Post on 29-Aug-2014
S I X
Transient Response Stability
SOLUTIONS TO CASE STUDIES CHALLENGES
Antenna Control: Stability Design via Gain
From the antenna control challenge of Chapter 5,
( ) 3 2
76.39151.32 198 76.39
KT ss s s K
=+ + +
Make a Routh table:
s3 1 198
s2 151.32 76.39K
s1 29961.36 76.39
151.32K−
0
s0 76.39K 0
From the s1 row, 392.2.K < From the s0 row, 0 K< . Therefore, 0 392.2.K< <
UFSS Vehicle: Stability Design via Gain
( )( )( )( )
( )
( )
( ) ( )( )
( )
1
12
1
2 4 3 2
3 1 2
13 4 3 2
3 14 3 2
3 1
0.125 0.437 22 1.29 0.193
10.25 0.10925
3.483 3.465 0.60719
0.25 0.109253.483 3.465 0.60719
0.25 0.109251 3.483 3.465 0.25 2.42
sG
s s s sGG
G ssG
s s s sG K G
s KG
s s s sG s s K
T sG s s s s K
− + ⋅=
+ + +
=+ −
− −=
+ + += −
+=
+ + ++
= =+ + + + +( ) 188 0.10925s K+
6-2 Chapter 6: Transient Response Stability
s4 1 3.465 0.10925K1
s3 3.483 ( )10.25 2.4288K + 0
s2 ( )11 45.844
3.483
K− − 0.10925K1 0
s1 ( )( )1 1
1
4.2141 26.420.25
45.84K K
K+ −
− 0 0
s0 0.10925K1 0 0
1For stability :0 26.42K< <
ANSWERS TO REVIEW QUESTIONS
1. Natural response
2. It grows without bound
3. It would destroy itself or hit limit stops
4. Sinusoidal inputs of the same frequency as the natural response yield unbounded responses
even though the sinusoidal input is bounded.
5. Poles must be in the left-half-plane or on the jω axis.
6. The number of poles of the closed-loop transfer function that are in the left-half-plane, the
right-half-plane, and on the jω axis.
7. If there is an even polynomial of second order and the original polynomial is of fourth order,
the original polynomial can be easily factored.
8. Just the way the arithmetic works out
9. The presence of an even polynomial that is a factor of the original polynomial
10. For the ease of finding coefficients below that row
11. It would affect the number of sign changes
12. Seven
13. No; it could have quadrantal poles.
14. None; the even polynomial has 2 right-half-plane poles and two left-half-plane poles.
15. Yes
16. ( )Det 0s − =I A
Solutions to Problems 6-3
SOLUTIONS TO PROBLEMS
1.
s5 1 5 1
s4 3 4 3
s3 3.667 0 0
s2 4 3 0
s1 2.75− 0 0
s0 3 0 0
2 rhp; 3 lhp
2.
The Routh array for ( ) 5 3 26 5 8 20P s s s s s= + + + + is:
s5 1 6 8
s4 θ ε 5 20
s3 5ε
− 20ε
−
s2 5 20
s 10θ
1 20
The auxiliary polynomial for row 4 is ( ) 25 20Q s s= + , with ( ) 10Q s′ = , so there are two roots
half-plane. The balance, one root must be in the left half-plane.
on the jω -axis. The first column shows two sign changes so there are two roots on the right
6-4 Chapter 6: Transient Response Stability
3.
s5 1 4 3
s4 1− 4− 2−
s3 ε 1 0
s2 1 4εε− 2− 0
s1 22 1 41 4
ε εε
+ −−
0 0
s0 2− 0 0
3 rhp, 2 lhp
4.
s5 1 3 2
s4 1− 3− 2−
s3 2− 3− ROZ
s2 3− 4−
s1 1/ 3−
s0 4−
( ) ( ) ( )Even 4 :4 ;Rest 1 :1rhp;Total 5 :1rhp;4j jω ω
5.
s4 1 5 6
s3 4 8 0
s2 3 6 0
s1 6 0 0 ROZ
s0 6 0 0
( ) ( )Even 2 :2 ;Rest 2 :2 lhp;Total:2 ;2 lhpj jω ω
Solutions to Problems 6-5
6.
s6 1 6− 1 6−
s5 1 0 1
s4 6− 0 6−
s3 24− 0 0 ROZ
s2 ε 6−
s1 144 /ε− 0
s0 6−
( ) ( )Even 4 :2 rhp;2lhp;Rest 2 :1rhp;1lhp;Total:3rhp;3lhp
7. Program: %–det ([si() si(); sj() sj()])/sj()
%Template for use in each cell.
syms e %Construct a symbolic object for
%epsilon.
%%%%%%%%%%%%%%%%%%%%%%%%%%$$$$$$$$$%%%%%%%%%%%
s5=[1 4 3 0 0] %Create s^5 row of Routh table.
%%%%%%%%%%%%%%%%%%%%%%%%%%%$$$$$$$$$%%%%%%%%%%
s4=[–1 –4 –2 0 0] %Create s^4 row of Routh table.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
if –det([s5(1) s5(2); s4(1) s4(2)])/s4(1)= =0
s3=[e...
–det([s5(1) s5(3); s4(1) s4(3)])/s4(1) 0 0];
%Create s^3 row of Routh table
%if 1st element is 0.
else
s3=[–det([s5(1) s5(2); s4(1) s4(2)])/s4(1). . .
–det([s5(1) s5(3); s4(1) s4(3)])/s4(1) 0 0];
%Create s^3 row of Routh table
%if 1st element is not zero.
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
if –det([s4(1) s4(2); s3(1) s3(2)])/s3(1)= =0
–det([s4(1) s4(3); s3(1) s3(3)])/s3(1) 0 0]; s2=[e...
6-6 Chapter 6: Transient Response Stability
%Create s^2 row of Routh table
%if 1st element is 0.
else
s2=[–det([s4(1) s4(2); s3(1) s3(2)])/s3(1) . . .
–det([s4(1) s4(3); s3(1) s3(3)])/s3(1) 0 0];
%Create s^2 row of Routh table
%if 1st element is not zero.
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
if –det([s3(1) s3(2); s2(1) s2(2)])/s2(1)= =0
s1=[e . . .
–det([s3(1) s3(3); s2(1) s2(3)])/s2(1) 0 0];
%Create s^1 row of Routh table
%if 1st element is 0.
else
s1=[–det([s3(1) s3(2); s2(1) s2(2)])/s2(1) . . .
–det([s3(1) s3(3); s2(1) s2(3)])/s2(1) 0 0];
%Create s^1 row of Routh table
%if 1st element is not zero
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
s0=[–det([s2(1) s2(2); s1(1) s1(2)])/s1(1) . . .
–det([s2(1) s2(3); s1(1) s1(3)])/s1(1) 0 0];
%Create s^0 row of Routh table.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
's3' %Display label.
s3=simplify(s3); %Simplify terms in s^3 row.
pretty(s3) %Pretty print s^3 row.
's2' %Display label.
s2=simplify(s2); %Simplify terms in s^2 row.
pretty(s2) %Pretty print s^2 row.
's1' %Display label.
s1=simplify(s1); %Simplify terms in s^1 row.
pretty(s1) %Pretty print s^1 row.
's0' %Display label.
s0=simplify(s0); %Simplify terms in s^0 row.
pretty(s0) %Pretty print s^0 row. Computer response: s5 =
1 4 3 0 0
s4 =
–1 –4 –2 0 0
ans =
s3
[e 1 0 0]
Solutions to Problems 6-7
ans =
s2
[ –1 + 4 e ]
[– -------------- –2 0 0]
[ e ]
ans =
s1
[ 2 ]
[ 2 e + 1 - 4 e ]
[- -------------- 0 0 0]
[ -1 + 4 e ]
ans =
s0
[–2 0 0 0]
8.
( ) 4 3 2
24010 35 50 264
T ss s s s
=+ + + +
s4 1 35 264
s3 10 50 0
s2 30 264 0
s1 –38 0 0
s0 264 0 0
2 rhp, 2 lhp
9.
( ) 4 2
14 4 1
T ss s
=+ +
s4 4 4 1
s3 16 8 0 ROZ
s2 2 1 0
s1 4 0 0 ROZ
s0 1 0 0
( )Even 4 :4 jω
6-8 Chapter 6: Transient Response Stability
10.
The characteristic equation is:
( )( )( )
21 0
1 3s
Ks s s
++ =
− + or
( )( ) ( )1 3 2 0s s s K s− + + + = or
( )3 22 3 2 0s s K s K+ + − + =
The Routh array is:
s3 1 3K −
s2 2 2K
s 3−
1 2K
The first column will always have a sign change regardless of the value of K. There is no
value of K that will stabilize this system.
11.
( ) 8 7 6 5 4 3 2
845 12 25 45 50 82 60 84
T ss s s s s s s s
=+ + + + + + + +
S8 1 12 45 82 84
S7 1 5 10 12
s6 1 5 10 12
s5 3 10 10 ROZ
s4 5 20 36
s3 5− 29−
s2 1− 4
s1 49−
s0 4
( ) ( )Even 6 :2 rhp, 2 lhp, 2 ;Rest 2 :0 rhp, 2 lhp,0 ;Total:2 rhp, 4 lhp, 2j j jω ω ω
Solutions to Problems 6-9
12.
( ) 4 3 2
12 5 2 1
T ss s s s
=+ + + +
s4 2 1 1
s3 5 2 0
s2 1 5
s1 23− 0
s0 5
Total: 2 lhp, 2 rhp
13.
( ) 7 6 5 4 3 2
82 2 4 8 4 8
T ss s s s s s s
=− − + + − − +
s7 1 1− 4 4−
s6 2− 2 8− 8
s5 12− 8 16− 0 ROZ
s4 0.6667 5.333− 8 0
s3 88− 128 0 0
s2 4.364− 8 0 0
s1 33.33− 0 0 0
s0 8 0 0 0
Even (6): 3 rhp, 3 lhp; Rest (1): 1 rhp; Total: 4 rhp, 3 lhp
6-10 Chapter 6: Transient Response Stability
14. Program: numg=8;
deng=[1 –2 –1 2 4 –8 –4 0];
'G(s)'
G=tf(numg,deng)
'T(s)'
T=feedback(G,1)
'Poles of T(s)'
pole(T)
Computer response: ans =
G(s)
Transfer function:
8
-----------------------------------------------
s^7 - 2 s^6 - s^5 + 2 s^4 + 4 s^3 - 8 s^2 - 4 s
ans =
T(s)
Transfer function:
8
---------------------------------------------------
s^7 - 2 s^6 - s^5 + 2 s^4 + 4 s^3 - 8 s^2 - 4 s + 8
ans =
Poles of T(s)
ans =
-1.0000 + 1.0000i
-1.0000 - 1.0000i
-1.0000
2.0000
1.0000 + 1.0000i
1.0000 - 1.0000i
1.0000
Solutions to Problems 6-11
15. Thus, there are 4 rhp poles and 3 lhp poles.
16. Even (6): 1 rhp, 1 lhp, 4 jω ; Rest (1): 1 lhp; Total: 1 rhp, 2 lhp, 4 jω
( ) 5 4 3 2
64 2 2 6
T ss s s s s
=+ + − + −
s5 1 2 1
s4 4 –2 –6
s3 2.5 2.5 0
s2 –6 –6 0
s1 –12 0 0 ROZ
s0 –6 0 0
Even (2): 2 jω ; Rest (3): 1 rhp, 2lhp; Total: 2 lhp, 1 rhp, 2 jω
17.
( ) ( )4 3 2
507 1;3 10 30 169
G s H ss s s s s
= =+ + + +
. Therefore,
( ) 5 4 3 2
5071 3 10 30 169 507
G sT sG H s s s s s
= =+ + + + + +
s5 1 10 169
s4 3 30 507
s3 12 60 0 ROZ
s2 15 507 0
s1 345.6− 0 0
s0 507 0 0
Even (4): 2 rhp, 2 lhp, 0 jω ; Rest (1): 0 rhp, 1 lhp, 0 jω ; Total (5): 2 rhp, 3 lhp, 0 jω
6-12 Chapter 6: Transient Response Stability
18.
( ) ( )( ) ( )
2
2
11 3 2
K sT s
K s s K+
=+ + + +
. For a second-order system, if all coefficients are
positive, the roots will be in the lhp. Thus, 1K > − .
19.
( ) ( )( )3 2
64 3 6
K sT s
s s K s K+
=+ + + +
s3 1 3 K+
s2 4 6K
s1 132
K− 0
s0 6K 0
Stable for 0 6K< <
20.
The characteristic equation for all cases is ( )( )
1 0K s as s b
−+ =
− or ( )2 0s K b s Ka+ − − = . The Routh
array is
s2 1 Ka−
s K b−
1 Ka−
a) 0, 0 , 0 0a b K b K K< < ⇒ > > ⇒ >
b) 0, 0 , 0a b K b K K b< > ⇒ > > ⇒ >
c) 0, 0 , 0 0a b K b K b K> < ⇒ > < ⇒ < <
d) 0, 0 , 0 Nosolutiona b K b K> > ⇒ > < ⇒
Solutions to Problems 6-13
21.
( ) ( )( )4 3 2
19 26 24
K sT s
s s s K s K+
=+ + + + +
s4 1 26 K
s3 9 0
s2 210 K− 9K 0
s1 2 105 5040
210K K
K− + +
− 0 0
s0 9K 0 0
Stable for 0 140.8K< <
22. Program: K=[0:0.2:200];
for i=1:length(K);
deng=poly ([0 –2 –3 –4]);
dent=deng+[0 0 0 K(i) K(i)];
R=roots (dent);
A=real (R);
B=max (A);
of B>0
R
K=K (i)
Break
end
end Computer response: R =
–8.0442
0.0000 + 4.2791i
0.0000 – 4.2791i
–0.9559
K =
140.8000
24 +K
6-14 Chapter 6: Transient Response Stability
23. Program: %-det([si() si();sj() sj()])/sj()
%Template for use in each cell.
syms K %Construct a symbolic object for
%gain, K.
s4=[1 26 K 0] %Create s^4 row of Routh table.
s3=[9 24+ K 0 0] %Create s^3 row of Routh table.
s2=[–det([s4(1) s4(2);s3(1) s3(2)])/s3(1). . .
–det([s4(1) s4(3);s3(1) s3(3)])/s3(1) 0 0];
%Create s^2 row of Routh table.
s1=[–det([s3(1) s3(2);s2(1) s2(2)])/s2(1). . .
%Create s^1 row of Routh table.
s0=[–det([s2(1) s2(2);s1(1) s1(2)])/s1(1). . .
–det([s2(1) s2(3);s2(1) s1(3)])/s1(1) 0 0];
%Create s^0 row of Routh table.
's2' %Display label.
s2=simplify(s2); %Simplify terms in s^1 row.
pretty(s2) %Pretty print s^1 row.
's1' %Display label.
s1=simplify(s1); %Simplify terms in s^1 row.
pretty(s1) %Pretty print s^1 row.
's0' %Display label.
s0=simplify(s0); %Simplify terms in s^0 row.
pretty(s0) %Pretty print s^0 row. Computer response: s4 =
[ 1, 26, K, 0]
s3 =
[ 9, 24+K, 0, 0]
ans =
s2
[70/3 – 1/9 K K 0 0]
ans =
s1
[ 2 ]
[–105 K – 5040 + K ]
[------------------------ 0 0 0 ]
[ –210 + K ]
–det([s3(1) s3(3);s2(1) s2(3)])/s2(1) 0 0];
Solutions to Problems 6-15
ans =
s0
[K 0 0 0]
Stable for 0 140.8K< <
24.
( ) ( )( )( ) ( )2
2 21 3 4
K s sT s
K s K+ −
=+ + −
. For positive coefficients in the denominator, 314
K− < < .
Hence marginal stability only for this range of K.
25.
( ) ( )5 4
12K s
T ss s Ks K
+=
+ + +. Always unstable since s3 and s2 terms are missing.
26.
( ) ( )( )( )( ) ( )3 2
2 4 57 1 2 3 40K s s s
T sKs K s Ks K
− + +=
+ + + + −
s3 K 2K
s1 254
7 1K KK
−+
0
s0 3 40K− 0
For stability, 1 354 40
K< <
6-16 Chapter 6: Transient Response Stability
27.
( ) ( )( ) ( )4 3 2
23 3 3 2 4
K sT s
s s s K s K+
=+ − + + + −
s4 1 3− 2 4K −
s3 3 3K + 0
s2 ( )123
K− + 2 4K − 0
s1 ( )3312
K KK
++
0 0
s0 2 4K − 0 0
Conditions state that 12, 2K K< − > , and 33K > − . These conditions cannot be met
simultaneously. System is not stable for any value of K.
28.
( ) ( )3 280 2001 15390KT s
s s s K=
+ + + +
s3 1 2001
s2 80 15390K +
s1 1 14469
80 8K− + 0
s0 15390K + 0
There will be a row of zeros at s1 row if 144690K = . The previous row, s2, yields the
auxiliary equation, ( )280 144690 15390 0s + + = . Thus, 44.73s j= ± . Hence, 144690K =
yields an oscillation of 44.73 rad/s.
Solutions to Problems 6-17
29.
( ) ( ) ( ) ( )4 2
2
2 21 2 1 2 1Ks Ks Ks KT s
K s K s K− + +
=+ + − + +
Since all coefficients must be positive for stability in a second-order polynomial,
1 K− < < ∞ ; 1; 1 2K K−∞ < < − < < ∞ . Hence, 1 12
K− < < .
30.
( ) ( )( )( ) ( )4 3 2
2 711 31 8 21 12
s sT s
s s K s K s K+ +
=+ + + + + +
Making a Routh table,
s4 1 31K + 12K
s3 11 8 21K + 0
s2 3 320
11K + 12K 0
s1 224 1171 6720
3 320K K
K+ +
+0 0
s0 12K 0 0
31.
s2 row says 106.7 K− < . s1 row says 42.15K < − and 6.64 K− < . s0 row says 0 K< .
( ) ( )( )3 2
43 2 4
K sT s
s s K s K+
=+ + + +
6-18 Chapter 6: Transient Response Stability
Making a Routh table,
s3 1 2 K+
s2 3 4K
s1 6 K− 0
s0 4K 0
a. For stability, 0 6K< < .
b. Oscillation for 6K = .
c. From previous row with 26,3 24 0K s= + = . Thus 8s j= ± , or 8ω = rad/s.
32.
a. ( ) ( )( )( )( )2
1 22 2 2
K s sG s
s s s− −
=+ + +
. Therefore, ( ) ( )( )( ) ( ) ( )3 2
2 14 6 3 2 2
s s KT s
s K s K s K− −
=+ + + − + +
.
Making a Routh table,
s3 1 6 3K−
s2 4 K+ 4 2K+
s1 ( )23 8 204
K KK
− + −
+ 0
s0 4 2K+ 0
From s1 row: 1.57, 4.24K = − ; From s2 row: 4 K− < ; From s0 row: 2 K− < . Therefore,
2 1.57K− < < .
b. If 1.57K = , the previous row is 25.57 7.14s + . Thus, 1.13s j= ± .
c. From part b, 1.13rad/sω = .
Solutions to Problems 6-19
33.
Applying the feedback formula on the inner loop and multiplying by K yields
( ) ( )2 5 7eKG s
s s s=
+ +
Thus,
( ) 3 25 7KT s
s s s K=
+ + +
Making a Routh table:
s3 1 7
s2 5 K
s1 35
5K− 0
s0 K 0
For oscillation, the s1 row must be a row of zeros. Thus, 35K = will make the system
oscillate. The previous row now becomes, 25 35s + . Thus, 2 7 0s + = , or 7s j= ± .
Hence, the frequency of oscillation is 7 rad/s .
34.
( ) ( ) ( )2
3 2
21 2 4 24Ks KsT s
s K s K s+
=+ − + − +
s3 1 2 4K −
s2 1K − 24
s1 22 6 20
1K K
K− −−
0
s0 24 0
For stability, 5K > ; Row of zeros if 5K = . Therefore, 24 24 0s + = . Hence, 6ω = for
oscillation.
6-20 Chapter 6: Transient Response Stability
35. Program:
K=[0:0.001:200];
for i=1:length(K);
deng=conv([1 –4 8],[1 3]);
numg=[0 K(i) 2*K(i) 0];
dent=numg+deng;
R=roots(dent);
A=real(R);
B=max(A);
if B<0
R
K=K(i)
break
end
end
Computer response:
R =
–4.0000
–0.0000 + 2.4495i
–0.0000 – 2.4495i
K =
5
a. From the computer response, (a) the range of K for stability is 0 5K< < .
b. The system oscillates at 5K = at a frequency of 2.4494 rad/s as seen from R, the poles
of the closed-loop system.
Solutions to Problems 6-21
36. ( ) ( )( ) ( )4 3 2
23 3 3 2 4
K sT s
s s s K s K+
=+ − + + + −
s4 1 3− 2 4K −
s3 3 3K + 0
s2 12
3K +
− 2 4K − 0
s1 ( )3312
K KK
++
0 0
s0 2 4K − 0 0
For 33 :1K < − sign change; For 33 12 :1K− < < − sign change; For 12 0 :1K− < < sign
change; For 0 2 : 3K< < sign changes; For 2 : 2K > sign changes. Therefore, 2K >
yields two right-half-plane poles.
37.
( ) ( )4 3 27 15 13 4KT s
s s s s K=
+ + + + +
s4 1 15 4K +
s3 7 13 0
s2 927
4K + 0
s1 1000 49
92K−
0 0
s0 4K + 0 0
a. System is stable for 4 20.41K− < < .
b. Row of zeros when 20.41K = . Therefore, 292 24.417
s + . Thus, 1.3628s j= ± , or
1.3628rad/sω = .
6-22 Chapter 6: Transient Response Stability
38.
( ) ( )3 214 45 50KT s
s s s K=
+ + + +
s3 1 45
s2 14 50K +
s1 580
14K− 0
s0 50K + 0
a. System is stable for 50 580K− < < .
b. Row of zeros when 580K = . Therefore, 214 630s + . Thus, 45s j= ± , or
6.71rad/sω = .
39.
( ) 4 3 28 17 10KT s
s s s s K=
+ + + +
s4 1 17 K
s3 8 10 0
s2 126
8 K 0
s1 32 1063
K− + 0 0
s0 K 0 0
a. For stability 0 19.69K< < .
Solutions to Problems 6-23
b. Row of zeros when 19.69K = . Therefore, 2126 19.698
s + . Thus, 1.25s j= ± , or
1.118rad/sω = .
c. Denominator of closed-loop transfer function is 4 3 28 17 10s s s s K+ + + + . Substituting
19.69K = and solving for the roots yield 1.118, 4.5s j= ± − , and 3.5− .
40.
( ) ( )( )
2
3 2
2 12 1K s s
T ss s K s K
+ +=
+ + + −
s3 1 1K +
s2 2 K−
s1 3 2
2K + 0
s0 K− 0
Stability if 2 03
K− < < .
41.
( ) ( )4 3 2
3 2
2 22
s K s KsT s
s s s K+ + +
=+ + +
s3 1 2
s2 1 K
s1 2 K− 0
s0 K 0
Row of zeros when 2K = . Therefore 2 2s + and 2s j= ± , or 1.414 rad/sω = . Thus
2K = will yield the even polynomial with 2 jω roots and no sign changes.
6-24 Chapter 6: Transient Response Stability
42.
1 K2 1
s3 K1 5 0
s2 1 2
1
5K KK− 1 0
s1 2
1 1 2
1 2
5 255
K K KK K
− +−
0 0
s0 1 0 0
For stability, 21 2 1 1 25; 25 5K K K K K> + < ; and 1 0K > . Thus 2
1 1 20 5 25K K K< < − , or
1 1 20 5 25K K K< < − .
43.
s4 1 1 1
s3 K1 K2 0
s2 1 2
1
K KK− 1 0
s1 2 2
1 1 2 2
2 1
K K K KK K− +
− 0 0
s0 1 0 0
For two 2 21 1 2 2poles, 0j K K K Kω − + = . However, there are no real roots. Therefore, there
is no relationship between K1 and K2 that will yield just two polesjω .
Solutions to Problems 6-25
44.
s8 1 1.18 03E + 2.15 03E + 1.06 04E− + 415−
s7 103 4.04 03E + 8.96 03E− + 1.55 03E− + 0
s6 1140.7767 2236.99029 10584.951− 415− 0
s5 3838.02357 8004.2915− 1512.5299− 0 0
s4 4616.10784 10135.382− 415− 0 0
s3 422.685462 1167.4817− 0 0 0
s2 2614.57505 415− 0 0 0
s1 1100.3907− 0 0 0 0
s0 415− 0 0 0 0
a. From the first column, 1rhp,7 lhp,0 jω .
b. G(s) is not stable because of 1 rhp.
45.
( ) 2 31 2 6K s s s s= + + +
s3 1 1
s2 6 2
s 23
1 2
No RHP roots
( ) 2 32 2 2 6K s s s s= + + +
s3 1 2
s2 6 2
s 53
1 2
No RHP roots
6-26 Chapter 6: Transient Response Stability
( ) 2 33 4 4K s s s s= + + +
s3 1 1
s2 4 4
s 8
1 4
Auxiliary equation ( ) 24 4Q s s= + , no roots in RHP, but two roots in axisjω .
( ) 2 34 4 2 4K s s s s= + + +
s3 1 2
s2 4 4
s 1
1 4
No RHP roots
The interval polynomial has no roots in the RHP.
46.
The characteristic equation for this system is:
( )2 2 2
4 2 20 002 2 2
0
1 0or 0T T T
s KK Ks a sm m ms s a
ω ωωω
⎛ ⎞++ = + + + =⎜ ⎟
+ ⎝ ⎠
The Routh array is:
s4 1 20
T
Kam
ω⎛ ⎞
+⎜ ⎟⎝ ⎠
20
T
Kmω
s3 4θ 202
T
Kam
θ ω⎛ ⎞
+⎜ ⎟⎝ ⎠
s2 201/ 2
T
Kam
ω⎛ ⎞
+⎜ ⎟⎝ ⎠
20
T
Kmω
s
22200
220
4r r
r
K Kam m
Kam
ω ω
ω
⎛ ⎞− + +⎜ ⎟
⎝ ⎠⎛ ⎞
+⎜ ⎟⎝ ⎠
1 20
T
Kmω
Solutions to Problems 6-27
The second row of zeros was substituted with the coefficients resulting from
differentiating the characteristic equation: ( )2
4 2 2 00a
T T
KKQ s s a sm m
ωω⎛ ⎞
= + + +⎜ ⎟⎝ ⎠
and
( ) 3 20' 4 2a
T
KQ s s am
ω⎛ ⎞
= + +⎜ ⎟⎝ ⎠
.
Since all the plant parameters are positive, there are two sign changes in the first column
of the Routh array. So there are two poles in the RHP, two must be in the LHP.
47.
The characteristic equation for the system is 21 0b
KI s
+ = or 2 0b
KsI
+ = . The system has two
complex conjugate poles at b
Ks jI
= ± . The arm will oscillate at a frequency rad/secb
KI
.
48.
( )( )( )
1 0 0 0 1 20 1 0 3 1 40 0 1 1 1 3
1.1427 2.5713 1.1245 2.5713 1.1245
ss
s s s i s i
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− −⎜ ⎟ ⎜ ⎟− = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− = + − − − +
I A
I A
Hence, eigenvalues are 1.1427,2.5713 1.1245j− ± . Therefore, 2 rhp, 1 lhp, 0 jω .
49. Program: A=[0 1 0;0 1 –4;–1 1 3];
eig(A) Computer response: ans =
1.0000
1.5000 + 1.3229i
1.5000 – 1.3229I
6-28 Chapter 6: Transient Response Stability
50.
Writing the open-loop state and output equations we get,
1 2
2 2 3
3 1 2 3
2 3
33 4 5
x xx x xx x x x uy x x
== += − − − += +
&
&
&
Drawing the signal-flow diagram and including the unity feedback path yields,
3x x 12x
1 1
1
s 3
1
s
1
s 1
-5 1
-3
-1
r c = y
1
1
-4
Writing the closed-loop state and output equations from the signal-flow diagram,
( )
1 2
2 2 3
3 1 2 3
1 2 3 2 3
1 2 3
2 3
33 4 53 4 53 5 6
x xx x xx x x x r c
x x x r x xx x x r
y x x
== += − − − + −= − − − + − += − − − += +
&
&
&
Solutions to Problems 6-29
In vector-matrix form,
[ ]
0 1 0 00 1 3 03 5 6 1
0 1 1
r
y
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦
=
x X
x
&
Now, find the characteristic equation.
( )( )
3 2
0 0 0 1 0 1 00 0 0 1 3 0 1 30 0 3 5 6 3 5` 6
5 9 9
s ss s s
s s
s s s
−⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥− = − = − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦
= + + +
I A
Forming a Routh table to determine stability
s3 1 9
s2 5 9
s1 365
0
s0 9 0
Since there are no sign changes, the closed-loop system is stable.
51. Program: A=[0,1,0;0,1,3;–3,–4,–5];
B=[0;0;1];
C=[0,1,1];
D=0;
'G'
G=ss(A,B,C,D)
'T'
T=feedback(G,1)
'Eigenvalues of T'
ssdata(T);
eig(T)
6-30 Chapter 6: Transient Response Stability
Computer response:
ans =
G
a =
x1 x2 x3
x1 0 1 0
x2 0 1 3
x3 –3 –4 –5
b =
u1
x1 0
x2 0
x3 1
c =
x1 x2 x3
y1 0 1 1
d =
u1
y1 0
Continuous-time model.
ans =
T
a =
x1 x2 x3
x1 0 1 0
x2 0 1 3
x3 -3 -5 -6
b =
u1
x1 0
x2 0
x3 1
Solutions to Problems 6-31
c =
x1 x2 x3
y1 0 1 1
d =
u1
y1 0
Continuous-time model.
ans =
Eigenvalues of T
ans =
–1.0000 + 1.4142i
–1.0000 – 1.4142i
–3.0000
52.
a. For ( )2
1 21, 1 0c
sn B sω
= = − = or 022 =+− cs ω . The Routh array is
s2 1− 2cω
s 2θ −
1 2cω
The auxiliary polynomial used in the second row is ( ) 2 2a cQ s s ω= − + , that row is
replaced with the coefficients of ( )' 2aQ s s= − .
The first column has one sign change, so there is one root I the RHP, one in the LHP.
b. For ( )4
2 42, 1 0c
sn B sω
= = + = or 4 4 0cs ω+ = .
The Routh array is
6-32 Chapter 6: Transient Response Stability
s4 1 0 4cω
s3 θ 4 0 0
s2 θ ε 4cω
s 44 cω
ε−
1 4cω
The second row was originally a row of zeros, the auxiliary equation used was
( ) 4 2a cQ s s ω= + , so its coefficients were substituted with the coefficients of
( ) 3' 4aQ s s= .
The first column in the array has two sign changes, so the polynomial has two roots in the
RHP and two must be in the LHP.
SOLUTIONS TO DESIGN PROBLEMS
53.
( ) ( )( )( ) ( ) ( )3 2
1 105.45 11.91 11 43.65 10
K s sT s
s K s K s K+ +
=+ + + + + +
s3 1 11.91 11K+
s2 5.45 K+ 43.65 10K+
s1 211 61.86 21.26
5.45K K
K+ +
+ 0
s0 43.65 10K+ 0
For stability, 0.36772 K− < < ∞ . Stable for all positive K.
Solutions to Problems 6-33
54.
( ) ( )( )4 3 2
0.7 0.12.2 1.14 0.193 0.07 0.01
K sT s
s s s s K+
=+ + + + +
s4 1 1.14 0.07 0.01K +
s3 2.2 0.193 0
s2 1.0523 0.07 0.01K + 0
s1 0.17209 0.14635K− 0 0
s0 0.07 0.01K + 0 0
For stability, 0.1429 1.1759K− < <
55.
( ) ( ) ( )2
5 4 3 2
0.6 10 60.1130 3229 10 2348 60.1 58000 0.6
K K s K sT ss s s K s K s K
+ +=
+ + + + + + +
S s5 1 3229 60.1 58000K +
s4 130 10 23480K + 0.6K
s3 10 396290K− + 7812.4 7540000K + 0
s2 2100 2712488 8.3247 9
10 396290K K E
K− + +
− + 0.6K 0
s1 4 3 2
3 2
7813 3 5.1401 11 7.2469 15 3.3213 19 2.4874 221000 66753880 9.9168 11 3.299 15
E K E K E K E K EK K E K E
− + + +− + +
0 0
s0 0.6K 0 0
6-34 Chapter 6: Transient Response Stability
Note: s3 row was multiplied by 130
From s1 row after canceling common roots:
( )( )( )( )( )( )( )
7813000 39629 967.31586571671 2776.9294183336 29908.0706151651000 39629 2783.405672635 29908.285672635
K K K KK K K
− + + −− + −
0
3
2
1
From row : 0From row : 39629From row : 29908.29;39629 From row : 29908.29 ,or 29908.07;
s Ks Ks K Ks K K
><< <
< <
Therefore, for stability, 0 29908.07K< <
56.
s5 1 1311.2 ( )1000 100 1K +
s4 112.1 10130 60000K
s3 1220.8 99465 1000K + 0
s2 10038 9133.4K− 60000K 0
s1 ( )( )
( )0.010841 1.0192
994651.0991
K KK
+ −−
0 0
s0 60000K 0 0
2
1
0
From row : 1.099From row : 0.010841 1.0192; 1.0991From row : 0
s Ks K Ks K
<− < < ><
Therefore, 0 1.0192K< <
Solutions to Problems 6-35
57.
Find the closed-loop transfer function.
( ) ( )( )( )
( ) ( )( ) ( ) ( )
6
6
3 2 6
63 1030 140 2.5
63 101 172.5 4625 10500 63 10
KG ss s s
G s KT sG s H s s s s K
×=
+ + +
×= =
+ + + + + ×
Make a Routh table.
s3 1 4625
s2 172.5 610500 63 10 K+ ×
s1 4564.13 365217.39K− 0
s0 610500 63 10 K+ × 0
The s1 line says 21.25 10K −< × for stability. The s0 line says 41.67 10K −> − × for
stability. Hence, 4 21.67 10 1.25 10K− −− × < < × for stability.
58.
Find the closed-loop transfer function.
( ) ( )( )( )( )
( ) ( )( ) ( )
( )( )( )3 2
7570 103 0.862.61 62.61
7570 103 0.81 7570 785766 3918.76 623768p p p
Kp s sG s
s s s
G s Kp s sT s
G s H s s K s K s K
+ +=
+ −
+ += =
+ + + − +
Make a Routh table:
s3 1 785766 3918.76pK −
s2 7570 623768Kp
s1 785766 4001.16pK − 0
s0 623768Kp 0
6-36 Chapter 6: Transient Response Stability
The s1 line says 35.09 10pK −> × for stability. The s0 line says 0pK > for stability.
Hence, 35.09 10pK −> × for stability.
59.
The characteristic equation is given by: 6 2 9 13
3 2 7 11
1 10 1.314 10 2.66 101 00.00163 5.272 10 3.538 10
s sKs s s
− − −
− −
× + × + ×+ =
+ + × + ×
Or
( ) ( ) ( )3 6 2 7 9 11 130.00163 1 10 5.272 10 1.314 10 3.538 10 2.66 10 0s K s K s K− − − − −+ + × + × + × + × + × =
The corresponding Routh array is:
s3 1 7 95.272 10 1.314 10 K− −× + ×
s2 60.00163 1 10 K−+ × 11 133.538 10 2.66 10 K− −× + ×
s ( )( )15
6
1.314 10 1371.6 457.80.00163 1 10
K KK
−
−
× + ++ ×
1 11 133.538 10 2.66 10 K− −× + ×
For stability row 2 requires 1630K > − and row 4 requires 133.008K > − . The dominant
requirement being the latter. It is clear also that when 133.008K > − , the first element on
row 3 is positive. So the overall requirement for stability is 133.008K > − .
60.
The characteristic equation of the system is given by:
2
11 01
C C
f
K Kms bs k K Ts
+ − =+ + +
or
( )( ) ( ) ( )2 21 1 0f f C CK ms bs k Ts K K Ts K ms bs k+ + + + + − + + = or
( ) ( )( ) ( ) ( )3 2 21 0f f C CK mTs bT m s kT b s k K K Ts K ms bs k+ + + + + + + − + + = or
( ) ( )3 2 0f f C f f C C f C CK mTs K bT m K m s K kT b K K T K b s k K K K k⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ + − + + + − + + − =⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Substituting numerical values the equation becomes:
Solutions to Problems 6-37
6 3 2 65.28 10 0.03444 1.8 1584.78 11700 36000 31.5 10 0f f f fK s K s K s K− ⎡ ⎤⎡ ⎤ ⎡ ⎤× + − + − + − × =⎣ ⎦ ⎣ ⎦ ⎣ ⎦
The Routh array is given by
s3 65.28 10 fK−× 1584.78 11700fK −
s2 0.03444 1.8fK − 622500 31.5 10fK − ×
s
6 65.28 10 36000 31.5 10 1584.78 11700 0.03444 1.8
0.03444 1.8f f f
f
K K K
K
− ⎡ ⎤ ⎡ ⎤ ⎡ ⎤× − × − − −⎣ ⎦ ⎣ ⎦⎣ ⎦⎡ ⎤− −⎣ ⎦
1 636000 31.5 10fK − ×
To obtain positive quantities on the first column it is required: 6
6
6 6
5.28 10 0 00.03444 1.8 0 52.2636000 31.5 10 0 875
1584.78 11700 0.03444 1.8 5.28 10 36000 31.5 10 0
f f
f f
f f
f f f f
K KK K
K K
K K K K
−
−
× > ⇒ >− > ⇒ >
− × > ⇒ >
⎡ ⎤⎡ ⎤ ⎡ ⎤− − − × − × >⎣ ⎦ ⎣ ⎦ ⎣ ⎦
or 2 254.5798 3255.552 21060 0.19008 166.32f f f fK K K K− + > −
or 254.3897 3421.872 21060 0f fK K− + >
or 2 62.9139 387.206 0f fK K− + >
or
( )( )6.914 55.999 0f fK K− − >
So either 6.914fK < and 55.999 6.914f fK K< ⇒ <
or 6.914fK > and 55.999 6.914f fK K> ⇒ >
The most dominant requirement is given by the fourth row. We conclude requiring
875fK > .
6-38 Chapter 6: Transient Response Stability
61.
a. The Mesh equations obtained by defining clockwise mesh currents are given by
I1: 1 2 11 R I RI V
sC⎛ ⎞+ − =⎜ ⎟⎝ ⎠
I2: 1 2 312 0RI R I RI
sC⎛ ⎞− + + − =⎜ ⎟⎝ ⎠
I3: 2 312 0RI R I
sC⎛ ⎞− + + =⎜ ⎟⎝ ⎠
Solving for I3,
1
21
3 2 23
1
12 0
0 0
1 1 1 20 2 3
12
10 2
R R VsC
R RsC
RR VI
RR R R R RsC sC sC sC
R R RsC
R RsC
+ −
− +
−−
= =⎛ ⎞⎛ ⎞+ − + + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
− + −
− +
3
12 3 2 2
31 1 22 3
R VV RIRR R R
sC sC sC
−= =
⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
22
1
11 1 21 2 3
VV
sRC sRC sRC
−=⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
b. The gain of the inverting amplifier is given by: 1 2
2 1
V RV R
= − or 2 1
1 2
1V RV R K
= − = − .
Equating to the transfer function obtained in Part a
Solutions to Problems 6-39
21 1
1 1 21 2 3K
sRC sRC sRC
−− =
⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
. Equivalently
21 01 1 21 2 3
K
sRC sRC sRC
− =⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
c. The characteristic equation can be written as: 21 1 21 2 3 0K
sRC sRC sRC⎛ ⎞⎛ ⎞+ + − − − =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
or
3 3 3 2 2 2
1 5 6 1 0Ks R C s R C sRC
+ + + − = or
( ) 3 3 3 2 2 21 6 5 1 0K s R C s R C sRC− + + + =
The Routh array is given by
s3 ( ) 3 31 K R C− 5RC
s2 6R2C2 1
s ( )298K RC−
1 1
So for oscillation it is required to have 29K = . The resulting auxiliary equation is
( ) 2 2 26 1Q s R C s= + . Solving the latter for the jω -axis poles we obtain 16
s jRC
= ± .
The oscillation frequency is 12 6
fRCπ
= .
6-40 Chapter 6: Transient Response Stability
62.
( ) ( )
11.7 6.8 61.6 7.73.5 24 66.9 8.40 1 01 0 10
24 66.9 8.4 3.5 66.9 8.4det 11.7 1 0 6.8 0 0
0 10 1 10
3.5 24 8.4 3.5 24 66.961.6 0 1 0 7.7 0 1
1 0 1 0 10
s K Ks K K
ss
s
s K K K Ks s s s
s s
s K s KK K s
s
+ − − −⎡ ⎤⎢ ⎥+ −⎢ ⎥− =⎢ ⎥−⎢ ⎥−⎣ ⎦
+ − −− = + − +
−
+ − +− − + −
− −
I A
I A
( ) ( )
( )
( )4 3 2
0 1 0 011.7 24 66.9 8.4
10 0 1 10
0 0 0 16.8 3.5 66.9 8.4
10 1 0 10
1 0 0 0 0 161.6 3.5 66.9 8.4
0 1 1 0
1 0 0 17.7 3.5 24 66.9
0 10 1 10 1 0
35.7 304.6 59.2 840.41
s ss s K K
s s
s sK K
s s
K K Ks s
s sK s K
s s K s
⎧ − ⎫= + + − −⎨ ⎬−⎩ ⎭
⎧ − ⎫+ − −⎨ ⎬−⎩ ⎭
⎧ − − ⎫− − −⎨ ⎬− −⎩ ⎭
⎧ − − ⎫+ − + +⎨ ⎬− −⎩ ⎭= + + + + 2713.3 1032.57 0Ks K K+ − =
The Routh array is:
s4 1 304.6 59.2K+ 713.3K
s3 35.7 21032.57K−
s2 35.66 304.6K + 228.92 713.3K K+
s 228936.58 230524.08
35.66 304.6K K
K++
1 228.92 713.3K K+
Row 3 is positive if 8.54K > −
Rows 4 and 5 are positive if 0K >
So the system is closed loop stable if 0K > .
840.4 1 K
Solutions to Problems 6-41
63.
Sensor
+
-Input
transducer
Desired force
Input voltage
Controller Actuator Pantograph dynamics
Spring
Fup
Yh-Ycat Spring
displacement
Fout1
100K 1
1000
0.7883( s + 53.85)
( s2 + 15.47 s + 9283 )( s2 + 8.119 s+ 376 .3)82300
1
100
+
-
Desired force
Controller Actuator Pantograph dynamics
Spring
Fup
Yh-Ycat Spring
displacement
Fout1
1000
0.7883( s + 53.85)
(s2 + 15.47s + 9283 )(s 2 + 8.119 s + 376 .3)82300
K
100
( ) ( ) ( )( )
( )( )( )
( ) ( ) ( ) ( )
( ) ( )( )( )
2 2
2 2
0.7883 53.8515.47 9283 8.119 376.3
/100 * 1/1000 * *82.3 30.6488 53.85
8.119 376.3 15.47 9283
h cat
up
e
e
Y s Y s sG s
F s s s s s
G s K G s eK s
G ss s s s
− += =
+ + + +
=
+=
+ + + +
( ) ( )
( )4 3 2
0.6488 53.85
23.589 9784.90093 0.6488 81190.038
K sT s
s s s K
+=
+ + + +
s + (34.94 K + 0.3493192910)
6-42 Chapter 6: Transient Response Stability
s4 1 9785 ( )0.3493 7 34.94e K+ +
s3 23.59 ( )0.6488 81190K + 0 +
s2 ( )0.0275 6343K− + ( )0.3493 7 34.94e K+ 0 230654K <
s1 20.0178 1058.7 432.59 6.0275 6343
K K eK
− + +− +
0 128966 188444K− < <
s0 ( )0.3493 7 34.94e K+ 0 99971 K− <
The last column evaluates the range of K for stability for each row. Therefore
99971 188444K− < < .
64.
The characteristic equation is given by
3 2
520 10.38441 02.6817 0.11 0.0126
sKs s s
− −+ =
+ + +
or
( )3 22.6817 0.11 0.0126 520 10.3844 0s s s K s+ + + − + =
or
( ) ( )3 22.6817 0.11 520 0.0126 10.3844 0s s K s K+ + − + − =
The Routh array is:
s3 1 0.11 520K−
s2 2.6817 0.0126 10.3844K−
s 0.2824 1384.1
2.6817K−
1 0.0126 10.3844K−
Solutions to Problems 6-43
Thus for stability
40.2824 1384.1 0or 2.04 102.6817
K K −−> < ×
and 30.0126 10.3844 0or 1.21 10K K −− > < ×
The intersection of both requirements gives 42.04 10K −< × .