system dynamics ogata
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B-2-1.
Note that
Solutions to B Problems
CHAPTER 2
& r t j = +- S
R e f erring to Egua tion (2-2 1, we obtain
Note tha t
3 sin (5t + 45 ") = 3 sin 5t cos 45' -+ 3 cos 5t sin 45'
- 3 - - 3
Jz s in 5t + - cos 5t
f l So we have
Note that
Referring to Equation (2-21, we obtain
B-24. f( t) = 0 t <U
= c o s 2 W t r o s 3 L L I t t 1 0
Noting that
cos 2 W t cos 3wt = + ( m s 5 d t + c Q s & J t )
w e have
F(S) = L[f( t )~ = [%ms 5 ~ t + r n s r ~ t ) ]
B-2-5. The function f I t ) can be written as
f(t) = (h - a) l(t - a )
The Laplace transform of f(t) is
-as ~ ( s ) = [~(t)] = J [(t - a) ~ ( t - a)] = +
6
B-2-6. - f[t) = c l[t - a ) - c I(t - b) The Laplace transform oE f ( t ) is
e-a S ~ ( a ) = c - - c - e-bs - - 2 (,-as - S S S
B-2-7. The Function f ( t ) can be written as
So the Laplace transform of f ( t ) becomes
As a approacl~es zero, t h e l imit ing value of F(s ) becomes as follows:
10 - 12.5 e-(a/5)s + 2 5 =-as l im F(s ) = lim a d 0 a+O 2 a s
( 2 . 5 e-(a/5)s - 2 . 5 .-as) = lim da
-0.5 s e-(a/5)s + 2.5 s e-as = lirn
a+O 2
B-2-8. The function f(t) can be wri t ten as
So the Laplace transform of f(t) becomes
The limiting ~ l u e 05 F(s) as a approaches zero is
24(1 - as e-*= l i m F(s) = l i r n a d 0 a 3 0 3 2 a s
3-2-11. Define y = x
Thw y(O+) = $(O+)
The i n i t i a l value of y can be obtained by use of the initial value theorem as follows:
Since y(s1 = J'+[y!t)l = d;rGct,1 = sX(S) - x((U)
we obtain
B-2-12. Note that
Referring to Problm A-2-12, we have
B-2-14.
f(t~ emst dt = e-st at = 1 0-
Referring to Problem A-2-12, we obtain
where
%(s) can thus be written as
and the inverse Laplace transform of Fl (s ) is
Cbl
where
FZ(s) can thus be written as
and the inverse Laplace transform of F Is) is 2
me inverse Laplace transform of F (s) is 1
(bl
where
E (s) can thus be written as 2
and the inverse Laplace transform of F,(s) is L-
i p ) = -3 .-t + B t .-2t . 3 .-2t
- 2 5 5 - 3 5 - 2 + - + - - - - 2 - + - s + l s s + L s + l s
Rre inverse Laplam transform of f (s ) is
The inverse Lapla- transform of F(S) is
f{t) = S ( t ) 'F 2 4- 4 t
H e n c e 1 f(t) = e-t cos 3t - - e-t sin 3t 3
Hence
The inverse Laplace transform of F(s) is
H e n c e
The inverse Laplace transform of Fl s ) is
The inverse Uplacre transform of ~ ( s ) is
1 1 (t - - sin L L J ~ ~ fit) = - ld2 W
B-2-23. C
~ ( s ) = - b (1 - e-as) , - e-as s2 S
a > O
The imrse Laplace transform of F(s) is
f(t) = ct - c ( t - a ) f ( t - a ) - b l(t - a)
3-2-24. A MATUG program to obtain prtial-fraction expansions of the given function F(s) is given b l e w .
From this eamputer output we obtain
I - - -0.25 + -0.75 0.5 F(s) = + +-
s4 + 333 + 2s2 s + 2 s + l s s2
The inverse Laplace transform of F(s) is
f(t) = -0.25 e-*t + e-t - 0.75 + 0.5t
B-2-25. A possible MATLIE! program to obtain partial-fraction expansions
of the given function F(s:l is given b l o w .
f I num = [O 0 3 4 I];
d e n = [ l 2 5 8 101; [r,p,kJ = residue(num,den)
r =
0.3661 - 0.4881i 0.3661 + 0.4881 i -0.3661 - 0.0006i -0.3661 + O.0006i
P =
0.2758 + 1 .go81 i 10.2758 - 1.9081 i
-1 -2758 + I.0309i - 1.2758 - 1.0309i
k =
0 L
From this cmputer output we obtain
Since the poles are complex quantities, we may rewrite F ( s ) as follows:
Then, the imrerse Laplace transform of F(s) is obtained as
f(tl = 0.7322 e092758t m s 1.9082t
+ 0.9762 e00275Bt sin 1.9081 t
- 0.7322 e-1-2758t cos 1.0309t
+ 0.001204 e-1*2758t sin 1.0309t
'Ihe Laplace transform of the given differential equation is
Substitution of the initial conditions into this l a s t equation gives
SoJ=ving for X(s ) , we obtain
The inverse Laplace transform oE K(s) is
x ( t ) = 5 COS 2t
This is the solution of the given differential equation.
--- - ---
.. B-2-27, X + L J 2 x = t , X ( O ) = 0, ;(a) = o
n
The Laplace transform of this differential equation is
Solving this q a t i o n for X ( s 1 , we obtain
The inverse Laplace transform of X ( s ] is
This is the solution of the given differential equation,
P - B-2-28 . d ; + & + x = l , x(0) = 0, 210) = 2
The Uplace transform of this differential equation is
1 2[s2x(s) - sx(0) - G(o)] + 2[sx(s) - ~(0)) + x ( s ) = - s
Substitution of the initial conditions into this equation gives
Solving this last equation for X ( s ) , we get
The inverse Laplace transform of ~ ( s ) giws
x(t ) = 4 e-005t s i n 0.5t + 1 - e cos 0.5t - e -0*5t sin 0.5k
= l + 3 e-0*5t sin 0.52 - a s 0.5t
Taking the Laplace transform of b h i s differential equation, we obtain
2[sZ~(s) - sxIO) - s(03 1 + 7 ~ a i s ) - x(0) 1 + 3X(s) = 0
By substituting the given initial conditions into this last equation,
~ [ s ~ x ( s ) - 3s] + 7[sX(s) - 31 + 3XCs) = 0
Solving for X [ s ) yields
Finallyf taking the inverse laplace tmnsform of X(s), we obtain
-0.5t x(t) = 3.6 e - 0.6 e-3t
.. - B-2-30, x * x = sin 3t, x(0) = 0, k ( 0 ) = 0
The Laplace transform of this differential equation is
Solving this equation for Xis), we get
The invezse. Laplace transform of X(s) gives
sin t - - x(t) = -- I s i n 3t '3 8
--
CHAPTER 3
B-3-2. Assume that the W y of known mmnt of i n e r t i a J is turned through a small angle @ akrout the ve r t i ca l axis and then ' released. The equation of motion far the oscillation is
where k is the torsfanal spring constant oE the string. This equation can be m i t t e n as
The period TO of this oscillation is
Next, we attach a rotating m y of unknown mmnt of inertia 9 and measure the period T of oscillation. The equation for the perid T is
By eliminating the trimown torsional spring constant Ir from Equations (1) and (2) , we obtain
The unknown moment of inertia J can therefore be determined hy measuring t h e perid af oscillation T and s u b s t i t t t t i n g it i n t o Equation ( 3 ) .
B-3-3. Define the vertical displacement of t h e b a l l as x( t ) with x(0) = 0. - The positive direction is downward. The equation of motion for the system is
with initial conditimsx(O) = O m a n d G(0) = 2 0 m / s . Sowe have
I\snrme that at t = tl the ball reaches the ground. Then
frm which w e obtain
The ball reaches the ground in 2,915 s.
B-3-4. Define the torque applied to tfie flywheel as T- The equation of motion for the system is
from which we obtain
%y substituting nmrical values into this equation, we have
Thus
T = 1256 N-III
** J Q = - T (T = braking torque)
Integrating this equation,
Substituting the given numerical values,
Solving for TJJ, we obtain
Hence, the deceleration given by the brake is 5.33 rad/s2.
The total angle rotated in 15-semnd period is obtained from
B-3-6. Assume tha t we apply force F to t he spring system. Then
F = k x + k ( x - y ) 1. 2
~liminating y f r a the preceding equations, we obtain
Hence the equivalent spring constant is given by
-3-7, The equations for the system are
Eliminating y from the two equations gives
The equivalent spring constant k is then obbined as @3
Next, consider the figure s h m below. Note that AABD and d C B E are similar. So we have
which can be rewritten as
-- -- K(OB + + OR) = OA(OB - 4 El
Solving for OC, we obtain
B-3-8,
( a ) The force f due to the dampers is
In terms of the ermivalent viscous friction coefficient beq, force f is given by
Hence
(b) The force f due to the dampers is
where z is the displacemnt of a point &tween damper bl and damper b2. (Note t ha t the same force is transmitted through the shaft.) From Equation (I), we have
In terrns of the equivalent viscous friction coefficient kt force f is given by
f = hq(y - ;) By substituting Equation (2) in to Equation (13, we have
Hence,
B-3-9. Since the sane force transmits the shaft, we have
where displacement z is defined in the figure Below.
In tern of the equitalent viscous friction coefficient, the force 5 is given by
f = - 4) (2 1
From Equation (1) we have
bl& + b2; + bgi = bl; + b2G + b3;
By substituting Equation (3) i n to Equation 1 1 ) ~ we have
Hence, by comparfng Equations (2) and (41 , we obtain
8-3-10. The equation For khe system is
$ + (lc + k + I c ) x = O 1 2 3
The natural frequency of the system is
B-3-ll. ?'he density p of the l i q u i d is
where A is the cross-sectio~lal area of the inside of the g l a s s tube. The mathernakical ~ d e l for the sysken is
The natural frequency is
B-3-12. For a mll displacement x, the torque balance equation for the system is
** I mx(2a) = - k(T x)a
or
Tire natural frequency is
-3-14. A modified diagram for the systim s h m in Figure 3-55 is given below.
A mathemtical model for the system is given by the following t y ~ equations :
13;~ = -k,e, - k2(e1 - 82) I*
J$32 = k2(91 - 82)
B-3-15. The following twa equations describe the nation of the system and they are a mathematical model of the system.
I.
mlx = - ~ ( x - ~3 - k(G - $1 + p(t)
m2$ = -kzy - kl(y - x ) - q(j. - i ) Rewriting, we obtain
B-3-15. A mathematical. m o d e l for the system is
B-3-17. The equations of motion for the system a r e
2 Noting that x = 2 y , R8 = x - y = y r and J = the three equations can b rewritten as
I 2" 1 - M R @ = - @ = 2 2 ( T ~ - T~)R
*. a = -
~ j ; * icy = TI + T*
Eliminating T2 f m the preceding equations gives
1 - M ~ + N ? + ] C Y = ml I - 2nrj; T2
2
By changing y into xt
The natural frequency 4s I
XE mass m is pullecl down a distance xo and released with zero i n i t i a l velmity, the motion of mass rn is
B-3-18. Referring to the figure below, we have
where T is the tension i n the wire. (Note that since x is masurd from the static equilibrium position, the term mq does not enter the equation.) For the rotational motion of the pulleyr WE have
Eliminating T frm Equetions (1) and (2), we obtain
Since x = Rlf l r we have
This Last equation is a mathematical model of the system. The natural f rquency of the system is
I l l
3-3-19. The equation of motion for the system is
Substituting the given numerical values for m, b, and k into th i s equation, we obtain
wherex(0) = 0.1 and ?(o) = 0. The respofise ta the given initial condition can be obtained by taking the Laplace transform of this equation, solving the resulting equation for ~ ( s ) , a n d finding the inverse Laplace transform of X(s) . The LapPaee transform of Equation (1 ) is
By subsbituting the given initial conditions ink0 this l a s t equation, we get
Solving t h i s equation for X l s ) gives
The inverse Laplace transform of this last equation gives
x( t ) = 0.1(J-- e-t s i n 3t + e l t - 3 t ) 3
--- . . - -- . - -.
B-3-20. The equation of notion for the system is
where Fk = &(mg - F sin 30'). ~ewritting this equation,
= 0.066 F - 0.3(mg - 0.5 F) For a constant speed motion, $ = 0 and the last equation becomes
1.016 F - 0.3 mg = b
or m%
3-3-21, Tfie equations of mtion for the system are rx .. M x = T -
kMs
0.
m = m g l - T
Elimination of T frm khese two equations gives
By substituting M = 2, m = I, and JLK= 0.2 i n t o this l a s t equation, we get
Noting that z ( 0 ) = 0, w e have
P s s w e that a t t = t . x(tl) - x(O) = 0.5 m. ?hen 1
Thus, the velocity of the block when it has maved 0.5 m can be found as
B-3-22. The equations of motion for the system are
2 where x = R8 and J = $ mR . So we obkain
The natural frequency of the system is
B-3-23. Assume that the direction of the static friction force F is to S
the left as shown in t h e diagram below. The equations for the system are
where J = $ m~'. Since the cylinder rolls without sliding, we have x = I?@. Consequently,
By eliminating from Equations (I) and (2), we have
Since < is found to be equal to -(1/3)F, the magnitude of Fs is one third
of F and its direction is opposite to that assumed in the solution.
B-3-24. The equation of motion for the system is **
ntx = F - q sin 30- - ppg cos 30°, x ( 0 ) = 0 , ; ( o ) = 0
By substituting the given numerical values into this equation, we obtain
* b x = F - 1 x 9.81 x 0.5 - 0.2 x 1 x 9.81 x 0.866
or
and
Assume that a t t = tl, x(t ) = 6 rn and i(tl) = 5 d s . T¶x?n - I
1 6 = (F - 6.604) - tI2 2
F m the last two equations, tl and F are fomd to h
Tfieref ore,
Work done by force F = F x 6 = 8.69 x 6 = 52.14 N-m
Work done by the grav3btional force =-mg s i n 30m x 6
- - - 9.81 x 0.5 x 6 = - 29-43 N-m
Work done by the sliding friction force
= - 0.2 x mg m s 30' x 6
8-3-26. The kinetic energy T is
1 2 '2 T = ~ H ~ ' e + d~ = + ( M + + ) I ~ ~ ~ -
The ptmtia l energy U is
U = t @ f ( 1 - cos 0 ) + - m s 0) d J
- - ( M + - + ) g j ? ( l - m s B)
Since the system f s conservative, we have
= constant
Noting that d(T + U)/dt = 0, we obtain
[(M + 1 2 ; + (M + +lg! sin 81 j) = 0
Since 0 is not identically zero, we have
Rewriting,
" + 8 + q s i n ~ = ~
M+m 1 3
For small values of 0 ,
So the natural frequency is
B-3-27. T"he ki ne t i c energy T of the system is
and the potential energy U of the system is
2 u = 'j r R + % k p l - e2) + )i k2o2Z 1 1
Using the law of conservation of energy, we have
Noting that d ( T + ~ ) / d " , = 0, we obtain
Since 6 and i) are not i d e n t i c a l l y zero, we must have 1 2
B-3-28. A t t = 0 (the instant the mss M is released' to m e ) the kinet energy T and the potential energy U of the system are
I 1
The potential energy I s measured relative to the floor.
A t the ins tant mass m h i t s the fnmr the kinetic energy T and the plkential energy U of the system are 2
2
&re v2 is the velocity of the hanging mass m and G2 is the angle of rota- - kion of the pulley both at the instant the mass hits the f loor; re2 - v2,
2 and J = % mpr . Using the law of consenration of energy, we obtain
By substituting ret = v2 into this las t equation,
, = + M V ~ ~ + + I T T V ~ . . ) ~ ~ ~ V ~ 2 ~2
Solving this equation for v 2
E-3-29. The force F necessary to move t he weight is
F = nl9 = 10OO x 9-81 = 1962 pJ 5 5
The p o w e r P is given by
where tJ = m~fx. So we obtain
-3-30. From the figure shmm to the right, we obtain -
To keep the bar AB horizontal when prlling the weight ng, the rnament a b u t p i n t P must balance. Thus,
Solving this equation for x, we obtain
B-3-31.
where
Note that
Gear ratio = 1/30
Since the power TLWl 0 2 the motor is
the torque T2 of the driven shaft is
B-3-32. Assume that the stiffness of the shafts of the gear t r a i n is i n f i - nite, t ha t there is neither backTash nor elastic defamation, and t h a t the nt.mber of teeth on each gear is proportional to the radius of gear. Ikf i n e the angular displacement of slraf t 1 and shaft 2. as 0 and Q2, respectively.
I
By applying Newton's second law to 'the system, we obtain far the mtbr shaft (shaft 1)
where Tm is the torque developed by the motor and TI is the laad torque an
5@ar 1 due to the rest of the gear t r a in . For shaf t 2 , we have
where T2 is the torque transmitted to gear 2. Since the work done hy gear 1 is equal to that of gear 2,
Sin- e2/Q1 = n /n , Equation ( 2 ) can be written as 1 2
Substituting Equation (3) into ~quatlon (I), we get
The equivalent moment of inertia of the gear t r a in referred to the motor sha f t is
Notice that if the ra t io n /n is very ttntch smaller than unity, then the 1 2
effect of J2 an the equivalent mwnent of inertia J is negligible. eq
&-3-33. The equivalent mament of inertia J of mass m referred to the motor shaft axis can be obtained f rcm m
where oE is the angular acceleration of the motor shaft and 5 is the linear acceleration of mass rn. Since d r = 2, we have
The equivalent mcment of inertia Jb of the belt is obtained from
Since there is no slippage between the belt and the pulleys, the work: done by the belt and the right-side pulley ( ~ ~ 0 ~ ) and that by the belt and the
left-side pulley (T2Q2) must he equal, or
where T is the load torque on the motor shaft and T is the torque trans- 1 2
rnitted to the left-side pulley s h a f t , is the angular d i s p l a c m n t of the
motor sha f t , and G2 is the amy~lar displacement of the left-side pulley shaft . Since the two pulleys are of the same size, we have BI = B2. Hence
For the motor shaf t , we have
Also, for the left-side pulley shaft, we have
Since TI = 5' we ham
Since 0 = B2. 1
this Last equat ion becomes
The equivalent moment of inertia J of the system with re- to the mtoi shaft axis is eq
CHAPTER 4
So we obtain
B-4-3. figure 4-55 can be r e a w n as s h below-
From the right side diagram we obtain the follaring equations:
10 i + 20(i - i3) = E 1 1
20 i2 + 10(i2 + ig) = E
100 i, + 1O(i + i 1 = ZO(i - i31 2 3 I
whi& yield
So w e obtain
Solving for i2, ve get
Figure ( a ) Figure (b) Figure Ic)
Fran Figure (a) abwe we find R 1 = ZR2. Referring to Figure (b) above, we set
So R is obtained as 1
By substituting R1 = 2R2 into this l a s t equation, we obtain
!t%en, frm Figure ( c ) a m , current i is obtained as
B-4-5. When switch S is open, resismm between points A and B is l60R and we have
When switch S is closed, resistanm between points A and B is
So we haw
Cansequen tly ,
Solving for Rj we obtain
R = 2 5 . R
M-6. F m the circuit diagram we obtain - di ,
Each of the p r d i n g two sets of equations constitutes a rrratharatifal model for the circuit.
3-4-7. Fran Figure 4-55, we have for t > 0
Taking Lapface transfom of Equations (1) and ( 2 ) , we get
1 ( 5 ) * R2[%(s) - T(s)I ' 0 1 1
where i -'(o) is g i m by 2
t = 0
F m Equation ( 3 ) we have
'men Equation (4 ) can be rewritten as
Substituting Equation (5) into Equation (6) we obtain
Tbe inwrse Laplam transform of this last equation gives
Referr ing to Equation 15 1 WE have
B-4-8, A t steady stake (t < 0) we haw -
and
For t 2 0 the equation for the circuit is
By d i f f e r en t i a t i q this l a s t quation, we obtain
Th-e Laplace transform of this q u a t i o n is
Rl[al(s) - i ( O ) ] + L I(s) = 0 C
where
H e n c e
or
The inverse Laplace ttansfonn of I (s) gives
B-4-9. The equations for t h e system are
where Jeq is the equi~lent moment of inertia of the system referred to the motor rotor axis and is the equivalent viscous friction coefficient of the system referred
By taking Idplace transform of Equations (1) and ( 2 ) , w e obtain
Elimination of Eb(s) from the above two equations yields
(R, + L,s)I,(s) + l(bsB1(5) = E , ( s )
The Laplace transform of Equation (3) is
J shl(s) + b SB ( 5 ) = KIa(s) eq 1
Hence
By substituting Equation (5) into Equation ( 4 ) , we obtain
The numerical values of the equivalent moment of inertia Jq and equivalent viscous friction coefficient b are, respectively,
eq
Substituting these numerical values into m a t i o n (61, we get
Since B2/B1= N /N = n = 0.1, ue have the transfer function B~(s)/E,(s) as 1 2
fallows :
R-4-10. From the circuit shown to t h e sight, we obtain
Since
we have
XSO, since
we obtain
Equation (1) san be written as
Upla- transforming this equation and simplifying, we get
Lapla- transforming Equation ( 2 ) , we obtain
fran which we obtain
By substituting Equation ( 4 ) i n to Equation (31, we have
frm which the transfer function Eo(s)/Ei (s) can be obtained as
334-11. From the circuit diagram shown, we obtain
Since
we have
Also, since
i3 = 14
we get
Thus,
( 2 )
Substituting Equation (2) i n t o Tquation (1) and simplifying, w obtain
Hence the transfer function Eo(s)/?Zi(s) is given by
3-4-12. The transfer function E,(s)/E~(s) can given in tern of camplw impedances Z1 and Z2 as follows :
where
Hence
-14.
H e n c e
ZL = Ls, I - - - RCs + 1 z 2
- R
Hence
Note that
Similarly,
The transfer function Eo(s)/Ei(s) can be given by
B-4-17. Define the voltage at point A as ek. Then
1
Define the voltage at point B as eg.
Noting that
and K 3> 1, we must have
EA(s) = EB(s) Hence
frm which we obtain
8-4-18. The voltage at point A is
1 e = - (e + @Dl ~ 2 i
or
The voltage a t point B is
Since
Thus, equating Equations (1) and (2) we obtain
W-19. Define the displacenent of midpoint between kg and b as x3. Then the equations for the system are
Using the force-voltage analogy, t h e preceding equations may k CcmVerted to
Rtle last three equations can be modified to
di ,
1 dt t i , - i 2 ) d t + - (i l - i )dt = e(t)
3 C3
d i 2
L 2 ~ + R(i2 - i3) + - (i2 - il)dt = 0
R(ig - i2) = - (i, - i g ) d t
From these thee equations we ean obtain the analogous electrical system as shown to the right,
B-s-20. Define the cyclic current in the left loop as il and that in t h e r l g n t Loop as i2. Then the equations for the circuit are
which can be rewritten as
Using the force-voltage analogy, we can convert the last two equations as follows:
From these equations an analogous mechanical system can be obtained as shuwn to the r ight . / / / / i = ~ / / / / /
CHAPTER 5
B-5-1. By substituting the given numerical va lues i n t o
WE! obtain
R-5-2. We sha l l solve t h i s problem by using ttm different approaches: - one based on the exact method and the other based on the use of an average resistance.
(1) Solution by the exact method. For t he liquid-level system we ha-
2 3 Py substituting C = 2 m , Pi = 0.05 n /s, and p = 0.02 lfi into t h i s last equation, we obtain
2 k t fi = x. Then H = x and dII = 2x &, So we have
As- that at t = t t k level reaches 2.5 rn. Then, t is obtained as 1 1
*, = \ dk ( * - & 1.. = - Z M ) x l y + 1 W O Jfi dx
5 - Z x I
1 = - 200( 6 - 1) t moo '"(5 - 2x)l /
(21 Solution by use of an averaqe resistance. Since Q = 0.02 5, the average resistance I? is obtained from
~ e f i n e h = H - 1. lhenqi = a i - 0.02 and qo=h/R. For the liquid-level system,
Which can be rewritten as
2 3 substituting C = 2 m , R = 129 s/m2, and qi = 0.05 - 0.02 = 0.03 m /s i n t o this l a s t equation yields
Taking Laplace transforms of both sides of this l a s t equation, w e obtain
3. B 7 258[sH(s) - h(O)] + H(s) = - 5
or
Solving t h i s q u a t i o n for H ( s ) ,
The inverse laplace transform of this last equation gives
Assume that at t = t h ( t ) = 1.5. The value of tl can be determined frm 1' 1
Rewriting,
So we have
his solution has $een obtained by use of an average resistance.
B-5-3. The equations for the Piquid-level system are
C dh = (6 + q - 6 - q,)dt 1 1
Since Rl = hl/ql and R 2 = h2/q2, the system equations can be rewritten as
From Equations (1) and (2) we obtain
By differentiating Equation ( 2 ) with r e s w t to t, we get
d S 2 +---- - 1 dh2 - I R2 d t RL d t
By eliminating dh /dt from Equations (3 ) and (41, we obtain 1
dZh2 R C R C - &2
t (RICI + R C - + h =liZq 1 1 2 2 &2 2 2 dt 2
Substitution of h2 = RZq2 into this last equation y i e l d s
dZq2 R C R C - dq2 + R C ) - 1 1 2 2 dt2 +(RIC1 2 2 dt + q 2 = 4
Hen= the transfer function of the system when q is the input and q2 is the output is g i m by
E-5-4. The equations for the system are
Thus, we have
From Equation (1) we obtain
ClsHl(s) = 2 R1
[H2(s1 - H~(s) 1
From Equation (3) we get
By adding Equatfons (I), (21, and (31, and taking the Laplace transform of the resulting equation, we obtain
By substituting Equations ( 4 ) and ( 5 ) in to Ekqaticm (6), we get
Since H3(s) = A 3 f ~ 0 ( ~ ) , this last equation can tR written as
This is k ? ~ transfer function relating Q O ~ S ) and Q~(s) =
21-5-5. For t h i s system
Hence
(+=27c Q1 = -0.005Jii dt
Assume that .the head moves d m frm H = 2m to x for the 60 sewrid mod. Then
5: H'.~ dR = 4.005 -
which can be rewritten as
Taking logarithm of both sides of this last qua t ian , we obtain
2.5 log x = 1wI0 3.5089 10
x = 1.652 m
B-5-6. Frm Figure 5-32 we obtain
Using the e1ect;rical-liquid-1-1 analogy given below, e q a t i d n s for an analogous electrical system can be obtained.
C (capacitance) R (resistance)
Analogaus equations for the electrical system are
Based on Equations ( 5 ) through (81, we obtain the analogous electrical system shuwn below,
B-5-7. The equations for the liquid-level. system of Figuse 5-20 are
Using t he table of electrical-liquid-level analogy shown in the s~lutian of Problem B-5-6, we can obtain an analogous electrical system. The analogous electrical equations are
Based on Equations 15) through ( 8 ) " r= 03tzin the analogous electrical system shown on n e t t page.
B-5-8. - pv = mR,irT
In this problem
p = 7 X 205 + 1.0133 x 105 = 8.0133 x l o 5 ~ / m ~ abs
The mass m of the air in the tank is
If ttre temperature of canpressed air is raised to 4bec, then T = 273 + 40 = 313 K and the pressure p beccanes
= 7.547 x 105 ~ / m 2 gage = 7.695 kgf/m2 gage
= 109.4 1bf/in.* gage
B-5-9. mte tha t
where q is the flow sate through the ~ 1 ~ x 3 and is given by
Hence
from which we obtain
PJs) - - 1
Pi l s ) RCS + 1
For the bell- and spring, we haw the following equation:
Ap, = lot
The transfer function x(s)/Pi (s) is then given by
B-5-10. Note t ha t
5 2 5 2 P1 = 0.5 x 10 NJm gage = 1.5133 x 10 N/m abs
2 5 2 P2
= 0 N/m gage = 1.0133 x 10 N/m abs
If p2 3 0.528pl, the speed of air flaw is subsonic. So the flaw throughout the system is subsonic. The flow rate through the inlet wive is
The flow rate through the outlet valve is
Since bokh valves have identical flow characteristics, we have TT = K = R. The equation for the system is 1 2
A t steady state, ve have d p i d t = 0 and t h i s last equation k c o m s
5 2 5 2 = 1.2633 x 10 N/m abs = 0.25 x 10 N/m gage
B-5-11. For the toggle j o in t shown in Figure 5-37, w e have
Hence
Neglecting the Mgher-clrder terms, a linearized equatlcm for the system can $e written as
C
Q - f(E) = a(H - A) where
f(E1 = f ( 4 ) = 0,2
Thus, a linearized m a t i o n becames
A l inearized equatian for the system is
- z - z = a(x - 2)
where 2 = 2, E = 20, and
Thus, a linearized equation becomes
A linearized mathematical d e l is
- where Z = 11, y = 5, z = 356, and
Thus* the l inea r l zd equation is
5 Define the radius and angle of rotation of the pinion as s and 8, respectively, Then, relative displacement between rack C and pinion B is re.
Relative displacement between rack A and rack C is 2re and this must qua1 displacement x, Therefore, we have
Since x = r0 + y, we obtain
B-5-17. The heat balance equations for the system are
c ae = (U - ¶,)at 1 1 (1
C2dG2 = (q, - s,)dt (2)
Noting that
Equations (1) and (2) can be modified to
from which we get
By eliminating Bl(s) frcm the preceding two equations, we obtain
Thus the transfer function EbZ(s)f l l (s) can be given by
B-6-1. &fine the current in the circuit as i (t) where t 2 0. The q u a t i a n for t h e circuit for t 9 0 is
Since the capacitor is not charged for t < 0, the laplace transform of this equation becaes
Since
we obtain
The i m f s e Laplace transform or E,(o) gives
B-6-2. The equation for the circuit is -
Since q(0) = 0 and i10) = {(o) = 0, the Laplace transform of this la* -a- tion gives
E LS~Q(S) + RsQ(s) + -1 Q ( s ) = - C S
or
Since khe'current itt) is d q ( t ) /a t r we have
The current i(t) w i l l be osc i l l a toq if t he t w o roots of the characteristic equation
are cmmple~ mnjugate. If two roots are real, then the current is not osci- Ilatory.
Case 1 (Two roots of the characteristic equation are complex conjugate):
For this case' define
Then
The inverse Laplace transform of I (s ) gives
The current i(t) approaches zero as t approaches infinity.
Case 2 (Two rmts of the characteristic equation are r e a l ) :
For this case define
Then
E Its) = - 1 L (S + + b)
The inverse Laplace transform OF I(s) gives
N o t i e that
Hence
E 1 - = - L b - a
The current i ( t) can thus be given by
B-6-3, Referring to the circuit diagram - shown to the right, we have Z1
Z1 = R1 " 7 Z = R 4.- C2s
Hence
Eo(s 1 - - Z2 - - KP2s + 1
Ei(5) '1 + '2 ( E ~ + R ~ ) C ~ S + 1
Next, we shall f ind the response eo( t ) when the input e. (t ) is the unit step Z
function of magnitude E. . Since 1
the Inverse Laplace transform of E0(s) is
which gives t b response to the s t e p input of magnitude E i '
~ 6 - 4 , The system equations are -
which can be rewritten as
0 t
bly + kly = k x + blxo 1 i
Noting that x,(O-) = O and y(O-) = 0, by t ak ing the b; - transform of these two equations we obtain
By eliminating Y ( s ) from the preceding two equations, we get
Simplifying ,
or
?he response of the systm to xi(t) = Xil(t) can be obtained by taking the
inverse Laplace transform of
as fellows:
m. The equations of motion for the system are
k ( x . - xo) = b (& - y ) 1 1 2 0
Rewriting these equations,
Noting that x(0-) = 0 and also yl(0-1 = 0, 2-transforms of these t w o equations becane
Elintinatfng Y ( s ) from the l a s t two equations, we obtain
which can be simplified as
Hence
Since the i npu t xi(t) is given as
we have
Tne response Xo(s) is t hen obtained as
Since
we have
Hence
B-6-6. - First note that
Since R2 = 1.5 Rlr C = C , and R C = 1, we obtain 2 1 1 1
and the transfer function %(s)/Ei(s) becomes
Since the input; ei(t) is given by
= 0 elsewhere
we have
Hence, the response eo(t) can be obtained as follow:
Since
we obtain the respanse eoIt) as follows:
e (t) = Ei (0.4 e-2t - 0.4 e- (L/3)t + 1) 0
'to lqarithmic d e c r e n t = In - = - xa l I n - - 5C.d T
Xl Xn n
Solving for 5 ,
By substituting n = 4 and xo/xq = 4 into this last equation, we obtain
Noting tha t Wn =@ and 2 5% = b/m. w e find
and
E-6-8. The system equation is
(rn+ 2 ) G + & + ) ~ = p
Substituting the given ntm#rical val~es m = 20 kg and p = 29 = 2 x 9.81 N into this l a s t equation, w e obtain
2 2 2 + b j t + h = 2 ~ 9 . 8 1
At steady state
l a S S = 2 x 9.81
From Figure 6-48 (b)p x,, = 0.08 m. 'Thus
- 2 x 9-81 = 0.08 Xss - k '
Solving for k, we obtain
Since
=& =@= 3.34 r a d / s , b R 2 2 U n -3
we obtain
E-5-9. FOP the x direction, the equation of motion is -
For the rotational motion of t k pendulum,
Rewriting the preceding tm mationst
+ rng (cos 0 iZ + sin % 51 1 s i n 8 = - m2g 1 sin 0
Simplifying,
.* 9 .e x + c o s e b 3 - m2 sin 8 b2 + 2k x = O
ml + m2 ml + m2 "1 + m2
Thus the equations of motion for the system are
mz 1 0. ;; + c o s Q 0 - m21 s i n O i2 + 2k x = O + m2 "1 + m2 "1 + m2
=.a The equation of mtian for t\e s y s t e m is
w e rn = 1 kg, k = 100 ~ / m , p ( t ) = 10 &(t) N, x(0-1 = 0.1 m, and G(0-) = 1 m/s. By substituting the given n m ~ i c a l valves into the system equation, we obtain
Taking the a- transfarm of this l a s t equation gives
Solving for X(s) gives
The inverse Laplace transform of X ( s ) gives
sin 10t + 0.1 cos 10t x(t1 =lo
L e t us def ine another impulse force to stop the nmtion as A 8 (t - T), *ere A is the undetermined magnitude of the impulse force and t = T is the un- determined i n s t a n t that this impulse is to be given to the system. 'fhm,
the equation of motion for the system when the t k f o impulse forces are given is
The L- transform of this last equation gives
[ m s 2 + k ) ~ F s ) - 1 + A e-ST
Solving for X(s) ,
me inverse taplace transform of X(s) is
If the mation of mass m is to be stopped at t = T, then x ( t ) mst be iden- tically zero for t k T.
Note t h a t x ( t ) can be made ident i ca l ly zero for t > T if we choose
Thus, t k Xmti0n of mass m can be stopped by anokher impulse force, such as
1 2 The system equation is
2 + b;t = d( t ) , X(O-) = 0, G(o-I = o
The &- transform 05 this q u a t i o n is
(ms2 + bs)X(s) = 1
Salving for X(S)~ we get
me response x( t ) of the system is
The velacity G(t ) is
I e-(b/m)t $(t) = - m
The initial velocity can also be obtainl;d by use of the initial value thwrm.
JJ-6-13. '1Zlc nwrnenL u1 itjestla of t . 1 ~ wndulum about the pivot is J = m i 2 . The angle of rotation of the pendulum is 8 rad. Def ine the force that a c t s on the pendulum a t the time of sudden stop as F(t ) . Then, k h e torque that acts on the pendulum due to the force F(t) is F ( t ) K cos 8. The equation for the pendulum system can be given by
We shall linearize this nonlinear ajvaiirn 4 erruning F(f) -3 1 angle 9 is small. (Although @ = 20" is not quite small, the resulting linearized q u a t i m i will give an approximate solution.) By approximating cos Cl f 1 and s i n 0 f 8, Equation (1) can be written as
J p9
Since the velocity of the car at t = 0- is 10 m / s and the car stops in 0.3 s, the deceleration is 33.3 rn/s2,
By a s d n g a constant acceleration of magnitude 33.3 m/s2 to act on the mass for 0.3 seconds, F ( t ) may & g i m by
Then, Equation ( 2 ) may be written as
Since 1 = 0.05 m, this l a s t equation becmes
Taking Laplace transforms of bath sides of this last qua t ion , we obtain
b
where we used the i n i t i a l conditions t h a t 8(0-) = 0 and B(0-) = 0. Solving Equation ( 3 ) for @ ( s ) ,
The irrverse Laplace transform of Qls) gives
(4) Note tha t l(t - 0-3) = O for 13 ,< t < 0.3.
Let us assme that at t = tl, 0 = 2 0 8 = 0.3491 rad. Tbenbyten- t a t i w l y assuming tha t tl occurs before t = 0.3, we solm the following equation for tl:
which can 3x simplified to
The result is
Since tl = 0.0327 < 0.3 our aasrenptim was correct. The terms involving l(t - 0 - 3 ) inEquat ion ( 4 ) d o n o t af fect t h e v a l u e o f tl. It takes approximately 33 ms for the pendulum to swing ZOO.
B-6-14. The equation of motion for t h i s system is
By substituting the numerical values for M, b, k, and g i n to this equation, we obtain
By taking the Laplace transform of th i s l a s t equation assuming zero initial conditions, we obtain
The inverse Laplace transf o m of X I S ) gives
xdt) = 0-04905(l - e-1*6666t cos 5.5277t - 0.3015 e-l-666fit s i n 5.5277t)
N e x t , we shall obtain the response clurve x(t ) versus t with MATUB. Note that: X(s] can be written as
Now define
n m = [ Q 0 1.5351
R possible MATLAB program to obtain the response curve is given belcrw.
% ***** MATLAB program to solve Problem B-6-14 *****
num = [O O 1.6351; den = [ I 3.3333 33.33331; step(num,den) grid title('Respoase ~(t)') Idabel('t sec') ylabel('x(t)')
The resulting response curve x(t ) versus t is sham kIm.
R-6-15. The equation of motion f o r the system is
By substituting the numerical values of m, kl, k2, and bl i n t o this equation, we obtain
Laplace transforming this equation, we get
Solving this equation for X(s),
X(s ) = sx(0) + $10) + 4x(O)
s2 + 4s 4 16
Since x(0) = 6-05 and g(0) = 1, X(s) becanes
The inverse Laplace transform of X ( s ) gives
f- x(t) = 0.05 e-2t m s 2 6 t + 0.3175 e-2t sin 2. 3 t
This equation gives the time xespQme x(t) . The response mme x(t1 versus t can be obtained easily by use of
NATLAE. Noting that X(s) can be written as
X(s) = 0.05~2 + 1.2s 2 s 2 + 4 s + 1 6 s
we may define
den = [I 4 163
and use a step camand. "ke following MRTLAB program will generate the reqmnse x(t) versus t as s h m below.
% ***** PvlATltA3 pro_nfam to solve Problem B-6-15 *****
mrn = L0.05 I .2 01; den=[I 4 161; st ep(num,den) ~d tit1 e(lriesponse x(t)3 xlabelrt secq> ylabel(k(t)')
B-6-17. In this system F is the input and x2 is the output. From Figure 6-57, we obtain the following equations:
Laplace transforming these tm -atimsr assuming zero i n i t i a l conditions, we obtain
By eliminating XI(s3 f r a these two equtltionsr we get
simplifying this last equation, we get
from which xe obtain
By substituting numerical ~ l u e s for ka, k2, blr and b2 into this l a s t equation, we obtain
Since the input F is a step formof ZM, we have ~ ( s ) = 21s. x2(s) can be obtained from
The inverse Laplace transform of X ~ ( S ) gives
The response curve x2(t ) versus t can be obtained with MATLAB as follows: F i r s t note tha t
Then, define
and use a step comnand. The following MATLAB program will yield the response curve x 2 ( t ) - The resulting response curve is shown in the figure below.
% ***** MATLAB program to solve Problem B-6-17 *****
num = [O 0 0.83; den=[l 6.4 81, step(num,den) grid titl$Response x( t )3 xlabel('t sect) ylabel('x(t)')
B-5-18. Referring to the figure showr to the right, we ham
1 = - J - + C 2 5 , Z2(s) = Rl + - I
Zl(d R2 C l S
H e n c e R2 Zr(s1 = Z ~ ( S ) = RrCls + 1
R ~ C ~ S + 1 CIS
Thus
By rmbstituting the given nrrmerical wlues for Rl, R 2 , C1, and C2, we obtain
men ei (t 1 = 5 V ( S*p input )l is applied to the system, we have
The reqmnse curve e,(t) versus t can be obtained by entering the follmj-lng MA'lTAEl program i n t o the mqmter.
The resulting response curve eo(t) versus t is shown below.
R e m e BdU
I Sec
Note that
Notice that the response curve is a sum of two exponential curves and a step function of magnitude 5.
CHAPTER 7
B-7-1. The equation of mtion for the system is
whesex1(0) = 0 and G(OE = O . The laplace transform of this equation is
By substituting the given numefical values of k, P, and c3 into this las t equation, we obtain
Solving for X(s) ,
X ( s > = 10 - 5 - - 2 - - 1 10 (s2 + 100)(s2 + 4) 96 s2 + 4 s2 + loo
The inverse Laplace tsansfarm of X(s) gives the response x ( t ) -
5 ' sin I&) x(t) = (sin 2t - - 5
B-7-2. The equation of motion for the system is -
By substituting the *ven nwrtlerlcal va3ues into this equation, we get
+ 24;; + 2 0 0 ~ = 5 sin 6t or
+ 123; + fOOx = 2.5 s i n 6t
73king the Laplace transform aE th is last equation, we o b i n
2 6 (S t 12s + 100)X(s) = 2.5 --
s - 1 - 6
Solving for ~ ( s ) ,
The i m s e laplace transfarm of X ( s ) gives
3 eat txe 8t + sin 6t - m s 6t x(t) = - e4t s i n 8t + - 1856 464
1 EJS) -
C s - - - 1 1 RCs + 1 E i ( d R + - Cs
Hence
For the input e. (t) = E sin L3 t, the steady-state output e (t) is g i m by 1 i 0
*-4. The equations of motion for the system are
which can be m i t t e n as I
Since we are interested in the steady state behavior of the system, we can assume that all initial conditions are zero. Thus, by assuming zero initial conditions and taking the Izlpkace transforms of the last two equations, we obtain
(m2s2 + bs + K)x*(s) = (h + k)x1(s)
Ran Equation (2) we have
Substituting Equation (3) into Equation (1) and simplifying, we obtain
The transfer function Gl(s) between X (3) and P(s) can thus b obtained as 1
H e n c e
fram Mch we obtain
The steady-state output i ( t ) a n , therefore, be given by
xl(t) = G ( j w ) P s i n dt + G ( j W ) 1 1 1 [ Ll
NgXt, referring to Equation 131 we have the transfer function G (s) between X ( 8 ) and P ( s l as fallows: 2
2
H e n c e
The magnitude and angle of G ( j W ) are given by 2
The steady-state output x (t) can be given by 2
xz(t) = I ~ , ( j w ) I P sin kt +&I
B-7-5. 2 2 tension = mu s = 0.1 X 6.28 x 1 = 3.94 M
Tne tension in the card is 3.94 N. 'rhe maximum angular speed can be ob- tained by salving the f o l l d n g q u a t ion for # .
?he maxfrmrm angular speed that can bE? attained without breaking the card is 1.59 Hz-
73-74. F m the d i a g m shown klokr , we obtain
centrifuqal force , nUJ2r - 0.15 gravitational force m9 0.2598 h
Solving for W , we obtain
B-7-7. - The equations of m t i o n for the system are
MZ + & + m = m t d r s i n u t :
Tf 10 % of the excitation form is to 1s transmitted to the foundation, the tranmnissibility must be qua1 to 0.1. Thus
Since 5 is desired to be 0.2, we substitute J = 0.2 into this last equation*
Solving for ' , we find I
86
Nating tha t & > O , velnuat have!* = 22.28. men
3 k = 17.7 x 10 N/m
The amplitude of force F transmitted to the foundation is t
Rewriting
By substituting x - y = z into this last equationr we obtain .. W
m z + & + k z = - m y
The Zaplace transform of this equation, assuming zero initial conditians, is
Far the sinusoidal input y = Y sin&) t r
The steady-state amplitude ratio of z to y is
Thus, the amplttude of sinusoidal dinplacetnent y of the base is equal to the amplitude of the relative displacemer~t z.
If w<< W , we have n
So the acceleration of the base is ptopartional. to z.
B-7-9. - Define the displacement of spring k as y. 'Then the eqmations of motion for the system are 2
The f o r e f(t) transmitted to the foundation is
By taking laplace transforrrrs of Equations (1) and (21, assuming zero ink t i a l conditions, WF obtain
By eliminating YCs] fran the last two equations amd simplifying, w e get
The Laplace transform af Equation (3 5 is
So w e have
and
The farce transmissibility TR is
The amplitude of the force transmitted to the foundation is
B-7-10. Define the d i ~ p l a - ~ k o f the t o p e n d of springk as z. Then the equations of mtim for the system ape 2
where p = P s inwt, Rewriting these equatfom,
Laplace transforming these two qtatims, assuming zero i n i t i a l m d i t i o n s , we obtain
Eliminating Z(s) fran the last tm equations and simplifying gives
So the transfer function ~(s)/P(s) is obtained as
WE motion transmissibility TR is
?RIP vibration amplitude I X( j W ) I of '.he machine is
B-7-11. The equations of matian for the system are
d + & + k x + ~ ( x - y ) = p = ~ s i n ~ t
Laplace tsansfoming these t w o equations, assuming zero i n i t i a l wnditicms, we obtain
Eliminating Y ( s ) E m the Last t w o equations and simplifying, we obtain
Note t b t if/-=#, then X ( j W ) = 0. Since
we have
2 By substituting %/m, = U Into this last equation, ve get
Hence
I p ( Y ( J W ) I - =-- p
ma LrlL ka ka "a-
m,
The amplitude of dbration 05 mass m, is P&.
8-7-12. Assuming ma11 angles Q1 and e2 the equations of nation for the system rriay be obtained as fo Llm:
Rewriting these equations, we obtain
which can be simplified to
To find the natural frequencies of the free yibntionr we assume the motion to be harmonic. That is, we assume
= A sin W t , €I2 = B s i n W t
u 0-
1 = -A& s i n t, e2 = -B&* s i n w t
Substituting the preceding expressions i n t o Equations (1) and (2), w e obtain
Since these equations must be satisfied at all times and s i n d t cannot be zero a t a l l times, the quantities in the brackets must be equal to zero. Thus
For constants A and B to be nonzero, ", determinant of the coefficients of Equations (3) and (4 ) must be equal t o zero, or
This determinant equation determines the natural frequencies of the system- Equation (5 ). can be rewritten as
This last equation can be factored as follows:
Thus
The first natural frequency is w l (first W e ) and the second natural fre- quency is L3 ( second mode) .
2 A t the first natura l freqwncy W =J=, we obtain from Equation
(3) the following expression:
kit2 -
[ ~ o t e that we obtain the same .result from Equation ( 4 ) . ] Thus, at the first nrode the amplitude ratio A D k a m e s unity, or A = B. This means that both masses move the same munt in the same direction. This mode is depicted in Figure ( a ) below.
A t the second natural frequency & = LJ2 we obtain from Equation (3)
ka - A - m - - m l l - 2
2 - - -
E3 ka* ka2 - 9 - + k a L L - - - - m 1 I mlt2 P m 1 t 2 m 2 1 2
[we obtain the same result fran Equation (4). 1 At the second mode, the amplitude ratio A D becomes - m /m or A = - (m2/ml)~. mis means that 2 1 masses mave in the opposite direction. This mode is depicted in Figure (b) shown b e l o w .
3-7-13. T h e ~ a t i o n s c E m t i m f o r thesystemare *.
mx = - (x + f l s i n O)kl - (x -I2 sin €Ilk2
For small angle B, we have sin 8 + 8 and cos +3 1 and the preceding two mat ions becnne
Notice that if flkl = [2k2t then the coupling terms become zero and ~I?Q
equations become independent. However, in this problem
Therefore, mupling exists between Equation (11 and Equation (2). To f i n d the natural frequencies for the system, assume the following
harmonic motion:
x = A s h u t r B =: T3 s i n W t
Then, frm Equations [I) and (2) we obtain
For amp1itudes A and 3 to be nonzero, the determinant of the coefficients of Equations 13) and (4) must be equal to zero, or
Thfs determinant equation determines t h e na tura l frequencies of the system. Equation (5) can ke rewritten as
which can be simplified to
Notice that this last: equation determines the natural frequencies of the system. By substituting the given numerical values f o r l l , 1 2 , kl, k2, m, and J into Equation (6), we ol~tain
2 a4 - 1 4 0 W + 3920 = O
Solving this qua t ion for W *, we get
H e n c e
The first natural frequency is Cdl = 6.2205 rad/s and the second natural frequency is u = 10.065 rad/s.
2 To determine the modes of vibration, notice tha t from Equations (3) and
(4) have
By substituting the given numerical values into this last equation, we obtain
For the first mode of vibration ( &J = 6.2205 rad/s) the amplitude ratio A h becomes as follows: 1
Notice that the ratio of the displacements of springs k and k are 1 2
'Ihe f i r s t mcde of vibration is sham in Figure (a ) on next page.
Figure (a)
For the second mode of vibration ( W = 10.065 rad/s) the ampl&ude ratio A/B k c m e s as 2
Hence the ratio of the displacements o= springs k and k becomes 1 2
The secand mode of vibration is sharn in Figure (b) below.
Figure (b)
B-7-14. The system shown in ? i p e 7-50 is a p i a l case of the system shown in Figure 7-32 (Problem A-7-14). By defining
k l = k , k2=2k, k 3 = k r m l = m , m 2 = m
Equations (740) and (7-41) become as follows:
Also, that satisfies Equations (7-38) and (7-39) bcmes as follows:
X)ef ine
By substituting Ld12 into Equation (I), we obtain
Similarly, by substituting uZ2 into Equation (13, get
[We get the sarrte result if we substitute dl2 or aZZ i n to Equation ( 2 ) , ] Hence we have
ses m he S a m n ( w i t 1 e same
t ion by ency w . Fig
the s: 2) r tht lures ( i
u n e m 2 msse 3 ) and
which means that: in the first mode of vibration (with frequenq dl), the ma5 E in t ? d i m In the second mod .brat io I E r e q in opposite d i r ; by th amount l u ) s r i m on next page deplct the first mode of vibration and second mode of vibration, respect- ively.
Figure ( a )
B-7-15. The equations of motion for the system are
Substituting the given numerical mlues into these two equations and simplifying, we have
To find the natural frequencies of the free vibration, assume that the motion is harmonic, or
Then .* .I
x = - A ~ * s i n a I t , y = - B w 2 s i n d t
If the preceding expressions are substituted into Equations (1) and (2) , we obtain
(-A a2 + II.A - B) sin = o
Since these two equations must be satisfied a t all times and s i n u t cannot be zero a t all times, w e must hsve
Rearranging,
For constants A and B to 'b? nonzero, the determinant of the coefficient matrix must be equal t;o zero, or
which yields
which can be rewritten as
( t~ 2 - 7.2985)C d2 - 13.7016) = 0
Hence r
The frequency of the first mode is 2.7016 rad/s and the frequency of the second mode is 3.7016 rad/s.
Frun Equations ( 3 ) and (4!, we obtain
By substituting uI2 = 7.2985 i n t o A D , we obtain
Similarly, by substftuting w 2Z = 13.7016 into A D , we get
Hence, at the first m o d e of vibration, two masses rnm in the same direction, while a t the second mode of vibration, two masses move i n opposite d i m - tions.
Next, we shaU obtain the vibrations x ( t ) and y(t) subjected to the given initial conditions. Laplace transforming Equations (1) and ( 2 ) ,
Substituting the given initial condition3 into the preceeding t w o equations we get
~liminating Y(S) frm Equations ( 5 ) and ( 6 ) ,
which can be simplified to
Similarly, we can obtain Y ( s ) as follows:
To obtain the responses x(t) and y ( t ) to the given initial conditions, we rewrite X(s) and Y ( s ) as follows:
Possible MATLAB programs to plot x( t ) and y( t ) , respectively, are given next. The resulting plots x( t 1 versus t and y ( t ) versus t are shown on next page.
% ***** MATLAB pro_mrn to obtain vibration x(t) *****
num = [0.05 0 0.5 0 01. den=[ l 0 21 0 1001; t = 0:0.05:30; x = step(num,den,t); plot(t,x) grid title('Vibration x(t) due to initial conditionds') xlabel('t sec') ylabd('x(t)')
- ****#I M A n A B progranl to obtain vibration y(t) *I*** ,
num = [0 0 0.5 0 01; den=[l 0 21 0 1001; t = 0:0.05:30; y = step(nurn,den,t); I
pIot(t,yl grjd title('Vibration y(t) dare to initial conditionds') xlabel('t sec') ylabel(ly(t)') t -
B-7-16. All necessary derivations of quatians for the system are given in Problem A-7-16. The e q u a t i ~ n s for the system are
Referring to Equation 17-54) we have
Case ( a ) : For the initial condition..;
x(0) = 0.2807, G(o) - O I y ( O ) = I, ~ C O ) = 0
Equation (33 becomes as fellows:
By substituting Equation ( 4 ) into Equation (23 and solving for ~ ( s ) , W obtain
S u b t i t u t i n g y ( 0 ) = 1 into the l a s t eya t ion and simplifying, we get
To obtain plots of x ( t ) versus t and y ( S ) versus t, w e may enter the following M4TLAE program into the m p u t e s . The resulting plots are shown in Figure (a) .
Case (b): For the i n i t i a l conditions
x(0) = 1.7808, G(03 = 0 , y ( 0 ) = -1, ;(o) = 0
% ***** MATLAB propo l to obtain x(t) and y(t), case (a) *****
numl = [0.2807 0 01; n u d = [ I 0 01; d e n = [ l 0 7.19221; step(num I ,den) hold Current plot held step(num2,den) text(2,-0.5,'x(t)') trn(3,0.3>'flt)') title(Responses x(t) and y ( t ] ~ due to initial conditions (a)') xlabel (It sec') ylabel (k(t) and y(t33
we obtain the following expressions for ~ ( s ) and ~(s):
I
k MATMI3 pragram for obtaining plots of x( t ) versus t and y(t) v;ersus t is 5- below. The resulting plots are shown in Figure (b) below.
*A **** * MATLAB program to obtain x(t) and fit), case (b) * * * **
numf =[1.7808 0 01; m = [ - l 0 01; den=[] 0 27.8081; step(num1 ,den) hold Current plot held step(num2, den) texf(l.7,l. S,"x(t)') text(3 -5,-1 .Slfi(t)? title(Responses x(t) and fit) due to initial conditions (b)") xlabel ('t sec') ylabel ('x(t) and fit)')
Case (c]: For the i n i t i a l conditions I
~ { o r = 0.5, ;f(o) = O, y ( o ) = - a s , + ( o ) = o
WE obtain the followhq expressions for XIS] and Y(s}:
A MATLAB program to obtain plots of x ( t ) versus t and y(t) versus t is s h u n below. The resulting plots are shown in f iqure (c] .
% ***** MATLAB progranr to obtain x(t) and fit), case (c) *****
nmml = [0.5 0 2.5 0 01 num2 = 1-0.5 O -7.5 0 Q]; den=[] 0 35 0 2003; t = 0:0.02:5; x = stq(mm 1 ,den,t); pfot(t,s'el) hold Current plot held y = step[nuM,den,t); P ~ o ~ ~ ~ , Y * ~ ~ ' ) text(1.6,0.5,'x(t)') text(l.1,-0.3,'y(t)3 title('Responses x(l) and y(t) due to initial conditions (c)? xlabel ('E sm') yfabel ('x(t) and y(t)')
Rsswmes: x(l) and dl) due to hRlal oondinions (c)
1
CHAPTER 8
B-8-1. Simplified b l w k diagrms - are s h m to the right, The transfer function ~(s)/R(s) is
B-8-2. Simplified block diagrams for the system are shown $elm.
The transfer function ~ ( s ) h ( s ) is
B-8-3, Define the i n p u t impedance and feedback impedance as Z and Z respxtively, as shown in the figure bel,>w. 1 2
Then
Since e l 0, Laplace transKorms af voltages e.(t) and e(t) are obtained as 1
E ~ ( s ) = ZII(s) = R I(s) 3.
Hence
USO r
Therefore,
The Mntrol acbion is propartianal plus integral.
-- -
108
B-8-5. Let us define d i s p l a m n t s e, x , and y as shown in the figure klow .
Fur relatively small angle 8 we can construct a block diagram as s h m below.
F r a the block diagram we obtain t h e transfer function Y(s)/B(s) as folloucts:
Since in such a systm 1 ~ a / [ o (a + b) 1 I is designed to be very large -red to I, ~(s)/B1sI may be simplified to
We see that the piston displa-nt y is proportional to d e f l e i o n angle 9 of the control lever. Also, from the system diagram we see t h a t fox each m L l Mlue of y , there is a corresponding value of angle dm Therefore, for each -12 angle e of the control lever, there is a mrresponding steady-state elevator angle 8.
B-8-6 If the engine spsed increases, the sleeve of the fly-ball -mar moves upward. This movement acts as the input to the hydraulic tontroller. A positive error signal (upward motion of the sleeve) causes the power p i s t a to move dawnward, reduces the fuel mlve apeningc and decreases the engine speed. Referring to Figure (a) shorn below, a block diagram for the system can be dram as shown i n figure (b) on n e e page.
Figure (a)
Figure Cb)
From Figure (b3 the transfer function Y(s)/E(s) is obtafned as
K - 8
al + a2 R al I + - bs s aI t a2 bs + k
Since such a speed mtruller is usually designed such that
the transfer f u n c t i o n ~ ( s ) @ ( s ) kcomes
Thus, the control action of this s p d controller is proportional-plus-intq- -1.
E-8-7. For the first-order system
the step response curve is an q n e n t i a l 50 the the constant T can k determined from such an exponential m ily. Erom Figure 8-99 the time cons tan t T is 2 s .
If this thermometer is placed in a bath, the temperature of f i c h is increasing at a rate of 10°C/rnin = 1/6 "C/s, or
where a is a constant, then the steady-state error can be determined as follows? Noting that
we obtain
where
Theref ere,
Thus, the steady-state error is 1/3*~. For a second-order system:
A typical response c u m , when this thermwneter is p l a d in a bath held at a constant temperature, is shown below.
B-8-B. The closed-loop transfer function of the system is
For the unit-step input, w e have
B-8-9. - Since M is specifimf as 0.05, we haw P
or
Rewriting,
Solving Ear the damping ratio 5 w e obtain
The s e t t l i n g time ts is specified as 2 seconds. So we have
or
Iheref ore,
-8-10. The closed-loop transfer f u n e i o n of the system is
c(s1 - - loo R l s ) s3 + 2s2 + 10s + 100
This system is unstable m u s e two m l e x pte closgd-loop poles are in the right half plane. To vi: tstable r e spnse , WE may enter the folrmring MATLAB program il .er . The resulting unstable response olrve is shown in L~IC r LYU e: UIL next page.
To make the system stable, it is necessary to reduce the gain o f the system or add an appropriate -nsator.
malize It0 the .h- .I=<-
-con juq the ur
cDmPut am-- --
num = [0 O 0 1001; den=[ l 2 10 lC101; t - 0:o. 1 5 ; s tv Inun ldqt ) €?id t i t l m i t - S t e p Response of Unstable System') AabelCt sec') yla bel('c(t )')
From the chara&eristic polynomial, we find
The rise t h w tr is obtained frm
t, = 7 t : - B
W d Since
Wa =w,JG= 4 JCTZ- 3.46
we have
?he peak time $ is obtained as
The maximum overshoot Mp is
The settling time t, is
B-8-12. The closed-loop transfer fun~tlon for the system is
From this quation, we obts in
Since the damping ratio 7 is specified as 0 .5 , we get
Th;lref ore, we have
0.5(1 + KKh) = JG The settling tine is specified as
Since the feedfomard transfer function G I s ) is
K
G I s ) = 2s + .L I 3t - = 1 - 1 + 3- S 2 8 + f , + m h S
2s - 1 the s ta t ic velocity error constalt Kv is
This value must be equal to or greater than 50, Hentx,
Thus, the-conditions to be satisfied can be surmaarized as follows:
O < R h < l
From W a t i a n s (1) and (2), wz gwk
Frm Equation (3 ) we obtain
If we choose K = 5000, then we get
Thus, we determined a set of values of K and Kh as follows:
R = 5000, Kh = 0.0198
With these values of K and Kh, a l l specifications are satisfied.
----------
Since R ( s ) = l/s2, the output C ( s ) is obtained as
Since the system is underdampecl, C(s) can be written as
The i nve r se Laplace transform of C(s) gives
The steady-state error e,, for s unit ramp input is
= l i m -
?he steady-state e r r o r can also 6e obbined by use of the f ina l value theorem. Since t h e error signal ~ ( s ) is
we obtain the steady-state error eSS as
B-8-14. The characteristic equation is
The Routh array for this equation is
3 5
6 R
For the systm to be stable, there should be no sign changes in the first column. This requires
30 - H > 0 , K > o
Hence, we get the mge of gain K fo r stability to be
3 O > R > O
8-8-15. Since the system is of higher order (5th order), it is easier to find the range of gain K fo r stability by first plotting the r m t laci and then finding critical p i n t s (for stability) on the soot loci. The open-loop transfer function G(s) can be written as
The MATLAB program given on next page wi l l generate a plot of the root loci for the system. The resulting root-locus plat is s h m also on n e page.
num= [O 0 Q I 2 43; den=[ ! 11.4 39 43.6 24 01; 1 rIocus(num, den)
Waning: Divide by zero v = [-8 2 -5 51; axis(v); (wis('squareq) gtid title(Root-Locus Plot (Problem B-8-15]?
Based on this plot, it can. be sem that the system is cbnditianally &able. Pill critical points for stability lie on the j# axis.
To deternine the crossing p i n t s of the r m t laci with the jd axis, substitute s = ju3 i n t o the characteristic equation w h i c h is
Then
(ju)5 + 11.4(ju)ld + 39(ju1)3 + (43.6 * ~) ( j ld )* + I24 + 2R)jkJ
This equation can be rewritten as
By equating the real part; and imaginary part equal to zero, respectively, we obtain
Equation (2) can be written as
From Equation (3) WE obtain
By substituting Equation ( 4 ) into Equat.ion (11, we get
which can IE simplified to
The roots of this l a s t q u a t i o n can be easily obtained by use of the Ft4ZZAB prcqram given below.
The rmt-locus branch in the upper half plane that goes to i n f i n i t y crosses the jul axis at MJ = 1.2130, icl = 2.1509, a n d w = 3.7553. The gain values at these crossing p i n t s are obtainel a s follows:
-1.2130~ + 39 x 1.2130~ - 2% = 15-61 K = for hJ = 1.2130 2
R = -2.1509~ + 39 X 2.1509~ - 24 = 67-51 for OJ = 2.1509 2
K = -3 . ~ 5 5 3 ~ + 39 x 3 - 7 5 ~ 3 ~ - 24 , 163.56 for W = 3.7553 2
Based or1 the K values above, we obtain t h e range of gain K for s t a b i l i t y as follows: me system is stable if
13-8-16. A MATLPlB program to plot the root loci and asymptotes for the following system
is given blow and the result5ng root-locus plot is shorn on next page-
N o t e that the equation for the asymptotes is
num=[O 0 0 0 I]; d e n = [ l 1.1 10.3 5 01; numa=[O 0 0 0 I]; I
dena = [ I 2.1 0.4538 0.083 19 0.00571 93; r = rlocus(num,den); plot(rsl-'1 hold Current plot 'held plot(r,'ol) rlocus(nurna,ctena);
v' [-5 5 -5 51; axis(v); axis('square7 titlerPlot ofRoot Loci and Asymptotes')
B-8-17. The open-loop transfer function G(s) is
The fallawing M A T L W pragram will generate a rmt-lms plot. The resulting plot is shown on n e page.
num=[O 0 0 I 11; den=[l 7 10 0 01; rlocus(num, den) v = 1-6 4 -5 51; axislv); &sCsquare') grid titlflmt-Locus Plot (Problem B-8-1713
Frm the plot we find that the critical Mlue of gain R for stability corresponds to the crossing pint of t;le rwt I m s branch that goes to infinity and the imaginary axis. Hence, we first f ind the crossing frquency and then find the mrrespding gain mlue .
The characteristic equation for t h i s systent is
By substituting s = ju) into the characteristic equation, we obtain
( j d 1 4 + 3 ( j # 1 ~ + 1 0 ( j d ) 2 + Z ~ ( j u ) + Z = O
which can be rewritten as
( d 4 - 10&12 3. 2K) 4 j U (-7&J2 + 2 ~ ) = 0
By equating the real part and imaginary part of this last equation to zero, respectively, we get
u4 - L O ~ J ~ + Z K = 0 C I. )
Equation ( 2 ) can be rewritten as
By suketitutimg Equation ( 2 ) in to Equation (11, we find
t ~ 4 - lo& + 7 k 1 2 = 0
which yields
Since W = 5 is the crossing frequency vith the j W ads, by substituting w = E i n t o Equation (3 ) we obtain the critical value of gain K for stability .
Hence tfie stability range for K is
R-8-18. The angle deficiency is
180* - 126' - 120° = - 6oe
A lead compensator can contribute 60*. L e t us chwse the zero of the l e a d compensator at s = -1. Then, to obtain phase Lead angle of 60 *, the pole of the cmpensator m u s t he locate3 at s = -4. Thus,
The gain K can be determined f ram t h e nlagnitude condition.
A m the lead rompensator beromes as f=rllows:
The f ed forward transfer function is
?"he following MATLAB program w i l l generate a root-locus plot. The resulting plot is shown on next page.
C
num = 10 0 8 81; den=[! 4 0 01; rloms(num,den) v = [-6 2 4 41; axis(v) axis('squarel) grid titlc('Root-Locus Plot (Problem B-8-18)')
h
4 - 5 4 3 - 2 - a 1 2 Real A*if
Note that the closed-loop trznsfer function is
The closed-loop poles are located at s = -1 2 j 5 and s = -2.
B-8-19. The MATUlB program a i m below generates a root-locus p lo t for the given system. The :ing plot is shorn on next page.
nutn = 60 0 O I]; den=[[ 5 4 01;
, rlocus(t~urn,den) v = [-6 4 -5 51; axislv); axis("square") grid tit le('Root-Locus Plot (Problem B-8-T9)y
Note t h a t constant-7 points (0 < 5<1) 1 straight line having angle 8 frm the j W axis as shown in the figure k l a w , From the figure we obtain . -
sin 8 = - Q n
Note also that = 0.6 line can be defined by
s = -0.7% + ja
where a is a variable (0 < a ( oo ) . To find the value of K such that the damping xatio J of the dominant closed-lcxn, mles is 0.6 .can b found by finding the intersection of the line s = -0.7Sa + j a and root locus, The intersection point can be determined by solving the following sim- taneaus equations far a.
By substituting Equa t ion (1) into Equation (21,
which can be rewritten as
~y equating t he real part and imaginary part of t h i s l a s t equation to zero, respctively, we obtain
Equation (43 can bs rewritten as
which can b written as
Of
Hence
From Equation (3 ) we f i n d
K = -1.8281a3 + 2. 1875a2 4 3a = 2.0535 for a = 0.5623
K = -1.8281a3 + 2.1875a2 + 3a = -1759.74 for a = 10.3468 . -
S i n ~ the K value is positive far a = 0.5623 and negative for a = 10.3468, - .
we choose a = 0.5623. The requixd gain R is 2,0535.
Since the characteristic equation with K = 2.0535 is
the closed-loop poles can be obtained by use of the following MATLdB prqram .
Thus, the closed-Imp poles are located at
The unit-step response of the systm with K = 2.0535 can be obtained by entering the following MRTLAE program into the caputer. The resulting unit-step response curve is s h m an next page.
- -
m m = [O 0 0 2.0535]; den= [ I 5 4 2.05351; step(num,den) &rid ticEe(Vnit-Step Response (Problem B-8-19]? xlabelrt Sect) ylabeE(Uu0utput9
,
B-8-20. The closed-loop transfer funct,ion of the system is
- - ~ ( s 2 c as + bs + ab)
s3 + s + ~ ( s 2 + as 4 bs + abl
Since the dominant closed-1- pales are located at s = -1 f jl, the characteristic equation must he divisib1.e by
Hence
where s = - W is the unknown t h i r d pole. Ey dividing the left side 05 this last equation by s2 + 2s + 2, we obtain
53 + K S ~ + (1 + aK + bR)s 4 abK = (s2 4 2s + 2 ) l s + K - 2 )
+ (aK + bK - 2R + 3)s + Kab - Z ( K - 2 )
?he remainder of divis ion must be zero. Hence we set
Kab - 2 ( K - 23 = 0
Since a is specified as 0 = 5 r hy s u k t i t ~ t i n g a = 0.5 into these two equations, we obtain
By substituting Equation (1) in to Equation (21, we haw
Then, by substituting K = 2 i n t o Eguation I l ) r we get
2 b = 1 . 5 X 2 - 3 - 0
The PFD controller with K = 2 and b = 0 kcoms
Thus, the controller beccxnes a P'D controller. The open-loop transfer funct ion bxanes
The closed-loop transfer fmction (with b = 0 and K = 2) becwnes as follows:
The root-lacus plat for the designed system can be obtained by enterring the following MATLAB program :.nto the computer.
num = 10 1 0.52; den= [ l 0 11; r!ocus(num,den) Y = [-2 1 -1.5 1-51; axistv); axis('sguare') &rjd title(Root-locus Plot (Problem B-8-20]?
- The resulting root-lmus plot i s shown on next page.
129
CHAPTER 9
8-9-3. A Bode diagram of the PI cont!t-oiler is shown in Figure (a ) . -
A Bode diagram of the ml controller is shown in Figure (b).
F i g u r e (b)
5 9 - 5 ,
Lead ne.t:wor-k
Lag network
B-9-6, The equation of motion for the q s k m is
'Ihe A_ transf o m of this equation, using zero i n i t i a l conditions, givfs
Hence
Notice that th i s system is a differentfating system. For the unit-step input X ( s ) = l/s, the output 8 ( s ) becomes
The inverse Laplace transform of 01s ) gives
1 ,-(k/bSt e ( t ) = - R
Note that if the mlue of l r b is large, the response B( t ) approaches a Fhlle signal as s h m in Figure ( a 3 b l o w -
Since
' W
we o m i n ""'I - 1
/ G ( j d ) = 9 0 O - tan-' Wb k
Figure (a )
The 6teady-state output es,(t) is therefore given by
1 e It] = - d X -1 W b SS
sin( Crl t + 90'- tan -) k
Next, substituting [ = 0.1 m, k = 2 ~ / m , and b = 0-2 N-si'm into G( jLJ ) gives
A Bode diagram of G ( j t d ) is shown below.
Bode Dlaqmn (Problem 0.96)
B-9-7. Noting that
we have
B-9-8. A possible MATLAB progmn for obtaining a Bode diagram of the given G C S 1 is s h m on next page. The resul t ing Bode diagram is ahown also art next page.
- num=[O 0 0 320 6401; den=[l 9 72 64 01; w = logspace(-2,3,IOO); bode(nupden,w) subplot(2,1,1); title('Rode D i a w (Problem B-9-8)')
3-9-9, A possible MBTLF_B program to obtain a Bode diagram of the given ~ ( s ) is sham below. The resul t ing Bode diagram is shown on next page.
num=[O 20 20 lo]; den=[l 11 10 03; w = lo_espace(-2,3,100]; bode(num,den,w) subplat{2,1,1); tit le(Sode Diagram (Problem 3-9-9)')
'3-9-10. A pssible WfTI.AB program for obtaining a Nyquist plot of the giwn ~ ( s ) is shown bzlow. Note Chat to plat G ( j W ) locus only fox w 0, we use the following command:
The resulting Nyquist plot is shown an n.ext gage.
C
num = [O 0 0 1 J; d e n = [ l 0.8 1 01; w = O.I:O.1:100; [re,im,w] = nyquist(num,dqw); plot(re,im) Y = 1-3 3 -4 23; ~ s ( v ) ; axis('squate? ~d titlewyquist Plot (Problem B-9-10$':1 XI abeI('Rea1 Axis'] ylabelC1Lmag h i s ' )
Since none of the open-loop poles lie in the right-half s plane and the G I j cc) 1 locus encircles the -1 + jO p i n t twice clockwise if G( j ul ) locus is plokted from W = - - to d = , the closed-loop system is unstable. ----I------ - --
B-9-11. A possible MATLAB program for obtaining a Nyquist plot of the given ~ ( s ) is sham below- The resulting N y q u i s t plot is shown on next page. Since none of the open-loop ples l i e in the right-haif s plane and frm the Nyquist plot it can be seen that the G { j W ) lorus daes not encircle the -1 + jO point, the system is stable.
num- [O 0 0 20 203 den= [ I 7 20 50 01; w = 0.1:0.1:100; [re,im,w] = nyquist(num,detr,w); plot(re,im) v=[-1.5 1.5 -2.5 0.5]:a~tis(v);axis~square') gtjd titfe('Nyquist Plot (Problem B-9-1 I)') d a b e l ( R d Axis') ylabel(7mag h i s ' )
B-9-12. A possible EaTLRB prqram for obtaining a Nyquist plot of the given GIs) is s h m below. me resulting N y q u i s t plot is shown un next page
num = [O 1 2 I]; den = [1 0.2 f 1 J; w = 0:0.005: 10; [re,im,w] = nyquist(nuqden,v~); plot(re,im) v = [-3 3 -3 31; aXis(v 1; isfsquare') grid title("Nycluist PIot (ProbIem B-9- 12)') xlabel('Real A x i s ' ) ylabel(7rnag Axis') -
Fran the plot, it is seen that the G ( j W ) lozus encircles the -1 + jO mint t w i c e a 8 a 1 f s =rid f r u n W = - w to W = 0 to W = & . Refer- ring to the Nyquist stability criterion (see page 497). we hare
N = n m k r of doclnrise encirclmmt of the -1 + jO p i n t = -2
P = n m h r of poles of G ( s ) In the right-half s plane = 2
Note that there are hn open-loop poles in the right-half s plane, because
s3. 0 . 2 ~ 2 + s + 1
Then Z = number of zeros of 1 + G ( s ) in the right-half s plane
Thus, there are no closd-loop p l e s in t h e right-half s plane and the closed-loop systan is stable.
B-9-13. A closed-loop systm with the following open-loop transfer function
K E ( S ) H ( S ) = ( T ~ > O )
s2(~is * 1) 11 3
is tmskable, w h i l e a closed-loop system with the follbwing open-laap transfer function is stable.
Nyquist plots of tkse two systems are shgvnm next page. Ndte that G(jw 1 H l j L . d ) laci s ta r t fram negative infinity on the real axis (&r = 0) and approach the origin I w = clo ) . The system with the o p n - l o o p transfer function given by Equation (1) encircles the -1 + jO point twice clockwise. The system is unstable. The system with open-loop transfer funckion given by equation (2 ) does not encircle the -1 + jO point* Hence, this system is s t a b l e ,
- C
! r T - - e 1 1 (Stable ) I
B-9-14. For t h i s system
~y setting K = 1, we draw a Nyquist diagram as shown below. Note that
The ~ ~ q i i s t locus crosses the negative real axis at CT = -0.442. Hence for stability, w e require
The same result can also €x obtained analytically. Since
by getting the imaginary past of G ( j w ) equal to zero, we obtain
Solv;ing this equation for the smallest positive =rue o f u r we obtain
L-d = 2.029
Substituting Cu = 2.029 in to G ( j w ) yields
K ~(j2-029) = ( m s 2.029 - 2.029 sin 2.029)
1 + 2.029~
The critical value of K for stability ran k obtained by letting G(jZ.029) = -I, or
0.4421 K = L
Thus, the range of gain K for stability is
The q u a d n t t e term kn the denominator has the undamped natural frqumcy of 2 raa/s and the damping ratio of 0.25. Define the frequency corres- ponding to the angle OF -130 to be
Solving this last equation for dl, we f ind W 1 = 1.491. nus , the phase a n g l e becomes equal to -130* at C J = 1.491 rad/s. A t t h i s fre- quency, themagnitudemust b e u n i t y , or IG(jdl)I = 1. The required gain K can ke determined from
Setting I~(j1.49111 = 0.2890K = I, we find
Note that the phase crossover frequency is at U = 2 rad/s, since
f ~ ( j 2 ) = - / j ~ - / - 0 . 2 5 x 22 + 0.25 :c j2 + 1 = -90'- 90.. - 1 8 0 ~
The magnitude IG(j2)1 with R = 3.46 becomes
Thus, the gain margin is 1.26 dB. The Bode diagram of G ( j W ) with R = 3.46 is shown below.
B-9-16. Note that
j Cc' + 0.1 TO - - ZK(10j 4 + f 3 K j ~ + o . 5 j W ( ~ j J + l l J j L J + l )
W e shall plot the W e diagram when 2R = 1. That is, w e plot the Bade
diagram o f ~ o j d + I
'(jd' = j w ( 2 j LO c l)(jd+ 1)
The diagram is shown below. me phase crurve shows that the phase angle is -130' at w = 1 ~d/s . Since we require the phase margin to be 50°, the magnitud j1.438 must be q a l to 1 or 0 dB- Since the Bode diagram irrdic .hat (G( j1.43811 is 5.48 dS, w e need to choose 2K = -5,48 dB, or
Since the phase rums lies above the -18~' line for all Ld , the gain margin is + ~ s dB.
B-9-17. Let us use the following lead compensator:
%?.+I I s + -
G,(s) = k O ( = R, T
d ~ s + 1 1 S f - e T
Since K, is specified as 4.0 s-1, we have
Ts + 1 IC, = l i m sK,d K
- = K , d K = 4 s+o, U T s + 1 s ( O . l s + l ) ( s + I )
n Le t us set K = 1 and define R c q = K. Then
N e x t , plot a Elode diagram of
The following MATLAB pryram prduces the W e diagram s h m klm.
num = I0 0 0 41: d e n = I O . f 1.1 1 01; bodelnum,den) subplot(2,l , I 1; title('8ode Diagram of G(s) = 4/{slO. Is+ J l(s + I)]')
From this plot, the phase and gain mrgins are 17 * and 8 ,7 dB, respectively.
Since the specifications call for a phase margin of 45 ', let us choose
(Tkis means that 12' has been added to canpensate for the shift in the gain crossover frequency.) The maximum phase lead is 40'. Since
sin & = 1 -u' (& = 40 O)
1 -FN
d i s d e l ; e n ~ i r ~ c d as 0.2174- f ~ t us choose, inskead o f 0.2174,M t o b e 0.21, or
Next s k p is to dctcrmi rkc t l ~ e mrncr rrcquencies @ = TJT and d = l/(d T) or tlie f pad mrp?tisalor. Notc? tlraI; the miximutn phase-lead angle &, occurs at ttie qemtr ic man or thc t w o carnor frequencies, or W = ~ / ( G T ) . 'I'l~e arwunt or tile nlorl l i icaLior~ i n tile rnagrlitude curve at tu'= l / ( & T ) due to t h e inclusion or the term (Ts + l}/(d Ts + 1) is
we need to find the frequency p i n t where, when the lead cmpnsator is a d d d , the total magnitude kmnes 0 dB. The magnitude I ~ ( j w ) I is -6.7778 dB corresponds to &J = 2.81 rad/s. we select this frequency to be the new gain crosswer f rquency Q , . Then we obtain
Hence
and
Thus
The open-loop transfer function becomes as
The CLosed-loop transfer function is
The following MRTLM!- program produces the unit-step response nwe as shown below,
I
num=[O 0 O 3.1064 41; den = [0.0163 I 0.2794 1.263 P 4.1064 41; step(nu m, d en) grid titIe('Ur Jt-Step Response of Compensated System (Problem B-9- 17)') KEabel('1 wc'] ylabeI('0utpu t c(t)')
I
Similarly, the following MATLA3 program produces the unit-ramp response cvrve as s h m on next page.
c = step(num,den,t);
ensated System problem B-9-17)')
B-9-18. To satisfy the xequirments, try a lead compensator Gc(s) of the form
D e f i n e
where K = Kc&. Since the s t a t i c v e l o c i t y error c o n s t a n t Kv isgiven as 50 s-l, we have
We shall now plot a W e diagrmn of
The following MATUB program produces the Bade diagram shom on next page.
C . num = [O 0 501; den=(I 1 01; w = lopspace(-1,2,100); bode(nrm,den,w); subpIot(Z,1, I); title(8odc Diagram of GI (5) (Problem B9-18)y
From this plot, the phase margin is found to & 7 . ~ ~ . The gain margin is + dB. Since the specifications call for a phase maruin of 50 O r
the additional phase lead angle nr lin requirement is 42 .ZO. We may as: 'ed to h 48'. This means tha t 5.8 ' Llt13 - t he shift in the gain crosswer frequency.
s i n &, = . - 1 +oC
3tisfy *
num pha: :d to a
se lead se m r g
I r q i ~ te for
Since
l - o l
F$, = 48 mrreqmnds to a = 0.114735. Note that ol = 0.15 -corresponds t o & = 4 7 . 6 5 J b . ) Whetherwe chmse8,= 4 8 " o r & , = 4 7 . 6 5 7 " d c e s n o t make much difference in t h e final solution. H e n c e , w e choose o( = 0.15.
The nc& step is to determine the corner frequencies W = 1SS and ti, = l/(o( T) of the lead compensator. Note that the m a x d phase lead angle &, m s at the geometric mean of the two comer frequencies, or W = 1 The m u n t o f the modification in themagnitude curve at # .= l / ( L ~ T ) due to the inclusion of the tern (Ts. + l ) J ( d T s + 2 ) is
Note that
We n e d to f ind the Frequency point where , when the lead rompensator is added, the to ta l magnitude beccmes O dB- The frequency at which
the magnitude of G ~ ( ~ L L ' S is q u a 1 to -8.239 dB -s kt- & = 10 and 100 r a d / s . From t h e Rode diagram we find the frequency p i n t where I G l ( j ~ ) I = -8.239 dB m s a t k 3 = 11..4 rad/s. Not ing tha t this fre- quency corresponds to 1 / ( K T ) , or
we obtain
The lead c a n p n s a t o r thus determined is
- "
T s -F 4.4152 cc(s) = yC = Kc
1 S f - s + 29.4347
where Kc is determined as
The folLowhg MATTAB program produces the W e diagram of the l e a d ccan- pensator just designed. It is s h m on next p g e .
num = [ I I .325 501; den = 1 11; w = lo, 1,3,100); bode(num,aen,w); subplot(2,2, I); titIeC'Bode Diagram of G4s) = SO(O.2265~ + 1)/(0.03397s + 1 )3
Tbe open-loop transfer function of the designed system is
150
The following MATTAR program produces the Bade diagram of G,(S)G(S 1 which is shown on next page.
From this Bode diagram, it is clearly seen that the phase margin is appro- ximately 52", the gain margin is + w dB, and ??, = 50 s-1; a13 specifications am met. ~ ~ I I S , t h e designed s v s t m is satisfactory.
e unit- and th
Next, we shall obtain thl ~d unit-- responses of the original uncompensated systm rnsated system, The original uncompensatd system has the following closed-loop transfer function:
The closed-loop transfer function of the cmpnsated system is
The closed-lmp ples of the campensated system are as follows:
The MATLkB proqsam given belw produces the wit-step responses of the uncompensated and campensated systems. The resulting r e s p n s e cumes are shown on next page.
-
nurn=[O 0 I]; den= 11 1 I]; nvmc=[O 0 1000 4415.21; denc = 13 97.3041 1088.3041 4415.21; t = 0:0.01:8; cl = step(num,den,t); c2 = step(nurnc,denc,t); plot(t,cl,t,c2) title("Unit-Step Responses of Uncompensated and Compensated Systems') xlabel('t S e c ' ) ylabel.IDutputs') text( 1, I .25,'Compensated system') tm(2,0.5,'Uncompensatd system?
The MITLAB prqram given &low produces the unit-ramp response of the uncanpensated system and mmpnsated system. The response c m s are shorn on next, page.
-
nurn=[O 0 0 I]; dm=[l 1 1 01; nurnc=[O 0 0 1000 4415.25; denc = [3 91.3041 1088.jD41 4415.2 01; t = 0:0.01:8; cl = step(num,degt); c2 = step(nurnc,denc,t); plat(t,c17t,c2,t,t) titlHVnit-Ramp Responses of Uncompensated and Compensated Systems') xlabel("t Sec') ylabel(lJnit-Ramp Input and System Outputs') text(l,5,'Campensated system') rext(4, I . 5,Vncompensated system')
Notice frm the unit-ramp respanse cu ?at the rompensated systm f o l l m the input ramp very closely. ;he compensaW system the error in fellowing t h e i npu t can be s : 0 < t < 0.5, but it is almost zero for 0 . 5 X t . (The steady state u r v r in themit-rampresponse is 0.C2. )
CHAPTER 10
R-10-1. The differential equation for t h e systm is
This is a s-nd-rder system. Therefore, WF need t w o state variables.
Define state variables xl and x2 as fullom:
Then we obtain
The output y for the system is simply xl. Thus,
Y = XI
'Ihe state space representation far this rptm is given by
B-16-2. The system q a t i o n s are
Then we obtain
The standard skate space representation for the system is
B-10-2. The equations for the system are
k (U - Z) = bl(k - 1
bl(i - j ) = k2y
By taking laplace transforms of these t w o equal:ions, assuming the zero i n i t i a l conditions, w e obtain
By eliminating Z(s1 from the l a s t two equations,
which can k simplified to
J - kAS 5
Yo- - - - k2
Using Equations 21) and (21, Z(s)/U(s) ran be obtaind as follows:
Define
where
Then, Equations (25 and (3) can b~ written as
Hence
Then
Equations 14) and ( 5 ) can be written as
Rewriting ,
from which we get:
1 a - i x 2 = - -
X2 aT U
aT
This is the state q u a t i o n .
F m Equation (6) we have
Frm Equation ( 7 ) we get
Hence
This is the output equation.
B-10-4. Note t ha t
Hence
which can be rewritten as
which is w i n l e n t to
This equation can be wri t t en as
where
13ef ine
men, m a t i o n (13 m s
from which the state equation can IE obtained as
Noting that y = eo = x + b u , the output equation can k given by 0
Equations (2) and (3) give a state space representation for the givlen system.
Method 1: The systm differential equation can bE? written as
Canparing t h i s equation with a standard third-order equation:
we f i n d
Referring t o Problem A-10-12, we obtain
T h ~ r the state equation and output equation are
Method 2: Referring to Problem A-10-13, ~(s?/U(s) can be written as
where
Then we have
... z + 18; + 192; + 6402 = u
Ikf ine
Then
Hence, the sta te space representation for the system by this approaeh fs
B-10-6. The equations for the circuit are
~ubstitution of ql = xlt $ - = x i n t o the last t w u equations yields - X ~ ' 9 2 3
H e n c e we have
B-10-7. A partial-fraction v s i o n of Y (s ]/u(s) gives
Notice that
Then, f r a the p d i n g equations for ~l(s). x ~ ( s ) , and x3(s) we obtain
Also,
By taking the inverse Laplace transforms of the last four quatims, we get
and
In the standard state space xepr@sentation, we have
B-10-8. The equations for the system are
Nuti- tht from Equations (1 1 and ( 2 )
Also, frm mat ion (3)
a .. = X I - 3(x1 - U) + 3(x1 - i - U) + O
Thus, q u a t i n g m a t i o n s ( 4 ) and (5) and simpl!:.fying, w e obtain
Substituting xl = y into this l a s t equation, we get I.. I.
y - 3 i ; + 3 j r - y = u - 2 G + u
['She same result can be obtained by use of Equation (10-51). ]
B-10-9. The eigenvector zi associated with an eigenvalue Xi is a -or
that sat is f ies the following equation :
which can be rewritten as
example of eigenvectors corresponding to eigerlvalue
[I:] - -
For Xi = A *
An example
B-10-lo. Wf ine
The eigenvedors for this matrix can be determined by solving the following equation for x. -
which can be rewritten as
which, i n turn, gives x = arbitrary constant 1
X* = 0
x3 = arbitrary ~ n s t a n t
H e n c e ,
where a and b are arbitrary nonzero constant. Notice that
Note that
a m linearly independent vectors. The eigerneors of matrix B involve two M l inearly independent eigenveetsrs.
N a , define
The eigemeckors for this matrix can be determined by solving the following equation for 2.
which can be s m i t t e n as
from which we obtain
x = arbitrary canstant 1
3 = arbitrary constant
x3 = arbitrary mnstant
Hence,
where a, b, and c are arbitrary nonzero m s t a n t , Notie that
Thus, the eigenvectors of matrix C involve three l inearly indepndent eigen- L
vectors.
A MATLAB pragram to obtain a state-space representation of this system is given on next page. Based on the MATLA13 cbutput we get the EoPloKing state spa- equations:
num = [6 6 25.04 5.0081; d e n = [ ] 5.03247 25.1026 5.008]; [A,B,C,D] = tf2ss(num,dm)
A =
-5.0325 -25.1026 -5.0080 1 .OOOO 0 0
0 1.0000 0
B =
I 0 0
C =
0 25.0400 5.0080
D =
0
B-10-12. Referring to Equation (10-511, WP have
where
Noking that
B-10-15. The solution of the state quat ion is
Since x(0) = 0 and u(t) - l(t). we obtain CH -
x ( t ~ = I f e ~ , ' ~ B d T
Note that Yb
where
Thus
At: R. = & -l[(s1 - A ) - ~ ]
Y b.
Hence
-TI B d T = a. b.m
Alternate solution Referring to Example 10-9, we have for the unit-step input the following solution:
Since x(O) = 0 in the present case, we have w. w
x ( t l = jI-'(ett - I J B W m .*I-
where
and
and x ( t ) is given by -
3-10-16. The solution to the state equation when u ( T ) = T can be g i m by
Note that
Hence
B-10-17. The equation of motion for the system is
By substituting the given numerical values for al, b, and k i n t o this last equation, we obtain
where u is the input and y is the output. By taking the Laplace transform of Equation (11, assuming the ~m
i n i t i a l conditions, we have
Since the input u ( t ) is specified as a step displacement of 0.5 m, we have
With this input, Equation (2) can be rewritten as
The following M A W program w i l l generate the desired r e p = . The resulting respbnse curve is s h m belaw.
I
num = [0 0.5 01; den=[[ 1 0.51; stepCnurqden) Erid titlerstep Response (Problem B- I 0-1 7)') xlabell't Sec') ylabel('y(t)")
Note that t h e respnse reaches zero as t increases. This is because the transfer function Y(s)/U(s) given by Equation (2 ) involves s in the nume- sator.
Onit-step response: The following MATLAB prqram y i e l d s the u n i t d t e p response of the given system. The resulting unit-step respnse curve is shown kllou.
r
A = [-5.03247 -25.1026 -5.008 1 0 0 0 E 01;
B = [l;b;O]; C=[O 25.04 5.0081; D = [O]; IY*%tl= ~W3CAB,C,D); pIot(t,y) srid titEe('Unit-Step Response (Problem B-10-IS)? xlabelrt sect) ylabel("Output y(t)')
Unit-ramp response: Referring to pages 596 - 599, we can obtain the unit-ramp response of the system by entering a MATLAB p q a m similar to M A W Program 10-7 into the computer. Note that
x4 is the butput of the system and is the unit-ramp respnse. The follow- ing M A W pregsam produces the unit-ramp response. The resulting res- ponse curve is shown on next page.
A = [-5.03247 -25.1626 -5.008 1 0 0 0 1 01 ;
B = [1;0;0]1; C = 10 25.04 5.0081; 0 = [O]; AA = [A zeros(3,I );C 01; BB = [B;O]; cc = [C 03; DD = [O]; t = 0:0.01:5; t4qtI = step(AtiBB,CCDD, Lt); x4= [O 0 0 I]*x'; pl~t(t,x4,5t) grid t i t l e ( V n i t - h p Response (Problem 33- 10-1 8)') xlabelrt Set') ylabel('lnput and Output')
CHAPTER 11
8-11-1. Since
sin(tclRT + 0) = sinlrlm m s 0 + cosd sin 8 we obtain
~ [ x ( r c r ) l = cm e [ s i n ~ ~ l + sin o 7, ~ Q ) S Y J ~ I
-1 + s in 8 1 - z mslFT
-1 1 - 22 msWT + z - ~
- - (ms O ) Z - ' I S ~ ~ W T ) + (sin 0 ) (1 - z'l msdT) -2
1 - 22-= cosLdT * z
- - sin 8 + 2-I s i n ( W T - 8)
1 - 2 f 1 m s W T + 2-2
By defining m = k - h, we obtain
Since x (KT - hT) = 0 for k < h, the limit of sumation in can be 1 00 k = h
&zinged to X . n u s k = O
ad d
!Zince X(z) = X l ( z ) X 2 ( z ) t by mmparing Eqwtims (1) and (21, we o m i n
Differentiating X(z) with xespect to z, ~e obtain
Multiplying lmth sides of the l a s t equation by -Tz gives ob
Thus ve have
Xo= (1 - e'aT 1 e -aT
3-~-
z -aT ( z - 1 ) I z - e s - 1 z - e -aT
Hence
The inverse z transform of X ( z ) gives
-aT -aw ~ ( k ) = L - Q- e
B-11-7. X(Z) can be rewritten as
Then
Hence
B-11-8. Since -T
x ( z > = II - e ]z- ( z - 1)(2 - eWT}
we have -T
X(z) = 1 - e - - - I I z - I - 2 - 1 z - e -T
Thus
B-11-9. Referring to Table 11-1, we f ind
By writing 6' = a, we obtain
H e n c e we have
%king the inverse z transform of each tern of this last equatizm, we obtain
3 2 1:-1 + .3*.k-1 a ( k ) = a k a
H e n c e
Since a = e'l, we ktave
-k-2 ~ ( k ) = k(k * 1) k = 0, I f 2, ...
B-11-10. The z transform of the given difference equation becomes
Suhstitwting. the initial data into this last equation, we get
Hence
The Ln-e z transform of X(z) gives
B-11-11 , If a # 0, then
xl l ) = x(0) + 1 = 1
x(2) = x(1) + a = 1 + a
x/31 = x(2) t a2 = 1 + a + a 2
+ ,k-1 = 1 - a k x ( k ) = l + a + a ' + - . - l - a
l a f 1, a #0)
= k ( a = 1)
If a = 0, then
~ ( k ) = 0 k = 0, I, Z r ... (a = 0 )
- 2 The pulse transfer function for G I s ) including the z e m r d e r hold is obtained as follaws?
P
- 3 . The pul se trans fer function forthe systemcanberemittenas
The inverse z transform of this l a s t equation gives
Clamparing this d i f f esence equation with the standard s m d 4 r d e r difference quation:
Define state variables as follows:
xl(k) = y ( l ) - h#(k)
h = b = O 0 0
h = b - alhO = 1
Also, define
h2 = b2 - alhl - aZhO = 2 - I = I
Then the state variables;
5 I k ) = y(k)
r ( k ) Z X (k + l) - u(L) 1.
Referring to Section 11-6, the state equation for the system can k given by
The output equation is
N o t e that a numkr of different state space represenbtions are possible far the system.
~1-u-14. Frm t;be given transfer function we obtain
Define a state variable x as
(me input Variable is u and the output ~ r i a b l e is y . ) Then we have t k following continuous-time state equation and a u t p u t equation:
x = - a x + u
The discretized state epuation is o b b i n d as
where
Hence the discrete-time state space representation for the system is given by
The pulse t r a n s f e r function for the discretized system is obtained from Equation (11-58) as follows:
8-11-15. Referring to Equation (11-501, the purse transfer functfm F(z ) is given by
- 6 The discretized state equation is
where
GCT) and HIT) can be obtained as follows: C .c
Hence
Since
we have
Thus, the discrete-time state equation is given by
where T = 0-1 s, Theoutput q a t i o n is
Referring to Equation (11-581, the pulse t.raflsfer function for t h i s discretized system is given by
- 7 The discretized s-te equation is
GITP and HIT) m be obtained as fellows: Since L. LI*
we have
Hence
Thus, the discretized state space representation of the system becames as f ollaws :
where T = 0.2 s.
Referring to Equation (11-583, the pulse transfer fmctim is obtain& as fbllows:
8-11-18 The given s y s t m has no input function, but is subjected to initial conditions. It is similar to free vibrations of a mechanical system consisting of a mass, danrper, and spring.
Cbnsider first. the following system:
a state space representation of the system is
w h e r e ~ ( k ) = m(k) + m(k)
Notice that the cneffirients bg, bl, and b2 affect only matrices C and D.
Next, consider the case where the system has no input (u = 0). For such a case, MATLAX? prduces a pulse transfer funckian of the fcllawing form :
If the original system bad an input function u such that
MATLAB prcduces the pulse transfer function
resenta 3 y ield
tion wl- t h e pc
The following FlA?ZAB program w i l l yield a discrete-time state-space rep1 Len the sampling period T is 0.1 s. This p r q m w i l l alsc ~Lse transfer fmction,
Eased an the MkTLAB output , the discrete-time state equation is
num = 10 0 03; den=[] 2 101; [AB,C,D J = tDss(num,den); [ G , q = c2d(A,B,O. 1)
G =
0.7753 -0.8913 0.08g1 0.9536
M =
0.089 1 0.0046
[n-denzl- sQtf(G,YC,D)
n u m =
0 0 0
denz =
1.0000 -1.7288 0.8 I87
I
I
The pulse transfer function obtained frm the M A W output is
If the original system had an input function U, the numerator becomes nonzero. The difference q u a t i o n carresponding to Equation (1) is
The i n i t i a l data y ( 0 ) and y(l) are g iwn by
Note that the original differential equation system given by n y + 2; + 10y = 0, y(0) = I, ;(o) = o
has the response curve (free vibration curve) as s h m in Figure ( a ) ,
Figure (a) F Sec
m e MRTLAB program used to obtain this c u r e is given below.
num=[l 2 01; den=[1 2 101; stepCnuqden) grid title(Tontinuous-Time System with y(O) = 1 and ydot(6) - 0 (Froblem B-I 1-18)') xlabel('t Sec') I
ylabel('0utput y(t) of Original Continuous-Time System')
The response curve (free vibration curve) of the discreti zed version of the system given by Equation ( 2 ) is shown in F i g w e { b ) .
k
The MATLA3 program used to obtain this curve is given blm.
rmml = [I -0.7752 01; den1 = [l -1 -7288 O.Sl87J; u = [ l zeros(I,50$]; k = 050; y = filter(mm1 ,den1 ,u); plotCky,"*,siy,'-l) u i d title@iscretised System with y(0) = 1 and y(35- 10.9536 (Prob B-11-18)') xlabel('k') yEabel('Output y(k) when SampIing Period is 0.1 srx')
r -
The pulse transfer function f o r this system m y be obtained by use of t he MATUB prqram s h m on n& page.
-
num = [O 0 21; dm= [i 3 23; [rlB,c,DI = tf2ss(num,den); , [G,W = c2d(&B,O I); [numz,demj = ssZtfIG,H,C,D)
nllm =
0 0.0091 0.0082
den%=
E .0000 - 1.7236 0.7408
me pulse transfer function obtained here is
n m - 0.00912-1 + 0,0082~'2 G(2) = -
denz 1 - 1.72362-1 + 0.74082-2
This result is identical w i t h the pulse transfer function obtained in P K Q ~ ~ B I I A-11-12.
Taking the z transform of this difference equation, we obtain
Hence
Of
The Fibonacci series can be generatd as the response of X(z) to the Kronecker delta input, where X(z) is given by Equation (1) and the ICro- necker d e l t a input u(k ) is defined by
following MATWIB prrqmm w i l l generate the Fibonacci series.
x = fiIter(num,den,u)
Columns 1 through 6
Columns 7 through 12
Columns 13 t h u g h 18
Columns I 9 through 24
2584 4181 6765 10946 17711 28657
Colums 25 through 30
46368 75025 121393 19H18 317811 514229
Note t ha t column 1 corresponds to k = 0 and c o l m 31 corresponds to k = 30. Thus, the Fibonacci series is givw by
The following MATLMl program will generate a plot of the mit-step response up to k = 50. The resulting unit-step response is s h m b l o w .
The following MATLAB program will generate a plot of the unit-- response of the system up to k = 16. (The sampling period is I s . ) The resulting plot is shown on next page.
I
- num = [O 0 0 0.3205 -0.18&5]; den=[1 -1.3679 0.3679 0.3205 -0.18851; u = ones(l,51); k = 0:50; y = filter(nwqden,u); ~lot(ky,"~r v = [O 50 0 1.21; axis(v) .@a titlewd-Step Response (P~*oblem B-11-21)> xtabel('k') ylabel('y(k)')
B-11-23. A MATLAB prQgram to discretize the given equation is presented below. (The sampling period T is 0.05 s. )
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~ l i s MAT& program produces the output as shown on next p a p .
mrm = [O 0.5 15 1 -0.1452 -0.2963 0,05283; den = [ 1 - 1.8528 1.5906 -0.6642 0.05281; k = 0:16; x=[k]; y = filter(nurqden,x); plotCky,'~',k~,'-',k,k'-3 1
grid titIcfUnit-Ramp Response (Ftoblem B- 1 1-22)') xlabel('kl) ylabel("y(k)3 I
From the MATLRB output, the discrete-time state equation is given &bow.
1.0259 0.0504 0
1.0389 k .0259 0
-0.0006 6.0000 1.10000 0 - 0560 q ( k + 13 6.0247 -0.0006 0 1.0000 0.0250
The END