/√

2 and solving for r gives r = 0.374.Thus the bandwidth corresponds to 0 ≤ r ≤ 0.374 or

0 ≤ ω ≤ 0.374RC



8.34 The circuit equation isLCv1 + RCv1 + v1 = vs

or10−7v1 + 10−6Rv1 + v1 = vs

Its natural frequency and damping ratio are

ωn =√

107 = 103√

10

ζ =10R

2√

107=

5R

1000√

10The resonant frequency is

ωr = ωn

√1 − 2ζ2 if ζ ≤ 1/

√2

Thus there is a resonant peak and a resonant frequency only if

5R

1000√

10≤ 1√

2

which implies that

R ≤ 1000√

105√

2= 447 Ω

If there is a resonant peak, its value is

Mr =1

2ζ√

1 − ζ2

A plot of Mr versus R for 0 ≤ R ≤ 447 shows that Mr = 3.2026 for R = 100. As R isincreased to 447, Mr decreases to Mr = 1.

For R > 447, the maximum value of M occurs at ω = 0. The plots on the next pageshow the variation of M versus r = ω/ωn, for R = 100, 250, 400, 500, 750, and 1000 Ω.The plots show that as R is increased, the system changes from a bandpass filter to a lowpass filter. For R ≤ 447, increasing R decreases the bandwidth. For R > 447, increasing Rdecreases the bandwidth.(continued on the next page)



0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

r

M

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

r

M

100

250

400

500 750

1000




8.35 The magnitude ratio is

M(ω) =1√

1 + R2C2ω2=

1√1 + 36 × 10−8ω2

and the phase angle is

φ(ω) = − tan−1 RCω = − tan−1 6 × 10−4ω

The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1667 rad/s. Thus the cos 720πt termand the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0, 240π

and 480π gives the following results.

M(0) = 1 M(240π) = 0.9111 M(480π) = 0.7415

φ(0) = 0 φ(240π) = −0.425 rad φ(480π) = −0.735 rad


voss =20π

− 0.9111403π

cos(240πt− 0.425)− 0.74154015π

cos(480πt− 0.735)

orvoss = 6.3662− 3.8664 cos(240πt− 0.425)− 0.6294 cos(480πt− 0.735)




M(ω) =1√

1 + R2C2ω2=

1√1 + 10−6ω2


φ(ω) = − tan−1 RCω = − tan−1 10−3ω

The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1000 rad/s. Thus the sin 360πt termand the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0 and 120π

gives the following results.

M(0) = 1 M(120π) = 0.9357

φ(0) = 0 φ(120π) = −0.361 rad


voss = 5 + 0.935720π

sin(120πt− 0.361)

orvoss = 5 + 5.9569 sin(120πt− 0.361)




mx + cx + kx = cy + ky

orx + 98x + 4900x = 98y + 4900y

The transfer function isX(s)Y (s)

=98s + 4900

s2 + 98s + 4900

The magnitude ratio is

M(ω) =√

49002 + 982ω2

√(4900− ω2)2 + 982ω2


φ(ω) = tan−1 98ω

4900− tan−1 98ω

4900− ω2

A plot of M versus ω shows that M has a peak of 1.2764 at ω = 55 rad/s. Noting that1.2764/

√2 = 0.9026 and that M(0) = 1, we see that the lower bandwidth frequency is 0.

The plot also shows that M(111) = 0.906 which is close to 0.9026. So the upper bandwidthfrequency is 111 rad/s. The circuit is a low pass filter with the bandwidth of 0 ≤ ω ≤ 111rad/s. So we see that only the terms up to and including the sin 30πt term lie within thebandwidth.

Evaluation of M and φ for ω = 0, 10π, 20π, and 30π gives the following results.

M(0) = 1 M(10π) = 1.1623 M(20π) = 1.263 M(30π) = 1.0395

φ(0) = 0 φ(10π) = −0.1057 rad φ(20π) = −0.5187 rad φ(30π) = 2.2467 rad


xss =120

−1.16231

10πsin(10πt−0.1057)−1.263

120π

sin(20πt−0.5187)−1.03951

30πsin(30πt+2.2467)

or

xss = 0.05− 0.037 sin(10πt− 0.1057)− 0.0201 sin(20πt− 0.5187)− 0.011 sin(30πt+2.2467)




T (s) =1

0.5s + 5=

0.20.1s + 1

The time constant is τ = 0.1 sec. and the bandwidth is 1/τ = 10 rad/sec.b) The magnitude ratio and phase angle are given by

M(ω) =0.2√

0.01ω2 + 1φ(ω) = − tan−1(1 + 0.1ωj)

The only components of the input that lie within the bandwidth are sin 4t and sin 8t. Thuswe evaluate M(ω) and φ(ω) at ω = 4 and ω = 8.

M(4) = 0.1856 φ(4) = −0.381 rad

M(8) = 0.1562 φ(8) = −0.675 rad

Thus,y(t) ≈ 0.1856 sin(4t − 0.381) + 4(0.1562) sin(8t − 0.675)

ory(t) ≈ 0.1856 sin(4t − 0.381) + 0.6248 sin(8t − 0.675)




T (s) =X(s)F (s)

=1

0.25s2 + 2s + 25=

0.040.01s2 + 0.08s + 1

From the frequency response plot we can determine that the bandwidth is from 0 to11.9 rad/sec.

b) Only the first two terms in the expansion have frequencies within the bandwidth.Thus

f(t) ≈ sin 3t +13

sin 9t

and at steady state:

x(t) ≈ M(3) sin[3t + φ(3)] + M(9) sin[9t + φ(9)]

whereM(3) = 0.0425 φ(3) = −0.2579 rad

M(9) = 0.0537 φ(9) = −1.3128 rad

Thusx(t) ≈ 0.0425 sin(3t − 0.2579) + 0.0179 sin(9t − 1.3128)



8.40 First normalize the vo data by dividing by the amplitude of the input. Let

v =vo

20

Then compute m = 20 log v dB and log ω, and plot m versus log ω. The plot looks likea first order system plot (see Figure 8.1.6). The peak value of m is −11.25 dB, and m isapproximately 3 dB below that value at ω = 0.8. Thus we estimate the time constant tobe τ = 1/0/8 = 1.25 s. Thus we surmise that the form of the transfer function is

Vo(s)Vs(s)

=K

1.25s + 1

To estimate K, note that for low frequencies (ω << 0.8), |vo| = K|v|. So from the firstdata point, we estimate that K = |vo|/20 = 5.48/20 = 0.274. Thus we estimate the transferfunction to be

Vo(s)Vs(s)

=0.274

1.25s + 1

The MATLAB code to generate this analysis is shown below.

omega = [0.1:0.1:1,1.5,2:7];vo = [5.48,5.34,5.15,4.92,4.67,4.4,4.14,3.89,3.67,...

3.2,2.59,2.05,1.42,1.08,0.87,0.73,0.63];v = vo/20;m = 20*log10(v);logom = log10(om);plot(logom,m)

This result can be checked by comparing the m plot of the transfer function with theplot of the data, by continuing the above session as follows:

[M, ph, w]=bode(sys);M = M(:);plot(log10(w),20*log10(M),logom,m,’o’)

The plots agree very well.



8.41 First convert the data to meters and then divide by 15, the amplitude of the input. Thisgives data for the magnitude ratio M(ω) = 10−3x/15. Then create the log magnitude ratioplot (see the following MATLAB program). The plot shows the presence of a denominatorterm s and a breakpoint frequency somewhere between 1 and 2 rad/s. This suggests atransfer function of the form

X(s)F (s)

=K

s(τs + 1)

This transfer function form agrees with the physical situation, because it can be due to amass attached to a damper, with no stiffness. This would have the following equation ofmotion: mx + cx = f(t) and a transfer function

X(s)F (s)

=1

s(ms + c)=

1/m

s(s + c/m)

The breakpoint frequency is at ω = c/m. From the data, we see that c/m is somewherebetween 1 and 2. The magnitude ratio is

M(ω) =1/m

ω√

ω2 + (c/m)2

To estimate the mass m, we can choose one of the data points. Choosing the first point,for which M = 0.0139 and ω = 0.1, we obtain

m =1

0.0139(0.1)√

ω2b + (0.1)2

where ωb is the estimate of the breakpoint frequency.The choice ωb = 2 gives a better fit to the data that ωb = 1. Thus the solution is

c/m = 2 and

m =1

0.0139(0.1)√

22 + (0.1)2= 359

from which we obtain c = 2m = 718. The estimated transfer function is

X(s)F (s)

=1

s(359s + 718)





The following program generates the plot, which shows good agreement with the data.

% File for problem 8.41x = (1e-3)*[209,52,28,19,7,2,1]/15; % Converted x dataw = [0.1,0.4,0.7,1,2,4,6]; % Frequency datam = 20*log10(x); % Decibel conversionwb = 2; % Estimated breakpoint frequencyM0 = x(1); % First converted data pointw0 = w(1); % First frequency data pointmass = 1/(M0*w0*sqrt(wb^2+w0^2)) % Estimated mass valuew1 = [0.1:0.01:6]; % Frequency vector for plottingm1 = 20*log10((1/mass)./(w1.*(sqrt(w1.^2+wb^2))));semilogx(w,m,’o’,w1,m1)

10−1

100

101

−85

−80

−75

−70

−65

−60

−55

−50

−45

−40

−35

ω (rad/s)

m (d

B)





mx + cx + kx = muεω2 sin ωt

The frequency transfer function for this system, with |x| normalized by m/muε is

m|x|muε

=r2

√(1− r2)2 + (2ζr)2

(1)

where

r = ω/ωn ωn =

√k

m

First plot the raw data after converting the frequency data to rad/sec and the displacementdata to meters. From this plot you can estimate that the resonant frequency is 19 rad/sec.Assuming that the damping is small enough so that the resonant frequency is close to ωn,we estimate that ωn = 19, and thus k = mω2

n = 100(192) = 36100 N/m.Next, normalize the frequency data to find r: r = ω/19. To plot the normalized

displacement m|x|/muε vs. r, we need to estimate muε. To do this we need to plot thetheoretical curve from (1), and this requires an estimate of ζ. After several attempts, usingdifferent values of muε and ζ, the estimates muε = 0.005 and ζ = 0.1 gave a good fit to thedata.

With ζ = 0.1 and k = 36100, we obtain c = 2√

mkζ = 2√

100(36100)(0.1) = 380 N·s/m.




m1x1 = −k1(x1 − x2)

m2x2 = k1(x1 − x2) + k2(y − x2)

Convert these to state variable form by dividing by m1 and m2. The state vector is[x1, x1, x2, x2]. This gives the following state and input matrices for the given parametervalues.

A =

0 1 0 0−64 0 64 00 0 0 164 0 −128 0

B =

00064

Since the output is given to be x1, the output matrices are

C =[

1 0 0 0]

D = [0]

The MATLAB program is the following.

A = [0,1,0,0;-64,0,64,0;0,0,0,1;64,0,-128,0];B = [0;0;0;64];C = [1,0,0,0]; D = 0;sys1 = ss(A,B,C,D);sys2 = tf(sys1)eig(A)bode(sys1)


X1(s)Y (s)

=4096

s4 + 192s2 + 4096

The roots are s = ±4.9443j and s = ±12.9443j. Thus the resonant frequencies in radianunits are 4.9443 and 12.9443. The Bode plots are shown on the next page. The resonantpeaks are very tight, so the bandwidths are very small. You can type bodemag(sys1) toobtain just the m plot, which can be expanded about the resonant peaks. After doing thisit is still difficult to identify the bandwidths, which appear to be about ±0.1 rad/unit timefor each peak.(continued on the next page)



−100

−50

0

50

100

150

Mag

nitu

de (

dB)

10−1

100

101

102

−720

−675

−630

−585

−540

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)





I1θ1 = kT (φ − θ1) − cT1(θ1 − θ2)

I2θ2 = cT1(θ1 − θ2) − cT2θ2

Convert these to state variable form by dividing by I1 and I2. The state vector is [θ1, θ1, θ2, θ2].This gives the following state and input matrices for the given parameter values.

A =

0 1 0 0−1 −0.1 0 0.10 0 0 10 0.1 0 −0.2

B =

0100

Since the output is given to be x1, the output matrices are

C =[

1 0 0 0]

D = [0]

The MATLAB program is the following.

A = [0,1,0,0;-1,-0.1,0,0.1;0,0,0,1;0,0.1,0,-0.2];B = [0;1;0;0];C = [1,0,0,0]; D = 0;sys1 = ss(A,B,C,D);sys2 = tf(sys1)eig(A)bode(sys1)


X1(s)Y (s)

=s + 0.2

s3 + 0.3s2 + 1.01s + 0.2

The roots are s = 0, s = −0.202, and s = −0.049± 0.9939j. Thus the resonant frequencyin radian units is 0.9939. The Bode plots are shown on the next page. You can typebodemag(sys1) to obtain just the m plot. The bandwidth can be computed by sliding thecursor along the plot and noting the frequencies corresponding to 3 dB below the peak.These are ω = 0.942 and ω = 1.04, so the bandwidth is 1.04− 0.942 = 0.098.(continued on the next page)



−40

−30

−20

−10

0

10

20

30

40

Mag

nitu

de (

dB)

10−1

100

101

−180

−135

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)





mx + cx + kx = ky(t)

Thus the transfer function is

X(s)Y (s)

=k

ms2 + cs + k=

500050s2 + 200s + 5000

=100

s2 + 4s + 100

The MATLAB code to generate the m, plot is

[mag,phase,w]=bode(100,[1,4,100]);mag = mag(:);plot(w,20*log10(mag)),axis([6 15 0 10]),ylabel(’Magnitude (db)’),...xlabel(’Frequency (rad/sec)’),grid

From the plot we can determine that the resonant frequency is approximately 9.6 rad/s,and the peak in m is approximately 8 db. Thus

Mr = 10mr/20 = 2.51

The bandwidth frequencies are found from the plot to be from 7.2 rad/sec to 11.5 rad/sec.



8.46 a) The MATLAB code is

KT = 0.04;Ke = 0.04;c = 7e-5;R = 0.6;L = 2e-3;Im = 2e-5;IL = 4e-5;I = Im + IL;den = [I*L,c*L+I*R,c*R+Ke*KT];num1 = [KT];sys1 = tf(num1,den);bode(sys1)[mag,phase,w] = bode(sys1);mag = mag(:);

To obtain the plots for the disturbance transfer function, type

num2 = [L,R];sys2 = tf(num2,den);bode(sys2)[mag,phase,w] = bode(sys2);mag = mag(:);

You can examine the data by typing [w,20*log10(mag)]. This enables you to determinethe resonant frequency and the bandwidth more precisely than from the plot.

The plots show that there is no resonant frequency. The bandwidth for the voltageinput is from 0 to 37 rad/s. The bandwidth for the disturbance torque input is from 0 to50 rad/s.

b) From the data generated by the above code, for the voltage input, m(0) = 27.733and φ(0) = 0. This m value corresponds to M = 1027.733/20 = 24.36. Thus the constantterm 10 in the input will produce a term 10(24.36) = 243.6 in the output. Also, from thedata, M(130) = 8.4853 and φ(130) = −94.7 = −1.653 rad. Thus the steady-state outputis

ωss(t) = 243.6 + 8.4853(2) sin(130t − 1.653) = 243.605 + 16.9706 sin(130t − 1.653)



8.47 The program is given below, along with the resulting plot. The resonant frequencyis 2.51 × 104 rad/s. By moving the cursor along the plot we can determine the points atwhich the curve is 3 dB below the peak. These frequencies are 2.39 × 104 and 2.61 × 104,so the bandwidth is (2.61− 2.39)× 104 = 2.2× 103 rad/sec.

m = 0.002;k = 1e6;Kf = 20;Kb = 15;R = 10;L = 0.001;sys = tf(Kf,[m*L,m*R,k*L+Kf*Kb,k*R])roots([m*L,m*R,k*L+Kf*Kb,k*R])bode(sys)

−220

−200

−180

−160

−140

−120

−100

Mag

nitu

de (

dB)

102

103

104

105

106

−270

−180

−90

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)




8.48 Transform the equations and arrange them as follows.

(m1s2 + c1s + k1)X1(s) − (c1s + k1)X2(s) = 0

−(c1s + k1)X1(s) + (m2s2 + c1s + k1 + k2)X2(s) = k2Y (s)

Use Cramer’s method to find the transfer functions. You can use MATLAB to perform thealgebra after inserting the given numerical values for the parameters. The general resultsare

X1(s)Y (s)

=k2(c1s + k1)

D(s)

X2(s)Y (s)

=k2(m1s

2 + c1s + k1)D(s)

where

D(s) = m1m2s4 + c1(m1 + m2)s3 + (k1m2 + k1m1 + k2m1)s2 + c1k2s + k1k2

The MATLAB code is

m1 = 250;m2 = 40;k1 = 15000;k2 = 150000;c1 = 1000;D = [m1*m2,c1*(m1+m2),k1*m2+k1*m1+k2*m1,c1*k2,k1*k2];num1 = k2*[c1,k1];num2 = k2*[m1,c1,k1];bode(num1,D)[mag,phase,w] = bode(num1,D,[5:.05:9]);mag = mag(:);plot(w,mag),axis([0 20 0 3]),xlabel(’Frequency (rad/sec)’),ylabel(’M’),grid

The plots show that the major peak has a resonance frequency of 7.3 rad/sec and a band-width of 5.3 to 8.8 rad/sec. The secondary resonance is at 62 rad/sec.

b) The peak amplitude occurs at ω = 7.3 rad/sec. With a period of 10 m, this frequencycorresponds to a speed of

v =7.3(1)

2π= 11.62 m/s

or 41.8 kilometers per hour.




to accompany


by

William J. Palm III


Solutions to Problems in Chapter Nine


CD1

Highlight

9.1 From (9.1-5),

v(t) =270024

(1 − e−4t

)= 112.5

(1 − e−4t

)

Because x = v,

x(t) =∫ t

0v(t) dt = 112.5t + 28.125

(e−4t − 1

)

Set x(t) = 2000.2000 = 112.5t + 28.125

(e−4t − 1

)

Combine terms and solve the following equation for t.

112.5t + 28.125e−4t − 2028.125 = 0

This can be solved by plotting or by iteration, for example. The answer is t = 17.8 sec.



9.2 τ = 0.25 sec. Thus the speed takes approximately 4(0.25) = 1 sec to reach constantspeed. Because 0.04 sec is small compared to 1 sec, the answer is yes.



9.3a) τ = 8/7, x(t) = 6e−7t/8

b) τ = 12/5, x(t) = 3e−5t/12

c) τ = 13/6, x(t) = −2e−6t/13

d) No time constant is defined because the model is unstable. x(t) = 9e5t/7



9.4 a) xss = 10, t = 4τ = 8b) xss = 10, t = 4τ = 8c) xss = 200, t = 4τ = 8



9.5 Note that the initial condition does not affect the steady-state response.a) xss = 4, t = 4τ = 24/5b) xss = 4, t = 4τ = 4/5c) No steady-state response exists because the model is unstable.



9.6 From (9.1-5),a) x(t) = 4

(1− e−5t

)

b) x(t) = e−5t + 4(1 − e−5t

)= 4− 3e−5t

c) x(t) = −2e6t/13 − 186

(1− e6t/13

)= −3 + e6t/13



9.7 The time constant of the rotational system is τ1 = 100/5 = 20 s. The time constant ofthe field circuit is τ2 = 0.002/4 = 5 × 10−4 s. The steady-state current is ifss = 12/4 = 3A, and thus the steady-state torque is Tss = KT ifss = 15(3) = 45 N·m. It takes about4τ2 = 2 × 10−3 s for the torque to reach steady state, and since this time is much smallerthan τ1, we may treat the torque as a step input to the rotational system. The steady-statespeed is 45/5 = 9 rad/s, and it takes about 4τ1 = 80 s to reach the steady-state speed.



9.8 RCv + v = vs, where RC = 3 s. From (9.1-5),

v(t) = 6e−t/3 + 12(1− e−t/3

)



9.9 (AR/g)h + h = (R/g)qv. With the given values, this becomes

93.1677h + h = 1.8634qv

From (9.1-5),h(t) = 2e−t/93.1677 + 18.634

(1 − e−t/93.1677

)

Set h(t) = 15 ft and solve for e−t/93.1677:

e−t/93.1677 = 0.2184

which gives t = 142 sec.



9.10 mcpRT + T = Tb, where mcpR = 100(500)0.09 = 4500 s. From (9.1-5),

T (t) = 20e−t/4500 + 80(1− e−t/4500

)



9.11 Transforming the first model, 2v + v = ˙ g(t) + g(t), gives

2[sV (s) − v(0)] + V (s) = (s + 1)G(s) =10(s + 1)

s

With v(0) = 5, this becomes

V (s) = 10s + 0.5

s(s + 0.5)=

10s

Thus v(t) = 10us(t).Transforming the second model, 2v + v = g(t), gives

2[sV (s) − v(0)] + V (s) = G(s) =10s

With v(0) = 5, this becomes

V (s) = 5s + 1

s(s + 0.5)=

10s

− 5s + 0.5

Thus v(t) = 10us(t) − 5e−0.05t.



9.12 Note that g = 0 for −∞ ≤ t ≤ ∞. Thus the model is equivalent to 5v + v = g. From(9.1-5),

v(t) = 10− 5e−0.2t

for both cases.



9.13 a) Transforming the equation gives

(6s + 3)V (s) = (s + 1)G(s) =s + 1

s

orV (s) =

s + 1s(6s + 3)

=16

s + 1s(s + 0.5)

=13

1s− 1

61

s + 0.5

Thusv(t) =

13− 1

6e−0.5t

Note that v(0+) = 1/6.b) Transforming the equation gives

(6s + 3)V (s) = (s + 1)G(s) = (s + 1)[1s− 1

s + 5

]= 5

s + 1s(s + 5)

orV (s) =

56

s + 1s(s + 5)(s + 0.5)

=13

1s− 5

271

s + 0.5− 4

271

s + 5

Thusv(t) =

13− 5

27e−0.5t − 4

27e−5t

Note that v(0+) = 0.



9.14 From (9.1.10),

v(t) =(

3 +52

)e−t/2 = 5.5e−t/2



9.15 Transforming the equation gives

(2s + 1)V (s) =5s2

orV (s) =

52s2(s + 0.5)

=5s2

− 10s

+10

s + 0.5

Thusv(t) = 5t − 10 + 10e−0.5t

The transient response is 10e−0.5t. The steady-state response is 5t − 10.



9.16 Transforming the equation gives

(9s + 3)V (s) =7s2

orV (s) =

79s2(s + 1/3)

=7

3s2− 7

s+

7s + 1/3

Thusv(t) =

73t − 7 + 7e−t/3

The steady-state response is 7t/3 − 7, which is not parallel to the input 7t.



9.17 The equation reduces to x + 4x = 0, whose roots are s = ±2j. Thus the oscillationfrequency is 2 rad/sec for both cases, and

x(t) = B sin(2t + φ)

andx(0) = B sin φ

x(0) = B cos φ

a) B sin φ = 5 and B cos φ = 0. Thus, φ = π/2 rad and the amplitude is

B =5

sin π/2= 5

b) B sin φ = 0 and B cos φ = 5. Thus, φ = 0 and the amplitude is

B =5

cos 0= 5



9.18 From Table 9.2.2,a) s = −2 ± 2j. x(t) = 0.5e−2t sin 2t.b) s = −6,−2. x(t) = 0.25e−2t − 0.25e−6t.c) s = −2, −2. x(t) = te−2t.



9.19a) The roots are s = −2 ± 2j, so we use the trial solution

x(t) = Be−2t sin(2t + φ) + C

At steady state, xss = 2/8 = 0.25, so C = 0.25. We obtain B and φ from the initialconditions. The solution is

x(t) =14

[√2e−2t sin

(2t +

5π

4

)+ 1

]

b) The roots are s = −6, −2, so we use the trial solution

x(t) = C1e−6t + C2e

−2t + C3

At steady state, xss = 2/12 = 1/6, so C3 = 1/6. We obtain C1 and C2 from the initialconditions. The solution is

x(t) =16

(12e−6t − 3

2e−2t + 1

)

c) The roots are s = −2, −2, so we use the trial solution


−2t + C3

At steady state, xss = 2/4 = 0.5, so C3 = 0.5. We obtain C1 and C2 from the initialconditions. The solution is

x(t) =12

[1 − (2t + 1)e−2t

]



9.20 Using Table 9.2.1,a) s = −2, −5.

x(t) =43e−2t − 1

3e−5t

b) s = −2, −2.x(t) = (1 + t)e−2t

c) s = −2 ± 5j.x(t) = 1.02e−2t sin (5t + 1.373)



9.21a) s = −2, −5.

x(t) =130

(23e−5t − 5

3e−2t + 1

)

b) s = −2, −2.

x(t) =120

[1− (2t + 1)e−2t

]

c) s = −2 ± 5j.

x(t) =158

[15

√29e−2t sin (5t + 4.33) + 1

]



9.22 a) The roots are s = −2, −5.

X(s) =4

3s2(s2 + 7s + 10)=

43

(0.1s2

− 0.07s

+0.0833s + 2

− 0.0133s + 5

)

Thusx(t) =

43

(0.1t − 0.07 + 0.0833e−2t − 0.0133e−5t

)

b) The roots are s = −2, −2.

X(s) =4

5s2(s2 + 4s + 4)=

45

(0.25s2

− 0.25s

+0.25

(s + 2)2+

0.25s + 2

)

Thusx(t) =

45

(0.25t− 0.25 + 0.25te−2t + 0.25e−2t

)

c) The roots are s = −2 ± 5j.

X(s) =5

2s2[(s + 2)2 + 25]=

52(292)

[29s2

− 4s− 21

(s + 2)2 + 25+ 4

s + 2(s + 2)2 + 25

]

Thusx(t) =

51682

(29t − 4 +

215


)



9.23X(s) =

a

s2(ms2 + cs + k)

Let e(t) be the difference between the input and the response: e(t) = f(t) − x(t). Then

E(s) = F (s) − X(s) =a

s2

(1 − 1

ms2 + cs + k

)=

a

s2

(ms2 + cs + k − 1

ms2 + cs + k

)

From the final value theorem,

ess = lims→0

sa

s2

(ms2 + cs + k − 1

ms2 + cs + k

)=

∞ if k 6= 1ac if k = 1

The response will be parallel to the input only if k = 1. In that case, the difference betweenthe input and the response is ac.



9.24 a) ζ = cos[tan−1(6/2)] = 0.316. τ = 1/2. ωd = 6. ωn =√

4 + 36 =√

40.b) Unstable so ζ and τ are not defined. ωd = 5. ωn =

√1 + 25 =

√26.

c) ζ = 1 because the roots are real and equal. τ = 1/10. ωd and ωn are not definedbecause the free response is not oscillatory.

d) ζ is not defined because the system is first order. τ = 1/10. ωd and ωn are notdefined because the free response is not oscillatory.



9.25 a) τ = 0.5, but ζ, ωn, and ωd do not apply because the dominant root is real.b) τ = 0.5, ζ = 0.707, ωn = 2

√2, ωd = 2.



9.26 The dominant root pair is s = −2 ± 4j. For this root, ζ = cos[tan−1(4/2)] = 0.447,τ = 1/2, and ωd = 4.



9.27 a) The Routh-Hurwitz criterion implies that the system is stable if and only if −(µ +2) > 0 and 2µ + 5 > 0, which gives −2.5 < µ < −2 for stability. Neutral stability occurs ifeither the s term or the constant term is missing in the characteristic equation. This occursif µ + 2 = 0 (the roots are s = ±j) or if 2µ + 5 = 0 (the roots are s = 0, −0.5).



9.28 We have 3x + cx + 27x = 0. Using Table 9.2.1,a) (c = 0)

x(t) = cos 3t

b) (c = 9)x(t) = e−1.5t cos 2.6t − 0.5774e−1.5t sin 2.6t

c) (c = 18)x(t) = −3te−3t + e−3t

d) (c = 22)x(t) = 1.3695e−5.774t − 0.3695e−1.558t



9.29 a) Stable if and only if 6d > 0 and 25d2 > 0 (from the Routh-Hurwitz criterion).Thus, stable if and only if d > 0.

b) The damping ratio is

ζ =6d

2√

100d2=

310

< 1 for all d > 0

Thus there will be damped oscillations for any d > 0.



9.30 a) Stable if and only if 6b > 0 and 5b − 10 > 0 (from the Routh-Hurwitz criterion).Thus, stable if and only if b > 2.

b) The roots ares = −3b ±

√9b2 − 5b + 10

There will be imaginary parts if f(b) = 9b2−5b+10 < 0. Solving df/db = 0 gives b = 5/18.At b = 5/18, df2/db2 = 18 > 0, and thus f(b) has a minimum value of f(5/18) = 9.305.Therefore there is no real value of b for which 9b2 − 5b + 10 < 0, and thus no decayingoscillations.



9.31 Transforming each equation gives

Ω(s) =T (s)

50s + 10

T (s) = 25If(s)

If (s)V (s)

=1

0.0001s + 5

Thus

Ω(s)V (s)

=25

50s + 101

0.001s + 5=

500(s + 0.2)(s + 5000)

=500

s2 + 5000.2s + 1000

The damping ratio is

ζ =5000.22√

1000= 79.06

The time constants are τ1 = 5 and τ2 = 1/5000 s. The undamped natural frequency isωn =

√1000/1 = 31.62 rad/s.



9.32 The characteristic equation becomes

24× 10−8s2 + (3.6× 10−5 + 4c× 10−3)s + 0.6c + 0.01 = 0


ζ =3.6× 10−5 + 4c × 10−3

2√

24× 10−8(0.6c + 0.010=

0.09 + 10c√6(0.6c + 0.01)

The undamped natural frequency is

ωn =

√0.6c + 0.0124× 10−8

= 5000√

0.6c + 0.016

If ζ < 1, the real part of the roots is

−3.6 × 10−5 + 4c × 10−3

24× 10−8= −900 + c × 105

6

Thus the time constant isτ =

6900 + c × 105



9.33 Transforming the equation and using Y (s) = 1/s, we obtain

X(s) =s + 3

s2 + 5s + 41s

=C1

s+

C2

s + 1+

C3

s + 4

where C1 = 3/4, C2 = −1, and C3 = 1/4. Thus

x(t) =34− e−t +

14e−4t for t > 0



9.34 Transforming the equation and using Y (s) = 1/s, we obtain

X(s) =s + 3

s2 + 4s + 851s

=C1

s+ C2

9(s + 2)2 + 81

+ C3s + 2

(s + 2)2 + 81

since the roots are s = −2 ± 9j. The coefficients are C1 = 3/85, C2 = 79/765, andC3 = −3/85. Thus

x(t) = C1 + C2e−2t sin 9t + C3e

−2t cos 9t =385

+79765

e−2t sin 9t − 385

e−2t cos 9t for t > 0



9.35 Transforming the equation and using Y (s) = 5, we obtain

X(s) =3(5)

s2 + 5s + 4=

C1

s + 1+

C2

s + 4

where C1 = 5 and C2 = −5. Thus

x(t) = 5e−t − 5e−4t for t > 0



9.36 ζ = 1/√

2 = 0.707 which implies from Figure 9.3.5a that the maximum percentovershoot is ≈ 5%. Because xss = 2/8 = 0.25, the overshoot is 0.05(0.25) ≈ 0.01.

For ζ = 0.707, Figure 9.3.5c shows that ωntr ≈ 3.2. Because ωn =√

8/1 = 2√

2,tr = 3.2/2

√2 = 1.1.

For ζ = 0.707, Figure 9.3.5a shows that ωntp ≈ 4.6. Because ωn = 2√

2, tp = 4.6/2√

2 =1.6.

For ζ = 0.707 and ωn = 2√

2, Table 9.3.2 gives td = 0.53.The roots are −2± 2j, so the time constant is τ = 0.5. The 2% settling time is 4τ = 2.



9.37 ωn =√

4/1 = 2. For tr ≤ 3, ωntr ≤ 2(3) = 6. From Figure 9.3.5c, we see that thisrequires that ζ ≤ 0.9. Thus ζ must be no greater than 0.9 if tr is to be no greater than 3.

Percent overshoot ≤ 20% implies from Figure 9.3.5a that ζ must be ≥ 0.43. Thus ζ mustbe in the range 0.43 ≤ ζ ≤ 0.9 to satisfy both specifications. To minimize the overshoot,Figure 9.3.5a shows that we should choose ζ as large as possible (within the above range).Thus we choose ζ = 0.9. The damping c is found from the definition of ζ: ζ = c/2

√4 = c/4.

Thus c = 0.9(4) = 3.6.



9.38 The approach is similar to that of Problem 9.37, but the two specifications cannot bothbe satisfied, so we give priority to the overshoot specification, as directed. The calculationsare as follows. ωn =

√4/9 = 2/3

For tr ≤ 3, ωntr ≤ 2(3)/3 = 2. From Figure 9.3.5c, we see that this requires thatζ ≤ 0.35. Thus ζ must be no greater than 0.35 if tr is to be no greater than 3.

Percent overshoot ≤ 20% implies from Figure 9.3.5a that ζ must be ≥ 0.43. Thus ζ

cannot satisfy both specifications. So we set ζ = 0.43, giving priority to the overshootspecification, as directed. (We could choose ζ > 0.43 to minimize the overshoot, but thiswould cause the rise time to become much larger than desired.)

Thus we choose ζ = 0.43. The damping c is found from the definition of ζ: ζ =c/2√

9(4) = c/12. Thus c = 0.43(12) = 5.16.



9.39tp =

π

ωn

√1 − ζ2

But the imaginary part of the roots is ωd = ωn

√1 − ζ2. Thus tp = π/ωd, and thus tp

depends only on the imaginary part of the roots.



9.40 For the system in part (a) of the figure,

mx + cx + kx = cy + ky

With the given values,3x + 18x + 10x = 18y + 10y

Note that Table 9.3.1 cannot be used because of the input derivative term y. Transformingthis equation with zero initial conditions gives

X(s) =18s + 10

3s2 + 18s + 10Y (s) =

6s + 10/3s(s + 0.6195)(s + 5.3805)

This expands to

X(s) =1s

+0.13

s + 0.6195− 1.13

s + 5.3805Thus

x(t) = 1 + 0.13e−0.6195t − 1.13e−5.3805t

For the system in part (b) of the figure,

mx + cx + kx = ky

With the given values,3x + 18x + 10x = 10y

Note that Table 9.3.1 can be used here. This result is

x(t) = 1 + 0.13e−0.5.3805 − 1.13e−0.6195t

The responses are shown in the plot on the following page. The two responses are verydifferent, due to the numerator dynamics of the model in part (a) of the figure.





0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

t

x(t)

(a)

(b)




9.41 For part (a) of the figure,

X(s)Y (s)

=18s + 10

3s2 + 18s + 10

In MATLAB, type

sysa = tf([18,10],[3,18,10]);

When the plot appears, right click on it and select “Characteristics”. From there you canselect the desired items to display on the plot. The results are: maximum per cent overshoot= 7%, peak time = 0.924, 2% settling time = 3.02, and rise time = 0.298. Note that youcannot use the formulas in Table 9.3.2 because this model has numerator dynamics.

For part (a) of the figure,

X(s)Y (s)

=10

3s2 + 18s + 10

In MATLAB, type

sysb = tf(10,[3,18,10]);

When the plot appears, right click on it and select “Characteristics”. From there youcan select the desired items to display on the plot. The results are: maximum per centovershoot = 0% (there is no overshoot), so the peak time does not apply. Also, 2% settlingtime = 6.51 and rise time = 3.6. Note how much larger the settling time at the rise timeare, as compared to the model having numerator dynamics. Thus numerator dynamics canreduce response time but can cause overshoot.



9.42 The free response of the circuit is

vC(t) = vC(0)e−t/τ

where τ = RC = 3 × 106C. A plot of ln vC(t) versus t gives a straight line whose slope−1/τ is computed from

−1τ

=ln 12 − ln 6.2

0 − 20= −0.033

Thus τ = 1/0.033 = 30.286 and C = τ/R = 10−5 F.



9.43 The free response is∆T (t) = ∆T (0)e−t/τ

A plot of ln ∆T (t) versus t gives a straight line whose slope −1/τ is computed from

−1τ

=ln(178− 68)− ln(82 − 68)

0 − 3000= −6.8714× 10−4

Thus τ = 1455 sec.Note that ∆T (0) = 178 − 68 = 110. The model is

T (t) = 68 + 110e−t/1455

For T = 135, we have135 = 68 + 110e−t/1455

ort = −1455 ln

135− 68110

= 721 sec



9.44 From the graph we can easily identify the following three characteristics: the steady-state response, xss = 0.025 m, the peak time, tp = 0.13 s, and the maximum per centovershoot,

M% =0.0395− 0.025

0.025100 = 58%

From Table 9.3.2,

R = ln100M%

= 0.5447

Thusζ =

R√π2 + R2

= 0.171

Thus√

1− ζ2 = 0.985.Because the applied force is 1000 N, xss = 1000/k, and thus k = 1000/0.025 = 40 000

N/m.From

tp =π

ωn

√1 − ζ2

we obtainωn =

π

tp√

1− ζ2= 24.53 rad/s

so thatm =

k

ω2n

=40 000(24.53)2

= 66.476 kg

Becauseζ =

c

2√

mk

we havec = 2ζ

√mk = 2(0.171)

√66.476(40 000) = 557.7 N · s/m



9.45 From the log decrement relations (9.4.12) and (9.4.14),

δ =1n

lnB1

Bn+1ζ =

δ√4π2 + δ2

Thusδ =

130

ln 5 = 0.0536 ζ = 0.0085

Also,ζ =

c

2√

mk=

c

2√

100k=

c

20√

k= 0.0085

Thus c = 0.17√

k.We are told the time to complete the 30 cycles is 60 s, so we can compute the period P

for one cycle from P = 60/30 = 2 s. Then,

k = mω2n =

mω2d

1 − ζ2=

m(2π/P )2

1 − ζ2=

100(2π/2)2

1− (0.0085)2= 987 N/m

and thus c = 0.17√

k = 5.34 N s/m.



9.46 From the diagram:

X(s) =1s6 [F (s) − X(s)] + 4X(s)

Solve for the ratio:X(s)F (s)

=6

s + 2



9.47 From the problem figure:

X(s) =1

s + 3G(s) + 10 [F (s) − X(s)]

Set G(s) = 0 and solve for the ratio:

X(s)F (s)

=10

s + 13



9.48 From the problem figure:

X(s) =1s

−G(s) − 8X(s) +

4s

[F (s) − 6X(s)]

Set G(s) = 0 and solve for the ratio:

X(s)F (s)

=4

s2 + 8s + 7



9.49 The following diagram is one of several possible ones:




9.50 The following diagram is one of several possible ones:




9.51 From the diagram,

C(s) =1

7s + 1

4s + 103s + 1

[R(s) − C(s)] − D(s)

Solve for C(s) to obtain

C(s) =4s + 10A(s)

R(s)− 3s + 1A(s)

D(s) (1)

whereA(s) = 21s2 + 14s + 11 (2)

From the diagram, and using (1),

E(s) = R(s) − C(s) = R(s) − 4s + 10A(s)

R(s) +3s + 1A(s)

D(s)

or

E(s) =21s2 + 10s + 1

A(s)R(s) +

3s + 1A(s)

D(s) (3)


M(s) =4s + 103s + 1

E(s) =(4s + 10)(7s + 1)

A(s)R(s) +

4s + 10A(s)

D(s)




C(s) =1

3s + 2

(1

4s + 1

6s

[R(s) − C(s)] − 10C(s)− D(s)

)

Solve for C(s) to obtain

C(s) =6

A(s)R(s) − 4s2 + s

A(s)D(s) (1)

whereA(s) = 12s3 + 11s2 + 12s + 6 (2)


E(s) = R(s) − C(s) = R(s) − 6A(s)

R(s) +4s2 + s

A(s)D(s)

or

E(s) =12s3 + 11s2 + 12s

A(s)R(s) +

4s2 + s

A(s)D(s) (3)

From the diagram,

C(s) =1

3s + 2[M(s) − D(s)]

Thus, using (1),

M(s) = (3s + 2)C(s) + D(s) =18s + 12

A(s)R(s) +

10s + 6A(s)

D(s)



9.53 For one loop,

v1 = Ldi1dt

+ Ri3 (1)

For the other loop:

Ri3 =1C

∫i2 dt + v2 (2)

From conservation of charge:i1 = i2 + i3 (3)

Solve (1) and (2) for i1 and i2:

Ldi1dt

= v1 − Ri3 (4)

1C

∫i2 dt = Ri3 − v2 (5)

Transform (3), (4), and (5), and solve for I1(s) and I2(s):

I1(s) =V1(s)− RI3(s)

Ls(6)

I2(s) =RI3(s) − V2(s)

Cs(7)

I2(s) = I3(s)− I1(s) (8)

The left side of the diagram on the following page is drawn from equation (6); the right sidefrom (7), and the middle summer from (8).



9.54 From the first loop:

v1 = R1i1 + L1di1dt

+ v3 (1)

From the second loop:

v3 = L2di2dt

+ v2 (2)

From conservation of charge:i1 = i2 + i3 (3)


(R + L1s)I1(s) + V3(s) = V1(s) (4)

L2sI2(s) − V3(s) = −V2(s) (5)

I1(s) = I2(s) + I3(s) (6)

Solve (4) for I1(s):

I1(s) =1

L1s + R[V1(s) − V3(s)]

This gives the left side of the diagram on the following page.Solve (5) for I2(s):

I2(s) =1

L2s[V3(s) − V2(s)]

This gives the top of the diagram.Solve (6) for I3(s):

I3(s) = I1(s) − I2(s)

This gives the summer in the middle of the diagram.Finally, use the fact that

v3 =1C

∫i3 dt

to obtainV3(s) =

1Cs

I3(s)

This gives the right side of the diagram.



9.55 See the following figure. The equivalent inertia and damping at the motor shaft are

Ie = Im +IL

N2ce =

cL

N2+ cm


Ωf(s)Vf(s)

=KT /N

(Lfs + Rf)(Ies + ce)

Ωf(s)TL(s)

=1/N2

Ies + ce




9.56 The Simulink digram is identical to that shown in Figure 9.8.6. Edit the Signal Builderand the Look-Up Table blocks to reflect the new data. Change the m-file damper.m to readas follows.

function f = damper(v)if v <= 0

f = -500*(abs(v)).^(1.2);else

f = 50*(abs(v)).^(1.2);end

A Stop Time of 15 seconds gives a suitable plot. The maximum overshoot is 0.522−0.3 =0.222 m at approximately t = 0.57 s. The maximum undershoot is 0− 0.1765 = −0.1765 mat approximately t = 4.73 s.



9.57 a) The state variable representation is much easier to do because it does not requirethe algebra needed to find the transfer functions. The state space matrices are

A =

0 1 0 0−5 −4 4 10 0 0 14 1 −4 −1

B =

0001

C =

[1 0 0 00 0 1 0

]D =

[00

]


X1(s)F (s)

=s + 4

s4 + 5s3 + 12s2 + 13s + 4

X2(s)F (s)

=s2 + 4s + 5

s4 + 5s3 + 12s2 + 13s + 4

b) The figure on the following page shows the Simulink model using both representations.For each you must create the following MATLAB function.

function f = func(t)if t <= 1

f = t;elseif t < 2

f = 2 - t;else

f = 0;end




to accompany


by

William J. Palm III


Solutions to Problems in Chapter Ten


CD1

Highlight

10.1 a) A traffic light may be either closed or open loop, depending on whether or not ituses a sensor to detect the presence of vehicles in the roadway.

b) Most washing machines are open loop, but some now have sensors that detect theamount of dirt in the wash water and adjust their cycle accordingly. These would beclosed-loop devices.

c) If the toaster uses a timer, it is open loop. If it uses a sensor to tell when the desiredtemperature has been reached, it is closed loop.

d) Cruise control is closed loop because it uses a measurement of the vehicle speed tocontrol the engine.

e) A aircraft autopilot is closed loop because it uses measurements of a variety of vari-ables (speed, altitude, angle, etc.) to adjust the control surfaces (ailerons, rudder, elevators,etc.).

f) Closed loop because it reacts to changes in the environment to keep the temperaturenear 98.6.



10.2 The diagram is like Figure 10.1.7 with

Gc(s) = KP

Ga(s) = 1

Gp(s) =1

4s2 + 6s + 3The command transfer function is found from the diagram as follows.

C(s) =1

4s2 + 6s + 3[Gf (s)R(s) + KP R(s) − KPC(s)]

This givesC(s)R(s)

=Gf (s) + KP

4s2 + 6s + 3 + KP

The system will give perfect response if C(s)/R(s) = 1. This requires that

Gf (s) = 4s2 + 6s + 3

The difficulty with implementing this design is that it requires the first and secondderivatives of the input, r and r, to be computed. This would be difficult to accomplish,especially for step inputs.



10.3 The diagram is a combination of Figures 10.1.7 and 10.1.8 with

Gc(s) = KP

Ga(s) = 1

Gp(s) =10s

Gf(s) = Kf

Gd(s) = Kd

The command transfer function is found from the diagram as follows.

C(s) =10s

[−D(s) + KdD(s) + KfR(s) + KP R(s)− KP C(s)]

orC(s) =

10(Kf + KP )s + 10KP

R(s) +10(Kd − 1)s + 10KP

D(s)

The disturbance will have no effect if Kd = 1. The command transfer function is

C(s)R(s)

=10(Kf + KP )

s + 10KP

If R(s) = 1/s, the steady-state response is

css =10(Kf + KP )

10KP

which will be 1 if Kf +KP = KP , which is true if Kf = 0. Thus Kf is not needed to satisfythe given specifications.

The time constant expression is

τ =1

10KP

Setting KP = 0.05 will give the desired time constant τ = 2. So the design is KP = 0.05,Kd = 1, and Kf = 0.




C(s) =6

15s + 2[M(s) − D(s)] (1)

M(s) =4K

3s + 1E(s) (2)

E(s) = R(s) − C(s) (3)

To obtain the output C(s) in terms of R(s) and D(s), eliminate M(s) and E(s) fromequations (1), (2), and (3) to obtain

C(s)R(s)

=24K

45s2 + 21s + 2 + 24K

C(s)D(s)

= − 6(3s + 1)45s2 + 21s + 2 + 24K

To obtain the error E(s) in terms of R(s) and D(s), eliminate C(s) and M(s) fromequations (1), (2), and (3) to obtain

E(s)R(s)

=45s2 + 21s + 2

45s2 + 21s + 2 + 24K

E(s)D(s)

=6(3s + 1)

45s2 + 21s + 2 + 24K

To obtain M(s) in terms of R(s) and D(s), eliminate C(s) and E(s) from equations (1),(2), and (3) to obtain

M(s)R(s)

=4K(45s2 + 21s + 2)

(3s + 1)(45s2 + 21s + 2 + 24K)=

4K(15s + 2)45s2 + 21s + 2 + 24K

M(s)D(s)

=24K

45s2 + 21s + 2 + 24K

The characteristic polynomial is the denominator of the transfer functions: 45s2 +21s+2 + 24K.



10.5 The command transfer function is

C(s)R(s)

=KP

5τas2 + (5 + τa)s + 1 + KP

If τa = 0.05, the characteristic roots are

s = −11.1 ±√

98.01− 4KP

The roots are real if KP ≤ 98.01/4 = 24.5. If KP > 24.5 the roots are complex and theclosed-loop time constant is 1/11.1 = 0.09.

If we neglect τa by setting it equal to zero, the characteristic equation becomes a first-order equation: 5s + 1 + KP = 0. Thus the root is always real and the closed-loop timeconstant is

τ =5

1 + KP

which goes to zero as we increase KP . So if we neglect the actuator time constant, we areled to believe that the step response will not oscillate and that we can make the closed-looptime constant as small as we like by increasing KP . However, in fact the time constant canbe made no smaller that 0.09 and the response will oscillate if KP > 24.5.



10.6 a) Summing forces parallel to the plane gives

mv = f − F − Wx

b) The block diagram is shown below. The control algorithm is Gc(s), and the motorconstants are KT and Rf . The second diagram is a simplified form where

G1(s) = Gc(s)KaKT

Rf

G2(s) =1

ms






c) From the diagram on the following page,

V (s) = G2(s)[T (s)R

− D(s)]

(1)

T (s) = G1(s)E(s) (2)

E(s) = Vr(s) − V (s) (3)

To obtain the output V (s) in terms of Vr(s) and D(s), eliminate T (s) and E(s) fromequations (1), (2), and (3) to obtain

V (s)Vr(s)

=G1(s)G2(s)

R + G1(s)G2(s)=

KaKTGc(s)mRfRs + KaKTGc(s)

V (s)D(s)

= − RG2(s)R + G1(s)G2(s)

= − RRf

mRfRs + KaKTGc(s)

To obtain the error E(s) in terms of Vr(s) and D(s), eliminate T (s) and V (s) fromequations (1), (2), and (3) to obtain

E(s)Vr(s)

=mRfRs

mRfRs + KaKTGc(s)

E(s)D(s)

=RRf

mRfRs + KaKTGc(s)

To obtain the torque T (s) in terms of Vr(s) and D(s), eliminate E(s) and V (s) fromequations (1), (2), and (3) to obtain

T (s)Vr(s)

=mKaKTRsGc(s)

mRfRs + KaKTGc(s)

T (s)D(s)

=KaKT RGc(s)

mRfRs + KaKTGc(s)





Figure : for Problem 10.6c



10.7 a) The diagram is shown in the following figure. Since no load torque is mentioned,we assume that TL = 0.

b)

Ie = I1 +122

I2 +1

[2(3)]2I3 = I1 +

14I2 +

136

I3

Ne = 2(3) = 6




10.8

KP +KI

s+ KDs =

KDs2 + KP s + KI

s

Thus KD = 15, KP = 6, and KI = 4, and

TI =KP

KI=

64

= 1.5

TD =KD

KP=

156

= 2.5



10.9 From Figure 10.4.4,

KP =Rf

Ri= 4

KI =1

RiC= 0.08

Using C = 10−6, we obtain

Ri =1

0.08× 10−6= 1.25× 107 Ω

Rf = 4Ri = 5 × 107 Ω



10.10 a) From Figure 10.4.6,

KP =R

R1 + R2= 2 (1)

TD = R2C = 2 (2)

α =R1

R1 + R2= 0.1 (3)

The denominator 1+αTDs causes the m curve to level off for ω > 1/αTD. So we choose

1αTD

= 5

which gives α = 0.2/TD = 0.2/2 = 0.1.Using C = 10−6, we obtain from (2): R2 = 2× 106 Ω. We then solve (3) for R1:

R1 =29× 106 Ω

and solve (1) for R:

R =409

× 106 Ω

b) The transfer function isVo(s)Vi(s)

= −2(1 + 2s)1 + 0.2s

The MATLAB code to obtain the frequency response plot is

sys = tf([4,2],[0.2,1]);bodemag(sys),grid

The plot is shown in the figure on the following page.





10−2

10−1

100

101

102

5

10

15

20

25

30

Mag

nitu

de (d

B)

Bode Diagram

Frequency (rad/sec)




10.11 From Figure 10.4.7,

KP = βRC + R2C1

R2C= 10 (1)

KI =β

R2C= 1.4 (2)

KD = βRC1 = 4 (3)

β =R2

R1 + R2(4)

From (2),

R2 =β

1.4C(5)

From (3),

R =4

βC1(6)

From (4),

R1 = R21 − β

β=

1 − β

1.4C(7)

Substituting (2), (5), and (6) into (1) gives

1.4(

4C

βC1+

βC1

1.4C

)= 10

This can be rearranged as follows:

(βC1)2 − 10C(βC1) + 5.6C2 = 0

Choosing C = 10−6 we obtain

(βC1)2 − 10−5(βC1) + 5.6× 10−12 = 0

which has the solutionsβC1 = 5.9546× 10−7, 9 × 10−6





Choosing the first solution we obtain

R = 6.7175× 106 Ω

R2 =0.4253

C1

R1 = 7.1429× 105 − 0.4253C1

(8)

From this, since we require that R1 > 0, we see that C1 must satisfy

C1 > 5.9546× 10−7 (9)

To limit the response above 100 rad/s, we require that

1βR1C1

≥ 100

or1

5.9546× 10−7R1≥ 100

This requires that R1 ≤ 1.679× 104. Combining this with (8) and (9) shows that C1 mustlie in the range

5.9546× 10−7 < C1 ≤ 6.3× 10−7

Choosing C1 = 6 × 10−7 F gives R1 = 5407 Ω, R2 = 7.089 × 105 Ω, and R = 4/βC1 =6.7175× 106 Ω.

The second solution, βC1 = 9 × 10−6, gives a much smaller allowable range for C1.



10.12 a)

Y (s) =6

7s + 314s

=84

(7s + 3)s

Thusyss = lim

s→0s

84(7s + 3)s

=843

= 28

b)

Y (s) =7s − 3

10s2 + 6s + 95s

Thusyss = lim

s→0s

7s − 310s2 + 6s + 9

5s

= −159

= −53

c)

Y (s) =3s + 5s2 − 9

12s

ThussY (s) = 12

3s + 5s2 − 9

The roots of the denominator are s = ±3. Since one root is positive, the final value theoremcannot be applied.

d)

Y (s) =4s + 3

s2 + 2s − 78s

ThussY (s) = 8

4s + 3s2 + 2s − 7

One of the roots of the denominator is positive, so the final value theorem cannot be applied.



10.13 For each case, the error transform is

E(s) = R(s) − C(s) = R(s) − T (s)R(s) = [1− T (s)]R(s)

whereT (s) =

C(s)R(s)

a)

E(s) =3s

3s + 16s2

=18

3s + 11s

ess = lims→0

s18s

3s + 11s2

= lims→0

183s + 1

= 18

b)

E(s) =3s − 43s + 1

6s2

ess = lims→0

s3s − 43s + 1

6s2

= ∞

c)

E(s) =3s2 + 5s

3s2 + 5s + 412s2

=3s + 5

3s2 + 5s + 412s2

ess = lims→0

s3s + 5

3s2 + 5s + 412s2

= lims→0

123s + 5

3s2 + 5s + 4= 15

d)

E(s) =2s2 + 4s − 52s2 + 4s + 5

8s2

ess = lims→0

s2s2 + 4s − 52s2 + 4s + 5

8s2

= lims→0

2s2 + 4s − 52s2 + 4s + 5

8s

= ∞



10.14 Divide the equation by 3 and obtain the characteristic equation:

s2 − (b + 2)s + 2b + 5 = 0

The Routh-Hurwitz criterion implies that the system is stable if and only if −(b+2) > 0and 2b + 5 > 0, which gives −2.5 < b < −2 for stability. Neutral stability occurs if eitherthe s term or the constant term is missing in the characteristic equation. This occurs ifb + 2 = 0 (the roots are s = ±j) or if 2b + 5 = 0 (the roots are s = 0,−0.5.).



10.15 From the Routh-Hurwitz criterion it is necessary and sufficient that K > 0 and9(26)− K > 0. Thus, the system is stable if and only if 0 < K < 234.



10.16 The time constant requirement means that no root can lie to the right of s = −2.Translate the origin of the s plane to s = −2 by substituting s = p−2 into the characteristicequation. This gives

(p− 2)3 + 9(p − 2)2 + 26(p− 2) + K = 0

Expanding and collecting terms gives

p3 + 3p2 + 2p + K − 24 = 0

We can apply the Routh-Hurwitz criterion to this equation to determine when all roots phave negative real parts (and thus when all s roots lie to the left of s = −2). This occurswhen K − 24 > 0 and 3(2)− (K − 24) > 0. Thus K must be in the range 24 < K < 30.



10.17 a) From the Routh-Hurwitz criterion,

a > 0 K > 0 b > 0

and 2aK > 2b, or aK > b.b) From the Routh-Hurwitz criterion,

a > 0 b > 0 K > 0

and 5aK > 25K, or ab > 5K.c) From the Routh-Hurwitz criterion,

4 + K > 0

and 12(12) > 4(4 + K). Thus−4 < K < 32



10.18 Referring to Example 10.3.5, we obtain the following transfer function.

Ω(s)Ω(s)

=KP KT

D(s)

where D(s) is given by equation (3) in that example. The equivalent inertia is

Ie = Im +IL + It

N2= 17.75× 10−4

The equivalent damping isce = cm +

cL

N2= 10−3

The characteristic polynomial is

D(s) = 1.42× 10−5s2 + 2.848× 10−3s + 0.0816 + 0.2KP

The steady-state requirement is

ωLss

ωr=

0.2KP

0.0816 + 0.2KP= 0.9

This gives KP = 3.672. The resulting roots are

s = −100.282± 217.735j

So the time constant is 1/100.282 = 0.00997 s and the damping ratio is 0.418.The disturbance transfer function is

ΩL(s)TL

= −(Las + Ra)/ND(s)

The steady-state deviation caused by the disturbance torque is

ωLss = − Ra/N

NRace + NKTKb + KP KTTL = −0.8/2

0.8160.2 = −0.49(0.2) = 0.098



10.19 The steady-state offset error is

offset error = ωr − ωss = 1 − KP

3 + KP= 0.2

if KP = 12. The time constant for this value of KP is

τ =I

c + KP=

23 + KP

=215

The steady-state response due to the disturbance is

disturbance response =−1

c + KP=

−115

= −0.067

If both inputs are applied, the actual steady-state speed is

ωss = command response + disturbance response = (1− 0.2) + (−0.067) = 0.733

Note that we can make the offset error, the disturbance response, and the time constantsmaller only by making KP larger than 12, but this increases the maximum required torqueand probably the cost of the system.



10.20 Substitute I = 2, c = 3, and KP = 12 into equations (1) and (2) of Example 10.6.1to obtain:

Ω(s)Ωr(s)

=12

2s + 15

Ω(s)Td(s)

= − 12s + 15

Using Td(s) = 1/s2 with the final value theorem, we find the steady-state disturbanceresponse to be

ωss = lims→0

s−1

2s + 151s2

= −∞

Thus the controller cannot keep the output near its desired value if the ramp disturbancelasts too long.

To investigate the command response, use Ωr(s) = 1/s2 with the first transfer functionto obtain

Ω(s) =12

2s + 151s2

=4

5s2− 8

75s+

875(s + 15/2)

Thus the response is

ω(t) =45t − 8

75+

875

e−15t/2

Since the slope is 4/5 and is less than the slope of the input, we can see that the speed ω(t)never catches up with the command input ωr(t), and that the steady-state error is infinite.We could have also obtained this result from the final value theorem, using the followingexpression for the error when no disturbance is present.

E(s) = Ωr(s) − Ω(s) = Ωr(s)(1 − 12

2s + 15

)= Ωr(s)

(2s + 32s + 15

)

Thus, with Ωr(s) = 1/s2,

ess = lims→0

s1s2

(2s + 32s + 15

)= ∞



10.21 The transfer functions are

Ω(s)Ωr(s)

=KI

20s2 + cs + KI

Ω(s)Td(s)

= − s

20s2 + cs + KI

The steady-state unit-step response for the command input is

ωss =KI

KI= 1

which is perfect. The steady-state deviation caused by a unit-step disturbance is

∆ωss = 0

which is also perfect.We also have

ζ =c

2√

20KI= 1

which gives

KI =c2

80Since ζ = 1 we can write

τ =2(20)

c=

40c

So, if c = 10,

KI =10080

= 1.25

τ =4010

= 4

If c = 0.2,

KI =0.0480

= 0.0005

τ =400.2

= 200



10.22 (a) The command transfer function is

Ω(s)Ωr(s)

=KP s + KI

4s2 + (4 + KP )s + KI(1)

For both cases (1) and (2), the time constant is

τ =8

4 + KP= 0.2

Therefore, KP = 36 for both cases. For case (1),

ζ =40

2√

4KI= 0.707

Thus, KI = 200. For case (2), the same procedure gives KI = 100.(b) Because ζ > 1 is required, there will be two real roots, and the dominant root must

be s = −1/τ = −5. For this to be the dominant root, the second root must be to the leftof s = −5. Choosing an arbitrary separation factor of 10, we place the secondary root ats = −50. The characteristic polynomial must therefore be

(s + 5)(s + 50) = s2 + 55s + 250 = 0

Multiply this equation by 4 before comparing it with the denominator of equation (1),so that the highest coefficients will be the same in each polynomial. Thus the requiredcharacteristic polynomial must be

4s2 + 220s + 1000

Comparing its coefficients with those of the transfer function denominator gives the equa-tions for the required gain values.

4 + KP = 220

KI = 1000

Thus, KP = 216.





For a unit-step command input, a MATLAB file to compute the unit-step response forthe first case, where KP = 216 and KI = 1000, is

KP = 216; KI = 1000;sys = tf(4*[KP,KI],[4,4+KP,KI]);

Right-click on the resulting plot, select “Characteristics”, then “Peak Response” to deter-mine the maximum percent overshoot, and the rise time. The following table summarizesthe important response characteristics. The overshoot decreases as ζ increases, as expected.But the 10-90% rise time is the smallest for ζ = 1.74. A common misconception is thatresponse is sluggish for ζ > 1, but this is clearly not the case here. The gains KP = 216,KI = 1000 give a good response with a very small rise time and a small overshoot. However,high gain values are required, and this might make the physical system expensive.

The fast response with ζ = 1.74 is due to the numerator dynamics, whose effect isincreased by the high gain values used in the third case. This effect can be seen by comparingthe actual overshoot and rise time with those predicted by the second-order model withoutnumerator dynamics. These values are given in the following table.

Without Numerator DynamicsOvershoot Rise Time Overshoot Rise Time

Case KP KI ζ % 10 − 90% % 10 − 90%1 9 50 0.707 16 0.14 4 0.362 9 25 1 8 0.18 0 0.663 54 250 1.74 5 0.03 0 0.44



10.23 With the disturbance torque absent, the motor torque expression can be found asfollows. From the block diagram,

T (s) =(

KPKI

s

)E(s) =

(KPs + KI

s

)[Ωr(s) − Ω(s)]

Using the transfer function Ω(s)/Ωr(s), we have

T (s) =(

KP s + KI

s

)Ωr(s)

(1 − KP s + KI

4s2 + (4 + KP )s + KI

)

or

T (s) = 4KP s2 + (KP + KI)s + KI

4s2 + (4 + KP )s + KIΩr(s) (1)

This can be used to obtain the unit-step responses for the various gain values computedabove.

For a unit-step command input, a MATLAB file to compute the actuator response forthe first case, where KP = 36 and KI = 200, is

KP = 36; KI = 200;sys = tf(4*[KP,KP+KI,KI],[4,4+KP,KI]);

Right-click on the resulting plot, select “Characteristics”, then “Peak Response” to deter-mine the peak value. For Case 1 it is 36. Repeating this for the other cases, we obtain peakvalues of 36 for case 2 and 216 for case 3.

Note that the maximum torque occurs at t = 0 for all three cases, that the maximumtorque equals KP , and thus is much greater for the case 3, which has the largest gain values.So we see that the fast rise time of case 3 is obtained at the expense of having to provide amotor with a larger torque capability.





In the absence of a disturbance, the maximum torque occurs at t = 0 for both theproportional control system and the PI control system, if the command is a step function.The maximum torque is KP M , where M is the step magnitude. This is because the stepinput is a sudden command, and the proportional term responds instantaneously to try toreduce the error, whereas the integral term takes time to build up. Thus the maximumerror, and maximum torque, occur at t = 0. At t = 0, the speed is zero and the torque is

T (0) = KP e(0) = KP [ωr(0)− ω(0)] = KP ωr(0) = KP M

This reasoning assumes that the maximum error occurs at t = 0. This is not true for somesystems, such as the system shown in Figure 10.6.7. For such systems, there is no simpleformula to compute the maximum torque.



10.24 The error equation is

E(s) =4s2 + 4s

4s2(4 + Kp)s + KIΩr(s) +

s

4s2(4 + Kp)s + KITd(s)

Apply the final value theorem using Ωr(s) = Td(s) = 1/s2. For the command error,

ess =4

KI

• For ζ = 0.707, KI = 200, and ess = 4/200

• For ζ = 1, KI = 100, and ess = 4/100

• For ζ = 1.74, KI = 1000, and ess = 4/1000

For the disturbance error,

ess =1

KI

• For ζ = 0.707, KI = 200, and ess = 1/200

• For ζ = 1, KI = 100, and ess = 1/100

• For ζ = 1.74, KI = 1000, and ess = 1/1000



10.25 From the figure given in the problem statement ,

Ω(s)Ωr(s)

=KP s + KI

2s2 + (2 + KP )s + KI

So the characteristic polynomial is

2s2 + (2 + KP )s + KI (1)

a) 1. The characteristic polynomial must be factored as

2(s + 10)(s + 8) = 2s2 + 36s + 160

Comparing coefficients with the characteristic polynomial (1), we obtain

KP = 34 KI = 160

2. The characteristic polynomial must be factored as

2(s + 10)(s + 20) = 2s2 + 60s + 400


KP = 58 KI = 400


2(s + 10)(s + 50) = 2s2 + 120s + 1000


KP = 118 KI = 1000




Ω(s)Ωr(s)

=KI

4s2 + (4 + K2)s + KI


4s2 + (4 + K2)s + KI

and the damping ratio is

ζ =4 + K2

2√

4KI=

4 + K2

4√

KI

a) 1. Since ζ ≤ 1, we can write an expression for the time constant:

τ =8

4 + K2= 0.2

which gives K2 = 36. Thus

ζ =4 + K2

4√

KI=

10√KI

= 0.707

which gives KI = 200.2. Since ζ ≤ 1, we can write an expression for the time constant:

τ =8

4 + K2= 0.2

which gives K2 = 36. Thus

ζ =4 + K2

4√

KI=

10√KI

= 1

which gives KI = 100.3. The characteristic polynomial must be factored as

4(s + 5)(s + 50) = 4s2 + 220s + 1000

Comparing coefficients with the characteristic polynomial, we obtain

K2 = 216 KI = 1000



10.27 The actuator equation is

T (s) =(4s + 4)KI

4s2 + (4 + K2)s + KIΩr(s) +

K2s + KI

4s2 + (4 + K2)s + KITd(s)

For a unit-step command input, a MATLAB file to compute the actuator response for thefirst case, where K2 = 36 and KI = 200, is

K2 = 36; KI = 200;sys = tf(KI*[4,4],[4,4+K2,KI]);

Right-click on the resulting plot, select “Characteristics”, then “Peak Response” to deter-mine the peak value. For Case 1 it is 14.5. Repeating this for the other cases, we obtainpeak values of 8.58 for case 2 and 16.1 for case 3.



10.28 The error equation is

E(s) =4s2 + (4 + K2)s

4s2(4 + K2)s + KIΩr(s) +

s

4s2(4 + K2)s + KITd(s)

Apply the final value theorem using Ωr(s) = Td(s) = 1/s2. For the command error,

ess =4 + K2

KI

• For ζ = 0.707, K2 = 36, KI = 200, and ess = 40/200

• For ζ = 1, K2 = 36, KI = 100, and ess = 40/100

• For ζ = 1.74, K2 = 216, KI = 1000, and ess = 220/1000

For the disturbance error,

ess =1

KI

• For ζ = 0.707, KI = 200, and ess = 1/200

• For ζ = 1, KI = 100, and ess = 1/100

• For ζ = 1.74, KI = 1000, and ess = 1/1000




Ω(s)Ωr(s)

=KI

2s2 + (2 + K2)s + KI


2s2 + (2 + K2)s + KI (1)

a) 1. The characteristic polynomial must be factored as

2(s + 10)(s + 8) = 2s2 + 36s + 160


K2 = 34 KI = 160


2(s + 10)(s + 20) = 2s2 + 60s + 400


K2 = 58 KI = 400


2(s + 10)(s + 50) = 2s2 + 120s + 1000


K2 = 118 KI = 1000



10.30 The compensated system diagram is shown in the following figure.


Command compensation cannot affect the characteristic roots or the disturbance re-sponse, so we set D(s) = 0 here. The error equation is

E(s) =s(Is + c− Kf)

Is2 + (c + KP )s + KI)R(s)

For a unit-step input, the final value theorem gives ess = 0 as long as the system is stable.Thus ess = 0 even if there is no command compensation (Kf = 0). For a unit-ramp input,the final value theorem gives

ess =c − Kf

KI= 0

if Kf = c. Thus the command compensation improves the ramp response without affectingthe step response.



10.31 The requirements are (for unit step inputs)

1. the steady-state command error must be zero.

2. the magnitude of the steady-state disturbance error must be ≤ 0.1

3. ζ = 0.707

Part (a). We can do P control and PD control at the same time by finding the errorequation for PD control, and then setting KD = 0 to obtain the results for P control. FromFigure 10.7.4 with I = 20 and c = 10, the error equation can be written as follows (afterdoing some algebra):

(1) E(s) = Θr(s)−Θ(s) =20s2 + 10s

20s2 + (10 + KD)s + KPΘr(s)+

120s2 + (10 + KD)s + KP

Td(s)

The characteristic equation is obtained from the denominator:

(2) 20s2 + (10 + KD)s + KP = 0

For a unit step command, the steady state error is found by setting Θr(s) = 1/s andTd(s) = 0 in equation (1), and using the final value theorem.

ess = lims→0

sE(s) = lims→0

s

[20s2 + 10s

20s2 + (10 + KD)s + KP

]1s

= 0

if the system is stable (that is, if 10 + KD > 0 and KP > 0).For a unit step disturbance, the steady state error is found by setting Td(s) = 1/s and

Θr(s) = 0 in equation (1), and using the final value theorem.

ess = lims→0

sE(s) = lims→0

s

[1

20s2 + (10 + KD)s + KP

]1s

=1

KP

if the system is stable.Thus the steady state errors are the same for P and PD control. So they both satisfy

specification #1, and specification #2 if KP ≥ 10. The difference between the two lies intheir transient performance.





P control: For P control, the characteristic equation is (set KD = 0 in equation (2)):

20s2 + 10s + KP = 0

Thusζ =

102√

20KP

To satisfy specification #3 (ζ = 0.707), KP must be 2.5, which is smaller than the valuerequired to satisfy specification #2. Thus P control cannot satisfy simultaneously bothspecification #2 and specification #3.

PD control. For PD control, since ζ < 1 we can write the following formula for the timeconstant [the real part of the roots of equation (1) is −(10 + KD)/40]:

(3) τ =40

10 + KD

The formula for ζ is

(4) ζ =10 + KD

2√

20KP= 0.707

A solution that exactly satisfies specification #3 is KP = 10. From equation (4) we findthat KD = 10. From equation (3) we see that τ = 40/20 = 2. Other solutions are possiblefor values of KP greater than 10.

Part (b). In summary

1. With P control, the requirements on damping ratio and disturbance error cannot bemet simultaneously.

2. With PD control, one solution is KP = 10, KD = 10

Thus P control does not work. PD control works but has numerator dynamics that increasesthe overshoot.



10.32 Because ζ = cos β, the value ζ = 0.707 corresponds to β = 45, which is a 45 line onthe complex plane. Thus ζ = 0.707 corresponds to a pair of roots whose real and imaginaryparts have the same magnitude; that is, s = −a ± ja. The time constant of these roots isτ = 1/a. Thus the roots must be s = −1±j if τ = 1 and ζ = 0.707. These roots correspondto the polynomial equation

(s + 1− j)(s + 1 + j) = (s + 1)2 + 1 = s2 + 2s + 2 = 0

or10s2 + 20s + 20 = 0

Compare this with the system’s characteristic equation obtained from the denominator ofthe transfer function:

10s2 + (3 + KD)s + KP = 0

Thus KP = 20 and 3 + KD = 20, or KD = 17.



10.33 From the figure given in the problem statement we obtain

Θ(s) =KP

20s2 + (10 + K2)s + KPΘr(s)−

120s2 + (10 + K2)s + KP

Td(s)


θss =KP

KP= 1

which is perfect. The steady-state deviation caused by a unit-step disturbance is

∆θss = − 1KP

Thus KP ≥ 10 is required to meet the specification.If ζ ≤ 1,

τ =40

10 + K2= 0.1

which gives K2 = 390.Thus

ζ =10 + K2

2√

20KP=

200√20KP

≤ 1

if KP ≥ 2000.So an acceptable design is KP = 2000, K2 = 390, which gives ζ = 1, τ = 0.1, and

∆θss = − 12000



10.34 (a) With the specific parameter and gain values used in Problem 10.33, the resultingcommand transfer function is

Θ(s)Θr(s)

=20 + 17s

10s2 + 20s + 20(1)

A MATLAB file to compute the unit-ramp response is the following.

systheta = tf([17,20],[10,20,20]);t = [0:0.001:4];input = t;lsim(systheta,input,t)

Even though the system is underdamped, we do not see oscillations in the response to a unitramp command because the oscillation period is 2π = 6.28, which is greater than 4τ = 1.Thus the oscillations die out before one period has occurred. The steady-state error canbe found using the final value theorem. From the block diagram in Figure 10.7.4, with thespecific parameter and gain values used here, we obtain

E(s) = Θr(s) − Θ(s) = Θr(s)(1 − 20 + 17s

10s2 + 20s + 20

)= Θr(s)

(10s2 + 3s

10s2 + 20s + 20

)

With Θr(s) = 1/s2, we have

ess = lims→0

sE(s) = lims→0

s1s2

(10s2 + 3s

10s2 + 20s + 20

)=

320

From Figure 10.7.4 with the specific parameter and gain values used here and Θr(s) =1/s2, we obtain the following actuator equation:

T (s) =170s3 + 251s2 + 60s

10s2 + 20s + 20Θr(s)

Note that the MATLAB tf function cannot be used to compute this transfer functionbecause the order of the numerator is greater than that of the denominator. We can,however, work around this difficulty as follows.





With Θr(s) = 1/s2, and canceling an s term from the numerator and denominator, weobtain

T (s) =170s2 + 251s + 6010s2 + 20s + 20

1s

This is equivalent to the unit-step response of the following transfer function

T (s)Ωr(s)

=170s2 + 251s + 6010s2 + 20s + 20

A MATLAB file to compute the unit-ramp response is the following.

systorque = tf([170,251,60],[10,20,20]);step(systorque)

The torque approaches the constant value of 3 needed to counteract the damping torque3θ = 3(1) = 3.





(b) Actual disturbance inputs are not always “clean” functions like steps, ramps, andsine waves. We often do not know the exact functional form of the disturbance. It isoften a random function, such as the disturbance torque due to wind gusts of a rotatingradar antenna. The analysis of random inputs is beyond the scope of this text. However,although frequency response plots strictly speaking describe only the steady-state responsefor periodic inputs, the plots are used to obtain a rough idea of the system’s transientresponse to fluctuating inputs that are not periodic. The disturbance transfer function forthe system in question is

Θ(s)Td(s)

=−1

10s2 + 20s + 20(1)

The frequency response plot can be obtained with the following MATLAB file.

sys=tf(1,[10, 20,20]);bodemag(sys)

Right-click on the plot, select “Characteristics”, then “Peak Response”, to see the peakvalue and the corresponding frequency. They are m = −26 dB at ω = 0 rad/s. The plotshows that the system responds more to slowly varying disturbances, and tends to rejectdisturbances whose frequencies lie above the frequency ω = 1.4 radians per unit time. Fordisturbance frequencies within the bandwidth of 0 ≤ ω ≤ 1.4, the system attenuates theinput by approximately m = −26 db, or by a multiplicative factor of M = 10−26/20 = 0.05.



10.35 a) Because this system is third order, we do not have formulas to use for the dampingratio and the time constant. In addition, we must interpret the specifications to apply tothe dominant root or root pair, and must choose the secondary root somewhat arbitrarily.The values ζ = 0.707 and τ = 1 correspond to the dominant root pair s = −1 ± j. Thethird root must be less than −1 so that it will not be dominant. We arbitrarily select thethird root to be s = −2. This choice can be investigated later if necessary. These threeroots correspond to the polynomial equation

(s + 2)(s + 1 − j)(s + 1 + j) = (s + 2)[(s + 1)2 + 1

]= s3 + 4s2 + 6s + 4 = 0

To compare this with the system’s characteristic equation, we multiply it by 10.

10s3 + 40s2 + 60s + 40 = 0

Compare this with the system’s characteristic equation:

10s3 + (3 + KD)s2 + KP s + KI

Thus KP = 60, KI = 40, and 3 + KD = 40, or KD = 37.b) The resulting disturbance transfer function for this system is

Θ(s)Td(s)

=−s

10s3 + 40s2 + 60s + 40

The frequency response plot can be obtained with the following MATLAB file.

sys=tf([1,0],[10, 40,60,40]);bodemag(sys)

Right-click on the plot, select “Characteristics”, then “Peak Response”, to see the peakvalue and the corresponding frequency. They are m = −33.7 dB at ω = 1.14 rad/s. Theplot shows that the system attenuates disturbance inputs by a factor of m = −33.7 dB ormore. This corresponds to an amplitude reduction of M = 10−33.7/20 = 0.0207. The systemrejects by an even greater amount any disturbances whose frequencies lie below or abovethe frequency ω = 1.14. Compare this performance with the PD control system of Problem10.35, which does not have as great an attenuation and responds more to low frequencydisturbances than high frequency ones.



10.36 The requirements are (for unit step inputs)

1. the steady-state command error must be zero.

2. the magnitude of the steady-state disturbance error must be ≤ 0.1

3. the time constant must be 0.1

We can do P control and PD control at the same time by finding the error equationfor PD control, and then setting KD = 0 to obtain the results for P control. From Figure10.7.4 with I = 20 and c = 10, the error equation can be written as follows (after doingsome algebra):

(1) E(s) = Θr(s)−Θ(s) =20s2 + 10s

20s2 + (10 + KD)s + KPΘr(s)+

120s2 + (10 + KD)s + KP

Td(s)

The characteristic equation is obtained from the denominator:

(2) 20s2 + (10 + KD)s + KP = 0

For a unit step command, the steady state error is found by setting Θr(s) = 1/s andTd(s) = 0 in equation (1), and using the final value theorem.

ess = lims→0

sE(s) = lims→0

s

[20s2 + 10s

20s2 + (10 + KD)s + KP

]1s

= 0

if the system is stable (that is, if 10 + KD > 0 and KP > 0).For a unit step disturbance, the steady state error is found by setting Td(s) = 1/s and


ess = lims→0

sE(s) = lims→0

s

[1

20s2 + (10 + KD)s + KP

]1s

=1

KP

if the system is stable.Thus the steady state errors are the same for P and PD control. So they both satisfy

specification #1, and specification #2 if KP ≥ 10. The difference between the two lies intheir transient performance.





P control. For P control, we have

ζ =10

2√

20KP

Thus, if KP ≥ 10, ζ ≤ 0.35. Because ζ < 1, the time constant gives

τ =4010

= 4

Thus specification #3 is not satisfied.

PD control. For PD control, specifications #1 and #2 are met if KP ≥ 10. There are twoways to meet specification #3: i) set ζ ≤ 1 or ii) set ζ > 1.

Case (i). For case (i), (ζ ≤ 1), the time constant is given by

τ =40

10 + KD= 0.1

if KD = 390. For this value, the damping ratio is found from equation (4) to be

ζ =10 + 3902√

20KP=

400√20KP

Thus ζ ≤ 1 only if KP ≥ 8000. Thus one solution is KP = 8000 and KD = 390. This givesa damping ratio ζ = 1.Case (ii). For case (ii), (ζ > 1), the dominant root must have a real part equal to −10 forthe time constant to be 0.1. The other root must be placed to the left of s = −10. Thusthere are an infinite number of solutions, depending on where the second root is placed.For example, if the second root is placed at s = −b, the characteristic equation will factoras 20(s + 10)(s + b) = 20s2 + (200 + 20b)s + 200b. Comparing this to the characteristicequation shows that KP = 200b and KD = 190 + 20b. Thus the larger b is, the greaterthe gains KP and KD. For example, using an arbitrary root separation factor of 10, onesolution is b = 100, KP = 20, 000 and KD = 2190. Generally, placing the second root far tothe left requires higher gains, which is not desirable. In this case, additional specificationsare needed to arrive at a unique solution.





Part (b). In summary,

1. Using P control, specifications #2 and #3 cannot be met simultaneously.

2. Using PD control, a solution with ζ ≤ 1 is: KP = 8000, KD = 390. A solution withζ > 1 is: KP = 20, 000, KD = 2190. Other solutions are possible.



10.37 From the following diagram with Td(s) = 0 we obtain

Θ(s) =1

s(Is + c)KfΘr + (KP + KDs) [Θr(s) − Θ(s)]

ThusΘ(s)Θr(s)

=Kf + KP + KDs

Is2 + (c + KD)s + KP


θss =Kf + KP

KP= 1

only if Kf = 0. The error equation is

E(s) = Θr(s) − Θ(s) =Is2 + cs − Kf

Is2 + (c + KD)s + KPΘr(s)

For a unit-ramp command,ess = lim

s→0sE(s) = ∞

regardless of the value of Kf . So Kf does not improve the steady-state response for a stepor a ramp input.




10.38 a) Try I-action: G(s) = KI/s. Thus

C(s)R(s)

=KI

20s2 + 0.2s + KI

and the error equation with a unit-ramp input is

E(s) = R(s)− C(s) =1s2

(20s2 + 0.2s

20s2 + 0.2s + KI

)

Thus ess = 0.2/KI = 0.01 if KI = 20. This gives ζ = 0.2/(2√

20KI) = 0.005 and τ =2(20)/0.2 = 200.

b) Try PI-action: G(s) = KP + KI/s. Thus

C(s)R(s)

=KP s + KI

20s2 + (0.2 + KP )s + KI


E(s) = R(s) − C(s) =1s2

(20s2 + 0.2s

20s2 + (0.2 + KP )s + KI

)

Thus ess = 0.2/KI = 0.01 if KI = 20. This gives

ζ =0.2 + KP

2√

20KI=

0.2 + KP

40= 1

if KP = 38.8. This gives τ = 2(20)/(0.2+ KP ) = 1.





c) Try PID-action: G(s) = KP + KI/s + KDs. Thus

C(s)R(s)

=KDs2 + KP s + KI

(20 + KD)s2 + (0.2 + KP )s + KI


E(s) = R(s)− C(s) =1s2

(20s2 + 0.2s

(20 + KD)s2 + (0.2 + KP )s + KI

)

Thus ess = 0.2/KI = 0.01 if KI = 20. The damping ratio is

ζ =0.2 + KP

2√

(20 + KD)KI=

0.2 + KP

2√

20(20 + KD)= 1

Because ζ = 1, the expression for the time constant is

τ =2(20 + KD)0.2 + KP

= 0.1

The solution is KP = 3.8 and KD = −19.8.



10.39 The error equation is:

(1) E(s) =s2 + s

s2 + (1 + KD)s + KPΘr(s) +

1s2 + (1 + KD)s + KP

Td(s)

For a unit ramp command, the steady state error is found by setting Θr(s) = 1/s2 andTd(s) = 0 in equation (1), and using the final value theorem.

ess = lims→0

sE(s) = lims→0

s

(s2 + s

s2 + (1 + KD)s + KP

)1s2

=1

KP

if the system is stable (that is, if 1 + KD > 0 and KP > 0).For a unit ramp disturbance, the steady state error is found by setting Td(s) = 1/s2 and


ess = lims→0

sE(s) = lims→0

s

(1

s2 + (1 + KD)s + KP

)1s2

= ∞

if the system is stable.



10.40 The system’s closed-loop transfer function can be found from the block diagram. Itis

C(s)R(s)

=KP + KDs

2s2 + (2 + KD)s + KP

Thus the characteristic equation is 2s2 + (2 + KD)s + KP = 0. Because ζ is specified to beless than 1, we can write the following formula for the time constant.

τ =2(2)

2 + KD= 1

This gives KD = 2. The damping ratio is given by

ζ =2 + KD

2√

2KP=

2 + 22√

2KP=

2√2KP

= 0.9

This gives KP = 2.469.If the command is a step input with a magnitude m, then R(s) = m/s, and the Final

Value Theorem gives

css = lims→0

sC(s) = lims→0

s

(KP + KDs

2s2 + (2 + KD)s + KP

)m

s= m

Thus the steady state error is zero, so all three specifications are satisfied.



10.41 a) The output equation is

Θ(s) =KDs2 + KP s + KI

10s3 + (2 + KD)s2 + KP s + KIΘr(s)−

s

10s3 + (2 + KD)s2 + KP s + KITd(s)

The characteristic equation

10s3 + (2 + KD)s2 + KP s + KI = 0 (1)

For case 1, the desired characteristic equation can be expressed as

10(s + 0.5)[(s + 5)2 + 25] = 10s3 + 105s2 + 550s + 250 = 0 (2)

Comparing coefficients in equations (1) and (2), we obtain

KP = 550 KI = 250 KD = 103

For case 2, the desired characteristic equation can be expressed as

10(s + 0.5)(s + 1)(s + 2) = 10s3 + 35s2 + 35s + 10 = 0 (3)

Comparing coefficients in equations (1) and (3), we obtain

KP = 35 KI = 10 KD = 33

b) The MATLAB file is

KP1 = 550; KD1 = 103; KI1 = 250;systheta1 = tf([KD1,KP1,KI1],[10,2+KD1,KP1,KI1]);KP2 = 35; KD2 = 33; KI2 = 10;systheta2 = tf([KD2,KP2,KI2],[10,2+KD2,KP2,KI2]);step(systheta1,systheta2)

Right-click on the plot, select “Characteristics”, then “Peak Response” to determine themaximum percent overshoot and the peak time. For Case 1, the maximum percent overshootis 22%; for Case 2, it is15%. For the gains given in Example 10.7.7, it is 10%. The separationfactors are 10, 2, and 10, respectively. On the basis of this example we would conclude thata large separation factor is better, provide that the dominant root is complex.





c) The disturbance transfer function is

Θ(s)Td(s)

= − s

10s3 + (2 + KD)s2 + KP s + KI

The MATLAB file, which is a continuation of the previous file, is

sysdist1 = tf([-1,0],[10,2+KD1,KP1,KI1]);sysdist2 = tf([-1,0],[10,2+KD2,KP2,KI2]);bodemag(sysdist1,sysdist2)

Right-click on the plot, select “Characteristics”, then “Peak Response” to determine thepeak response and the corresponding frequency. For Case 1, the maximum response ism = −54.2 dB at ω = 2.44; for Case 2, it is −30 dB at ω = 0.585. For the gains given inExample 10.7.7, it is m = −34.1 dB at ω = 0.7. The separation factors are 10, 2, and 10,respectively. On the basis of this example we would conclude that a large separation factorgives more attenuation, provided that the dominant root is real.



10.42 Combining equations (1) and (2) in Example 10.7.5, with Td(s) = 0, we obtain

T (s) =KIs(10s + 2)

10s3 + (2 + K2)s2 + K1s + KIΘr(s)

Using the values KI = 25, K1 = 55, and K2 = 58 from the example, we can use theMATLAB code shown below to obtain the plot of T (t).

KI=25;K1=55;K2=58;sys=tf([10*KI,2*KI,0],[10,2+K2,K1,KI])



10.43 With I action only, the transfer function is

C(s)R(s)

=KIK

τs2 + s + KIK

andζ =

12√

τKIK

With DI action,C(s)R(s)

=K(KDs2 + KI)

(τ + KKD)s2 + s + KIK

andζ =

12√

(τ + KKD)KIK

If KD > 0 the derivative action makes ζ smaller, and thus the response is more oscillatory.So D action does not improve the response.



10.44 The compensated system diagram is shown in the following figure.


Command compensation cannot affect the characteristic roots or the disturbance re-sponse, so we set D(s) = 0 here. The error equation is

E(s) =Is2 + cs − Kf

Is2 + cs + KPR(s)

For a unit-step input, the final value theorem gives ess = −Kf/KP as long as the system isstable. Thus ess = 0 only if there is no command compensation (Kf = 0)! For a unit-rampinput, the final value theorem gives ess = ∞. Thus the command compensation does notimprove the ramp response. Therefore, command compensation for this system degradesthe step response, and does not improve the ramp response.



10.45 a) The closed-loop transfer function is

C(s)R(s)

=3KP + 3KDs

s2 + 3KDs + 3KP − 4


ζ =3KD

2√

3KP − 4= 0.707

The time constant isτ =

23KD

= 0.1

These two conditions give KP = 68 and KD = 20/3.b) The closed-loop transfer function with the negative rate feedback gain K1 is

C(s)R(s)

=3KP

s2 + 3K1s + 3KP − 4

This denominator has the same form as the transfer function in part (a) with KD replacedby K1. Thus KP = 68 and K1 = 20/3.

c) For part (a),C(s)R(s)

=20s + 204

s2 + 20s + 200

This has numerator dynamics and will overshoot more than the design in part (b), for which

C(s)R(s)

=204

s2 + 20s + 200



10.46 Since the desired value of θ is 0, the error signal is e = θr − θ = 0 − θ = −θ. TryingPD control we use f = KP e + KDe = −KP θ −KDθ. Substituting this into the equation ofmotion we obtain

MLθ − (M + m)gθ = f = −KP θ − KDθ


MLs2 + KDs + KP − (M + m)g = 0

Using M = 40, m = 8, L = 20, and g = 32.2 this becomes

800s2 + KDs + KP − 1545.6 = 0

A 2% settling time of 10 seconds corresponds to 4τ = 10 or τ = 2.5 seconds. Because ζ < 1,we can use the following formula for the time constant.

τ =2(800)KD

= 2.5

This gives KD = 640 lb/rad/sec.The formula for the damping ratio is

ζ =KD

2√

800(KP − 1545.6)=

6402√

800(KP − 1545.6)= 0.707

Solve this for KP to obtain KP = 1801.6 lb/rad.



10.47 The closed-loop transfer function is

Θ(s)Θr(s)

=2412 + 1200s

1500τs3 + 1500s2 + (1200− 1932τ)s + 480


1500τs3 + 1500s2 + (1200− 1932τ)s + 480 = 0

From the Routh-Hurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498,so the case where τ = 1 is unstable.

If τ = 0.1, the roots ares = −0.3418± 0.4761j

The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specificationscall for a dominant time constant of 2.5 and a damping ratio of 0.707.




Θ(s)Θr(s)

=(2412 + 1200s)(τs + 1)

1500τs3 + 1500s2 + (1200− 1932τ)s + 480


1500τs3 + 1500s2 + (1200− 1932τ)s + 480 = 0

From the Routh-Hurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498,so the case where τ = 1 is unstable.

If τ = 0.1, the roots ares = −0.3418± 0.4761j

The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specificationscall for a dominant time constant of 2.5 and a damping ratio of 0.707.



10.49 a) From Figure P10.49,

(ms2+cs+k)X(s) = Fd(s)+(ces+ke)X(s)+me

s2Xr(s) +

KDs2 + KP s + KI

s[Xr(s) − X(s)]

If me = m, ce = c, and ke = k, this equation becomes

(ms3 + mKDs2 + mKP s + mKI)X(s) = sFd(s) + (ms3 + mKDs2 + mKP s + mKI)Xr(s)

ThusX(s)Xr(s)

= 1

which is perfect, and

X(s)Fd(s)

=s

ms3 + mKDs2 + mKPs + mKI

For a step disturbance, Fd(s) = 1/s, Xss = 0. For a unit-ramp disturbance, Fd(s) = 1/s2,Xss = 1/mKI .

Since the characteristic equation is third order, there is no expression for the dampingratio or time constant. We can, however, obtain a dominant root pair that meets therequirements that ζ = 1 and τ = τd. This means that the dominant root pair must be twoidentical roots at s = −1/τd. Thus the third root will be real. Denote it by s = −b. Thenthe characteristic equation can be factored as

ms3+mKDs2+mKP s+mKI = m

(2 +

1τd

)2

(s+b) = m

[s3 +

(b +

2τd

)s2 +

(1τ2d

+2b

τd

)s +

b

τ2d

]

Comparing coefficients, we see that the gains must be chosen so that

KP = b +2τd

KP =1τ2d

+2b

τd

KI =b

τ2d

The value of b must be selected such that the root at s = −b is not the dominant root. Thisrequires that b > 1/τd.





b) In addition to requiring accurate estimates of m, c, and k, this scheme requires thatthe second derivative, xr, of the command input must be computed in real time. This canbe difficult to do for noisy inputs or inputs having discontinuities in xr or xr, such as stepsand ramps.



10.50 a) From the diagram,

ΩL(s) =1

N(Ies + ce)KT

Las + Ra[Vm(s) − KbΩm(s)]

and Ωm(s) = NΩL(s). Thus

ΩL(s)Vm(s)

=KT

NIeLas2 + N(ceLa + RaIe)s + ceRaN + NKTKb

With the given values,

ΩL(s)Vm(s)

=1.4 × 105

4.368s2 + 894.4s + 2.8015× 104=

4.9971.5592× 10−4s2 + 0.0319s + 1

Since the two roots are real, this can be expressed as

ΩL(s)Vm(s)

=K

(τ1s + 1)(τ2s + 1)

where K = 4.997, τ1 = 2.591× 10−2 and τ2 = 6.02× 10−3.





b) Using PI control with the above transfer function results in a third-order system,for which there are no convenient design formulas. Since τ2 τ1, we will neglect τ2 andexpress the transfer function as

ΩL(s)Vm(s)

=K

τ1s + 1

With PI control, the closed-loop transfer function is

ΩL(s)Ωr(s)

=K(KPs + KI)

τ1s2 + (1 + KKP )s + KKI

Since we will set ζ < 1, we can express the closed-loop time constant as

τ =2τ1

1 + KKP= 0.05

ThusKP =

2τ1 − 0.050.05K

= 7.284× 10−3


ζ =1 + KKP

2√

τ1KKI=

20τ1√τ1KKI

Thus

KI =400τ2

1

ζ2τ1K

Choosing ζ = 0.707, we obtainKI = 4.1481

The resulting third-order transfer function is

ΩL(s)Ωr(s)

=0.0364s + 20.73

0.0001559s3 + 0.03193s2 + 1.036s + 20.73

The roots of the third-order system are s = −170.39 and s = −17.21 ± 22j. Thus thedominant time constant is 1/17.21 = 0.058, which is close to the desired value. The dampingratio of the dominant root pair is ζ = cos[tan−1(22/17.21)] = 0.616, which is within therequired range.



10.51 The value from Problem 10.19 is KP = 3.6772. From Example 10.8.6,

Tm(s)Ωr(s)

=NKP KT (Ies + ce)

D(s)

and thus, since ia = TM/KT ,

Ia(s)Ωr(s)

=NKP (Ies + ce)

D(s)

whereD(s) = N

[LaIes

2 + (RaIe + ceLa)s + Race + KT Kb + KP KT

]

Note that the term N in D(s) cancels N in the numerator of Ia(s)/Ωr(s).The MATLAB program is the following.

KT = 0.2; Kb = 0.2;Ie = 1.775e-3;ce = 1e-3;Ra = 0.8;La = 4e-3;KP = 3.672;num = KP*[Ie, ce];den = [La*Ie, Ra*Ie+ce*La, Ra*ce+KT*Kb+KP*KT];sys = tf(num, den);step(209.4*sys)

Note that we can use the step function by multiplying the transfer function by magnitudeof the step input. The plot is shown on the next page. The maximum current is 389 A!This unattainable value can be reduced by using a modified step input that increases slowly(see Problem 10.53 for a discussion of this type of input).





0 0.01 0.02 0.03 0.04 0.05 0.06−200

−100

0

100

200

300

400

Step Response

Time (sec)

Cur

rent

i a (A

)




10.52 The new transfer functions are

Ω(s)Ωr(s)

=KP s + KI

0.5s3 + 5.2s2 + (2 + KP )s + KI

T (s)Ωr(s)

=KP s + KI

0.1s + 10.5s2 + 5.2s + 2

0.5s3 + 5.2s2 + (2 + KP )s + KI

Ω(s)Td(s)

=−s(0.1s + 1)

0.5s3 + 5.2s2 + (2 + KP )s + KI

The three cases are:

1. KP = 18, KI = 40;

2. KP = 18, KI = 20;

3. KP = 108, KI = 200;

a) The MATLAB program is

KP = [18, 18, 108];KI = [40, 20, 200];% Command transfer functionk = 1; sysa1 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 2; sysa2 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 3; sysa3 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]);% Actuator transfer functionk = 1; numb1 = conv([KP(k), KI(k)],[0.5, 5.2, 2]);denb1 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]);sysb1 = tf(numb1, denb1)k = 2; numb2 = conv([KP(k), KI(k)],[0.5, 5.2, 2]);denb2 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]);sysb2 = tf(numb2, denb2)k = 3; numb3 = conv([KP(k), KI(k)],[0.5, 5.2, 2]);denb3 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]);sysb3 = tf(numb3, denb3)





subplot(2,1,1), step(sysa1, sysa2, sysa3)subplot(2,1,2), step(sysb1, sysb2, sysb3)

The characteristic roots are as follows:KP KI Intended ζ Intended Roots Actual Roots18 40 0.707 −2 ± 2j (dominant) −2.2353± 2.9147j (ζ = 0.6086), −5.929418 20 1 −2, −2 (dominant) −1.5016, −4.4492± 2.6159j108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168± 13.6248j

The plots are shown in the following figure. The peak actuator values are 16.2, 14.2,and 52.

0 0.5 1 1.50

0.5

1

1.5

0 0.5 1 1.5−20

0

20

40

60

Step Response

Time (sec)

Spe

ed (r

ad/s

ec)

Step Response

Time (sec)

Act

uato

t Res

pons

e

1 2

3

1 2

3

Figure : for Problem 10.52a.





b) The MATLAB program is

KP = [18, 18, 108];KI = [40, 20, 200];% Disturbance transfer functionk = 1; sysc1 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 2; sysc2 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 3; sysc3 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]);bodemag(sysc1, sysc2, sysc3)

The plots are shown in the figure on the following page. The peak values are −23.3,−24.7, and −30.8 dB.

c) The peak actuator values are less than in Example 10.6.4 because the actuator re-sponds more slowly in this problem (it has a time constant of 0.1, whereas in the examplethe actuator responded instantaneously). The peak disturbance response values are slightlyhigher than in the example.



10−1

100

101

102

−70

−65

−60

−55

−50

−45

−40

−35

−30

−25

−20

Mag

nitu

de (d

B)

Bode Diagram

Frequency (rad/sec)

1

2

3

Figure : for Problem 10.52b.



10.53 a) The transfer functions are

Ω(s)Ωr(s)

=KP s + KI

5s2 + (2 + KP )s + KI

T (s)Ωr(s)

=(KP s + KI)(5s + 2)

5s3 + (2 + KP )s + KI


1. KP = 18, KI = 40;

2. KP = 18, KI = 20;

3. KP = 108, KI = 200;


KP = [18, 18, 108];KI = [40, 20, 200];% Command transfer functionk = 1; sysa1 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]);k = 2; sysa2 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]);k = 3; sysa3 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]);% Actuator transfer functionk = 1; sysb1 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)])k = 2; sysb2 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)])k = 3; sysb3 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)])tc = [0:.001:2];rc = 1 - exp(-20*tc);subplot(2,1,1),lsim(sysa1,sysa2,sysa3,rc,tc)ta = [0:.001:0.8];ra = 1 - exp(-20*ta);subplot(2,1,2),lsim(sysb1,sysb2,sysb3,ra,ta)





The plots are shown in the following figure. The peak actuator values are 14.5, 13.4,and 40.1. The peak actuator values are less than in Example 10.6.4 because the commandinput does not act as fast as the step input used in the example. This gives the systemmore time to respond, and thus makes less demand on the actuator.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

10

20

30

40

Linear Simulation Results

Time (sec)

Am

plitu

de

Linear Simulation Results

Time (sec)

Am

plitu

de

Command

1

2

3

1

2

3




10.54 The new transfer functions are

Ω(s)Ωr(s)

=KI

0.5s3 + 5.2s2 + (2 + K2)s + KI

T (s)Ωr(s)

=KI(5s + 2)

0.5s3 + 5.2s2 + (2 + K2)s + KI

Ω(s)Td(s)

=−s(0.1s + 1)

0.5s3 + 5.2s2 + (2 + K2)s + KI


1. K2 = 18, KI = 40, ζ = 0.707;

2. K2 = 18, KI = 20, ζ = 1;

3. K2 = 108, KI = 200, ζ = 1.74;


K2 = [18, 18, 108];KI = [40, 20, 200];% Command transfer functionk = 1; sysa1 = tf(KI(k),[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysa2 = tf(KI(k),[0.5, 5.2, 2 + K2(k), KI(k)]);k = 3; sysa3 = tf(KI(k),[0.5, 5.2, 2 + K2(k), KI(k)]);% Actuator transfer functionk = 1; sysb1 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysb2 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysb3 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + K2(k), KI(k)]);subplot(2,1,1), step(sysa1, sysa2, sysa3)subplot(2,1,2), step(sysb1, sysb2, sysb3)





The characteristic roots are as follows:K2 KI Intended ζ Intended Roots Actual Roots18 40 0.707 −2 ± 2j (dominant) −2.2353± 2.9147j (ζ = 0.6086), −5.929418 20 1 −2, −2 (dominant) −1.5016, −4.4492± 2.6159j108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168± 13.6248j

The plots are shown in the following figure. The peak actuator values are 8.64, 4.87,and 11.3.

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.5 1 1.5 2 2.5 3 3.50

2

4

6

8

10

12

Step Response

Time (sec)

ω(t)

Step Response

Time (sec)

T(t)

ζ = 0.707

ζ = 1.74

ζ = 1

ζ = 0.707

ζ = 1.74

ζ = 1






b) The MATLAB program is

K2 = [18, 18, 108];KI = [40, 20, 200];% Disturbance transfer functionk = 1; sysc1 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysc2 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 3; sysc3 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]);bodemag(sysc1, sysc2, sysc3)

The plots are shown in the following figure. The peak values are −23.3, −24.7, and−30.8 dB.

c) The peak actuator values are greater than in Example 10.6.5. The peak disturbanceresponse values are slightly higher than in the example.

10−2

10−1

100

101

102

−110

−100

−90

−80

−70

−60

−50

−40

−30

−20

Mag

nitu

de (d

B)

Bode Diagram

Frequency (rad/sec)

ζ = 1.74

ζ = 0.707

ζ = 1




10.55 LetK =

KpotKaKT

6Ie

whereIe = I1 +

14

[I2 +

19I3

]= 0.0157

and K = 12.7547.The transfer function is

Θ(s)Θr(s)

=K(KDs + KP )

Ls3 + Rs2 + KKDs + KKP

a) For L = 0,Θ(s)Θr(s)

=K(KDs + KP )

Rs2 + KKDs + KKP

To obtain τ = 0.5 and ζ = 1,

τ = 0.5 =2R

KKD

andζ = 1 =

KKD

2√

RKKP

These give

KD = KP =4R

K= 0.0941





Parts (b), (c), and (d) are done with the following MATLAB program.

I1 = 0.01; I2 = 5e-4; I3 = 0.2;Ka = 1; KT = 0.6; Kpot = 2;R = 0.3;Ie = I1+(1/4)*(I2+(1/9)*I3)K = Kpot*Ka*KT/(6*Ie)KD = 4*R/KKP = 4*R/Ksys1 = tf([K*KD,K*KP],[R,K*KD,K*KP])L = 0.015;sys2 = tf([K*KD,K*KP],[L,R,K*KD,K*KP])step(sys1,sys2)

The resulting plot is shown below. With L = 0, the overshoot is 13.5% and the settlingtime is 2.7 s. With L = 0.015, the overshoot is 16.2% and the settling time is 2.59 s. Sothe response predicted by the second order model, which makes the calculation of the gainsKP and KD much easier, is close to that of the third order system. This will not be true ifL is much larger than 0.015.

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Ampl

itude

L = 0.015

L = 0




10.56 Since the acceleration time is 4 s, we set τ = 1 so that the controllers can follow theramp input. The time constant expressions for the three controllers are:

1. τ = I/(c + KP ) = 10/(3 + KP ) for P control.

2. τ = I/(c + KP ) = 10/(3 + KP ) for PI control, if ζ ≤ 1.

3. τ = I/(c + K2) = 10/(3 + K2) for Modified I control, if ζ ≤ 1.

Requiring, for example, a maximum steady state error of 10% of the slew speed meansthat ess ≤ 0.1. The slope of the input is 1/4, and the steady state errors for a rampcommand of slope 1/4 is

1. ess = 0.25/(c + KP ) = 0.25/(3 + KP ) for P control.

2. ess = 0.25c/KI = (0.25)3/KI for PI control.

3. ess = 0.25(c + K2)/KI = 0.25(3 + K2)/KI for Modified I control.

Combining these requirements and satisfying the requirement on τ exactly, we obtainthe following gains:

1. KP = 7 for P control.

2. KP = 7 , KI = 7.5 for PI control.

3. K2 = 8 , KI = 110 for Modified I control.





The following program is a modification of the programs command_tf.m, actuator_tf.m,and trap1.m given in Section 10.9. The program calls on the files named plot_commandand plot_actuator listed in Table 10.9.3.

KPa = 120;KPb = 7; KIb = 7.5;K2 = 8;KIc = 110;I = 10;c = 3;% Create the command transfer functions.sysa = tf(KPa,[I,c+KPa]);sysb = tf([KPb,KIb],[I,c+KPb,KIb]);sysc = tf(KIc,[I,c+K2,KIc]);% Create the actuator transfer functionssysaACT = tf(KPa*[I,c],[I,c+KPa]);sysbACT = tf(conv([KPb,KIb],[I,c]),[I,c+KPb,KIb]);syscACT = tf(KIc*[I,c],[I,c+K2,KIc]);% A specific trapezoidal profilet = [0:0.01:19];for k = 1:length(t)if t(k) <= 4r(k) = (1/4)*t(k);elseif t(k) <= 10r(k) = 1;elseif t(k) <= 14r(k) = 14/4-(1/4)*t(k);elser(k) = 0;endendsubplot(2,1,1),plot_commandsubplot(2,1,2),plot_actuator

The value KP = 7 for P control was found to be too small, so KP was increased to 120for P control only. This improved response is shown in the figure on the following page.





0 2 4 6 8 10 12 14 16 18 20−0.2

0

0.2

0.4

0.6

0.8

1

1.2

Time

Com

man

d R

espo

nse

0 2 4 6 8 10 12 14 16 18 20−4

−2

0

2

4

6

Time

Act

uato

r O

utpu

t

P

PI I Modified

I Modified

P

PI




10.57 a) First we reflect the load and tachometer inertias back to the motor shaft to obtainthe equivalent inertia Ie:

Ie = Im +1

N2(IL + It)

= 2 × 10−5 +1

1.52(4× 10−3 + 10−5) = 1.802× 10−3 kg · m2

Next do the same for the load damping coefficient:

ce = cm +cL

N2= 0 +

10−3

1.52= 4.444× 10−4 N · m · s/rad

The transfer functions derived in Example 10.3-5 and are repeated here.

ΩL(s)Ωr(s)

=KPKT

D(s)(1)

ΩL(s)TL(s)

= −(Las + Ra)/ND(s)

(2)

where the denominator is

D(s) = NLaIes2 + N(RaIe + ceLa)s + NRace + NKTKb + KP KT (3)

Thus the characteristic equation is

D(s) = 5.407× 10−6s2 + 1.623× 10−3s + 2.8× 10−3 + 0.04KP = 0

The Routh-Hurwitz criterion shows that the system is stable if and only if KP > 0.The desired speed ωr is a constant and so can be modeled as a step function of magnitude

1000(2π)/60 = 104.7 rad/s. Applying the Final Value Theorem to equation (1) we have

ωLss =KP KT

NRace + NKTKb + KP KTωr =

0.04KP

2.8× 10−3 + 0.04KPωr (4)

Solve for KP as follows.

KP =0.07ωLss/ωr

1 − ωLss/ωr(5)

For the steady-state speed ωLss to be within 10% of ωr, ωLss/ωr = 0.9 or 1.1. These valuesgive KP = 0.63 and KP = −0.77 respectively.





Choosing the positive solution for stability, we obtain the roots s = −282 and −18.4.The system is stable and the result obtained from the Final Value Theorem is thus valid.Thus the solution is KP = 0.63 for which ωLss = 900 rpm. The largest time constant of theclosed-loop system is 1/18.4 = 0.054 s. The speed will reach its steady-state value of 900rpm after approximately 4(0.054) = 0.217 s. It will not oscillate because the roots are real.With KP = 0.63, we can then determine the value of the gain K1, which is given by

K1 =KP

KtachKa=

0.6310(5)

= 0.0126

as discussed in Example 10.3-5.b) Applying the Final Value Theorem to equation (2), with KP = 0.63 and TL = 1 N-m,

we obtain the steady-state deviation caused by the load torque.

ωLss = − Ra/N

NRace + NKTKb + KP KTTL = −14.3 rad/s (6)

or −136 rpm. This is not the actual load speed, but only the deviation in the load speedcaused by the load torque. Thus the load torque causes the steady-state speed to decreaseby 136 rpm.

The actual load speed is the sum of the speeds obtained from equations (1) and (2), forthe given inputs. Here, at steady-state, the actual load speed is ωLss = 900 − 136 = 764rpm.

You should note that because the model is linear, we can deduce that for KP = 0.63,the steady-state speed will be 90% of the desired speed, regardless of the magnitude of ωr,as long as it is a step function and if the load torque is zero. Similarly, linearity enables usto see from equation (6) that the steady-state deviation caused by a constant load torqueis proportional to TL; that is, ωLss = −14.3TL rad/s.



10.58 The maximum speed error occurs at t = 0 and is 1000 rpm, or 104.7 rad/s.From (6.5.7), the energy consumption from t = 0 to t = T is

E =∫ T

0Rai

2a(t) dt +

∫ T

0ceω

2L(t) dt (1)

The rms current is

irms =

√1T

∫ T

0i2a(t) dt (2)

The rms speed error is

ωrms =

√1T

∫ T

0[104.7− ωL(t)]2 (t) dt (3)

We can perform these computations either 1) by obtaining the step response numericallyand then performing the integrations numerically, or 2) by first obtaining the closed-formexpressions for the current and speed, and then evaluating the integrals. These integralscan be evaluated in closed form.

Both methods require the transfer functions. From the example, we have the expressionfor the speed:

ΩL(s)Ωr(s)

=KP KT

D(s)

whereD(s) = NLaIes

2 + N(RaIe + ceLa)s + NRace + NKTKb + KP KT

The expression for the current is

Ia(s)Ωr(s)

=KP (Ies + ce)

D(s)





The following MATLAB program does the computations.

N = 1.5;La = 2e-3;Ra = 0.6;Ie = 1.802e-3;ce = 4.444e-4;KP = 0.63;KT = 0.04;Kb = 0.04;D = [N*La*Ie,N*(Ra*Ie+ce*La),N*Ra*ce+N*KT*Kb+KP*KT];sysia = tf(KP*[Ie,ce],D);syswL = tf(KP*KT,D);[ia,t1] = step(104.7*sysia);[wL,t2] = step(104.7*syswL);integrand1 = Ra*ia.^2;integrand2 = ce*wL.^2;E = trapz(t1,integrand1) + trapz(t2,integrand2)max_current = max(ia)rms_current = sqrt((1/max(t1))*trapz(t1,ia.^2))rms_speed_error = sqrt((1/max(t2))*trapz(t2,(104.7-wL).^2))

The results are:Energy consumed = E = 97.0283 J

The maximum current is 64.2584 A, and the rms current is 23.1163 A. The rms speed erroris 34.9 rad/s, or 333.271 rpm.

For those not wishing to use MATLAB, we give the following expressions for the speedand current.

ωL(t) = 94.041 + 6.5643e−282t − 100.6053e−18.4t

ia(t) = 1.0448− 83.3224e−282t + 82.2776e−18.4t

The integrals in equations (1), (2), and (3) can be evaluated in closed form or numeri-cally.



10.59 From Figure 10.6.4, we see that the only case where the saturation block with anupper limit of 20 will have any effect is the case for which ζ = 1.74. This corresponds toKP = 108 and KI = 200.

The Simulink model is shown in the following figure. In the To Workspace block,specify the Save format as Array. Type KP = 108; KI = 200; in MATLAB and then runthe model. Then in MATLAB type

subplot(2,1,1),plot(tout,simout(:,2)),subplot(2,1,2),plot(tout,simout(:,1))

The resulting plot, which has been edited, is shown on the next page. As compared to thecase with no limit on the actuator, the overshoot in ω(t) is larger (18% vs approximately5%).






0 0.5 1 1.50

0.2

0.4

0.6

0.8

1

1.2

t

ω(t)

0 0.5 1 1.50

5

10

15

20

25

t

T(t)




10.60 The Simulink model is shown in the following figure. We must use the PID Controllerblock instead of a Transfer Function block, because the latter does not allow the denominatorto have an order greater than the numerator. In the Block Parameters window of the PIDController block, enter 55 for Proportional, 25 for Integral, and 58 for Derivative.

We must use the Transfer Function (with initial outputs) block because the initial po-sition is not zero. In the To Workspace block, specify the Save format as Array.


Run the model and then in MATLAB type

plot(tout,simout)

The resulting plot, which has been edited, is shown on the following page.





0 1 2 3 4 5 6 7 8 9 100.5

1

1.5

2

2.5

3

t

θ(t)




10.61 a) The Simulink model and the responses are shown in the following figures.






0 5 10 15 20 25 300

2

4

6

t

θ(t)

0 5 10 15 20 25 30

0

10

20

t

T(t

)

0 5 10 15 20 25 30−5000

0

5000

10000

t

T(t

)






b) Comparison with Figure 10.7.4 shows that the response with a limited actuator ismuch slower. However, the response shown in Figure 10.7.7 is misleading because it iscaused by an actuator response that contains an impulse, something that is not physicallypossible. To see this, derive the expression for the actuator output, as follows. From thediagram in Figure 10.7.3, we can derive the following expression.

T (s) =IKDs4 + (IKP + cKD)s3 + (IKI + cKP )s2 + cKIs

Is3 + (c + KD)s2 + KP s + KIΘr(s)

If the command is a unit step, then Θr(s) = 1/s, and, after canceling the s terms in thenumerator and denominator, we obtain

T (s) =IKDs3 + (IKP + cKD)s2 + (IKI + cKP )s + cKI

Is3 + (c + KD)s2 + KP s + KI

Using synthetic division, this can be expressed as

T (s) = KD +N(s)

Is3 + (c + KD)s2 + KP s + KI

where N(s) is a second-order polynomial. Thus the constant term KD will produce animpulse of strength KD in T (t).

The top graph shows the command response, which is much slower than when theactuator output is unlimited. The second of the three graphs shows the limited actuatoroutput. The third of the three graphs shows the actuator output if unlimited. It has avery large positive value initially and then goes negative. So the limiter causes the veryslow command response. Note that Simulink cannot deal with an impulse in the actuatorresponse, but rather computes the response numerically. This effect causes the simulated,unlimited actuator response to have some approximation error.



10.62 a) The Simulink model and the responses are shown in the following figures. Wemust use the PID Controller block instead of a Transfer Function block, because the latterdoes not allow the denominator to have an order greater than the numerator. In the BlockParameters window of the PID Controller block, enter 55 for Proportional, 25 for Integral,and 58 for Derivative.






0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

t

θ(t)

0 0.5 1 1.5 2 2.5 3−200

−100

0

100

200

300

t

T(t

)


b) The maximum overshoot in the command response is 42%, as compared to 10% inthe example. The 2% settling time is 2.5 as compared with 2.89 in the example.



10.63 a) The Simulink model is shown in the following figure. Note that since N = 1, wedo not need a gain block for the gain 1/N .






In MATLAB, type

Kb = 0.199; KT = 0.14;Ie = 2.08e-3; ce = 3.6e-4;Ra = 0.43; La=2.1e-3;KP = 7.284e-3; KI = 4.1481;

Then run the model and type the following in MATLAB.


This produces the following plots, which have been edited.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−10

−8

−6

−4

−2

0

2

t (sec)

ω L(t) (r

ad/s

ec)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−2

0

2

4

6

8

10

t (sec)

i a(t) (A

)






b) The Simulink model is shown in the following figure. Note that since N = 1, we donot need a gain block for the gain 1/N .


In MATLAB, type




This produces the plots shown on the following page, which have been edited.





0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

t (sec)

Spe

ed E

rror

, ωL(t

) −

ω(t

) (

rad/

sec)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.005

0.01

0.015

0.02

t (sec)

i a(t)

(A)




10.64 Modify the Simulink model from Problem 10.63a as shown below.






In MATLAB, type




This produces the following plots, which have been edited.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

50

100

150

t (s)

ωL(t

) (r

ad/s

ec)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

−10

−5

0

5

10

t (sec)

i a(t)

(A)





to accompany


by

William J. Palm III


Solutions to Problems in Chapter Eleven


CD1

Highlight

11.1 From the quadratic formula,

s =−12 ±

√144− 12k

6= −2 ±

√4− k/3

For k ≤ 12 the roots are real, the dominant root lies between 0 and −2, and the dominanttime constant lies between 1/2 and ∞. For k > 12 the roots are complex, and the real partis −2. Thus the dominant time constant is 1/2. The root locus plot is identical to Figure11.1.2. The smallest possible dominant time constant is 1/2 and is obtained for any k ≥ 12.



11.2 From the quadratic formula,

s =−c ±

√c2 − 1446

For c ≥ 12 the roots are real, the dominant root lies between 0 and −2, and the dominanttime constant lies between 1/2 and ∞. For c < 12 the roots are complex, and the undampednatural frequency is ωn =

√12/3 = 2. This means that the complex roots lie on a semicircle

of radius 2 centered at the origin. The root locus plot is identical to Figure 11.1.3. Thesmallest possible dominant time constant is 1/2 and is obtained for c = 12.



11.3 Dividing by the highest coefficient and factoring out c gives the standard form (11.1.4):

s2 + 666.7s + 2.78× 104 + 8333c(s + 666.7) = 0

where α = β = 666.7, γ = 2.78 × 104, and µ = 8333c. From the results of (11.1.4) and(11.1.5) we see that the root locus is a circle centered at s = −666.7 with a radius of 166.7,and is similar to Figure 11.1.4(a) with the ×’s at s = −44.7 and s = −622, and the © ats = −667. The starting points with c = 0 are at the ×’s. As c → ∞ one root approachess = −667 and the other root approaches s = −∞.

The plot illustrates the sometimes counterintuitive behavior of dynamic systems. Onewould think that increasing the damping c would increase the time constant and thus slowthe response. However, as c is increased up to the value where the two roots meet ats = −833, the dominant time constant decreases, and we see that the smallest possibledominant time constant is τ = 1/833 s.



11.4 Let µ = (m− 2)/2. Thus m = 2µ + 2 and the characteristic equation becomes

(2µ + 2)s2 + 12s + 10 = 0

Divide by 2:(µ + 1)s2 + 6s + 5 = 0

From the quadratic formula,

s =−3 ±

√4− 5µ

µ + 1

For 0 ≤ µ ≤ 4/5, the roots are real. For µ = 0, the roots are s = −1 and s = −5. Forµ = 4/5, the roots are equal at s = −1.67. For µ > 4/5 the roots are complex and approachs = 0 as µ → ∞. The plot is similar to that in Figure 11.1.5 with the ×’s located at s = −1and s = −5, and with the circle intersecting the real axis at s = −1.67.



11.5 a) The equation can be expressed as

1 +3p

61

s(s + 4/3)= 0

so K = 3p/6 = p/2.b) The equation can be expressed as

1 +p

3s + 2

s2 + 2s + 5/3= 0

so K = p/3.c) The equation can be expressed as

1 +4p

4s2 + 1/4

s(s2 + 0.5)= 0

so K = 4p/4 = p.



11.6 a) Write Gc(s)Gp(s) as

Gc(s)Gp(s) = KP TD

(s2 + 1

TDs + 1

TDTI

)(s + 10)

s(s + 1)(s + 2)= 0.5KP

(s2 + 2s + 10)(s + 10)s(s + 1)(s + 2)

The poles are s = 0, −1, and −2. The zeros are s = −10 and −1 ± 3j.b) K = TDKP = 0.5KP

c) The closed-loop transfer function is

C(s)R(s)

=Gc(s)Gp(s)

1 + Gc(s)Gp(s)

When KP = 10, this becomes

C(s)R(s)

=5s3 + 60s2 + 150s + 5006s3 + 63s2 + 152s + 500

The closed-loop poles are the roots of the denominator, and are s = −8.688 and −0.9059±2.962j. The closed-loop zeros are the roots of the numerator, and are s = −10 and −1±3j.



11.7Part (a)

• Poles: s = 0 and s = −5

• Zeros: None

• Two paths must leave the plot, so there will be two asymptotes.

• The locus exists on the real axis in the following interval: −5 < Re(s) < 0.

• To find the breakaway and breakin points:

K = −(s2 + 5s)

dK

ds= −(2s + 5) = 0

The solution is s = −2.5, which must be a breakaway point. The value of K at thebreakaway point is

K = − (s2 + 5s)∣∣∣s=−2.5

= 6.25

• Angles of the asymptotes:

θ =(2n + 1)180

2 − 0= 90, 270

• Intersection point of the asymptotes:

σ =∑

sp −∑

sz

2 − 0=

0 − 5 − 02

= −2.5

• Crossover point: None for K ≥ 0

• The root locus plot is shown in the figure on the following page.



−5 −4.5 −4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

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11.7 Part (b)

• Poles: s = 0, s = −7, and s = −9. Zeros: None

• Three paths must leave the plot, so there will be three asymptotes.

• The locus exists on the real axis in the following intervals:

Re(s) < −9 − 7 < Re(s) < 0

• To find the breakaway and breakin points:K = −(s3 + 16s2 + 63s) and

dK

ds= −(3s2 + 32s + 63) = 0

The solutions are s = −8.0618 and s = −2.6049. The point s = −8.0618 correspondsto a breakaway or breakin point for K < 0. The value of K at the breakaway points = −2.6049 is

K = − (s3 + 16s2 + 63s)∣∣∣s=−2.6049

= 73.22


θ =(2n + 1)180

3 − 0= 60, 180, 300


σ =∑

sp −∑

sz

3 − 0=

0 − 7− 9 − 03

= −163

= −5.333

• The crossover point is found as follows. Substitute s = jω into the characteristicequation s3 + 16s2 + 63s + K = 0.

−jω3 − 16ω2 + 63ωj + K = 0

which gives(63− ω2)ω = 0 and K − 16ω2 = 0

The first equation gives ω = 0, which is not of interest, and ω = ±√

63. From thesecond equation, K = 16(63) = 1008 at the crossover points. The root locus plot isshown in the figure on the following page.



−35 −30 −25 −20 −15 −10 −5 0 5 10 15−20

−15

−10

−5

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5

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20

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11.7 Part (c)

• Poles: s = −1.5± 1.6583j

• Zero: s = −3

• One path must leave the plot, so there will be one asymptote, which is 180.

• The locus exists on the real axis in the following interval: Re(s) < −3


K = −s2 + 3s + 5s + 3

dK

ds= −(s + 3)(2s + 3) − (s2 + 3s + 5)

(s + 3)2=

s2 + 6s + 4(s + 3)2

= 0

The solutions are s = −5.2361 and s = −0.7639. The point s = −0.7639 correspondsto a breakaway or breakin point for K < 0. The value of K at the breakaway points = −5.2361 is

K = − s2 + 3s + 5s + 3

∣∣∣∣∣s=−5.2361

= 7.47

• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s2 + 3s + 5 + K(s + 3) = 0.

−ω2 + 3jω + 5 + K(jω + 3) = 0

which gives(K + 3)ω = 0 and 5 + 3K − ω2 = 0

The first equation gives two possibilities: 1) ω = 0, which is not of interest becauseit does not lie on the root locus for K > 0, and 2) K = −3. With K = −3, thesecond equation gives ω = ±

√5− 9 = ±4j, which means that s = −4 and is thus not

a solution of interest.




−9 −8 −7 −6 −5 −4 −3 −2 −1 0

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11.7 Part (d)

• Poles: s = 0 and s = −4. Zero: s = −5

• One path must leave the plot, so there will be one asymptote, which is 180.

• The locus exists on the real axis in the following intervals

Re(s) < −5 and − 4 < Re(s) < 0


K = −s2 + 4s

s + 5

dK

ds= −(s + 5)(2s + 4)− (s2 + 4s)

(s + 5)2=

s2 + 10s + 20(s + 5)2

= 0

The solutions are s = −7.2361 and s = −2.7639. The point s = −2.7639 correspondsto the breakaway point, and s = −7.2361 corresponds to the breakin point. Becauseeach point is symmetrically placed a distance 2.2361 from the zero, this indicates thatthe locus is a circle of radius 2.2361 off the real axis.

• The value of K at the breakaway point is

K = − s2 + 4s

s + 5

∣∣∣∣∣s=−2.7639

= 1.5279

The value of K at the breakin point is

K = − s2 + 4s

s + 5

∣∣∣∣∣s=−7.2361

= 10.4721




Problem 11.7d continued:

• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s2 + 4s + K(s + 5) = 0.

−ω2 + 4jω + K(jω + 5) = 0

which gives(K + 4)ω = 0 and 5K − ω2 = 0

The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K > 0, and 2) K = −4. With K = −4, the secondequation gives ω = ±j

√20, which means that s = −

√20 and is thus not a solution of

interest. The root locus plot is shown in the following figure.

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0

−3

−2

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1

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3

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Figure : for Problem 11.7d



11.7 Part (e)

• Poles: s = 0 and s = −1.5± 1.6583j

• Zeros: None


• The locus exists on the real axis in the following interval: Re(s) < 0


K = −(s3 + 3s2 + 5s)

dK

ds= −(3s2 + 6s + 5) = 0

The solutions s = −1 ± 0.8165j are complex, so there are no breakaway or breakinpoints.


θ =(2n + 1)180

3 − 0= 60, 180, 300


σ =∑

sp −∑

sz

3 − 0=

0− 1.5− 1.6583j − 1.5 + 1.6583j − 03

= −33

= −1

• The crossover point is found as follows. Substitute s = jω into the characteristicequation s3 + 3s2 + 5s + K = 0.

−jω3 − 3ω2 + 5ωj + K = 0

which gives(5 − ω2)ω = 0 and K − 3ω2 = 0

The first equation gives ω = 0, which is not of interest, and ω = ±√

5. From thesecond equation, K = 3(5) = 15 at the crossover points.




−8 −6 −4 −2 0 2 4−5

−4

−3

−2

−1

0

1

2

3

4

5

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Figure : for Problem 11.7e



11.7 Part (f)

• Poles: s = 0, s = −3 and s = −7. Zero: s = −4

• Two paths must leave the plot, so there will be two asymptotes. Angles of the asymp-totes:

θ =(2n + 1)180

3 − 1= 90, 270


σ =∑

sp −∑

sz

3 − 1=

0 − 3 − 7 + 42

= −3


−7 < Re(s) < −4 and − 3 < Re(s) < 0


K = −s3 + 10s2 + 21s

s + 4

dK

ds= −(s + 4)(3s2 + 20s + 21)− (s3 + 10s2 + 21s)

(s + 4)2= −2s3 + 22s2 + 80s + 84

(s + 4)2= 0

The only real solution is s = −1.7812. This point corresponds to a breakaway point.The value of K at the breakaway point is

K = − s3 + 10s2 + 21s

s + 4

∣∣∣∣∣s=−1.7812

= 5.1062




Problem 11.7f continued:

• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 10s2 + 21s + K(s + 4) = 0.

−ω3 − 10ω2 + 21jω + K(jω + 4) = 0

which gives(K + 21− ω2)ω = 0 and 4K − 10ω2 = 0

The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K > 0, and 2) K = ω2 − 21. Substituting this intothe second equation gives ω = ±j

√7/3, which means that s = −

√7/3 and is thus

not a solution of interest. The root locus plot is shown in the following figure.

−15 −10 −5 0 5−10

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−4

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Figure : for Problem 11.7f



11.8 The root locus equation is

1 + Gc(s)Gp(s) = 1 + KP

(1 +

1TIs

+ TDs

)s + 10

(s + 2)(s + 5)= 0

or

1 + KP TD

s2 + 1TD

s + 1TITD

s

s + 10(s + 2)(s + 5)

= 0

Substitute the given values to obtain

1 + 0.5KP(s2 + 2s + 10)(s + 10)

s(s + 2)(s + 5)= 0

The root locus parameter is K = 0.5KP .

• Poles: s = 0, s = −2 and s = −5

• Zeros: s = −10 and s = −1 ± 3j

• No paths must leave the plot, so there will be no asymptotes.


−10 < Re(s) < −5 and − 2 < Re(s) < 0


K = − s3 + 7s2 + 10s

s3 + 12s2 + 30s + 100

dK

ds= −5s4 + 60s3 + 390s2 + 1400s + 1000

(s3 + 12s2 + 30s + 100)2= 0

The only relevant solution is s = −0.9187, which is a breakin point for K > 0. Thevalue of K at this breakin point is

K = − s3 + 7s2 + 10s

s3 + 12s2 + 30s + 100

∣∣∣∣∣s=−0.9187

= 0.0496





• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 10s2 + 21s + K(s + 4) = 0.

−ω3 − 10ω2 + 21jω + K(jω + 4) = 0

which gives(K + 21− ω2)ω = 0 and 4K − 10ω2 = 0

The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K > 0, and 2) K = ω2 − 21. Substituting this intothe second equation gives ω = ±j

√7/3, which means that s = −sqrt7/3 and is thus

not a solution of interest. The root locus plot is shown in the following figure.

−15 −10 −5 0 5−10

−8

−6

−4

−2

0

2

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8

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11.9 a) The equation can be expressed as

1 +5p

4s2 + 6s + 8

s[s2 + (25/4)s + 4]= 0

so K = 5p/4.b) The poles are the roots of s[s2 + (25/4)s + 4] = 0, which are s = 0, −5.5262, and

−0.7238. The zeros are the roots of s2 + 6s + 8 = 0, which are s = −2 and −4.One path must leave the plot, so there will be one asymptote, which is 180.The locus exists on the real axis in the following intervals:

Re(s) < −5.5262 − 4 < Re(s) < −2 − 0.7238 < Re(s) < 0

To find the breakaway and breakin points:

K = −s[s2 + (25/4)s + 4]s2 + 6s + 8

= −s3 + (25/4)s2 + 4s

s2 + 6s + 8

dK

ds= −(s2 + 6s + 8)(3s2 + (25/2)s + 4)− (s3 + (25/4)s2 + 4s)(2s + 6)

(s2 + 6s + 8)2

=s4 + 12s3 + 57.5s2 + 100s + 32

(s2 + 6s + 8)2= 0

The solutions are s = −4.4115 ± 2.9797j, s = −2.7693, and s = −0.4077. The breakawaypoint is at s = −0.4077 and the breakin point is at s = −2.7693. The value of K at thebreakaway point is

K = − s3 + (25/4)s2 + 4s

s2 + 6s + 8

∣∣∣∣∣s=−0.4077

= 0.1153

The value of K at the breakin point is

K = − s3 + (25/4)s2 + 4s

s2 + 6s + 8

∣∣∣∣∣s=−2.7693

= 16.49

There is no crossover point for K ≥ 0.The root locus plot is shown in the figure on the following page.



−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0

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11.10 a) The poles are s = 0, 0, and −9. The zero is s = −1. The plot is shown in thefollowing figure.

−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1

−3

−2

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1

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4

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11.10 b)

K = −s2(s + 9)s + 1

dK

ds= −(s + 1)(3s2 + 18s) − (s3 + 9s2)

(s + 1)2= 0

which gives s = 0 and s = −3 for the breakaway points. For the breakaway point of interest,s = −3,

K = − s2(s + 9)s + 1

∣∣∣∣∣s=−3

= 27

From Guide 10, ∑roots = −3 − 3 + r3 = −9

Thus the third root is r3 = −3. So when K = 27, all three roots are at s = −3.c) The smallest possible dominant time constant occurs at the breakaway point s = −3

and is τ = 1/3.



11.11 a) The root locus equation is

1 + Ks(s + 0.5)(s + 1.5)

(s + 2.5)(s2 + 7s + 24.5)

The poles are s = −2.5, and −3.5± 3.5j. The zeros are s = 0, −0.5, and −1.5. The plot isshown in the figure on the following page.

b) The two points on the root locus that have a time constant of τ = 0.5 can be read offthe plot. They are s = −2 and s = −2 ± 3.35j. You can use the graphical method to findthe value of K at each point, and then use Guide 10 to determine the locations of the otherroots. For the first point, s = −2, K = 4.92, and the other roots are s = −0.64 ± 2.19j,which are the dominant roots. Thus the dominant time constant is not 0.5. For the secondpoint, s = −2 ± 3.35j, K = 0.73, and the third root is s = −2.32, which is not dominant.Thus the solution is K = 0.73.

You can solve this problem very easily by using the rlocus and rlocfind functions ofMATLAB to find the values of K and the root locations.

−6 −5 −4 −3 −2 −1 0 1 2 3−4

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11.12 a) Separate the equation as follows:

9s3 + 6s2 + 2 − 5ps = 0

The root locus form of the equation is

1 +(−5p

9

)s

s3 + (2/3)s2 + 2/9= 0

The root locus parameter is K = −5p/9 ≤ 0.b) Separate the equation as follows:

4s3 + 2s + 7 − ps2 = 0


1 +(−p

4

)s2

s3 + (1/2)s2 + 7/4= 0

The root locus parameter is K = −p/4 ≤ 0.c) Separate the equation as follows:

s2 + 3s + 4 − p(s − 4) = 0


1 + (−p)s − 4

s2 + 3s + 4= 0

The root locus parameter is K = −p ≤ 0.



11.13Part (a)

• Poles: s = 0 and s = −5

• Zeros: None

• Two paths must leave the plot, so there will be two asymptotes.


Re(s) < −5 and Re(s) > 0


θ =m360

0 − 2= ±180

• The two previous results lead us to conclude that there are no breakaway or breakinpoints.

• Crossover point: None for K ≤ 0




−12 −10 −8 −6 −4 −2 0 2 4 6 8

−6

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6

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11.13 (b)

• Poles: s = −1.5± 0.866j

• Zero: s = −3

• One path must leave the plot, so there will be one asymptote.

• The locus exists on the real axis in the following interval: Re(s) > −3

• Angles of the asymptote:

θ =m360

1 − 2= 0

• The breakaway and breakin points are found as follows.

K = −s2 + 3s + 3s + 3

dK

ds= −(s + 3)(2s + 3)− (s2 + 3s + 3

(s + 3)2=

s2 + 6s + 6(s + 3)2

= 0

The solutions are s = −4.7321 and s = −1.2679. The point is at s = −4.7321 is abreakaway or breakin point for K > 0. The point s = −1.2679 is a breakin point forK < 0. The value of K at the breakaway point is

K = − s2 + 3s + 3s + 3

∣∣∣∣∣s=−1.2679

= −0.4641

• The crossover point obviously occurs at s = 0, at which K = −1.




−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5

−1.5

−1

−0.5

0

0.5

1

1.5

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11.13 (c)

• Poles: s = 0, s = −1.5± 0.866j

• Zeros: None


• The locus exists on the real axis in the following interval: Re(s) > 0


θ =m360

0 − 3= 0, −120, −240

which are equivalent to θ = 0, −120, 60.

σ =∑

sp −∑

sz

2 − 0=

0− 1.5− 0.866j − 1.5 + 0.866j − 03

= −1


K = −(s3 + 3s2 + 3s)

dK

ds= −(3s2 + 6s + 3) = 0

The solutions are s = −1 and s = −1, which do not lie on the locus for K < 0. Thusthere are no breakaway or breakin points for K < 0.

• There is no crossover point for K ≤ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 3s2 + 3s + K = 0.

−jω3 − 3ω2 + 3jω + K = 0

which gives(3 − ω2)ω = 0 and K − 3ω2 = 0

The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K < 0, and 2) ω = ±

√3. Substituting this into the

second equation gives K = 9, which corresponds to the locus for K > 0.




−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

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11.13 (d)

• Poles: s = 0, s = −5, s = −7. Zeros: None



−7 < Re(s) < −5 and Re(s) > 0


θ =m360

0 − 3= 0, −120, −240

which are equivalent to θ = 0, −120, 60.


σ =∑

sp −∑

sz

2 − 0=

0 − 5 − 7 − 03

= −4

• The breakaway and breakin points are found as follows: K = −(s3 + 12s2 + 35s) and

dK

ds= −(3s2 + 24s + 35) = 0

The solutions are s = −6.0817 and s = −1.9183, which does not lie on the locus forK < 0. The value of K at the breakaway point is

K = − (s3 + 12s2 + 35s)∣∣∣s=−6.0817

= −6.0411




Problem 11.13d continued:

• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 12s2 + 35s + K = 0.

−jω3 − 12ω2 + 35jω + K = 0

which gives(35− ω2)ω = 0 and K − 12ω2 = 0

The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K < 0, and 2) ω = ±

√35. Substituting this into the

second equation gives K = 12(35), which corresponds to the locus for K > 0. Theroot locus plot is shown in the following figure.

−25 −20 −15 −10 −5 0 5 10 15 20 25−20

−15

−10

−5

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5

10

15

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Figure : for Problem 11.13d



11.13 (e)

• Poles: s = 0, s = −3

• Zero: s = −4

• One path must leave the plot, so there will be one asymptote.


−4 < Re(s) < −3 and Re(s) > 0

• Angle of the asymptote:

θ =m360

1 − 2= 0

• From the two previous items, it is obvious that there are no breakaway or breakinpoints for K < 0.

• There is no crossover point for K ≤ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s2 + 3s + K = 0.

−ω2 + 3jω + K = 0

which givesω = 0 and K − ω2 = 0

The first equation gives ω = 0, which is not of interest because it does not lie on theroot locus for K < 0. The second equation gives K = 0.




−4 −3 −2 −1 0 1 2 3 4 5 6

−3

−2

−1

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3

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Figure : for Problem 11.13e



11.13 (f)

• Poles: s = and s = −6. Zero: s = 4. One path must leave the plot, so therewill be one asymptote. The locus exists on the real axis in the following intervals:−6 < Re(s) < 0 and Re(s) > 4. The angle of the asymptote is θ = m360/(1−2) = 0


K = −s2 + 6s

s − 4

dK

ds= −(s − 4)(2s + 6)− (s2 + 6s

(s − 4)2=

s2 − 8s − 24(s − 4)2

= 0

The solutions are s = 10.325 and s = −2.325. The point is at s = 10.325 is a breakinpoint. The point s = −2.325 is a breakaway point. The value of K at the breakinpoint is

K = − s2 + 6s

s − 4

∣∣∣∣∣s=10.325

= −26.65

The value of K at the breakaway point is

K = − s2 + 6s

s − 4

∣∣∣∣∣s=−2.325

= −1.35

The breakin and breakaway points are symmetrically located each a distance 6.325from the zero at s = 4. The locus off the real axis is a circle of radius 6.325 centeredat the zero.

• The crossover points can be found as follows. Substitute s = jω into the characteristicequation s2 + 6s + K(s − 4) = 0.

−ω2 + 6jω + K(jω − 4) = 0

which gives(6 + K)ω = 0 and 4K + ω2 = 0

The first equation gives 1) ω = 0, which is not of interest because it does not lie onthe root locus for K < 0, and 2) K = −6. Substituting K = −6 into the secondequation gives ω = ±

√24 = ±2

√6. The root locus plot is shown in the figure on the

following page.



−6 −4 −2 0 2 4 6 8 10 12−8

−6

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−2

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6

8

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Figure : for Problem 11.13f




C(s)R(s)

=KP (8− s)

s2 + 2s + 3 + KP (8− s)=

KP (8− s)s2 + (2 − KP )s + 3 + 8KP

The root locus equation is

1 + KP8 − s

s2 + 2s + 3= 1 + K

s − 8s2 + 2s + 3

= 0

where K = −KP ≤ 0. The poles are s = −1± 1.414j, and the zero is s = 8. The root locusplot is shown in the figure on the following page.

The breakaway and breakin points are found as follows.

K = −s2 + 2s + 3s − 8

dK

ds= −(s − 8)(2s + 2)− (s2 + 2s + 3)

(s − 8)2= −s2 − 16s− 19

(s − 8)2= 0

The solutions are s = 17.11, which is the breakin point, and s = −1.11, which is a breakawayor breakin point for K > 0. From the location of the breakin point we can tell that the rootlocus off the real axis is a circle of radius 17.11− 8 = 9.11 centered on the zero at s = 8.

From the Routh-Hurwitz criterion applied to the characteristic equation

s2 + (2− KP )s + 3 + 8KP = 0

the system is stable if 2 − KP > 0 and 3 + 8KP > 0; that is, if −3/8 < KP < 2. Thereforethe crossover points correspond to KP = 2 and can be located by solving the characteristicequation for this value of KP . The result is

s2 + 19 = 0

or s = ±√

19.



−5 0 5 10 15 20−10

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y A

xis




11.14 b) To achieve τ = 2/3, either 1) the two roots must be real with the dominant rootat s = −3/2, or 2) the roots must be complex with the real part equal to −3/2. Since thelocus does not exist on the real axis for Re(s) < 0, it is impossible for case (1) to occur.Case (2) is also impossible because the circular locus lies entirely to the right of the verticalline passing through s = −3/2. Thus we see that the smallest time constant possible forthis system is 1/1 = 1, which corresponds to KP = 0.

c) With KP = 1 the closed-loop transfer function becomes

C(s)R(s)

=4 − s

s2 + s + 3 + 11

The characteristic roots are s = −0.5± 3.2787j. The unit step response can be found withthe Laplace transform.

C(s) =4 − s

(ss2 + s + 3 + 11)=

C1

s+ C2

3.2787s2 + s + 3 + 11

+ C3s + 0.5

s2 + s + 3 + 11

whereC1 + C3 = 0 C1 + 3.2787C2 + 0.5C3 = −1 11C1 = 8

The solution is C1 = 8/11 = 0.7273, C2 = −0.4159, and C3 = −0.7273. Thus

c(t) = 0.7273− 0.4149e−0.5t sin 3.2787t− 0.7273e−0.5t cos 3.2787t

The response is plotted in the figure on the following page. The negative sign in thenumerator causes the negative initial slope in the response.



0 2 4 6 8 10 12−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Step Response

Time (sec)

Am

plitu

de

Figure : For Problem 11.14c



11.15 a) The root locus equation is

1 + KP−2s2 + s + 26s(s + 2)(s + 3)

= 1 + Ks2 − 0.5s − 13s(s + 2)(s + 3)

= 0

where K = −2KP ≤ 0. The poles are s = 0, −2, −3, and the zeros are s = 3.86 and −3.36.The root locus plot is shown in the figure on the following page.


s(s + 2)(s + 3) + K(s2 − 0.5s − 13) = s3 + (5 + K)s2 + (6− 0.5K)s− 13K = 0

From the Routh-Hurwitz criterion, the system is stable if

5 + K > 0 6 − 0.5K > 0 − 13K > 0

and(5 + K)(6− 0.5K) > −13K

These reduce to −1.7277 < K < 0, or 0 < KP < 0.8639.At a breakaway or breakin point, ζ = 1. These points are found as follows.

K = − s3 + 5s2 + 6s

s2 − 0.5s− 13

dK

ds= −(s2 − 0.5s − 13)(3s2 + 10s + 6)− (s3 + 5s2 + 6s)(2s − 0.5)

(s2 − 0.5s − 13)2= 0

This is true if s4 − s3 − 4.75s2 − 130s − 78 = 0, which gives s = 8.5, −3.96, −2.68, and−0.8628. The first solution is an unstable point, the second and third solutions correspondto KP < 0, which is unstable, and the fourth is the desired solution. At this point,

K|s=−0.8628 =

(− s3 + 5s2 + 6s

s2 − 0.5s− 13

)∣∣∣∣∣s=−0.8628

= −0.177

which gives KP = 0.177/2 = 0.088.



−4 −2 0 2 4 6 8 10

−5

−4

−3

−2

−1

0

1

2

3

4

5

Root Locus

Real Axis

Imag

inar

y A

xis




11.15 b) The closed-loop transfer function is

C(s)R(s)

=KP (−2s2 + 1s + 26)

s3 + (5 − 2KP )s2 + (KP + 6)s + 26KP

The step response with KP = 0.088 is shown in the figure on the following page. Thenegative sign in the numerator causes the negative initial slope in the response.

0 1 2 3 4 5 6 7 8−0.2

0

0.2

0.4

0.6

0.8

1

1.2

Step Response

Time (sec)

Am

plitu

de




11.16 For ∆ = 0, the closed-loop transfer function is

C(s)R(s)

=KP + KDs

s2 + KDs + KP − 5

Assuming the system is stable, the steady-state response for a unit-step input is

css = lims→0

KP + KDs

s2 + KDs + KP − 51s

=KP

KP − 56= 1

Thus we cannot achieve the specification that css = 1.For ζ < 1, the time constant expression is

τ =2

KD= 0.1

Thus KD = 20.For the damping ratio,

ζ =KD

2√

KP − 5= 0.707

which gives KP = 205. Thus the steady-state response will be css = 205/200 = 1.025,which is close to the desired value.

With KP = 205 and KD = 20, for ∆ 6= 0, the closed-loop transfer function is

C(s)R(s)

=205 + 20s

s2 + 20s + 200− ∆


1 + K1

s2 + 20s + 200= 0

where K = −∆ and −1 ≤ K ≤ 0. The root locus for K ≤ 0 is shown in the figure onthe following page. The breakin point is at s = −0.1 when K = −100. It shows that thetime constant remains fixed at τ = 0.1 for −100 ≤ K ≤ 0, that is for 0 ≤ ∆ ≤ 100. Thedamping ratio variation as K varies from 0 to −1 is too small to be seen on the plot, butwe can calculate the sensitivity of ζ to changes in K as follows.

ζ =20

2√

200 + K

From this we see that ζ varies from ζ = 0.707 to ζ = 0.709 as K varies from 0 to −1. Thusthe system is insensitive to K and therefore to ∆.



−20 −15 −10 −5 0−10

−8

−6

−4

−2

0

2

4

6

8

10

Root Locus

Real Axis

Imag

inar

y A

xis




11.17 a) With proportional control, the root locus equation is

1 + KP6

s(2s + 2)(3s + 24)= 0

or1 + K

1s(s + 1)(s + 8)

= 0

where K = KP . The root locus is similar in shape to that shown in Figure 11.1.20.The breakaway point is found as follows.

K = −s(s + 1)(s + 8) = −(s3 + 9s2 + 8s

)

dK

ds= −(3s2 + 18s + 8) = 0

which gives s = −5.517 and s = −0.4833. The first solution corresponds to a breakin pointfor K < 0. The second solution is the breakaway point for K > 0. For this plot, the farthestto the left the dominant root can lie is at the breakaway point. Thus the smallest possibledominant time constant is 1/0.4833 = 2.07.



11.17 b) PD action will add a zero that will pull the locus to the left, stabilize the system,and allow a smaller time constant to be obtained, if the zero is placed properly. With PDaction, the open-loop transfer function becomes

KP (1 + TDs)s(s + 1)(s + 8)

=KP TD

(s + 1

TD

)

s(s + 1)(s + 8)

The poles are still at s = 0, s = −1, and s = −8, and the zero is at s = −1/TD. The rootlocus parameter is K = KP TD.

We want τ = 0.5. Thus the zero at s = −1/TD must be placed far enough to the left topull the locus to the left of s = −2. Thus we choose TD = 0.5 to place the zero at s = −2.The resulting root locus is shown in the figure on the following page.

Next we choose ζ = 0.707, which along with τ = 0.5, specifies the point s = −2 ± 2j.From the plot we can see that the line corresponding to ζ = 0.707 passes through thelocus, so we know that a solution exists. At s = −2 ± 2j, we find that K = 19.7. ThusKP = 19.7/TD = 2(19.7) = 39.4. The third root lies at s = −5 and thus is not dominant.Therefore, one solution is TD = 0.5, KP = 39.4. Other solutions are possible, depending onthe choices made for TD and ζ.

−5 −4 −3 −2 −1 0 1 2 3 4

−2

−1

0

1

2

3

4

0.707

0.707

Root Locus

Real Axis

Imag

inar

y A

xis

s = −2 + 2j





1 + KPs + 10

(s + 2)(s + 3)= 0

The root locus plot is shown in the following figure.The line tangent to the circle gives the smallest ζ. This line has an angle of 49. Thus

the smallest ζ is ζ = cos 49 = 0.66.The 45 line corresponds to ζ = 0.707. This line intersects the circle at two points. The

left-most point gives the smallest time constant. This point, which can be found graphicallyor analytically, is s = −6.73 + 6.73j, which corresponds to

KP = − (s + 2)(s + 3)s + 10

∣∣∣∣s=−6.73+6.73j

= 8.46

The time constant is τ = 1/6.73 = 0.149.

−18 −16 −14 −12 −10 −8 −6 −4 −2 0−8

−6

−4

−2

0

2

4

6

8

Root Locus

Real Axis

Imag

inar

y A

xis

ζ = 0.66 ζ = 0.707




11.19 The characteristic equation is found from

1 + KP Gp(s) = 0

which becomes1 + 2

5(s + 4)(s + 3)(s + p)

= 0

ors2 + 13s + 40 + p(s + 3) = 0

Let K = p − 7. Then the characteristic equation becomes

s2 + 20s + 61 + K(s + 3) = 0

where K ≥ 0. The poles are s = −3.76 and s = −16.2. The zero is s = −3. The root locusplot is sketched in the figure on the following page. As p increases above 7, the dominantroot moves from −3.76 to −3, and the other root moves from −16.2 to −∞. The roots arealways real, so there will be no oscillations in the free response.

The dominant time constant varies from τ = 1/3.76 = 0.266 to τ = 0.333 as p increasesabove p = 7.

−25 −20 −15 −10 −5 0

−8

−6

−4

−2

0

2

4

6

8

Root Locus

Real Axis

Imag

inar

y A

xis




11.20 Express the controller transfer function as

Gc(s) =KP (s + KI

KP)

s=

KP (s + b)s

where b = KI/KP . The root locus equation is

1 + KPs + b

s(s + 1)(s + 2)= 0

The poles are s = 0, s = −1, and s = −2. The zero is at s = −b. The asymptotic anglesare θ = ±90. The asymptotes intersect at

σ =b − 3

2

Sketches of the root locus plots for three cases are shown in the figure on the followingpages. These three cases correspond to

a) 0 < b < 1, using b = 0.5b) 1 < b < 2, using b = 1.5, andc) b > 2, using b = 3.Case (a) does not satisfy the specification that the dominant root have a damping ratio

of ζ = 0.707. Case (b) allows a smaller dominant time constant than Case (c).



−6 −5 −4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

Root Locus

Real Axis

Imag

inar

y A

xis

Figure : for Problem 11.20, case (a)



−6 −4 −2 0 2 4−5

−4

−3

−2

−1

0

1

2

3

4

5

Root Locus

Real Axis

Imag

inar

y A

xis

Figure : for Problem 11.20, case (b)



−8 −6 −4 −2 0 2 4−5

−4

−3

−2

−1

0

1

2

3

4

5

Root Locus

Real Axis

Imag

inar

y A

xis

Figure : for Problem 11.20, case (c)





s3 + 3s2 + (KP + 2)s + bKP = 0

Because we require that ζ = 0.707, the dominant root pair has the form s = −c ± cj. Thecharacteristic equation can be factored as

[(s + c)2 + c2](s− s3) = s3 + (2c− s3)s2 + (2c2 − 2cs3)s − 2c2s3 = 0

where the third root is s3 and must be real. Comparing coefficients, we see that

2c− s3 = 3 2c2 − 2cs3 = KP + 2 − 2c2s3 = bKP

Thus the third root is s3 = 2c− 3. In order for s = −c ± cj to be dominant, s3 < −c, andthus we must choose c < 1. To minimize the dominant time constant τ = 1/c, we shouldchoose c as large as possible, subject to the restriction that c < 1. However, the closer c

is to 1, the closer the secondary root s3 is to the dominant root. If s3 is too close to thedominant root, the response expected for ζ = 0.707 might not be obtained. Thus we shouldtry a value for c and check the effects of the secondary root by analysis or simulation.

For example, one solution is c = 0.9. This gives s3 = −1.2, b = 1.09, and KP = 1.78.



11.21 The characteristic equation is

1 + Gc(s)Gp(s) = 0

which becomes1 +

(KP +

KI

s+ KDs

)4

3s2 + 3= 0

or3s3 + 4KDs2 + (3 + 4KP )s + 4KI = 0 (1)

To achieve τ = 1 and ζ = 0.5, the desired dominant roots are s = −1 ± j√

3. The thirdroot is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will notbe the dominant root). The polynomial corresponding to these three roots is

[(s + 1)2 + 3](s + b) = 0

ors3 + (b + 2)s2 + (4 + 2b)s + 4b = 0

To compare this with equation (1), we must multiply by 3:

3s3 + 3(b + 2)s2 + 3(4 + 2b)s + 12b = 0 (2)

Comparing the coefficients of (1) and (2), we obtain

KP =9 + 6b

4

KI = 3b

KD =3(b + 2)

4The gains can be computed once a value for b > 1 has been selected.



11.22 a) Assuming that sin φ ≈ φ, cosφ ≈ 1, and sinφ φ2 ≈ 0, we obtain from (2.4.8) and(2.4.11),

(IG + mL2)φ− mLx = T + mgLφ (1)

(m + M)x− mLφ = −f (2)

Solve (2) for x and substitute into (1) to obtain[(m + M)IG + mML2

]φ − (m + M)mgLφ = (m + M)T − mLf (3)

Note that (3) implies that the arm dynamics (φ) are independent of the base dynamics (x),unless either T or f are dependent on x through feedback control.

b) Substituting the given values into (3) gives

500φ− 29 430φ = 60T − 50f (4)

c) For the personal transporter discussed in Chapter 2, T = 0 since the device is con-trolled entirely with the base force f . So we will imitate this approach. Noting that (4)represents an unstable system because the φ term is missing and because the coefficient ofφ is negative, we try a feedback control law of the form

f = KP φ + KDφ (5)

This is a PD control algorithm.Setting T = 0 in (4), substituting (5) into (4), and cleaning up, we obtain

φ − 58.86φ = −0.1KPφ − 0.1KDφ

orφ + 0.1KDφ + (0.1KP − 58.86)φ = 0

A settling time of 4 s implies a time constant of 1 s. If ζ ≤ 1, the time constant is givenby

τ =2

0.1KD= 1

which gives KD = 20.Choosing ζ = 1 gives

ζ =0.1KD

2√

0.1KP − 58.86=

1√0.1KP − 58.86

= 1

which gives KP = 598.6.



11.23 From Example 11.3.1,

LIs3 + I(R + K3)s2 + +KT K2s + KTK1 = 0

Here,IL = 1.2 × 10−7 IR = 3.6× 10−6

Thus

1.2× 10−7s3 + (3.6× 10−6 + 6 × 10−5K3)s2 + 0.04K2s + 0.04K1 = 0 (1)

We wants = − 1

0.5= −2 and s =

10.05

(−1 ± j) = −20 ± 20j

This requires a characteristic equation of the form

(s + 2)[(s + 20)2 + 202

]= 0

ors3 + 42s2 + 880s + 1600 = 0

Multiply by the leading coefficient of Eq. (1) to obtain

1.2×10−7(s3+42s2+880s+1600) = 1.2×10−7s3+50.4×10−7s2+1056×10−7s+1920×10−7 = 0 (2)

Comparing the coefficients of Eqs. (1) and (2) gives

3.6× 10−6 + 6× 10−5K3 = 50.4× 10−7

0.04K2 = 1056× 10−7

0.04K1 = 1920× 10−7

These giveK1 = 0.0048 V/rad

K2 = 0.00264 V · rad

K3 = 0.024 V/rad



11.24 Let x1 = θ and x2 = θ. Thenx1 = x2

x2 =1

mL(mgθ − mu) =

g

Lx1 −

1L

u

But u = K1x1 + K2x2, sox1 = x2

x2 =g

Lx1 −

1L

(K1x1 + K2x2)

Taking the Laplace transform of each equation with zero initial conditions and collectingterms, we obtain

sX1(s)− X2(s) = 0

(K1 − g)X1(s) + (Ls + K2)X2(s) = 0

The determinant of these equations must be zero. This gives∣∣∣∣∣

s −1(K1 − g) Ls + K2

∣∣∣∣∣ = 0

This givess(Ls + K2) + (K1 − g) = 0

orLs2 + K2s + K1 − g = 0

The control system will be stable if and only if K2 > 0 and K1 > g.



11.25 a) Let the state variables be

z1 = x1 z2 = ˙ x1 z3 = x2 z4 = ˙ x2

The input is f . The state equations are

z1 = z2 z2 =1

m1[k1y − (k1 + k2)z1 − cz2 + k2z3 + cz4 − f ]

z3 = z4 z4 =1

m2[k2z1 + cz2 − k2z3 − cz4 + f ]

where the disturbance input is the road displacement y. The matrices are

A =

0 1 0 0−k1+k2

m1− c

m1

k2m1

cm1

0 0 0 1k2m2

cm2

− k2m2

− cm2

=

0 1 0 0−3240 −2 240 2

0 0 0 148 0.2 −48 −0.2

B =

0− 1

m1

01

m2

=

0−0.02

00.004

b) The MATLAB script file is

A = [0,1,0,0;-3240,-2,240,2;0,0,0,1;48,0.2,-48,-.2];B = [0;-.02;0;0.004];p = [-1.397+69.94j,-1.397-69.94j,-0.168+7.779j,-0.168-7.779j];K = acker(A,B,p)

The result is K = [-80 794, -22, 12 688, 124, 124]. For this application, r(t) =0, and the control algorithm is f(t) = K1(0 − x1) − K2x2 − K3x3 − K4x4 = 80 794x1 +22x2 − 12 688x3 − 124x4.



11.26 There are two circuits and one inertia in this system, and we must write an equationfor each. For the inertia I , Newton’s law gives:

Idω

dt= T − cω − TL

where T = KT ia. For the generator circuit, Kirchhoff’s voltage law gives

vf = Rf if + Lfdifdt

For the motor circuit, Kirchhoff’s voltage law gives

va = Raia + Ladiadt

+ Kbω

where va = Kf if . Substitute for va and T , and rearrange to obtain:

Idω

dt+ cω = KT ia − TL

Lfdifdt

+ Rf if = vf

Ladiadt

+ Raia = Kf if − Kbω

Let the state variable vector be x = [ω, if , ia]. Let the input vector be u = [vf , TL]. Thestate equations are

x = Ax + Bu

where

A =

−c/I 0 KT/I

0 −Rf/Lf 0−Kb/La Kf/La −Ra/La

=

−2 0 0.050 −10 0

−2.5 250 −5

B =

0 −1/I−1/Lf 0

0 0

=

0 −0.1−5 00 0





State variable feedback with proportional control has the form

vf = KP (ωr − ω) − K2if − K3ia

To determine the gain vector K = [KP , K2, K3] we use the acker function with the inputmatrix B1 consisting of the first column of B: B1 = [0, 1/Lf , 0]T .

First we must specify the three desired roots. Since the speed time constant is I/c = 0.5,the field time constant is Lf/Rf = 0.1, and the armature time constant is La/Ra = 0.2, wemay reasonably choose a dominant closed-loop time constant of 1 s. Choosing ζ = 0.707,gives the two dominant roots to be s = −1± j. Choosing the third root to be s = −2 givesthe root vector p = [-1 + j, -1-j, -2]. In MATLAB type

A=[-2, 0, 0.05; 0, -10, 0; -2.5, 250, -5];B1 = [0;5;0];p = [-1 + j,-1-j,-2];K = acker(A,B1,p)K =

0.0100 -2.6000 0.0135

Thus the control algorithm is

vf = 0.01(ωr − ω) + 2.6if − 0.0135ia

The closed-loop dynamics are described by

x = (A− B1K)x +

0 −1/IKP 00 0

[

ωr

TL

]

Let

B2 =

0 −1/IKP 00 0





To find the transfer functions, first create the state space model Mss as follows:

KP = K(1);B2 = [0,-1/10; KP,0;0,0]C = [1, 0, 0]; D = [0];Mss = ss(A -B1*K,B2,C,D)

Then find the two transfer functions as follows:

Mtf = tf(Mss)

The first transfer function displayed on the screen is Ω(s)/Ωr(s), which is

Ω(s)Ωr(s)

=0.125

s3 + 4s2 + 6s + 4

The second transfer function displayed on the screen is Ω(s)/TL(s), which is

Ω(s)TL(s)

= −0.1s2 + 0.2s + 0.1875s3 + 4s2 + 6s + 4

For a unit-step command ωr, the steady-state response is

ωss = 0.0313

which is a 97% error. For a unit-step disturbance TL, the steady-state response is

ωss = −0.0469



11.27 With b = 10, τ = 1, and u = −K1δ − K2ω − K3φ, the equations become

δ = u = −K1δ − K2ω − K3φ

ω = −ω + 10δ

φ = ω

Taking the Laplace transform of each equation with zero initial conditions and collectingterms, we obtain

(s + K1)∆(s) + K2Ω(s) + K3Φ(s) = 0

−10∆(s) + (S + 1)Ω(s) + k3Φ(s) = 0

0∆(s)− Ω(s) + sΦ(s) = 0

The determinant of these equations must be zero. This gives∣∣∣∣∣∣∣

(s + K1) K2 K3

−10 (s + 1) 00 −1 s

∣∣∣∣∣∣∣= 0

This gives

(s + K1)

∣∣∣∣∣(s + 1) 0−1 s

∣∣∣∣∣+ 10

∣∣∣∣∣K2 K3

−1 s

∣∣∣∣∣ = 0

This reduces tos3 + (K1 + 1)s2 + (K1 + 10K2)s + 10K3 = 0 (1)

The desired roots are s = −20 and s = −10 ± 10j. These correspond to the factoredcharacteristic equation:

(s + 20)[(s + 10)2 + 102

]= s3 + 40s2 + 600s + 4000 = 0 (2)

Comparing the coefficents of Eqs. (1) and (2), we obtain

K1 + 1 = 40

K1 + 10K2 = 600

10K3 = 4000

The solution isK1 = 39 K2 = 56.1 K3 = 400



11.28 The basic equations for the motor circuit is

e = Raia + Ladiadt

+ Keωm (1)

Tm = KT ia (2)

The dynamics of the windup roll give

d(Iωr)dt

= nTm − cωr − TR (3)

The rate of change of paper tension is given as

dT

dt= k(vr − vp) (4)

The linear and rotational roll velocities are related as

Rvr = ωr (5)

and the effect of the gear pair isωm = nωr (6)

From calculus,d(Iωr)

dt= Iωr + Iωr (7)

andI =

12ρπW (4RR3) = 2ρπWR3

(12dW

)= ρπdW 2R3 (8)

a) If R ≈ 0, then I ≈ 0 andd(Iωr)

dt= Iωr

The three state variables are ia, ωm, and T . Thus the state equations are

diadt

=1La

(e − Raia − Keωm) (9)

dωm

dt=

1I

(n2KT ia − cωm − nRT

)(10)

dT

dt=

k

nRωm − kvp (11)





b) In this case, I 6= 0 and

R =d

2Wt + R(0)

From (3), (7), and (8),

ρπdW 2R3

nωm +

12

ρπWR4

nωm = nKT ia −

c

nωm − RT

or

ωm =2n

ρπWR4

[−ρπdW 2R3

nωm + nKT ia −

c

nωm − RT

](12)

The state model consists of (9), (11), and (12).



11.29 Change a) For the given values,

Y (s)Ic(s)

=KaK1K2

s(τs + 1)(s2 + 2ζωns + ω2n)

=3× 106

s4 + 260s3 + 26000s2 + 106s

which can be expressed as

Y (s)Ic(s)

=3 × 106/s4

1 + 260/s + 26000/s2 + 106/s3

Let x1 = y and u = ic. Then the state model is

x =

−260 1 0 0−2600 0 1 0−106 0 0 1

0 0 0 0

x +

000

3 × 106

u





b) Note that if Gc(s) = KP , then

ic = KPe = KP (K3yr − K3y) = 20KPyr − 20KPy

Substitute this into the matrix equation in part (a) to obtain

x = Ax + Bu ==

−260 1 0 0−2600 0 1 0−106 0 0 1

−60 × 106KP 0 0 0

x +

000

60× 106KP

u

The charactersitic polynomial of A is found from the detrminant |sI− A| = 0, and is

s4 + 260s3 + 2600s2 + 106s + 6 × 107KP = 0


1 + K1

s4 + 260s3 + 2600s2 + 106s= 0

where K = 6 × 107KP . The root locus plot is shown in the figure on the following page.Since two root always lie in the right-half plane, it is impossible for the proportional controlsystem to be stable.(continued on the next page)



−1000 −800 −600 −400 −200 0 200 400 600 800−800

−600

−400

−200

0

200

400

600

800

Root Locus

Real Axis

Imag

inar

y A

xis

Figure : For Problem 11.29





(m + M)x− mLφ = −f (2)


]φ − (m + M)mgLφ = (m + M)T − mLf (3)



500φ− 29 430φ = 60T − 50f (4)

c) For the personal transporter discussed in Chapter 2, T = 0 since the device is con-trolled entirely with the base force f . So we will imitate this approach. Set T = 0 anddivide (4) by 500 to obtain

φ − 58.86φ = −0.1f (5)

Let x1 = φ and x2 = φ to put the model into the following state variable form:

x1 = x2

x2 = 58.86x1 − 0.1f

For linear state variable feedback, f = K1x1 + K2x2, and

x1 = x2

x2 = (58.86− 0.1K1)x1 − 0.1K2x2

Transforming both equations and rearranging gives

sX1(s) − X2(s) = 0 (6)

(0.1K1 − 58.86)X1(s) + (s + 0.1K2)X2(s) = 0 (7)





Solve (6) for X2(s) and substitute into (7) to obtain

(0.1K1 − 58.86)X1(s) + (s + 0.1K2)sX1(s) = 0 (7)

or (s2 + 0.1K2s + 0.1K1 − 58.86

)X1(s) = 0

Thus the characteristic equation is

s2 + 0.1K2s + 0.1K1 − 58.86 = 0 (8)


τ =2

0.1K2= 1

which gives K2 = 20.Choosing ζ = 1 gives

ζ =0.1K2

2√

0.1K1 − 58.86=

1√0.1K1 − 58.86

= 1

which gives K1 = 598.6.



11.31 a) Using P action, the closed-loop transfer function is

T (s) = T (s) =0.5KP

s3 + 1.5s2 + 1.5s + 0.5 + 0.5KP

Substitute s = jω into the characteristic equation:

j(−ω3 + 1.5ω) + (−1.5ω2 + 0.5 + 0.5KP ) = 0

Thus ωu = 1.225, KPu = 3.5, Pu = 2π/ωu = 5.129.From Table 11.4-1,

For P action : KP = 0.5KPu = 1.75

For PI action : KP = 0.45KPu = 1.575 TI = 0.83Pu = 4.2572

KI =KP

TI= 0.37

For PID action : KP = 0.6KPu = 2.1 TI = 0.5Pu = 2.5646

KI =KP

TI= 0.8189

TD = 0.125Pu = 0.6411 KD = KP TD = 1.3464

b) The responses are shown in the figure on the following page.



0 5 10 15 20 25 300

0.5

1

t

θProportional

0 5 10 15 20 25 300

0.5

1

1.5

t

θ

PI control

0 5 10 15 20 25 300

0.5

1

1.5

t

θ

PID control

Figure : For Problem 11.31. Responses of the P, PI, and PID controllers.





The overshoot in the PID response can be reduced by using KP = 1, and reducingthe other gains by a factor of KP = 1.75, that KP = 1, KI = 0.8189/1.75, and KD =1.3464/1.75. The value of KP = 1 was found by simulation. The response of the improvedPID is shown in the following figure.

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4

t

θ

PID

Improved PID

Figure : For Problem 11.31. Response of the improved PID controller.




T (s) =5KP

2s3 + 10s2 + 2s + 4 + 5KP

Substituting s = jω into the characteristic equation gives ωu = 1, KPu = 1.2, Pu = 2π/ωu =2π.

From Table 11.4-1,For P action : KP = 0.5KPu = 0.6

For PI action : KP = 0.45KPu = 0.54 TI = 0.83Pu = 5.215

KI =KP

TI= 0.1035


KI =KP

TI= 0.2292

TD = 0.125Pu = 0.7854 KD = KP TD = 0.5655

b) The responses are shown in the figure on the following page.



0 10 20 30 40 50 60 700

0.5

1Proportional

t

y

0 10 20 30 40 50 60 700

0.5

1PI control

y

t

0 10 20 30 40 50 60 700

1

2

t

y

PID control

Figure : For Problem 11.32. Responses of the P, PI, and PID controllers.





The oscillations in the PID response can be reduced by increasing the derivative gain.The value of KD = 2 was found by simulation. The response of the improved PID is shownin the following figure.

0 5 10 15 20 25 30 35 40 45 500

0.2

0.4

0.6

0.8

1

1.2

1.4

t

y

PIDImproved PID

Figure : For Problem 11.32. Response of the improved PID controller.



11.33 a) Set p = 1 and ζ = 0.5. Using P action, the closed-loop transfer function is

C(s)R(s)

=4KP

s3 + 3s2 + 6s + 4 + 4KP

Substituting s = jω into the characteristic equation gives ωu =√

6 = 2.45, KPu = 3.5,Pu = 2π/ωu = 2.57.

From Table 11.4-1,


KI =KP

TI= 1.6374

TD = 0.125Pu = 0.3206 KD = KP TD = 0.6733

The PID control law is

G(s) = 2.1(1 +

11.285s

+ 0.3206s

)=

0.6733s2 + 2.1s + 1.637s

and the resulting closed-loop transfer function is

C(s)R(s)

=2.693s2 + 8.4s + 6.55

s4 + 3s3 + 8.693s2 + 12.4s + 6.55

The roots are s = −0.4848± 2.3038j and s = −1.0152± 0.3886j.In terms of p and ζ, the closed-loop transfer function is

C(s)R(s)

=4p(0.6733s2 + 2.1s + 1.637)

s4 + (4ζ + p)s3 + (4 + pζ + 2.6932p)s2 + 12.4ps + 6.548p

b) With p = 1 the root locus equation in terms of ζ is

1 + 4ζs3 + s2

s4 + s3 + 6.932s2 + 12.4s + 6.548= 0

So the root locus parameter is K = 4ζ. The plot is shown in the figure on the followingpage. For 0.4 ≤ ζ ≤ 0.6, the dominant roots vary from s = −0.313 ± 2.42j, which have adamping ratio of 0.129 and a time constant of 3.19, to s = −0.641 ± 2.16j, which have adamping ratio of 0.284 and a time constant of 1.56.



−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−3

−2

−1

0

1

2

3

Root Locus

Real Axis

Imag

inar

y A

xis

Figure : For Problem 11.33b.





c) With ζ = 0.5 the root locus equation in terms of p is

1 + ps3 + 4.6932s2 + 12.4s + 6.548

s4 + 2s3 + 4s2= 0

So the root locus parameter is K = p. The plot is shown in the following figure. For0.5 ≤ p ≤ 1.5, the dominant roots vary from s = −0.589±1.9j, which have a damping ratioof 0.296 and a time constant of 1.69, to s = −0.542± 2.6j, which have a damping ratio of0.204 and a time constant of 1.845.

Root Locus

Real Axis

Imag

inar

y A

xis

−4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−4

−3

−2

−1

0

1

2

3

4

Figure : For Problem 11.33c.




C(s)R(s)

=KP

s3 + 6s2 + 11s + 6 + KP

Substituting s = jω into the characteristic equation gives ωu =√

11 = 3.317, KPu = 60,Pu = 2π/ωu = 1.8945. From Table 11.4-1,

For PID action : KP = 0.6KPu = 36 TI = 0.5Pu = 0.9472

KI =KP

TI= 38.006

TD = 0.125Pu = 0.2368 KD = KP TD = 8.525

b) The response is shown in the figure on the following page, and is labeled ”OriginalPID”. The overshoot can be lowered by decreasing KP , for example, from 36 to 5, anddecreasing KI and KD by the same ratio. The response labeled ”Improved PID” wasobtained with the gain values: KP = 5, KI = (5/36)38.0065.2786, and KD = (5/36)8.525 =1.184.



0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

t

Res

pons

e

Original PID

Improved PID

Figure : For Problem 11.34.



11.35 In the linear region,C(s)R(s)

=KP s + KI

5s2 + KP s + KI

To avoid saturation,

KP ≤ mmax

rmax=

202

= 10

If we set ζ ≤ 1,

τ =10KP

= 0.1 which gives KP = 100 > 10

If we set KP = 10, the characteristic equation becomes 5s2 + 10s + KI = 0. If ζ ≤ 1, thenτ = 10/10 = 1 > 0.1. If ζ > 1 and τ = 0.1, the two roots will be real and the equation canbe factored as 5(s + 10)(s + a) = 5s2 + (50 + 4a)s + 50a = 5s2 + 10s + KI . Thus a = −8and KI = 50a = −40, which gives an unstable system.

Thus all the possibilities have been exhausted, and we cannot achieve τ = 0.1 becauseof the limit on KP .

So one possible solution is KP = 10 with ζ = 1. This gives

ζ =KP

2√

5KI= 1 which gives KI =

K2P

20= 5

but then τ = 10/5 = 2 > 0.1, so all the specifications have not been achieved.Using the Saturation block with Simulink, simulation with a step input of magnitude

2 shows that the actuator is not saturated except at t = 0+, and that the steady stateresponse is css = 2 but it takes about 8 seconds to reach steady state.




C(s)R(s)

=KP s + KI

10s2 + (20 + KP )s + KI


ζ =20 + KP

2√

10KI= 1

The time constant isτ =

2020 + KP

= 0.1

These two conditions give KP = 180 and KI = 1000.b) Because ζ ≤ 1,

τ =20

20 + KP= 0.1

Thus, to minimize τ , we maximize KP . The largest we can make KP without causingsaturation is

KP =mmax

rmax=

11

= 1

Thus the damping ratio is

ζ =21

2√

10KI

To obtain ζ = 1 requires that KI = 11.025. The solution is KP = 1, KI = 11.025.c) The easiest way to compare these two designs is to use the Saturation block with

Simulink. The unit-step responses show that saturation occurs for both designs and thatthe steady state response is css = 0.05. Examining the plant’s equation of motion, whichis 10c + 20c = m(t), we see that if the maximum value of m is 1, then the steady stateresponse will be css = 1/20 = 0.05. So with the given value of mmax = 1, it is impossibleto achieve a steady state response of css = 1.



11.37 In the linear region,

C(s)R(s)

=KP s + KI

7s2 + (5 + KP )s + KI

To avoid saturation,

KP ≤ mmax

rmax=

205

= 4

If ζ > 1 and τ = 0.2, the two roots will be real and the equation can be factored as7(s + 5)(s + a) = 7s2 + (35 + 5a)s + 35a. Thus 5 + KP = 35 + 5a and KI = 35a. ThusKI = 5KP − 150 and the system is stable KI > 0, which is true only if KP > 30.

So the only choice is ζ ≤ 1, which gives

τ =14

5 + KP≥ 14

9> 0.2 because KP ≤ 4

With KP = 4, τ = 14/9, and KI can be selected to achieve any value of ζ ≤ 1. However,the time constant specification cannot be satisfied because of the limit on KP .

For example, KI = 2.8929 gives ζ = 1. Using the Saturation block with Simulink,simulation with a step input of magnitude 5 shows that the actuator is always saturated,and that the response reaches c = 4.5 at t = 4 and takes more than 10 seconds to reachsteady state at css = 5. Examining the plant’s equation of motion, which is 7c+4c = m(t),we see that since mmax is 20, then the steady state response will be css = 20/4 = 5. So withthe given value of mmax = 20, it is possible to achieve a steady state response of css = 5,but with a time constant much larger than the desired value of 0.2.



11.38 a) For PI control with ζ = 1

KP =mmax

rmax=

63

= 2

KI =1

4Iζ2

(mmax

rmax

)2

=1

4ζ2= 0.25

For modified I controlK2 = 5.44

mmax

rmax= 10.88

KI =29.64I

(mmax

rmax

)2

= 7.4

b) For a unit-step command, the PI controller has an overshoot of about 15% andthe modified I controller has no overshoot. Neither controller saturates the actuator. Ifthe magnitude of the step command is rmax = 3, the PI controller has an overshoot ofabout 13% and the modified I controller has no overshoot. Neither controller saturates theactuator.

c) For a unit-ramp command, neither controller saturates the actuator. The responseof the modified I controller lags behind that of the PI controller by about 1.5 seconds atsteady-state.



11.39 The closed-loop transfer functions for the PI controller are

C(s)R(s)

=KP s + KI

Is2 + KP s + KI

C(s)D(s)

=−s

Is2 + KP s + KI

Assuming the controller is critically damped (ζ = 1), the roots are repeated at s = −KP /2I .The closed-loop transfer functions for the modified I controller are

C(s)R(s)

=KI

Is2 + K2s + KI

C(s)D(s)

=−s

Is2 + K2s + KI

Assuming the controller is critically damped (ζ = 1), the roots are repeated at s = −K2/2I .a) For a unit ramp disturbance, the response of the PI controller is

c(t) =4I

K2P

[1 −

(1 +

KP

2It

)e−KP t/2I

]

and the response of the modified I controller is

c(t) =4I

K22

[1 −

(1 +

K2

2It

)e−K2t/2I

]

For both controllers, css = 1/KI , but note that the value of KI for the modified I controlleris 29.6 times lager than for PI. Thus the steady-state response of the modified I controlleris 29.6 times smaller than that of the PI controller.

b) For a sinusoidal disturbance of unit amplitude and frequency ω, D(s) = ω/(s2 +ω2),and the magnitude ratio of the PI controller is

M(ω) =∣∣∣∣C(iω)D(iω)

∣∣∣∣ =ω√

(KI − Iω2)2 + (KPω)2

Its maximum value occurs at ω =√

KI/I and is 1/KP .





The magnitude ratio of the modified I controller is

M(ω) =∣∣∣∣C(iω)D(iω)

∣∣∣∣ =ω√

(KI − Iω2)2 + (K2ω)2

Its maximum value occurs at ω =√

KI/I and is 1/K2.Because K2 = 5.44KP (see (11.5.3)), the maximum magnitude ratio of the PI controller

is 5.44 times larger than that of the modified I controller. This means that the deviationcaused by the disturbance is 5.44 times greater for the PI controller.

c) For a sinusoidal command of unit amplitude and frequency ω, R(s) = ω/(s2 + ω2),and the magnitude ratio of the PI controller is

M(ω) =∣∣∣∣C(iω)R(iω)

∣∣∣∣ =KI√

(KI − Iω2)2 + (KP ω)2

The magnitude ratio of the modified I controller is

M(ω) =∣∣∣∣C(iω)R(iω)

∣∣∣∣ =

√K2

I + (K2ω)2√

(KI − Iω2)2 + (K2ω)2

The ratio for PI is greater than 1 at low frequencies. The ratio for modified I is alwaysless than 1. The PI controller has a high- frequency slope of -20 db/decade, whereas themodified I controller has a slope of -40 db/decade.



11.40 A plot of the response data yields an eyeball estimate of R = 58 and L = 1. If othersalso estimate this line by eye, some variations can be expected in the values of R and L. Abetter way is the use a least squares fit. A Matlab file to do this is shown below.

% Times at which measurements were made.tm=[0:0.5:6,7];% Measured Response.resp=[0,4,20,32,56,84,116,140,160,172,184,190,194,196];% Fit a 5th order polynomial to the data.coef=polyfit(tm,resp,5);% Evaluate the polynomial at a large number of points.dt = .01;t=[0:dt:7];resppoly=polyval(coef,t)’;% Find the coefficients of the polynomial’s derivative.coef2=[5*coef(1),4*coef(2),3*coef(3),2*coef(4),coef(5)];% der is the derivative polynomial.der=polyval(coef2,t)’;% Find the maximum slope R, and the time tdmax at which it occurs.[R,i]=max(der);tdmax = i*dt;% Compute the response where the slope is the maximum.slopemax=resppoly(i);% Compute the intercept L on the time axis.L=(R*tdmax-slopemax)/R;% Compute the points on the straight line having the slope R% and intercept L.stline=R*(t-L);plot(t,resppoly,t,stline,tm,resp,’+’), ...axis([0 7 0 200]),ylabel(’Response’),xlabel(’Time (min)’)RL% Gains For P Control:KP=1/(R*L)





% Gains For PI Control:KP=0.9/(R*L)KI=KP/(3.3*L)% Gains For PID Control:KP=1.2/(R*L)KI=KP/(2*L)KD=.5*L*KP

The resulting plot is shown in the figure on the following page. The tangent line and thedata are shown along with the polynomial for visual checking. The results are R = 54.5612and L = 0.9422. The resulting values of the gains are

For P action : KP =1

RL= 0.0195

For PI action : KP =0.9RL

= 0.0175 KI =KP

TI= 0.0056

For PID action : KP =1.2RL

= 0.0233 KI =KP

TI= 0.0124 KD = KP TD = 0.0110



0 1 2 3 4 5 6 70

20

40

60

80

100

120

140

160

180

200

Re

spo

nse

Time (min)




11.41 A MATLAB file to fit a 5th degree polynomial to the data is given below. Thepolynomial is then used to compute R, L, and the PID gains.

% Raw Data.% Times at which measurements were made.tm=[0:1:12];% Measured Temperatures (deg C).tempm=[156,157,159,162,167,172,175,179,181,182,183,184,184];% Change in steam pressure.deltap = 3;% Shift the data to indicate relative temperature change.tstart = tempm(1);deltatempm = tempm-tstart;% Scale the data for a unit step input.scaledtempm=deltatempm/deltap;% Fit a 5th order polynomial to the data.coef=polyfit(tm,scaledtempm,5);% Evaluate the polynomial at a large number of points.dt = .01;t=[0:dt:12];scaledtemp=polyval(coef,t)’;% Find the coefficients of the polynomial’s derivative.coef2=[5*coef(1),4*coef(2),3*coef(3),2*coef(4),coef(5)];% der is the derivative polynomial.der=polyval(coef2,t)’;% Find the maximum slope R, and the time tdmax at which it occurs.[R,i]=max(der);tdmax = i*dt;% Compute the scaled temperature where the slope is the maximum.scaledtdmax=scaledtemp(i);% Compute the intercept L on the time axis.L=(R*tdmax-scaledtdmax)/R;% Compute the points on the straight line having the slope R% and intercept L.





stline=R*(t-L);plot(t,scaledtemp,t,stline,tm,scaledtempm,’+’), ...axis([0 12 0 15]), title(’Unit Step Response’),...ylabel(’Change in Temperature, deg C’),xlabel(’Time (min)’)RL% Gains For PID Control:KP=1.2/(R*L)KI=KP/(2*L)KD=0.5*L*KP

The resulting plot is shown in Figure 11.41. The tangent line and the data are shown alongwith the polynomial for visual checking. The computed values are R = 1.4842, L = 1.5396,KP = 0.5251, KI = 0.1705, and KD = 0.4043.

0 2 4 6 8 10 120

1

2

3

4

5

6

7

8

9

10Unit Step Response

Cha

nge

in T

empe

ratu

re, d

eg C

Time (min)




11.42 The standard form is1 +

c

5s

s2 + 9= 0

where the locus parameter is K = c/5. Type rlocus([1,0],[1,0,9]) to plot the rootlocus.




4x + 8x + kx = f(t)

and the characteristic equation is

4s2 + 8s + k = 0

The standard form is1 +

k

41

s2 + 2s= 0

where the locus parameter is K = k/4. Type rlocus(1,[1,2,0]) to plot the root locus.




4x + cx + 64x = f(t)


4s2 + cs + 64 = 0


c

4s

s2 + 16= 0

where the locus parameter is K = c/4. Type rlocus([1,0],[1,0,16]) to plot the rootlocus.




2x + 8x + (26 + k2)x = k2y(t)


2s2 + 8s + 26 + k2 = 0


k2

21

s2 + 4s + 13= 0

where the locus parameter is K = k2/2. Type rlocus(1,[1,4,13]) to plot the root locus.The plot shows that the largest attainable value of ζ for k2 > 0 is ζ = cos[tan−1(3/2)] = 0.55.




2x + (8 + c2)x + 26x = c2vi(t)


2s2 + (8 + c2)s + 26 = 0


c2

2s

s2 + 4s + 13= 0

where the locus parameter is K = c2/2. Type rlocus([1,0],[1,4,13]) to plot the rootlocus. The plot shows that the smallest attainable value of τ for c2 > 0 occurs whenK = 3.21 at s = −3.61. This gives τ = 1/3.61 and c2 = 2K = 6.42.



11.47 The characteristic equation in standard form is

1 + K1

s3 + 13s2 + 52s + 60= 0

Type

rlocus(1,[1,13,52,60]),sgrid([0.5,0.707],[3,5])

to plot the root locus with a (ζ, ωn) grid. The plot shows that the specifications are achiev-able with K in the range 38.6 ≤ K ≤ 83.8.



11.48 a) The characteristic equation in standard form is

1 +K

21

s3 + 6s2 + 8s= 0

where the locus parameter is K/2. Type rlocus(1,[1,6,8,0]),sgrid(0.707,[]) to plotthe root locus with a (ζ, ωn) grid.

b) The plot shows that the specifications are achievable with K/2 ≈ 5.18; that is, withK = 10.36.

c) The step response with K = 10.36 is obtained by typing

sys = tf(10.36,[2,12,16,10.36]);step(sys)

Right click on the plot and select “Characteristics”, then “Peak response”. The maximumovershoot is 4.03% at t = 4.4.

With K = 10.36, the roots are s = −4.4694 and s = −0.7653± 0.7572j. The dominanttime constant is 1/0.7653 = 1.31. The secondary time constant is 1/4.4694 = 0.2237. It ismuch smaller than the dominant time constant and thus has little effect on the response.This is illustrated by the fact that the overshoot predicted from the dominant roots withζ = 0.707 is 4.33%, which is almost identical to that of the third-order system.



11.49 Isolate the parameter c as follows.

LaIs2 + RaIs + KbKT + c(Las + +Ra) = 0

Put this into the form of (11.6.1):

1 +c

I

s + Ra/La

s2 + (Ra/La)s + KbKT/LaI= 0

where K = c/I . The zero is s = −Ra/La = −666.67. The poles are the roots ofs2 + (Ra/La)s + KbKT/LaI = s2 + 666.67s + 8.3333 × 104 = 0 and are s = −500 ands = −166.7. Type rlocus([1, 666.67], [1, 666.67, 8.3333e+4]), axis equal toplot the root locus. The locus is a circle centered at s = −666.67 with a radius of 288.5.The smallest possible time constant occurs when both roots are repeated at s = −955. Thisoccurs when K = 1240, or when c = 1240I = 0.0496.



11.50 a) The charactersitic polynomial is

D(s) = s4 + (16 + c1)s3 + (9 + 8c1)s2 + (8 + 4c1)s + 4 = 0

Isolate the parameter c1 as follows.

s4 + 16s3 + 9s2 + 8s + 4 + c1(s3 + 8s2 + 8s) = 0

Put this into the standard root locus form:

1 + c1s3 + 8s2 + 8s

s4 + 16s3 + 9s2 + 8s + 4= 0

where K = c1. The zeros are the roots of s3 +8s2 +8s = 0 and are s = 0, s = −6.8284, ands = −1.1716. The poles are the roots of s4 + 16s3 + 9s2 + 8s + 4 = 0 and are s = −15.4499,s = −0.0107 ± 0.6997j, and s = −0.5287. Type rlocus([1, 8,8,0], [1,16,9,8,4]),sgrid(0.707,[]) to plot the root locus with a gridline corresponding to ζ = 0.707.

b) By moving the cursor along the plot we can determine that a root exists at s =−0.973 + 0.966j, which gives ζ = 0.71, which is close enough to 0.707. The value of c1

required to give this root is determined as we move the cursor, and is the “gain”, which is3.69. However, this root is not the dominant root. This can be seen by typing K = 3.69;roots([1,(16+K),(9+8*K),(8+8*K), 4]). The roots are s = −17.6245, s = −0.9725 ±0.9673j, and s = −0.1206. The latter is the dominant root.

So it is not possible to achieve a dominant root with a damping ratio of 0.707.





c) Type rlocus([1,8,8,0], [1,16,9,8,4]),sgrid([],0.25) to plot the root locuswith a gridline corresponding to ωn = 0.25. This will give a circular gridline that in-tersects the real axis at the point corresponding to τ = 4. By moving the cursor alongthe plot we can determine that a root exists at s = −0.252, which gives τ close enoughto 4. The value of c1 required to give this root is determined as we move the cur-sor, and is the “gain”, which is 1.51. To see if this is the dominant root, type K =1.51; roots([1,(16+K),(9+8*K),(8+8*K), 4]). The roots are s = −16.2908, s =−0.4835 ± 0.8601j, and s = −0.2522. The latter is the dominant root. So it is possi-ble to achieve a dominant real root with a time constant of 4.

d) With c1 = 1.51 the two transfer functions are:

X1(s)D(s)

=8s + 4

s4 + (16 + c1)s3 + (9 + 8c1)s2 + (8 + 4c1)s + 4=

8s + 4s4 + 17.51s3 + 21.08s2 + 14.04s + 4

X2(s)D(s)

=s2 + 9.51s + 5

s4 + 17.51s3 + 21.08s2 + 14.04s + 4

The unit-step response plot for x1 is found by typing

sys1 = tf([8,4],[1,17.51,21.08,14.04,4]);step(sys1)

The unit-step response plot for x2 is found by typing

sys2 = tf([1,9.51,5],[1,17.51,21.08,14.04,4]);step(sys2)



11.51 The root locus plots are shown in the solutions to Problem 11.7.



11.52 The root locus plots are shown in the solutions to Problem 11.13.



11.53 a) K = 27.7. The three roots are s = −1.37± 1.37j, and s = −7.2.b) K = 16.9. The three roots are s = −1.5, s = −1.65, and s = −6.87.



11.54 a) K = 1.27. The three roots are s = −0.569, s = −0.569, and s = −7.86.b) When K = 1.14, the dominant root is s = −0.402. When K = 1.4, the dominant

roots are s = −0.576±).156j.



11.55 It is possible. With K = 3.96, the dominant roots are s = −0.724± 0.724j.



11.56 The two candidate solutions are on the boundaries of the ζ-ωn lines. One candidateis ζ = 0.5, which gives ωn = 2.39 and ζωn = 1.7. The other candidate is ωn = 2, which givesζ = 0.677 and ζωn = 1.35. So the solution is ζ = 0.5 and ωn = 2.39, for which K = 43.6.



Problem 11.57 The characteristic equation is 10s3 + (2 + KD)s2 + KP s + KI = 0. WithKP = 55 and KD = 58, we have 10s3 + 60s2 + 55s + KI = 0, or

1 +KI

101

s3 + 6s2 + 5.5s= 0

The root locus gain is K = KI/10. In MATLAB type

sys = tf(1, [1, 6, 5.5, 0]);rlocus(sys), axis equal

The plot follows. In Example 10.7.4, KI = 25. This gives ζ = 0.707 and τ = 2. The plotshows that to reduce the error by increasing KI , the dominant time constant will becomelarger and the damping ratio will decrease, making the system slower and more oscillatory.Moving the cursor along the plot shows that if we decrease the error by half by doublingKI to 50, the new damping ratio will be approximately 0.44 and the new time constant willbe approximately 1/0.43 = 2.3.

−20 −15 −10 −5 0 5 10−15

−10

−5

0

5

10

15

Root Locus

Real Axis

Ima

gin

ary

Axi

s




11.58 The characteristic equation is s3 +KDs2 +(KP − 4)s+KI = 0. With KP = 604 andKD = 40, we have s3 + 40s2 + 600s + KI = 0 or

1 + KI1

s3 + 40s2 + 600s= 0

The root locus gain is K = KI . In MATLAB type

sys = tf(1, [1, 40, 600, 0]);rlocus(sys)

The plot follows. In Example 10.8.3, KI = 4000. This gives ζ = 0.707 and τ = 0.1. The plotshows that to reduce the error by increasing KI , the dominant time constant will becomelarger and the damping ratio will decrease, making the system slower and more oscillatory.If we decrease the error by half by doubling KI to 8000, the plot shows that the newdamping ratio will be approximately 0.3 and the new time constant will be approximately1/5.68 = 0.18.

−60 −50 −40 −30 −20 −10 0 10−40

−30

−20

−10

0

10

20

30

40

Root Locus

Real Axis

Imag

inar

y A

xis




11.59 a) The model is shown in the figure. The outputs of the proportional and integralterms are stored in the variables simout(:,1) and simout(:,2). To plot them, typeplot(tout,simout(:,1),tout,simout(:,2)






b) The model is shown in the figure. With KA = 0, the overshoot id about 18% andthe actuator remains saturated until abot t = 0.6. There are a range of values for KA thatwill reduce the overshoot. For example, with KA = 50, the overshoot is about 3% and theactuator remains saturated until about t = 0.3.




11.60 The model is shown in the figure. Without the rate limiter, the overshoot is about9%, but with the rate limiter, the overshoot is about 50%.




11.61 a) By reducing the loops in the block diagram we obtain the following transferfunction:

Θ(s)Θr(s)

=KT KI

ILs4 + (IR + IK3)s3 + K2KTs2 + K1KTs + KIKT

To place the four roots at s = −2, −20, and −20 ± 20j, requires the following polynomial:

s4 + 62s3 + 1720s2 + 19 200s + 32 000

This polynomial was obtained using the MATLAB conv function as follows. Note that theroot pair −20 ± 20j can be written as (s + 20)2 + 400 = s2 + 40s + 400.

step1 = conv([1,2],[1,40,800];)step2 = conv(step1,[1,20])step2 =

1 62 1720 19200 32000

Comparing the denominator of the transfer function with this polynomial, and account-ing for the leading coefficient IL, we obtain

KI =32 000IL

KT= 0.0960

K1 =19 200IL

KT= 0.0576

K2 =1720IL

KT= 0.0052

K3 = 62L− R = −0.4760





b) The Simulink model is shown on the next page, using a unit-step command startingat t = 0 and a step disturbance of magnitude 0.1 starting at t = 1. Before running thismodel, in the Command window enter the values of the parameters as follows.

L=0.002;R=0.6;KT=0.04;I=6e-5;KI=32000*I*L/KT;K1=19200*I*L/KT;K2=1720*I*L/KT;K3=62*L-R;

The second figure shows the response.





Figure : for Problem 11.61b. The Simulink model.

0 0.5 1 1.5 2 2.5 3 3.5 4−5

−4

−3

−2

−1

0

1

Time (s)

Posit

ion (r

ad)

Figure : for Problem 11.61b. The response.

c) The model is shown on the following page, using a unit-step command starting att = 0 and a step disturbance of magnitude 0.1 starting at t = 1. Be sure to first enter theparameters as discussed in part (b). The second figure shows the response. Obviously thesaturation limit prevents the system from counteracting the disturbance.





Figure : for Problem 11.61c. The Simulink model.

0 0.5 1 1.5 2 2.5 3 3.5 4−1800

−1600

−1400

−1200

−1000

−800

−600

−400

−200

0

200

Time (s)

Posit

ion (r

ad)

Figure : for Problem 11.61c. The response.



11.62 The model without rate limiting is shown in the figure on the following page. Thecorresponding response is shown in the following plot. The disturbance acts at t = 20.Saturation blocks 1 and 2 were added to prevent the liquid heights from becoming negative,which of course is physically impossible.





Figure : for Problem 11.62. Model without rate limiting.

0 10 20 30 40 50 60 700

5

10

15

20

25

30

35

Time

Flow

Rat

e

With no limit on rate

0 10 20 30 40 50 60 700

1

2

3

4

5

6

7

Time

Heigh

t h2

Figure : for Problem 11.62. Response with no rate limiting.




The model with rate limiting is shown in the figure on the following page. The cor-responding response is shown in the following plot. As we would expect, the flow rateresponds slower with rate limiting, and the liquid height also responds slower. The steady-state response is the same.





Figure : for Problem 11.62. Model with rate limiting.

0 10 20 30 40 50 60 700

10

20

30

40

50

60

Time

Flow

Rat

e

With rate limited

0 10 20 30 40 50 60 700

2

4

6

8

10

12

14

16

Time

Heigh

t h2

Figure : for Problem 11.62. Response with rate limiting.




to accompany


by

William J. Palm III


Solutions to Problems in Chapter Twelve


CD1

Highlight


C(s)R(s)

=KP + KDs

s2 + KDs + KP − 5

For ωn = 0.5,

ωn =

√KP − 6

5= 0.5

Thus KP = 6.25.For the damping ratio,

ζ =KD

2√

5(KP − 6=

KD

2√

5(0.25= 0.707

which gives KD = 2.5.



12.2 a) For K = 4 the roots are s = −1 ±√

3j. For ωn = 4 and ζ = 0.5, the roots mustbe s = −2 ± 2

√3j. A simple gain adjustment will not give the desired roots. The angle

deficiency is6

4s(s + 2)

∣∣∣∣s=−2+2

√3j

= 6 4 − 6 s − 6 (s + 2) = −210

Thus the compensator must add (210 − 180) = 30 to the system. This implies a leadcompensator. Place the compensator’s pole and zero to the left of s = −2. Using the sameprocedure as in Example 12.1.2, with µ = 10, we obtain T = 0.0138 and T = 0.293. ChooseT = 0.0138 because the second value would place the zero to the right of s = −2, and thusviolate the assumed geometry. The compensator’s pole is s = −1/T = −72.6, and its zerois s = −1/aT = −7.26. The compensated open-loop transfer function is

Gc(s)G(s) =s + 7.26s + 72.6

K

s(s + 2)

The value of K required to place the root at s = −2± 2√

3j is K = 155.6. Other solutionsare possible, depending on the choice for µ.

For C = 1µF ,

T = 0.0138 =R1R2

R1 + R210−6





b) A ramp error of 0.05 implies that Cv = 20. With K = 4, s = −1 ± j√

3 and Cv = 2.To obtain Cv = 20 we need to increase the gain by a factor of 10, but without changing theroot locations appreciably. Therefore we use a lag compensator. Because the PM and GMare not specified, we use a root locus design method.

Step 1: With no compensation, K = 4 will place the roots at the desired locations.Thus KP1 = 4.

Step 2: To obtain Cv = 20, K must be increased from 4 to 40. Thus KP2 = 40 andµ = KP1KP2 = 0.1.

Step 3: Choose T large. Note that the plant has a pole at s = −2. Select the compen-sator’s pole and zero to be well to the right of s = −2. So try s = −0.02 and s = −0.2.This gives T = 100.

Step 4: The open-loop compensated transfer function is

Gc(s)G(s) =110

2 + 0.2s + 0.02

K

s(s + 2)

The precise value of the desired roots s = −1±j√

3 probably will not lie exactly on the newroot locus, but if T is chosen large enough, they should be close. The two specifications areζ = 0.5 and ωn = 2, so we can look for a root that satisfies one specification exactly, andhope that the other is close to being satisfied.

For T = 100, the root locus shows that s = −0.8989+1.56j gives ζ = 0.5, but ωn = 1.8,when K = 36. If ωn is close enough to 2, we can stop. Otherwise a larger value of T canbe chosen. Note that the roots s = −0.8989± 1.56j are “dominant” only if we consider thepole at s = −0.2 to be canceled by the zero at s = −0.2.



12.3 Try a PID compensator. The transfer function is

C(s)R(s)

=5(KDs2 + KP s + KI)

s3 + 5KDs2 + 5KP s + 5KI

The steady-state error for a step command input will be zero, as required. The characteristicequation is

s3 + 5KDs2 + 5KPs + 5KI = 0 (1)

To achieve τ = 1 and ζ = 0.45, the desired dominant roots are s = −1 ± j1.985. The thirdroot is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will notbe the dominant root). The polynomial corresponding to these three roots is

[(s + 1)2 + (1.985)2](s + b) = 0

ors3 + (b + 2)s2 + (4.94 + 2b)s + 4.94b = 0 (2)

Comparing the coefficients of (1) and (2), we obtain

KP =4.94 + 2b

5

KI =4.94b

5

KD =b + 2

5The gains can be computed once a value for b > 1 has been selected.



12.4 Because the desired roots are obtainable with a simple gain selection, we need a lagcompensator to achieve the desired Cv value. The lag compensator has the form

Gc(s) = Ks + z

s + p

ThusCv =

Kz

2p

To achieve Cv = 5 with K = 1, we need z/p = 10. Choosing z = 0.05 and p = 0.005satisfies this requirement, and places the compensator’s pole and zero near the origin, faraway from the desired root locations of s = −0.338 ± 0.562j. Using the root locus plot,we can determine the value of K required to place the closed-loop roots near the desiredlocation. This value is K = 1.024. Thus one solution is the compensator transfer function

Gc(s) = 1.024s + 0.05s + 0.005

The resulting closed-loop roots are s = −2.326, −0.055, and −0.312 ± 0.55j, which areclose to the desired locations. The closed-loop zero at s = 0.05 approximately cancels theclosed-loop pole at s = −0.055. The resulting Cv value is Cv = 5.12, which is close to thedesired value.



12.5 The values ζ = 0.5 and ωn = 2 correspond to a dominant root location of s =−0.5 ± j

√3. These roots can be obtained by setting K = 1. So the gain KP required to

achieve the desired transient performance has already been established as KP1 = K = 1.A lag compensator is indicated, because the steady-state error is too large. The second

step is to determine the value of the parameter µ. For this system, the coefficient Cv is

Cv = lims→0

sKP

s(s + 1)= KP

and KP2 is the value of KP that gives Cv = 10. Thus, KP2 = 10, and the parameterµ = KP /KP2 = 1/10. The compensator’s pole and zero must be placed close to theimaginary axis, with the ratio of their distances being 1/10. Noting that the plant has apole at s = −1, we select locations well to the right of this pole, say, at s = −0.01 ands = −0.1 for the pole and zero, respectively. This gives T = 100.

The open-loop transfer function of the compensated system is thus

Gc(s)G(s)H(s) =0.1Kc(s + 0.1)

s(s + 1)(s + 0.01)

The root locus is shown in the figure on the following page.(continued on the next page)




−1 −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 0−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Root Locus

Real Axis

Imag

inar

y A

xis


For the desired damping ratio of 0.5, the locus essentially lies on the asymptote thatpasses through

σ =∑

sp −∑

sz

P − Z=

0 − 1 − 0.01 + 0.13 − 1

= −0.455

So the real part of the root is approximately −0.455, and the imaginary part will be0.455 tan 60 = 0.788 to obtain ζ = 0.5. Thus the root is approximately s = −0.455 +0.788j. The value of Kc required to obtain this root is found from

0.1Kc = −∣∣∣∣s(s + 1)(s + 0.01)

s + 0.1

∣∣∣∣s=−0.455+0.788j

= 0.9128

or Kc = 9.128. This is only a tentative estimate because the root does not lie exactly on thelocus. Using this value of Kc in the characteristic equation, we find that the actual rootsare s = −0.4496 ± 0.788j and s = −0.1109. So our estimate turns out to be very close tothe true value.(continued on the next page)




The error coefficient is thus Cv = 9.128 and less than the desired value of 10. Also,the dominant roots differ somewhat from the desired locations at s = −0.5 + j0.866. Ifthese differences are too large, the compensator’s pole and zero can be placed closer to theimaginary axis, say, at s = −0.01 and s = −0.001, respectively, with T = 1000. This willdecrease the compensator’s influence on the locus near the desired root locations.



12.6Step 1: Adjust KP to meet Cv = 20. Cv = K/2 = 20. Thus KP = 40. Use KP = 40 in

the following steps.Step 2: Check phase margin and gain margin with KP = 40, for

G(s) =20

s(0.5s + 1)

The phase margin is 17 and the gain margin is infinite. Thus we must add 40 − 17 = 23

to obtain a phase margin of 40. This requires a lead compensator.Following the method of Example 12.2.3 we add a 5 safety factor and choose φm to be

23 + 5 = 28. Then

µ =1 + sin 28

1 − sin 28= 2.7698

Now find the frequency at which the uncompensated gain equals −20 log√

µ = −4.42 db.This occurs at about ω = 8.2 rad/sec. Thus choose ωm = 8.2 and ωm = 1/T

√µ = 8.2,

which gives T = 0.073. Thus the compensator’s parameters are µ = 2.7698 and T = 0.073.The pole and zero are s = −1/T = −13.699 and s = −4.946. The open-loop transferfunction of the compensated system is

Gc(s)G(s) =s + 4.946s + 13.699

40s(s + 2)

As s → 0, the compensator’s gain is seen to introduce a gain factor of 4.946/13/699 =0.361 = 1/µ. Thus, in order to preserve Cv = 20, we must increase KP by a factor of1/0.361 = 2.7698. Therefore we set KP = 40(2.7698) = 110.79. The compensated open-loop transfer function is

Gc(s)G(s) = 110.79s + 4.946s + 13.699

1s(s + 2)

=20(0.2022s + 1)

s(0.073s + 1)(0.5s + 1)

The frequency response plots show that the phase margin is 42, the gain crossoverfrequency is ωg = 8, and the gain margin is infinite. Thus the specifications are satisfied.The closed-loop transfer function is

Gc(s)G(s)1 + Gc(s)G(s)

=20(0.2022s + 1)

0.0365s3 + 0.573s2 + 5.044s + 20

The roots are s = −7.104 and s = −4.297±7.659j, with a dominant time constant of 0.233,a dominant damping ratio of 0.489, and natural frequency of 8.78.



12.7 With no compensation, Cv = 0.625, which is much less than the desired value ofCv = 80. Thus a lag compensator is needed to increase the gain by 80/0.625 = 128. Leadcompensation is needed because it is not possible to place the roots at the desired locationwith a simple gain adjustment. Thus a simple lead or a simple lag compensator will notwork.

We therefore will design a lag-lead compensator. The angle deficiency is found from

6 G(s)|s=−2+2√

3j = 610

s(s + 2)(s + 8)

∣∣∣∣s=−2+2

√3j

= 120− 90− 30 = −240

Thus, the angle deficiency is 180− 240 = 60. For the lead compensator G1(s),

6 G1(s)|s=−2+2√

3j = 61 + µT1s

1 + T1s

∣∣∣∣s=−2+2

√3j

= 60

This is satisfied if the lead compensator has a pole at s = −53 and a zero at s = −3.65.Thus 1/µT1 = 3.65 and 1/T1 = 53. These give T1 = 0.0189 and µ = 14.5.

For the lag compensator, the choice of T2 = 100 gives∣∣∣∣∣s + µ

T2

s + 1T2

∣∣∣∣∣ =∣∣∣∣s + 0.145s + 0.01

∣∣∣∣s=−2+2

√3j

= 0.967

which we take to be close enough to 1 to indicate adequate pole-zero cancelation.The lag-lead compensator is thus given by

Gc(s) = 128s + 3.65s + 53

s + 0.145s + 0.01



12.8 We try a compensator of the form

Gc(s) = Ks + a

s + b

The characteristic equation is found from

1 + Gc(s)Gp(s) = 0

which becomes1 + K

s + a

s + b

10s2(0.1s + 1)

= 0

or0.1s4 + (1 + 0.1b)s3 + bs2 + 10Ks + 10Ka = 0 (1)

For τ = 1 and ζ = 0.5, the required dominant root pair is s = −1 ± 1.732j. So thecharacteristic equation must be factored as

0.1[(s + 1)2 + (1.732)2](s2 + cs + d) = 0

or

0.1(s2+2s+4)(s2+cs+d) = 0.1s4+0.1(2+c)s3+0.1(4+2c+d)s2+0.1(4c+2d)s+0.4d = 0 (2)

Comparing the coefficients of equations (1) and (2), we obtain

1 + 0.1b = 0.1(2 + c) (3)

b = 0.1(4 + 2c + d) (4)

10K = 0.1(4c + 2d) (5)

10Ka = 0.4d (6)

We can choose c and solve (3) for b and (4) for d. The choice of c = 14 gives b = 6 andd = 28. So the secondary roots from s2 + cs + d = 0 are s = −2.417 and s = −11.58.

Then solve (5) for K to obtain K = 1.12. Finally, solve (6) for a to obtain a = 1. Theresulting compensator is

Gc(s) = 1.12s + 1s + 6




1 +K

5s + a

s2(s + b)= 0

The poles are s = 0, s = 0, and s = −b. The zero is s = −a. The asymptotic angles areθ = ±90. The asymptotes intersect at

σ =a − b

2

For stability, the intersection point should be negative; thus, we should select b > a.For a 5% overshoot, the damping ratio of the dominant root must be ζ = 0.69, which

corresponds to a complex root. For a time constant of τ = 0.5, the dominant roots mustbe s = −2 ± 2.096j to achieve ζ = 0.69. The corresponding factor is (s + 2)2 + (2.096)2 =s2 + 4s + 8.393. Thus the characteristic equation can be factored as

(s2 + 4s + 8.393)(s− s3) = s3 + (4− s3)s2 + (8.393− 4s3)s − 8.393s3 = 0


s3 + bs2 +K

5s + a

K

5= 0

Comparing coefficients, we see that

s3 = 4 − bK

5= 8.393− 4s3 a = −8.393s3

K/5

The third root must lie to the left of the dominant root, whose real part is −2. Therefore,b must be chosen so that b > 6. The other restriction is that b > a.

One solution is b = 10, which gives s3 = −6, K = 161.965, and a = 1.55.If we try to cancel one of the poles at s = 0 by letting a → 0, the root locus equation

becomes1 +

K

51

s(s + b)= 0


s2 + bs +K

5= 0

The specifications that ζ = 0.69 and τ = 0.5 require that s = −2± 2.096j. This is achievedwith b = 4 and K = 41.95.



12.10 The open-loop poles are s = 0 and −2 ± 3j.a) The root locus equation for the lead compensator is

1 +s + 1

µT

s + 1T

Kc

s(s2 + 4s + 13)= 0

where µ > 1 and the root locus parameter is K = Kc ≥ 0.b) The root locus equation for the lag compensator is

1 +s + 1

µT

s + 1T

µKc

s(s2 + 4s + 13)= 0

where µ < 1 and the root locus parameter is K = µKc ≥ 0.c) The root locus equation for the reverse-reaction compensator is

1 +s − 1

T1

s + 1T2

−T1T2

Kc

s(s2 + 4s + 13)= 0

where the root locus parameter is K = −T1Kc/T2 ≤ 0.The root-locus plots are shown in the following figures. All three compensators can

produce an unstable system if the gain Kc is too large. However, only the reverse-reactioncompensator can pull the complex-root paths starting at s = −2±3j to the left. This enablesa dominant root to be obtained that is not close to the imaginary axis. The optimum rootlocations are shown on the plot on the next page.(continued on the next page)





(m + M)x− mLφ = −f (2)


]φ − (m + M)mgLφ = (m + M)T − mLf (3)



500φ− 29 430φ = 60T − 50f (4)

c) For the personal transporter discussed in Chapter 2, T = 0 since the device is con-trolled entirely with the base force f . So we will imitate this approach. Noting that (4)represents an unstable system because the φ term is missing and because the coefficient ofφ is negative, we try a feedback control law of the form

f = KP φ + KDφ (5)

This is equivalent to a series compensator of the PD type.Setting T = 0 in (4), substituting (5) into (4), and cleaning up, we obtain

φ − 58.86φ = −0.1KPφ − 0.1KDφ

orφ + 0.1KDφ + (0.1KP − 58.86)φ = 0


τ =2

0.1KD= 1

which gives KD = 20.Choosing ζ = 1 gives

ζ =0.1KD

2√

0.1KP − 58.86=

1√0.1KP − 58.86

= 1

which gives KP = 598.6.



12.12 a) The plot for K = 2 is shown in figure (a). It can be obtained with the followingMATLAB file.

K = 2;sys = tf(5*K,[1,6,5,0])margin(sys)

The results are that the gain margin is 9.54 dB; the phase crossover frequency is 2.24rad/sec; the phase margin is 25.4; and the gain crossover frequency is 1.23 rad/sec.

b) The plot for K = 20 is shown in figure (b). It can be obtained in a manner similar tothat used for part (a). The results are that the gain margin is −10.5 dB (which means thatthe system is unstable); the phase crossover frequency is 2.24 rad/sec; the phase margin is−23.7 (another indication of instability); and the gain crossover frequency is 3.91 rad/sec.

c) For K = 6 both the gain and phase margins are 0. Thus they both are a limitingfactor for stability.



12.13 The open-loop transfer function is

G(s) = Gc(s)Gp(s) =2 + 19s

100s2 + s

The Bode plots are shown in following figure. They can be obtained with the MATLABmargin function. The gain margin is infinite because the phase curve is always above the−180 line. The phase margin is 66. The system is stable.

−50

0

50

100

150

Mag

nitu

de (d

B)

10−4

10−3

10−2

10−1

100

101

−150

−120

−90

Pha

se (d

eg)

Bode DiagramGm = Inf , Pm = 66.3 deg (at 0.212 rad/sec)

Frequency (rad/sec)





G(s) = Gc(s)Gp(s) =25(7s + 64)

5s3 + 6s2 + 5s=

175s + 16005s3 + 6s2 + 5s

The Bode plots are shown in following figure. They can be obtained with the MATLABmargin function. Both the gain and phase margins are negative, so the system is unstable.

−100

−50

0

50

100

150

Mag

nitu

de (d

B)

10−2

10−1

100

101

102

103

−270

−225

−180

−135

−90

Pha

se (d

eg)

Bode DiagramGm = −47.3 dB (at 1.07 rad/sec) , Pm = −41.5 deg (at 7.46 rad/sec)

Frequency (rad/sec)





G(s) =0.1KP (10)

(0.2s + 1)(s2 + s + 10)=

KP

0.2s3 + 1.2s2 + 3s + 10

The Bode plots, phase margins, and gain margins can be obtained with the MATLABmargin function. The results are as follows:

a) KP = 1: Gain margin = 20.9 dB, Phase margin = 88.3

b) KP = 10: Gain margin = 0.916 dB, Phase margin = 70.4

c) KP = 100: Gain margin = −19.1 dB, Phase margin = −115

Cases (a) and (b) are stable, but Case (c) is unstable. This means that any initial rollangle will continue to increase.



12.16 A solution is K = 0.376.



12.17 The answers are given in the following table.

Case Type No. Cp Cv Ca Step Error Ramp Error(a) 1 ∞ 20 0 0 1/20(b) 0 20 0 0 1/21 ∞(c) 2 ∞ ∞ 7 0 0




G(s) = KP e−(D1+D2)s 1100s + 1

= KP e−100s 1100s + 1

The delay of 100 s lowers the phase curve by 100ω rad, or 100ω(180/π) degrees. A plotof m and φ for KP = 1 shows that instability is caused by the negative gain margin.The phase crossover frequency is approximately 0.02 rad/sec, and m at this frequency isapproximately −7 dB. To achieve a positive gain margin, we must therefore increase KP toat least 107/20 = 2.239. For this value of KP both the phase margin and the gain marginare zero.

The calculations for KP = 1 can be done in MATLAB as follows.

sys = tf(1,[100,1]);w = [0.001:0.001:0.02]’;[mag, phase] = bode(sys,w);m = 20*log10(mag(:));phasetotal = phase(:)-100*w*(180/pi);semilogx(w,m,w,phasetotal),grid



12.19 With D = 0, the open-loop frequency response plot of 10/(0.1s + 1) has a gaincrossover at ω ≈ 100, and a phase margin of 95. For PM = 40, the dead time D canreduce the phase curve at ω = 100 by no more than 95− 40 = 55. Because

6 P (iω)e−iωD = 6 P (iω) − ωD

we have100D ≤ 55

π

180rad

or D ≤ 0.0096.



12.20 One solution is the compensator transfer function

Gc(s) =s + 0.1s + 0.01

This gives a gain margin of 14.3 dB and a phase margin of 41.6.




Gc(s) = 42s + 4.4s + 18

This gives an infinite gain margin and a phase margin of approximately 50.




Gc(s) = 206.66s + 166.6s + 1





Gc(s) = 200.34s + 10.07s + 1





Gc(s) = 50(s + 1)(s + 0.16)(s + 0.01)(s + 16)

The closed-loop transfer function is

C(s)R(s)

=50s2 + 58s + 8

0.2s5 + 4.402s4 + 20.24s3 + 66.2s2 + 58.16s + 8

There is pole-zero cancelation of the factor s + 1, so the transfer function becomes

C(s)R(s)

= 250s + 0.16

(s + 17.185)(s + 0.168)(s2 + 3.656s + 13.846)

The roots are s = −17.1837, s = −0.1681, and s = −1.8291 ± 3.2401j. If we take theroots s = −0.1681 to be canceled approximately by the compensator zero at s = −0.16,the root pair s = −1.8291 ± 3.2401j is the dominant root. It has a damping ratio ofζ = cos(tan−1(3.2401/1.8291)) = 0.4916, which is approximately the desired value of 0.5.




Gc(s) = 20(s + 0.7)(s + 0.15)(s + 7)(s + 0.015)




M =ω2

n

ω√

4ζ2ω2n + ω2

A gain margin of 0 implies that M = 1. So, setting M = 1 in the above equation results inthe following polynomial:

ω4 + 4ζ2ω2nω2 − ω4

n = 0

The positive solution is

ω = ωn

√−2ζ2 +

√4ζ4 + 1 (1)

This is the gain crossover frequency.The phase angle is

φ = − 6 jω − 6 (2ζωn + jω) = −90 − tan−1 ω

2ζωn

The phase margin is

PM = φ(ω) − (−180) = φ(ω) + 180 (2)

where ω is given by (1). With this substitution, we obtain

PM = 90 − tan−1 ω

2ζωn= 90 − tan−1

√−2ζ2 +

√4ζ4 + 1

2ζ(3)

Using the identitiescot−1 y

x= tan−1 x

y

tan−1 y

x+ cot−1 y

x= 90

we can show that90 − tan−1 y

x= tan−1 x

y(4)

Letting y =√−2ζ2 +

√4ζ4 + 1 and x = 2ζ in (3), identity (4) gives

PM = tan−1 2ζ√−2ζ2 +

√4ζ4 + 1



12.27 Proportional control is sufficient here, with KP = 912.01. The resulting maximumpercent overshoot is 1.03% and a 2% settling time of 0.866 s.



12.28 A closed-loop damping ratio of ζ = 0.4559 corresponds to an overshoot of 20%,and a phase margin of 48.1. Try PD control action with Gc(s) = KP (TDs + 1), with aguessed value of TD = 0.2. Using the margin function with different values of KP givesGc(s) = KP (TDs + 1) = 605(0.2s + 1), which gives an infinite gain margin and a phasemargin of 48.

So the open-loop transfer function is

2.42s + 12.10.02s3 + 0.3s2 + s

The closed-loop transfer function is

2.42s + 12.10.02s3 + 0.3s2 + 3.42s + 12.1

The roots are s = −5 and −5.0000 ± 9.7980j. So the time constant is 0.2, which gives asettling time of 0.8, less than the required value of 1 s.

The error transfer function is

0.02s3 + 0.3s2 + s

0.02s3 + 0.3s2 + 3.42s + 12.1

Using the final value theorem with a unit-ramp input, we find that the steady-state erroris 1/12.1 = 0.0826, which is less than 0.1 as required.

So one solution is Gc(s) = KP (TDs + 1) = 605(0.2s + 1). Other solutions are possible.



12.29 a) Since the open-loop transfer function is

KP s + KI

s(s + 4)

the margins are calculated in MATLAB as follows:

KP = 6;KI = 50;num = [KP,KI];den = [1,4,0];sys = tf(num,den);[GM, PM, Wg, Wp] = margin(sys)

The results are GM = ∞ and PM = 70.26.b) In MATLAB, type

sysd = tf(num,den,’iodelay’,0.1);[GM, PM, Wg, Wp] = margin(sysd)

The results are GM = 1.9068 and PM = 25.5. So the dead time has reduced the stabilityof the system quite a bit.

To use the feedback function, first convert the continuous-time function sysd to thediscrete-time function discrete using the c2d function with a sampling time equal to 1/100of the time constant. Then use the feedback and step functions.

discrete = c2d(sysd, 0.002);sysCL = feedback(discrete,1);step(sysCL)

The step response is very oscillatory.c) Iterate using the following commands, increasing the dead time until the margin

function indicates an unstable system.

sysd = tf(num,den,’iodelay’,0.157);[GM, PM, Wg, Wp]=margin(sysd)

To three decimal places, the upper limit on the dead time is 0.157. Any value greater thanthat results in a negative phase margin.



12.30 The new open-loop transfer function with KP = 101 is

101(s + 3)(s + 0.0578)(s + 3.2)(s + 0.01)(s + 11.9)(s(s + 5)

=101s2 + 308.8s + 17.51

s5 + 20.11s4 + 113.8s3 + 191.5s2 + 1.904s

The new closed-loop transfer function is

101s2 + 308.8s + 17.51s5 + 20.11s4 + 113.8s3 + 292.5s2 + 310.7s + 17.51

which has the roots −13.2, −3.66± 3.25j, −0.056, and 0.206± 0.973j, which indicate thatthe system is unstable.



12.31 The Simulink diagram is shown below. The PID controller block has the gains:KP = KI = 16, KD = 6. Set the command step to start at t = 0 and the disturbance stepto start at t = 5. You will see an overshoot of about 27% for the command response, andan undershoot of 5% for the disturbance response. In both cases the steady-state error iszero.





to accompany


by

William J. Palm III


Solutions to Problems in Chapter Thirteen


CD1

Highlight

CD1

Highlight

13.1 The equation of motion ismx + kx = ky

where y(t) = Y sin 6πt = 4× 10−3 sin 6πt. Also,

ω2n =

k

m=

5000.5

= 1000 rad/s

From (13.1.9), with ζ = 0 and r2 = (6π)2/1000,

X = Y

∣∣∣∣1

1 − r2

∣∣∣∣ = 6.2× 10−3 m or 6.2 mm



13.2 For a period of 20 feet and a vehicle speed of v (mph), the frequency ω is

ω =(

528020

)(1

3600

)(2π) = 0.4608v rad/sec

Thus ω = 0.4608(20) = 9.216 rad/sec for v = 20 mph, and ω = 0.4608(50) = 23.04 rad/secfor v = 50 mph.

For a car weighing 2000 lb, the quarter-car mass is m = 500/32.2 slugs. Its naturalfrequency is ωn =

√k/m =

√2000/(500/32.2) = 11.35 rad/sec. Its frequency ratio at 20

mph is r = ω/ωn = 9.216/11.35 = 0.812, and at 50 mph it is r = 23.04/11.35 = 2.03. Itsdamping ratio is

ζ =360

2√

2000(500/32.2)= 1.02

Now substitute these values of r and ζ and Y = 0.03 ft into the following expressions,obtained from (13.1.9) and (13.1.12).

X = Y

√1 + 4ζ2r2

(1 − r2)2 + 4ζ2r2

Ft = r2kX

This gives the following table.

2000 lb Car(ζ = 1.02)

v (mph) r X (ft) Ft (lb)20 0.812 0.034 45.1250 2.03 0.025 203.4



13.3 We are given m = 1500 kg, k = 20, 000 N/m, ζ = 0.04, and Y = 0.01 m. At resonance,

r =ωr

ωn=

ωn

√1 − 2ζ2

ωn= 0.998

and

Ft = r2kY

√4ζ2r2 + 1

(1 − r2)2 + 4ζ2r2= 2500 N



13.4 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also,

ω = 40 Hz = 215 rad/s ωn =

√k

m=

√66 667

(200/9.81)= 57.2 rad/s

r2 =(

ω

ωn

)2

=(

25157.2

)2

= 19.26

From (13.1.9) with c = 0,X

Y=

1|1 − r2|

= 0.055

Thus 5.5% of the airframe motion is transmitted to the module.



13.5 a) Neglect damping in the isolator, and determine its required stiffness k. From(13.1.9),

X

Y=

1|1− r2| = 0.1

which gives r2 = 11. Thus

ω2n =

ω2

r2=

[3000(2π)/60]2

11=

(314)2

11

and

k = mω2n =

232.2

(314)2

11= 556.7lb/ft

b) r = ω/ωn and ωn = 314/√

11. Thus

r1 =2500(2π)/60

314/√

11= 2.77

r2 =3500(2π)/60

314/√

11= 3.87

From (13.1.9)X

Y=

1|1 − r2|

Thus the highest percentage of motion will be transmitted at the lowest r value, whichcorresponds to 2500 rpm. For 2500 rpm,

X

Y=

1|1− r2

1|= 0.15

For 3500 rpm,X

Y=

1|1− r2

2|= 0.07

Thus at most, 15% of the crane motion will be transmitted to the module.



13.6 We are given m = 5/32.2 slug and

r2 =(

ω

ωn

)2

=(30(2π)/60)2

32.2k/5

From (13.1.9) with ζ = 0,X

Y=

1|1− r2|

= 0.1

which gives r2 = 11. Thus(30(2π)/60)2

32.2k/5= 11

Solve for k to obtain k = 0.139 lb/ft. The transmitted force is

Ft =(r2 X

Y

)kY = 11(0.1)0.139(0.003) = 4.59× 10−4 lb



13.7 a) From Newton’s law,mx = c(y − x) − kx

ormx + cx + kx = cy

T (s) =X(s)Y (s)

=cs

ms2 + cs + k

ThusT (jω) =

cωj

−mω2 + cωj + k=

cωj/k

1 − r2 + cωj/k

This gives

T (jω) =cωmmk j

1 − r2 + cωmj/mk=

2ζωnωj/ω2n

1 − r2 + 2ζωnωj/ω2n

=2ζrj

1 − r2 + 2ζrj

where we have used the fact that ωn =√

k/m, r = ω/ωn, and c/m = 2ζωn. The magnitudeis

X =2ζr√

(1− r2)2 + (2ζr)2Y

b) Thus

Ft = kX =2ζrk√

(1− r2)2 + (2ζr)2Y



13.8 The displacement transmissibility with r = 2 is

X

Y=

√4ζ2r2 + 1

(1 − r2)2 + 4ζ2r2=

√1 + 16ζ2

9 + 16ζ2

This has a minimum of 1/3 when ζ = 0. This is the best choice for ζ if the velocity increases(r > 2). However, if the velocity decreases (so that r → 1, and the system approachesresonance), an non-zero value of ζ would be a better choice to limit the resonant response.

With ζ = 0 and a 20% increase in r (r = 2.4), we have X/Y = 0.21. With ζ = 0 and a20% decrease in r (r = 1.6), we have X/Y = 0.64.



13.9 The relations areFt = r2kX

X = Y

√1 + 4ζ2r2

(1 − r2)2 + 4ζ2r2

whereζ =

c

2√

mkr =

ω

ωn

At 40 mph, the forcing frequency is

ω =(

528020

)(1

3600

)2π(4) = 18.432 rad/sec

and r = 18.432/ωn, where

ωn =

√k

m

To minimize Ft, choose either

1. r small (with ζ near 1 and ωn large; this means k large and c large), or

2. r > 2 (this means ωn small).

For example, with m = 800/32.2, choosing ζ = 1 and r = 0.5, implies that

ωn = 2(18.432) = 36.864 k = m(36.864)2 = 33 763 lb/ft

c = 2√

mkζ = 1832 lb sec/ft

This gives

X = Y

√4ζ2r2 + 1

(1 − r2)2 + 4ζ2r2= 0.0589 ft

andFt = r2kX = 497 lb





On the other hand, choosing ζ = 1 and r = 3 for example, we obtain

k = mω2n =

80032.2

(18.432

3

)2

= 937.9 lb/ft

c = 2√

80032.2

937.9 = 305 lb sec/ft

This gives

X = Y

√4ζ2r2 + 1

(1 − r2)2 + 4ζ2r2= 0.0317 ft

andFt = r2kX = 268 lb

Note that the plot in Figure 13.1.3 is a plot of Ft/kY . Thus the curves in the plot mustbe interpreted with the value of k kept in mind. Even though the curve for r = 0.5 is lowerthat the curve for r = 3, for ζ = 1, the corresponding values of k are different, and thus thecase r = 3 gives a lower value of Ft.



13.10 For the lumped-mass equivalent system, we have the following spring constant andequivalent mass (after converting inches to feet):

k =Ewh3

4L3=

(4.32× 109)(0.333)(0.03125)3

4(0.5)3

= 8.78× 104 lb/ft

We compute the equivalent mass of the beam as follows. Using 15.2 slug/ft3 for the densityof steel, and including 23% of the beam’s mass, we obtain

me =20

32.17+ 0.23(15.2)(0.333)(0.5)(0.03125) = 0.640 slug

The unbalanced mass is m = 1/32.17 = 0.0311 slug.The model for the system is

mex + cx + kx = f(t) = muεω2 sin ωt

This gives the transfer function

T (s) =X(s)F (s)

=1

mes2 + cs + k

=1/k

mek s2 + c

ks + 1

Thus,

T (jω) =1/k

1 −(

ωωn

)2+ 2ζω

ωnj

where 1/k = 1.139×10−5 and ωn =√

k/me = 370 rad/sec = 3537 rpm. The log magnituderatio is.

m(ω) = 20 log(1/k)− 10 log

(

1 − ω2

ω2n

)2

+(

2ζω

ωn

)2

The motor speed of 1750 rpm gives a forcing frequency of ω = 183 rad/sec. Usingζ = 0.1 and ω/ωn = 183/370, we see that m = 2.36 − 98.9 = −96.5 db, which correspondsto a magnitude ratio of 1.5 × 10−5. The amplitude of the forcing function is muεω2 =0.0311(0.01)(183)2 = 10.4 lb. Thus, the steady-state amplitude is 10.4(1.5 × 10−5) =15.6×10−5 ft. On the downward oscillation, the total amplitude as measured from horizontalis the preceding value plus the static deflection, or 15.6× 10−5 +20/87800 = 3.84× 10−4 ft.(continued on the next page)




Because ω is not close to ωn in this problem, the preceding results are not very sensitiveto the assumed value of ζ = 0.1. For example, the calculated amplitudes of vibration forζ = 0.05 and 0.2 are close to the amplitude for ζ = 0.1 (the amplitudes are 15.7 × 10−5

ft and 15.2 × 10−5 ft, respectively). However, if we had used a motor with a speed of3500 rpm = 366 rad/sec, this choice would put the forcing frequency very close to thenatural frequency. In this region, the assumed value of ζ would be critical in the amplitudecalculation. In practice, such a design would be avoided. That is, in vibration analysis,the most important quantity to know is the natural frequency ωn. If the damping is slight,the resonant frequency is near ωn. If ωn is designed so that it is not close to the forcingfrequency, it is often not necessary to know the precise amount of damping.



13.11 We are given that the natural frequency is ωn1 = 900 rpm. Thus, at 1750 rpm,

r1 =1750900

= 1.94

If we decrease the stiffness k by 1/2, then the new natural frequency will be

ωn2 =

√k/2m

=1√2ωn1 =

900√2

rpm

Therfore,

r2 =1750

900/√

2= 2.75

From the rotating unbalance equation (13.2.6) with ζ = 0,

X =muε

m

r2

|1− r2| (1)

We are given that at r1, X1 = 8 mm. We need to compute X2. From (1),

X2

X1=(

r2

r1

)2 1 − r21

1 − r22

= 2−2.7636−6.5625

= 0.842

ThusX2 = 0.842X1 = 6.73 mm



13.12 We are given that

k =500

0.25/12= 24, 000 lb/ft

Thus

ωn1 =

√k

m=

√24, 000

500/32.2= 39.31 rad/sec

From the rotating unbalance equation (13.2.6) with ζ = 0,

X =muε

m

r2

|1 − r2|

and thusX2

X1=(

r2

r1

)2 1 − r21

1 − r22

wherer1 =

1750(2π)/6039.31

= 4.66

After the block is added, the new mass is m2 = 4m1 and the new natural frequency is

ωn2 =

√k

4m1=

12ωn1

Therefore,r2 =

ω

ωn2= 2

ω

ωn1= 2r1

Thus,X2

X1= (2)2

1 − r21

1 − r22

= 4−20.7156−85.8624

= 0.9651

and given that X1 = 0.1 in, we have

X2 = 0.9651(0.1) = 0.0965 in



13.13 The transmitted force is given by

Ft = muεω2

√1 + 4ζ2r2

(1− r2)2 + 4ζ2r2

We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. R = 0.1/12 =0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus

r =ω

ωn=

104.7√500/1.55

= 5.83

Thus

Ft = (0.00155)(0.0083)(104.7)2√

1 + 136ζ2

1088 + 136ζ2= 0.141

√1 + 136ζ2

1088 + 136ζ2

a) For ζ = 0.05, Ft = 0.0049 lb.b) For ζ = 0.7, Ft = 0.034 lb.



13.14 We haveFt = muεω2Tr

where Ft = 15, ω = 200(2π)/60 = 20.9 rad/s, and because c = 0,

Tr =1

r2 − 1

We haver =

ω

ωn=

20.9√k/M

=20.9√

2500/75= 3.62

Thus Tr = 0.0826 andFt = muε(20.9)2(0.0826) = 15

Solve for muε to obtain muε = 0.416 kg m.



13.15 We want Tr = 0.1. Neglecting damping, we have from (13.2.12)

r2 =1 + Tr

Tr= 11

Butr =

3000(2π)/60ωn

=314ωn

=√

11

Thus ωn = 314/√

11 = 94.7. But

ωn =

√k

m=

√k

3

from which we obtain k = 3ω2n = 3(94.7)2 = 26 904 N/m.



13.16 Damping is assumed to be negligible. Thus for the vertical motion,

Ft

F= Tr =

1r2 − 1

wherer =

ω

ωn=

ω√4k/m

=1750(2π)/60√

8000/25= 10.24

Thus Tr = 0.0096.The equation of motion for rotation is

Iθ = −(4kRθ)D/2

Thus the natural frequency for rotation is ωn =√

4kD2/4I =√

80/0.2 = 20. For therotational motion,

Tt

T= Tr =

1r2 − 1

wherer =

ω

ωn=

1750(2π)/6020

= 9.16

Thus Tr = 0.012.



13.17 We are given m = 20/g slug, mu = 1/g slug, ε = 0.01 ft, and ω = 3500(2π)/60 =366.5 rad/sec. Thus the unbalance force amplitude is

muεω2 =(

132.2

)(0.01)(366.5)2 = 41.719 lb

The beam stiffness is

k =Ewh3

4L3=

4.32× 109(1/3)(3/96)3

4(1/2)3= 8.79× 104 lb/ft

The beam mass is m = ρV = 15.2(1/3)(3/96)(1/2) = 0.079 slug.The first design equation for the absorber is r2 = 1, or

r2 =ω

ωn2=

366.5√k2/m2

= 1

This implies that k2/m2 = (366.5)2. The second design equation is X2 = 0.25/12 = 1/48ft, where

X2 =1k2

F =41.719

k2

Thus k2 = 41.719(48) = 2002.5 lb/ft. Substitute this value into the first design equation toobtain

m2 =k2

(366.5)22002.5

(366.5)2= 0.0149 slug



13.18 We are given that X2 ≤ 0.08/12 ft, ω = 6000(2π)/60 = 628.3 rad/sec, and that theunbalance force amplitude is muεω2 = 60 lb. Thus

muε =60ω2

=60

628.3= 0.0955

The first design equation for the absorber is

k2 =F

X2=

600.08/12

= 9000 lb/ft

The second design equation is √k2

m2= ω = 628.3

Thusm2 =

k2

(628.3)2= 0.023 slug



13.19 We are given that ω = 200(2π)/60 = 20.94 rad/sec, the amplitude of the unbalanceforce is muεω2 = 4 lb, and that X2 ≤ 1/12 ft. Assume that the table legs are rigid.

The first design equation for the absorber is√

k2

m2= ω = 20.94

The second design equation is

k2 =F

X2=

41/12

= 48 lb/ft

Thusm2 =

48(20.94)2

= 0.109 slug



13.20 The machine mass is m1 = 8 kg. a) We are given that ωn = 2π(6) = 12π rad/sec,muεω2 = 50 N, ω = 4(2π) = 8π rad/sec, and X2 ≤ 0.1 m. The design equation for theabsorber is √

k2

m2= ω = 8π

Thus k2 = 50/0.1 = 500 N/m, and

m2 =k2

(8π)2=

500(8π)2

= 0.792 kg

b) We have that k1 = m1ω2n = (12π)2m1 = 1421m1 = 11368 N/m. From (13.3.7),

T1(jω) =X1(jω)F (jω)

=1

11368

∣∣∣∣∣1 − r4

2b2r4

2 − [1 + (1 + µ)b2] r22 + 1

∣∣∣∣∣

whereb =

ωn2

ωn1=

8π

12π=

23

µ =m2

m1=

0.7928

= 6.336

1 + (1 + µ)b2 =139

+3.168

8= 1.84

The amplitude of F (jω) is muεω2, where r2 = ωωn2 = ω/8π. Thus

X1

muε=

111368

∣∣∣∣∣ω2(1 − r2

2

)

49r4

2 − 1.84r22 + 1

∣∣∣∣∣

orX1

muε=

111368

∣∣∣∣∣∣

ω2(1 − ω2

64π2

)

[49

ω4

4096π4 − 1.84 ω2

64π2 + 1]

∣∣∣∣∣∣

The plot is shown in the figure on the following page.(continued on the next page)




0 10 20 30 40 50 60 70 80 90 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Forcing frequency ω (rad/s)

X1/m

uε (s

lug−1

)




13.21 An absorber tuned to 2100 rpm requires that√

k2

m2=

√k1

m1=

2100(2π)60

= 219.9 rad/sec

Because we are given that m2 = 5/32.2 = 0.155, we have that k2 = (219.9)2m2 = 9747.8.Also,

k1 = (219.9)2m1 (1)

The characteristic equation of the combined system is given by the denominator of (13.3.1),and is

m1m2s4 + (m2k1 + m2k2 + m1k2)s2 + k1k2 = 0 (1)

One of the observed resonances of the combined system is 2850 rpm, or 2850(2π)/60 =298.45 rad/sec. Thus substituting s = 298.45j and relation (1) into (2), along with thevalues m2 = 0.155 and k2 = 9747, we can determine the value of m1, which is m1 = 0.4slug. Thus gives k1 = 19, 344 lb/ft.

With m1 and k1 determined, we now calculate the required values of m2 and k2. Supposewe choose to put the resonances just outside the operating range, say at 1400 and 3100 rpm(146.6 and 324.6 rad/sec). Let λ = s2, and let λ1 and λ2 denote the desired values of s2.We can factor the characteristic equation (2) as follows:

m1m2(λ− λ1)(λ− λ2) = m1m2λ2 − m1m2(λ1 + λ2)λ + m1m2λ1λ2 = 0 (3)

Comparing the coefficients of (3) with (2), we see that

(m2k1 + m2k2 + m1k2 = −m1m2(λ1 + λ2) (4)

andm1m2λ1λ2 = k1k2 (5)

We can solve (5) for k2 as follows.

k2 =m1m2λ1λ2

k1= Dm2 (6)

whereD =

m1λ1λ2

k1





Substitute (6) into (4) and solve for m2.

m2 = −m1(λ1 + λ2) + k1 + m1D

D

The desired values are λ1 = −(146.6)2 and λ2 = −(324.6)2. These give the absorber valuesm2 = 0.2706 slug and k2 = 12, 672 lb/ft.



13.22 a) From Newton’s law,

m1x1 = f − kx1 + c(x2 − x1)

m2x2 = −c(x2 − x1)

Transform these equations with zero initial conditions to obtain

(m1s2 + cs + k)X1(s) − csX2(s) = F (s)

−csX1(s) + (m2s2 + cs)X2(s) = 0

The solutions obtained with Cramer’s rule are

X1(s) =m2s

2 + cs

D(s)F (s)

X2(s) =cs

D(s)F (s)

where Cramer’s determinant is

D(s) = s(m1m2s3 + c(m1 + m2)s2 + km2s + ck)

Define the following parameters:

µ =m2

m1ω2

1 =k

m1

ζ =c

2√

m1kr =

ω

ω1

Then D(jω) can be written as

D(jω) =∣∣∣ωj[−m1m2ω

3j − c(m1 + m2)ω2 + km2ωj + kc]∣∣∣

orD(jω) = m2

1ω41r√

(2ζr)2[1 − (1 + µ)r2]2 + µ2r2(1 − r2)2





The numerator of kX1(jω)/F (jω) is | − m2ω2 + cωj| = m1rω

21

√4ζ2 + µ2r2. The nu-

merator of X2(jω)/F (jω) is |cωj|. Thus,

kX(jω)F (jω)

=√

4ζ2 + µ2r2

√(2ζ)2[1− (1 + µ)r2]2 + µ2r2(1− r2)2

andX2(jω)F (jω)

=1ω2

1

2ζ√(2ζ)2[1− (1 + µ)r2]2 + µ2r2(1− r2)2



13.23 a) From Newton’s law,

m1x1 = f − kx1 + k2(x2 − x1) + c(x2 − x1)

m2x2 = −k2(x2 − x1) − c(x2 − x1)

Transform these equations with zero initial conditions to obtain

(m1s2 + cs + k1 + k2)X1(s)− (cs + k2)X2(s) = F (s)

−(cs + k2)X1(s) + (m2s2 + cs + k2)X2(s) = 0

The solutions obtained with Cramer’s rule are

X1(s) =m2s

2 + cs + k2

D(s)F (s)

X2(s) =cs + k2

D(s)F (s)

where Cramer’s determinant is

D(s) = m1m2s4 + c(m1 + m2)s3 + (m2k1 + m2k2 + m1k2)s2 + ck1s + k1k2

Define the following parameters:

µ =m2

m1ω2

1 =k1

m1ω2

2 =k2

m2

α =ω1

ω2r =

ω

ω1

ζ =c

2√

m1k1λ =

k2

k1

Then D(jω) can be written as

D(jω) =∣∣∣m1m2ω

4 − (m2k1 + m2k2 + m1k2)ω2 + k1k2 + [ck1ω − (m1 + m2)cω3]j∣∣∣

orD(jω) =

∣∣∣m21ω

4[µr4 − (1 + λ + µλ)r2 + µα2] + [2ζr − 2(1 + µ)r3]j∣∣∣





The numerator of k1X1(jω)/F (jω) is |k2 − m2ω2 + cωj| = k1

∣∣[λ − µr2 + 2ζri]∣∣. The

numerator of k1X2(jω)/F (jω) is |cωj + k2| = k1|(λ + 2ζrj)|. Thus,

k1X(jω)F (jω)

=√

(λ − µr2)2 + (2ζr)2√[µr4 − (1 + λ + µλ)r2 + µα2]2 + (2ζr)2 [1− (1 + µ)r2]2

andk1X2(jω)

F (jω)=

√λ2 + (2ζr)2√

[µr4 − (1 + λ + µλ)r2 + µα2]2 + (2ζr)2 [1− (1 + µ)r2]2




m1x1 = −k1x1 − k2(x1 − x2)

m2x2 = k2(x1 − x2)

From these equations we can write the modal amplitude equations by substituting x1 =A1e

st and x2 = A2est, and using the given parameter values. The result is

(10s2 + 30, 000)A1 − 20, 000A2 = 0

−20, 000A1 + (30, 000s2 + 20, 000)A2 = 0

These give the solution

A2 =s2 + 3000

2000A1 (1)

The roots are found from Cramer’s determinant of the modal equations, which is

3s4 + 11, 000s2 + 2 × 106 = 0

The roots are s2 = −3475 and s2 = −192. Substitute s2 = −3475 into (1) to obtainA2 = −0.2375A1. Substitute s2 = −192 into (1) to obtain A2 = 1.404A1.

In the first mode, the masses oscillate in opposite directions with a radian frequencyof

√3475. The displacement amplitude of mass 2 is 0.2375 times that of mass 1. In the

second mode, the masses oscillate in the same direction with a radian frequency of√

192.The displacement amplitude of mass 2 is 1.404 times that of mass 1.




m1L2θ1 = −mg1Lθ1 − kd(dθ1 − dθ2)

m2L2θ2 = −m2gLθ2 + kd(dθ1 − dθ2)

From these equations we can write the modal amplitude equations by substituting θ1 =A1e

st and θ2 = A2est. The result is

(m1L2s2 + m1gL + kd2)A1 − kd2A2 = 0

−kd2A1 + (m2L2s2 + m2gL + kd2)A2 = 0

Using the given parameter values and g = 9.81 m/s2, these equations give the solution

A2 = (3.13s2 + 7.13)A1 (1)


2500s4 + 10810s2 + 11 586 = 0

The roots are s2 = −1.96 and s2 = −2.36. Substitute s2 = −1.96 into (1) to obtainA2 = 0.995A1. Substitute s2 = −2.36 into (1) to obtain A2 = −0.257A1.

In the first mode, the masses oscillate in the same direction with a radian frequency of√1.96. The displacement amplitude of mass 2 is 0.995 times that of mass 1. In the second

mode, the masses oscillate in the opposite direction with a radian frequency of√

2.36. Thedisplacement amplitude of mass 2 is 0.257 times that of mass 1.




I1θ1 = k2(θ2 − θ1) − k1θ1

I2θ2 = −k2(θ2 − θ1)

From these equations we can write the modal amplitude equations by substituting θ1 =A1e

st and θ2 = A2est. The result is

(I1s2 + k1 + k2)A1 − k2A2 = 0

−k2A1 + (I2s2 + k2)A2 = 0

Using the given parameter values, these equations give

A2 =I1s

2 + k1 + k2

k2θ1 =

s2 + 43

A1 (1)


5s4 + 23s2 + 3 = 0

The roots are s2 = −0.134 and s2 = −4.47. Substitute s2 = −0.134 into (1) to obtainA2 = 1.29A1. Substitute s2 = −4.47 into (1) to obtain A2 = −0.157A1.

In the first mode, the masses oscillate in the same direction with a radian frequency of√0.134. The displacement amplitude of mass 2 is 1.29 times that of mass 1. In the second

mode, the masses oscillate in the opposite direction with a radian frequency of√

4.47. Thedisplacement amplitude of mass 2 is 0.157 times that of mass 1.




mx = −2k sin 45 x = −1.41kx

my = −ky − 2k sin 45 x = −2.41kx

In the first mode, the mass oscillates in the x direction with a radian frequency of√

1.41k/m.In the second mode, the mass oscillates in the y direction with a radian frequency of√

2.41k/m.



13.28 From (13.4.8):

(750)(1350)s4 +730[(1.95× 104)(1.5)2 + 2.3× 104(1.1)2] + 1350(4.25× 104)

s2

+ 1.95(2.3)× 108(2.6) = 0

ors4 + 111.329s2 + 1.166× 104 = 0

This givess2 = −99.43 and s2 = −11.9

ors = ±9.971j and s = ±3.45j

These correspond to frequencies of 1.587 Hz and 0.549 Hz.

A1

A2=

x

θ=

k1L1 − k2L2

ms2 + k1 + k2=

3.95× 103

730s2 + 4.25× 104=

5.411s2 + 58.2192

For mode 1 (s2 = −99.43),

x

θ= −0.131 m ahead of the mass center

For mode 2 (s2 = −11.9),

x

θ= 0.1168 m behind the mass center



13.29 Referring to Figure 13.4.4, we are given that m1g = 1000 lb, c1 = 0, and k2 = 1300lb/in. The ride rate should be

ke =m1g

∆=

10009.8

= 102 lb/in

Thus the suspension stiffness should be

k1 =kek2

k2 − ke=

102(1300)1300− 102

= 111 lb/in



13.30 Define the following (refer to Figure 13.4.2):k1 = rear quarter-car suspension stiffness.k2 = front quarter-car suspension stiffness.kr = total rear suspension stiffness = 2k1.kf = total front suspension stiffness = 2k2.ker = total rear ride rate (including suspension and tire stiffness).kef = total front ride rate (including suspension and tire stiffness).ke1 =quarter-car rear ride rate.ke2 =quarter-car front ride rate.kt = individual tire stiffness.Referring to the guidelines on page 851, and using equations (4) and (5) of Example

13.4.3 as approximations for the bounce and pitch dynamics, we have

kef = 0.7ker (1)

ωbounce

2π=

12π

√kf + kr

m≤ 1.3 Hz (2)

ωpitch

2π=

12π

√kfL2

1 + krL22

IG≤ 1.3 Hz (3)

ke1 =(m/4)g

∆1(4a)

ke2 =(m/4)g

∆2(4b)

From the above definitions,

kef = 2ke2 ker = 2ke1 (5)

Because the tire stiffness is in parallel with the suspension stiffness,

ke1 =k1kt

k1 + ktke2 =

k2kt

k2 + kt(6)





These relations must be satisfied by k1, k2, and kt. Using the given values m =4800/32.2, IG = 1800, L1 = 3.5, and L2 = 2.5, these equations can be rearranged asfollows:

0.02019√

k1 + k2 ≤ 1.3 (7)

0.0375√

24.5k1 + 12.5k2 ≤ 1.3 (8)

k1 =1200∆1

kt

kt − 1200∆1

(9)

k2 =1200∆2

kt

kt − 1200∆2

(10)

∆1 = 0.7∆2 (11)

The procedure is to select suitable values for kt and ∆2, solve (9) and (10) for k1 and k2,and see if (7) and (8) are satisfied. Trying ∆2 = 9.8/12 ft, as suggested on page 851, andkt = 1200(12) = 14400 lb/ft, we obtain from (9) and (10) k1 = 2457 lb/ft and k2 = 1636lb/ft. With these values, the left-hand sides of (7) and (8) are 1.29 and 1.07. Thus therequirements have been met.



Problem 13.31 The equation of motion with the controller is

125x + (5000 + KD)x + (7 × 106 + KP )x = 100 sin ωt

where ω = 2500(2π)/60 = 261.8. It is desired that

ωn = 100 =

√7 × 106 + KP

125

which gives KP = (1.25)2 × 108 − 7 × 106 = 1.4925× 108 N/m.Also

ζ =5000 + KD

2√

125[(125)2 × 108]= 0.5

This gives KD = 6.25√

5 × 104 − 5000 = 1.3475× 105 N·s/m.Substituting these values into the equation of motion and simplifying, we obtain

x + 500√

5x + 1.25× 106x = 0.8 sin ωt

The steady state amplitude is

X(ω) =0.8∣∣∣−ω2 + 1.25× 106 + 500

√5ωi

∣∣∣



Problem 13.32 From the given information, the stiffness is

k = mω2n = 20(13.2)2 = 3485

and the damping isc = 2(0.28)

√20(3485) = 147.8

It is desired that ωn = 141. Thus

ωn = 141 =

√k + KP

20

andKP = (141)220− k = 3.9414× 105 N/m


ζ =c + KD

2√

20(k + KP )= 0.707

which givesKD = 2(0.707)

√20(k + KP ) = 3987 N · s/m



Problem 13.33 Assuming that xs, xw, and xr are measured from the equilibrium positions,the equation of motion for the sprung mass is

msxs = c(xw − xs) + ks(xw − xs) + f

where f is the actuator force. Since we are neglecting the tire mass, a force balance at thatpoint gives

mwxw = 0 = kt(xr − xw)− ks(xw − xs)− c(xw − xs) − f

The actuator force isf = −Kp(xs − xw) − KD(xs − xw)

Taking the Laplace transform of these equations, using zero initial conditions, eliminat-ing the variables f and xw , and reverting back to the time domain, we obtain

ms(c + KD)d3xs

dt3+ ms(ks + kt + KP )

d2xs

dt2+ kt(c + KD)

dxs

dt+ kt(ks + KP )xs

= kt(c + KD)dxr

dt+ kt(ks + KP )xr



Problem 13.34 Set the derivatives to zero to find the equilibrium solution ye.

1700y3 + 900y − 5g = 0

which has the roots y = 0.0542 and s = −0.027 ± 0.729i. Discard the complex valuedsolution. Near y = 0.0542,

y3 ≈ (0.0542)3 + 3y2e (y − ye) = (0.0542)2 + 0.00881x

where x = y − 0.0542. The linearized model is 5x = −900x − 1200(0.000881)x, or 5x +914.977x = 0. The roots are s = ±13.5i, and the natural frequency is ωn = 13.5 rad/sec.

b) Because the roots are s = ±13.5i, the solution of 5x + 914.977x = 0 has the formx(t) = A sin 13.5t + B cos 13.5t. Note that x(0) = B = 0.002 and x(0) = 13.5A = 0.005.Thus A = 3.704× 10−4, and

x(t) = 3.704× 10−4 sin 13.5t + 0.002 cos 13.5t



Problem 13.35dv

dt= − c

mv3

∫ v(t)

v(0)

dv

v3= − c

m

∫ t

0dt = − c

mt

− v−2

2

∣∣∣∣∣

v(t)

v(0)

= − c

mt

v−2(t) − v−2(0) = 2c

mt

Thus

v(t) = v(0)

√1

1 + 2cv2(0)t/m



Problem 13.36 The equilibria are the roots of

16ye − 4y3e = 0

and are ye = 0 and ye = ±2.The linearized model is

x + 12x + 16x − 12(ye)2x = 0

For ye = 0,x + 12x + 16x = 0

which is locally stable.For ye = ±2,

x + 12x− 32x = 0

which is locally unstable.The state variable form of the model is

y1 = y2

y2 = −16y1 + 4y31 − 12y2

where y1 = y and y2 = ˙ y.Create the following function file.

function ydot = prob13p36(t,y)ydot = [y(2); -16*y(1)+4*y(1)^3-12*y(2)];

For y(0) = 1, type the following in MATLAB.

y0 = 1;[t, y] = ode45(′prob13p36′, [0, 1], [y0, 0]);plot(t, y(:,1))

Do the same for the other initial conditions. The simulations show that for y(0) = ±2.1,the model is unstable (|y(t)| → ∞). For y(0) = ±1 and y(0) = ±1.9, y(t) → 0. Thus theequilibria at ye = ±2 are globally unstable, and the equilibrium at ye = ±0 is locally stable.



Problem 13.37 The state variable form of the model is

y1 = y2

y2 = −2y1 − 0.1y31


function ydot = prob13p37(t,y)ydot = [y(2); -2*y(1)-0.1*y(1)^3];

For y(0) = 10, type the following in MATLAB.

y0 = 10;[t, y] = ode45(′prob13p37′, [0, 10], [y0, 0]);plot(t, y(:,1))

Do the same for the other initial condition. The simulations show that for y(0) = 10, theresponse oscillates with a frequency of about 3 rad/s. For y(0) = 40, the response oscillateswith a frequency of about 11 rad/s. So the oscillation frequency depends greatly on theinitial condition.



Problem 13.38 The three cases are of the form

x + cx + 2x = 0

The state variable form isx1 = x2

x2 = −2x1 − cx2

In MATLAB create the following file.

function xdot = prob13p38(t,y)global cxdot = [x(2); -2*x(1)-c*x(2)];

For part (a), where c = 0.1, type the following in MATLAB.

global cc = 0.1;[t, x] = ode45(′prob13p38′, [0, 50], [1, 0]);plot(x(:,1), x(:,2))

The phase plane plot is a tight spiral ending at (0, 0).For parts (b) and (c) change the second line to c = 2 and c = 4, respectively. The

phase plane plots head almost directly toward the point (0, 0) without much of a spiral.




y1 = y2

y2 = −2y1 − 3y31 − 2y2


function ydot = prob13p39(t,y)ydot = [y(2); -2*y(1)-3*y(1)^3-2*y(2)];

Then type the following in MATLAB.

[t, y] = ode45(′prob13p39′, [0, 10], [1, 0]);plot(y(:,1), y(:,2))

The phase plane plot heads almost directly toward the point (0, 0) without much of aspiral.




y1 = y2

y2 = −y1 + 5(1− y21)y2


function ydot = prob13p40(t,y)ydot = [y(2); -y(1)+5*(1-y(1)^2)*y(2)];

Then type the following in MATLAB.

[t, y] = ode45(′prob13p40′, [0, 50], [1, 0]);plot(y(:,1), y(:,2))

The phase plane plot is a limit cycle.




to accompany


by

William J. Palm III


Solutions to Problems in Appendix C


CD1

Highlight

C.1 The session is:

a = 1.12; b = 2.34; c = 0.72;d = 0.81;f = 19.83;x = 1 + a/b + c/f^2x =

1.4805s = (b-a)/(d-c)s =

13.5556r = 1/(1/a + 1/b + 1/c + 1/d)r =

0.2536y = a*b/c*f^2/2y =

715.6766



C.2 The session is

x = -7-5i;y = 4+3i;x+yans =-3.0000 - 2.0000ix*yans =-13.0000 -41.0000ix/yans =-1.7200 + 0.0400i(3/2)*jans =0 + 1.5000i3/(2j)ans =0 - 1.5000i



C.3 The session is:

x=[1:0.2:5];y = 7*sin(4*x);length(y)ans =

21y(3)ans =

-4.4189

There are 21 elements. The third element is −4.4189.



C.4 The session is

wage = [5,5.5,6.5,6,6.25]; hours = [40,43,37,50,45];output = [1000,1100,1000,1200,1100];earnings = wage.*hoursearnings =

200.0000 236.5000 240.5000 300.0000 281.2500total_salary = sum(earnings)total_salary =

1.2582e+003total_widgets = sum(output)total_widgets =

5400average_cost = total_salary/total_widgetsaverage_cost =

0.2330average_hours = sum(hours)/total_widgetsaverage_hours =

0.0398[maximum,most_efficient] = max(output./earnings)maximum =

5most_efficient =

1[minimum,least_efficient] = min(output./earnings)minimum =

3.9111least_efficient =

5

The workers earned $200, $236.50, $240.50, $300, and $281.25 respectively. The total salarypaid out was $1258.20, and 5400 widgets were made. The average cost to produce one widgetwas 23.3 cents, and it took an average of 0.0398 hr to produce one widget. The first worker,who produced 5 widgets per dollar of earnings, was the most efficient. The fifth worker,who produced 3.911 widgets per dollar of earnings, was the least efficient.



C.5 The MATLAB expressions are:

f = 1./sqrt(2*pi*c./x)E = (x + w./(y + z))./(x + w./(y - z))A = exp(-c./(2*x))./(log(y).*sqrt(d*z))S = x.*(2.15 + 0.35*y).^1.8./(z.*(1-x).^y)



C.6 For part (a) note that the first quarter material cost is computed by

7(16) + 3(12) + 9(8) + 2(14) + 5(13) = 326

and the second quarter material cost is computed by

7(14) + 3(15) + 9(9) + 2(13) + 6(16) = 346

and so on. Thus the quarterly costs can be computed by multiplying the transpose of thematrix of unit costs by the matrix of quarterly production volumes. The resulting 3 × 4matrix is quarterly_costs. Its first row contains the material costs, its second row containsthe labor costs, and the third row contains the transportation costs. The four columnsof quarterly_costs correspond to the four quarters. For part (b) the yearly costs formaterials, labor, and transportation are found by summing the rows of quarterly_costs,or equivalently, by summing the columns of the transpose of quarterly_costs. For part(c) the total quarterly costs are found by summing the columns of quarterly_costs. Thesession is

unit_cost = [7,3,2;3,1,3;9,4,5;2,5,4;6,2,1];quarterly_volume = [16,14,10,12;12,15,11,13;8,9,7,11;14,13,15,17;13,16,12,18];quarterly_costs = unit_cost′*quarterly_volumequarterly_costs =

326 346 268 364188 190 168 214177 186 160 204

yearly_costs = sum(quarterly_costs′)yearly_costs =

1304 760 727total_quarter_cost = sum(quarterly_costs)total_quarter_cost =

691 722 596 782




Problem C.6 continued:

The table C.6 was created from the matrix quarterly_costs. All costs are in thousandsof dollars.

Table C.6Quarterly Costs

Category Quarter 1 Quarter 2 Quarter 3 Quarter 4Materials 326 346 268 364

Labor 188 190 168 214Transportation 177 186 160 204

From the vector yearly_costs we obtain the following information: yearly materialscost = $1,304,000, yearly labor cost = $760,000, and yearly transportation cost = $727,000.From the vector total_quarter_cost we find that the total costs in each quarter are$691,000, $722,000, $596,000, and $782,000 respectively.



C.7 The session is

roots([13,182,-184,2503])ans =

-15.68500.8425 + 3.4008i0.8425 - 3.4008i

poly(ans)ans =

1.0000 14.0000 -14.1538 192.538513*ans

ans =1.0e+003 *0.0130 0.1820 -0.1840 2.5030

This gives the original coefficients.



C.8 The session is

roots([36,12,-5,10])ans =

-0.86510.2659 + 0.5004i0.2659 - 0.5004i

polyval([36,12,-5,10],ans)ans =

1.0e-013 *0.01780.1243 + 0.0178i0.1243 - 0.0178i

Since the last result is essentially zero, the polynomial evaluates to zero, showing that theroots are correct.



C.9 The session is

poly([3+6j,3-6j,8,8,20])ans =

1 -42 645 -5204 24960 -57600roots(ans)ans =

20.00003.0000 + 6.0000i3.0000 - 6.0000i8.00008.0000



C.10 The session is

conv([10,-9,-6,12],[5,-4,-12,8])ans =

50 -85 -114 272 -48 -192 96

The answer is 50s6 − 85s5 − 114s4 + 272s3 − 48s2 − 192s + 96.



C.11 The session is

[q,r] = deconv([14,-6,3,9],[5,7,-4])q =

2.8000 -5.1200r =

0.0000 0.0000 50.0400 -11.4800

The answer is 2.8s− 5.12 with a remainder of 50.04s− 11.48.



C.12 The session is:

x = [0:0.01:2];u = 2*log10(60*x+1);v = 3*cos(6*x);plot(x,u,x,v,′--′),ylabel(′Speed (mi/hr)′),...xlabel(′Distance x (mi)′)



C.13 The session is

p1 = [3,-6,8,4,90];p2 = [3, 5, -8, 70];x=[-3:0.01:3];y = polyval(p1,x);z = polyval(p2,x);plot(x,y,x,z,′--′),xlabel(′x′),ylabel(′y and z′),gtext(′y′),gtext(′z′)



C.14 The script file is:

x = -5;if x < -1

y = exp(x + 1)elseif x < 5

y = 2 + cos(pi*x)else

y = 10*(x - 5) + 1end

The answer for x = −5 is y = 0.0183. Change the first line to x = 3 to obtain y = 1. Thenchange the first line to x = 15 to obtain y = 101.



C.15 a) z = 0 1 0 0 1b) z = 1 0 1 0 0c) z = 1 1 1 0 1d) z = 0 0 0 1 0e) z = 0 1 0 1 1



C.16 (a) The session is

price_A = [19,18,22,21,25,19,17,21,27,29];price_B = [22,17,20,19,24,18,16,25,28,27];price_C = [17,13,22,23,19,17,20,21,24,28];length(find(price_A>price_B&price_A>price_C))ans =

4

Thus the price of stock A was above both B and C on four days.(b) Replace the fourth line in the above session with

length(find(price_A>price_B|price_A>price_C))ans =

9

The answer is nine days.(c) Replace the fourth line in the session in part (a) with

length(find(xor(price_A>price_B,price_A>price_C)))ans =

5

The answer is five days.



C.17 The script file is

price = [19,18,22,21,25,19,17,21,27,29];cost_new_shares = 100*sum(price.*(price<20))income_selling = 100*sum(price.*(price>25))shares_change = 100*(sum(price<20)-sum(price>25));total_shares = 1000 + shares_changenet_increase = price(10)*total_shares - price(1)*1000

The results are: cost_new_shares = 7300, income_selling = 5600, total_shares =1200, and net_increase = 15800. Thus you spent $7300 in buying shares. You received$5600 from the sale of shares. After the tenth day you own 1200 shares. The net increasein the worth of your portfolio is $15,800.




sum = 0;for k = 1:10

sum = sum + 5*k^3;endsum

The answer is sum = 15125.




sum = 0;k = 0;while sum <= 2000

k = k + 1;sum = sum + 2^k;

endksum

The answers are k = 10 and sum = 2046.




amt1 = 1000;amt2 = 1000;k1 = 0;k2 = 0;while amt1 < 50000

k1 = k1 + 1;amt1 = amt1*1.055 + 1000;

endwhile amt2 < 50000

k2 = k2 + 1;amt2 = amt2*1.045 + 1000;enddiff = k2 - k1

The answer is diff = 2. Thus it takes 2 more years in the second bank.



C.21 The function is

function theta = asine(x,q)if abs(x) <= 1

if q==1theta=asin(x);

elseif q==2theta = asin(x)+pi;

elseif q==3theta = asin(x)+pi;

elsetheta = asin(x);

endtheta = theta*(180/pi)

elsedisp(’|x| >1’)

end

The function does not protect against incorrect values of the quadrant number q, but theproblem statement did not require this.



C.22 The function file is

function y = f6(x)y = 1 + exp(-0.2*x).*sin(x+2);

You can plot the function to obtain solution estimates to use with fminbnd, or you cansimply try values of x between 0 and 10. The session is

fminbnd(′f6′,0)ans =2.5150f6(ans)ans =

0.4070fminbnd(′f6′,10)ans =

8.7982f6(ans)ans =

0.8312

So the solutions are (x, y) = (2.5150, 0.4070) and (x, y) = (8.7982, 0.8312).



C.23 The function file is

function t = time(h,v0,g)% Computes time t to reach a specified height h, with initial speed v0.roots([0.5*g,-v0,h])

A test session follows.

time(100,50,9.81)ans =

7.46122.7324

The smaller value is the time to reach the height while ascending; the larger value is thetime to reach the height while descending.




to accompany


by

William J. Palm III


Solutions to Problems in Appendix D


CD1

Highlight

D.1 The closed-form solution isy(t) = 6e−5t

After five steps, t = 0.1 and the exact solution is

y(0.1) = 3.6392



D.2 The closed-form solution isy(t) = 6 + sin t


y(1.5) = 6.9975



D.3 The closed-form solution is

y(t) = 12 − 2 cos 3t


y(0.5) = 11.8585




y(t) = 3.25− 12.5e−4t


y(0.125) = −4.3316




y(t) =657

e−3t +57e4t


y(0.05) = 8.8647





y(0.1) = 3.6392



D.7 The closed-form solution isy(t) = 6 + sin t


y(1.5) = 6.9975




y(t) = 12 − 2 cos 3t


y(0.5) = 11.8585




y(t) = 3.25− 12.5e−4t


y(0.125) = −4.3316




y(t) =657

e−3t +57e4t


y(0.05) = 8.8647



D.11 The equations of motion are given by equations (3) and (4) in Example 2.2.4. Theseequations are programmed in the following m file.

function ydot = mast(t,y)Q=sqrt(2020+1650*cos(1.33+y(1)));ydot(1) = y(2);ydot(2) = (1/25400)*(-17500*cos(y(1))+(626000./Q).*sin(1.33+y(1)));ydot = [ydot(1);ydot(2)];

This file is called as follows. Since we do not know how long it will take for the mast to reach90, we must experiment with the stop time. Since the geometry on which the equationsare based breaks down if θ ≥ 90, we start with a low value of the stop time, say 2 seconds,and gradually increase the stop time uintil the plot shows that 90 has been reached. Totwo decimal places, the answer is 8.62 seconds.

[t,y] = ode45(@mast,[0, 8.62],[pi/6,0]);plot(t,(180/pi)*y(:,1))





y(0.1) = 3.6392




y(t) = 6 + sin t


y(1.5) = 6.9975




y(t) = 3.25− 12.5e−4t


y(0.125) = −4.3316





y(0.1) = 3.6392




y(t) = 6 + sin t


y(1.5) = 6.9975




y(t) = 12 − 2 cos 3t


y(0.5) = 11.8585




y(t) = 6 + sin t


y(1.5) = 6.9975




y(t) = 12 − 2 cos 3t


y(0.5) = 11.8585



System Dynamics 2nd Edition Solution manual

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