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Unformatted text preview: NINE Design via Root Locus
SOLUTIONS TO CASE STUDIES CHALLENGES
Antenna Control: LagLead Compensation
76.39K
a. Uncompensated: From the Chapter 8 Case Study Challenge, G(s) = s(s+150)(s+1.32) =
7194.23
1 6.9
s(s+150)(s+1.32) with the dominant poles at  0.5 ± j6.9. Hence, ζ = cos (tan 0.5 ) = 0.0723, or
%OS = 79.63% and Ts = 4
4
7194.23
= 0.5 = 8 seconds. Also, Kv = 150 x 1.32 = 36.33.
ζωn b. LeadCompensated: Reducing the percent overshoot by a factor of 4 yields, %OS = 79.63
=
4 8
19.91%, or ζ = 0.457. Reducing the settling time by a factor of 2 yields, Ts = 2 = 4. Improving
Kv by 2 yields Kv = 72.66. Using Ts = 4
= 4, ζωn = 1, from which ωn = 2.188 rad/s. Thus, the
ζωn design point equals ζωn + j ωn 1ζ2 = 1 + j1.946. Using the system's original poles and
assuming a lead compensator zero at 1.5, the summation of the system's poles and the lead
compensator zero to the design point is 123.017o . Thus, the compensator pole must contribute
1.946
123.017o180o = 56.98o. Using the geometry below, p  1 = tan 56.98o, or pc = 2.26.
c Adding this pole to the system poles and the compensator zero yields 76.39K = 741.88 at 1+j1.946.
Hence the leadcompensated openloop transfer function is GLeadcomp(s) = 318 Chapter 9: Design Via Root Locus 741.88(s + 1.5)
. Searching the real axis segments of the root locus yields higherorder
s(s + 150)(s + 1.32)(s + 2.26)
poles at greater than 150 and at 1.55. The response should be simulated since there may not be
pole/zero cancellation. The leadcompensated step response is shown below. Since the settling time and percent overshoot meet the transient requirements, proceed with the lag
741.88 x 1.5
compensator. The leadcompensated system has Kv = 150 x 1.32 x 2.26 = 2.487. Since we want Kv
s+0.002922
72.66
= 72.66, an improvement of 2.487 = 29.22 is required. Select G(s)Lag = s+0.0001 to improve the
steadystate error by 29.22. A simulation of the laglead compensated system,
741.88(s+1.5)(s+0.002922)
GLagleadcomp(s) = s(s+150)(s+1.32)(s+2.26)(s+0.0001) is shown below. Answers to Review Questions 319 UFSS Vehicle: Lead and Feedback Compensation
0.25K2(s+0.437)
Minor loop: Openloop transfer function G(s)H(s) = (s+2)(s+1.29)(s+0.193) ; Closedloop transfer
function: TML (s) = 0.25K 2 (s + 0.437)
. Searching along the 126.87o line (ζ = 0.6), find the
s(s3 + ...) dominant secondorder poles at 1.554 ± j2.072 with 0.25K2 = 4.7. Thus K2 = 18.8. Searching the
real axis segment of the root locus for a gain of 4.7 yields a 3rd pole at 0.379.
Major loop: The unity feedback, openloop transfer function found by using the minorloop closed0.25K1(s+0.437)
loop poles is GML(s) = s(s+0.379)(s+1.554+j2.072)(s+1.554j2.072) . Searching along the 120o line
(ζ = 0.5), find the dominant secondorder poles at 1.069±j1.85 with 0.25K1 = 4.55. Thus K1 = 18.2.
Searching the real axis segment of the root locus for a gain of 4.55 yields a 3rd pole at 0.53 and a 4th
pole at 0.815. ANSWERS TO REVIEW QUESTIONS
1. Chapter 8: Design via gain adjustment. Chapter 9: Design via cascaded or feedback filters
2. A. Permits design for transient responses not on original root locus and unattainable through simple gain
adjustments. B. Transient response and steadystate error specifications can be met separately and
independently without the need for tradeoffs
3. PI or lag compensation
4. PD or lead compensation
5. PID or laglead compensation
6. A pole is placed on or near the origin to increase or nearly increase the system type, and the zero is
placed near the pole in order not to change the transient response.
7. The zero is placed closer to the imaginary axis than the pole. The total contribution of the pole and zero
along with the previous poles and zeros must yield 1800 at the design point. Placing the zero closer to the
imaginary axis tends to speed up a slow response.
8. A PD controller yields a single zero, while a lead network yields a zero and a pole. The zero is closer to
the imaginary axis.
9. Further out along the same radial line drawn from the origin to the uncompensated poles
10. The PI controller places a pole right at the origin, thus increasing the system type and driving the error
to zero. A lag network places the pole only close to the origin yielding improvement but no zero error.
11. The transient response is approximately the same as the uncompensated system, except after the
original settling time has passed. A slow movement toward the new final value is noticed.
12. 25 times; the improvement equals the ratio of the zero location to the pole location. 320 Chapter 9: Design Via Root Locus 13. No; the feedback compensator's zero is not a zero of the closedloop system.
14. A. Response of inner loops can be separately designed; B. Faster responses possible; C. Amplification
may not be necessary since signal goes from high amplitude to low. SOLUTIONS TO PROBLEMS
1. Uncompensated system: Search along the ζ = 0.5 line and find the operating point is at 1.5356 ±
j2.6598 with K = 73.09. Hence, = %OS = e −ζπ / 1 −ζ 2 x100 = 16.3%; Ts = 4
= 2.6 seconds; Kp
1.5356 73.09
=2.44. A higherorder pole is located at 10.9285.
30 Compensated: Add a pole at the origin and a zero at 0.1 to form a PI controller. Search along the ζ =
0.5 line and find the operating point is at 1.5072 ± j2.6106 with K = 72.23. Hence, the estimated
performance specifications for the compensated system are: %OS = e −ζπ / 1 −ζ 2 x100 = 16.3%; Ts = 4
= 2.65 seconds; Kp = ∞. Higherorder poles are located at 0.0728 and 10.9125. The
1.5072
compensated system should be simulated to ensure effective pole/zero cancellation.
2.
a. Insert a cascade compensator, such as Gc (s ) = s + 0.01
.
s b.
Program:
K=1
G1=zpk(,[0,2,5],K)
Gc=zpk([0.01],[0],1)
G=G1*Gc
rlocus(G)
T=feedback(G,1)
T1=tf(1,[1,0])
T2=T*T1
t=0:0.1:200;
step(T1,T2,t) Computer response:
K=
1
Zero/pole/gain:
1
s (s+2) (s+5) %G1=1/s(s+2)(s+5)
%Gc=(s+0.01)/s %Form 1/s to integrate step input
%Show input ramp and ramp response Solutions to Problems 321 Zero/pole/gain:
(s+0.01)
s
Zero/pole/gain:
(s+0.01)
s^2 (s+2) (s+5)
Zero/pole/gain:
(s+0.01)
(s+5.064) (s+1.829) (s+0.09593)
(s+0.01126) Transfer function:
1
s
Zero/pole/gain:
(s+0.01)
s (s+5.064) (s+1.829) (s+0.09593)
(s+0.01126) 3.
a. Searching along the 126.16o line (10% overshoot, ζ = 0.59), find the operating point at 322 Chapter 9: Design Via Root Locus
20
1.4 + j1.92 with K = 20. Hence, Kp = 1 x 5 x 3 = 1.333.
s+0.3
b. A 3x improvement will yield Kp = 4. Use a lag compensator, Gc(s) = s+0.1 .
c. 4.
a. Searching along the 126.16o line (10% overshoot, ζ = 0.59), find the operating point at
1.009 + j1.381 with K = 17.5. Hence, Kv = 17.5
= 1.1667.
5x3 b. A 3.429x improvement will yield Kv = 4. Use a lag compensator, Gc(s) = s + 0.3429
.
s + 0.1 c.
Program:
K=17.5
G=zpk(,[0,3,5],K)
Gc=zpk([0.3429],[0.1],1)
Ge=G*Gc;
T1=feedback(G,1);
T2=feedback(Ge,1);
T3=tf(1,[1,0]);
%Form 1/s to integrate step input
T4=T1*T3;
T5=T2*T3;
t=0:0.1:20;
step(T3,T4,T5,t)
%Show input ramp and ramp responses Solutions to Problems 323 Computer response:
K=
17.5000
Zero/pole/gain:
17.5
s (s+3) (s+5)
Zero/pole/gain:
(s+0.3429)
(s+0.1) 5.
a. Uncompensated: Searching along the 126.16o line (10% overshoot, ζ = 0.59), find the operating
45.72
20
point at 2.03 + j2.77 with K = 45.72. Hence, Kp = 2 x 4 x 6 = 0.9525. An improvement of 0.9525
0.201
= 20.1 is required. Let Gc(s) = 0.01 . Compensated: Searching along the 126.16o line (10%
overshoot, ζ = 0.59), find the operating point at  1.99+j2.72 with K = 46.05. Hence, Kp =
46.05 x 0.201
2 x 4 x 6 x 0.01 = 19.28. 324 Chapter 9: Design Via Root Locus b. c. From (b), about 28 seconds 6.
Uncompensated: Searching along the 135o line (ζ = 0.707), find the operating point at
2.32 + j2.32 with K = 4.6045. Hence, Kp = π
2.32
ωn = = 1.354 seconds; %OS = e −ζπ / 1−ζ 2 4.6045
4
= 0.153; Ts =
= 1.724 seconds; Tp =
30
2.32 x100 = 4.33%; 2.322 + 2.322 = 3.28 rad/s; higherorder pole at 5.366. Compensated: To reduce the settling time by a factor of 2, the closedloop poles should be – 4.64 ±
j4.64. The summation of angles to this point is 119o . Hence, the contribution of the compensating
zero should be 180o 119o =61o . Using the geometry shown below, 4.64
= tan (61o). Or, zc = 7.21.
zc − 4.64 Solutions to Problems 325 After adding the compensator zero, the gain at 4.64+j4.64 is K = 4.77. Hence, Kp = 4.77 x 6 x 7.21
4
π
= 6.88 . Ts =
= 0.86 second; Tp =
= 0.677 second;
4.64
2 x3 x5
4.64 %OS = e −ζπ / 1−ζ 2 x100 = 4.33%; ωn = 4.642 + 4.642 = 6.56 rad/s; higherorder pole at 5.49. The problem with the design is that there is steadystate error, and no effective pole/zero
cancellation. The design should be simulated to be sure the transient requirements are met.
7.
Program:
clf
'Uncompensated System'
numg=[1 6];
deng=poly([2 3 5]);
'G(s)'
G=tf(numg,deng);
Gzpk=zpk(G)
rlocus(G,0:1:100)
z=0.707;
pos=exp(pi*z/sqrt(1z^2))*100;
sgrid(z,0)
title(['Uncompensated Root Locus with ' , num2str(z), ' Damping Ratio
Line'])
[K,p]=rlocfind(G); %Allows input by selecting point on graphic
'Closedloop poles = '
p
i=input('Give pole number that is operating point
');
'Summary of estimated specifications'
operatingpoint=p(i)
gain=K
estimated_settling_time=4/abs(real(p(i)))
estimated_peak_time=pi/abs(imag(p(i)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(i))^2+imag(p(i))^2)
Kp=dcgain(K*G)
'T(s)'
T=feedback(K*G,1)
'Press any key to continue and obtain the step response'
pause
step(T)
title(['Step Response for Uncompensated System with ' , num2str(z),...
' Damping Ratio'])
'Press any key to go to PD compensation'
pause
'Compensated system' 326 Chapter 9: Design Via Root Locus done=1;
while done>0
a=input('Enter a Test PD Compensator, (s+a). a =
')
numc=[1 a];
'Gc(s)'
GGc=tf(conv(numg,numc),deng);
GGczpk=zpk(GGc)
wn=4/[(estimated_settling_time/2)*z];
rlocus(GGc)
sgrid(z,wn)
title(['PD Compensated Root Locus with ' , num2str(z),...
' Damping Ratio Line', 'PD Zero at ', num2str(a), ', and Required Wn'])
done=input('Are you done? (y=0,n=1) ');
end
[K,p]=rlocfind(GGc); %Allows input by selecting point on graphic
'Closedloop poles = '
p
i=input('Give pole number that is operating point
');
'Summary of estimated specifications'
operatingpoint=p(i)
gain=K
estimated_settling_time=4/abs(real(p(i)))
estimated_peak_time=pi/abs(imag(p(i)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(i))^2+imag(p(i))^2)
Kp=dcgain(K*GGc)
'T(s)'
T=feedback(K*GGc,1)
'Press any key to continue and obtain the step response'
pause
step(T)
title(['Step Response for Compensated System with ' , num2str(z),...
' Damping Ratio']) Computer response:
ans =
Uncompensated System
ans =
G(s)
Zero/pole/gain:
(s+6)
(s+5) (s+3) (s+2)
Select a point in the graphics window
selected_point =
2.3104 + 2.2826i
ans =
Closedloop poles =
p=
5.3603
2.3199 + 2.2835i Solutions to Problems 327 2.3199  2.2835i
Give pole number that is operating point 2 ans =
Summary of estimated specifications
operatingpoint =
2.3199 + 2.2835i
gain =
4.4662
estimated_settling_time =
1.7242
estimated_peak_time =
1.3758
estimated_percent_overshoot =
4.3255
estimated_damping_ratio =
0.7070
estimated_natural_frequency =
3.2552
Kp =
0.8932
ans =
T(s)
Transfer function:
4.466 s + 26.8
s^3 + 10 s^2 + 35.47 s + 56.8
ans =
Press any key to continue and obtain the step response
ans = 328 Chapter 9: Design Via Root Locus Press any key to go to PD compensation
ans =
Compensated system
Enter a Test PD Compensator, (s+a). a = 6 a=
6
ans =
Gc(s)
Zero/pole/gain:
(s+6)^2
(s+5) (s+3) (s+2)
Are you done? (y=0,n=1) 1
Enter a Test PD Compensator, (s+a). a = 7.1 a=
7.1000
ans =
Gc(s)
Zero/pole/gain:
(s+7.1) (s+6)
(s+5) (s+3) (s+2)
Are you done? (y=0,n=1) 0
Select a point in the graphics window
selected_point =
4.6607 + 4.5423i
ans =
Closedloop poles =
p=
4.6381 + 4.5755i
4.6381  4.5755i
5.4735
Give pole number that is operating point
ans =
Summary of estimated specifications 1 Solutions to Problems 329 operatingpoint =
4.6381 + 4.5755i
gain =
4.7496
estimated_settling_time =
0.8624
estimated_peak_time =
0.6866
estimated_percent_overshoot =
4.3255
estimated_damping_ratio =
0.7070
estimated_natural_frequency =
6.5151
Kp =
6.7444
ans =
T(s)
Transfer function:
4.75 s^2 + 62.22 s + 202.3
s^3 + 14.75 s^2 + 93.22 s + 232.3
ans =
Press any key to continue and obtain the step response 330 Chapter 9: Design Via Root Locus Solutions to Problems 331 332 Chapter 9: Design Via Root Locus 8.
The uncompensated system performance is summarized in Table 9.8 in the text. To improve settling
time by 4, the dominant poles need to be at 7.236 ± j14.123. Summing the angles from the openloop
poles to the design point yields 277.326o. Thus, the zero must contribute 277.326o  180o = 97.326o.
Using the geometry below, jω j14.123 splane
97.236o 7.236 zc σ 14.123
7.236  zc = tan(18097.326). Thus, zc = 5.42. Adding the zero and evaluating the gain at the design
point yields K = 256.819. Summarizing results: Solutions to Problems 333 9.
4
a. ζωn = T = 2.5; ζ =
s %OS
 ln ( 100 )
= 0.404. Thus, ωn = 6.188 rad/s and the operating
%OS
π2 + ln2 ( 100 ) point is  2.5 ± j5.67.
b. Summation of angles including the compensating zero is 150.06o. Therefore, the compensator
pole must contribute 150.06o  180o = 29.94o.
5.67
c. Using the geometry shown below, p  2.5
c = tan 29.94o. Thus, pc = 12.34. 334 Chapter 9: Design Via Root Locus d. Adding the compensator pole and using 2.5 + j5.67 as the test point, K = 357.09.
e. Searching the real axis segments for K = 1049.41, we find higherorder poles at 15.15, and 1.186.
f. Pole at 15.15 is more than 5 times further from the imaginary axis than the dominant poles. Pole at
1.186 may not cancel the zero at 1
g. A simulation of the system shows a percent overshoot of 37.5% and a settling time of 2.12 seconds.
Thus, the specifications were not met because polezero cancellation was not achieved. A redesign is
required.
10.
4
a. ζωn = T = 2.4; ζ =
s %OS
 ln ( 100 )
= 0.5. Thus, ωn = 4.799 rad/s and the operating point is
2 + ln2 (%OS)
π
100 2.4 ± j4.16.
b. Summation of angles including the compensating zero is 131.36o. Therefore, the compensator
4.16
pole must contribute 180o  131.36o = 48.64o. Using the geometry shown below, p  2.4
c = Solutions to Problems 335
tan 48.64o. Thus, pc = 6.06. c. Adding the compensator pole and using 2.4 + j4.16 as the test point, K = 29.117.
d. Searching the real axis segments for K = 29.117, we find a higherorder pole at 1.263.
e. Pole at 1.263 is near the zero at 1. Simulate to ensure accuracy of results.
29.117
f. Ka = 6.06 = 4.8
g. From the plot, Ts = 1.4 seconds; Tp = 0.68 seconds; %OS = 35%. 336 Chapter 9: Design Via Root Locus 11.
a.
Uncompensated Root Locus with 0.8 Damping Line 10 8 6 4 Imag Axis 2 0 2 4 6 8 10 10 8 6 4
Real Axis 2 0 2 b. and c. Searching along the ζ = 0.8 line (143.13o), find the operating point at
–2.682 + j2.012 with K = 35.66.
4
d. Since ζωn = T , the real part of the compensated dominant pole is 4. The imaginary part is
s 4 tan (180o143.13o) = 3. Using the uncompensated system's poles and zeros along with the
compensator zero at  4.5, the summation of angles to the design point, 4 + j3 is –158.71o. Thus, the
contribution of the compensator pole must be 158.71o  180o = 21.290. Using the following
geometry, 3
= tan 21.290, or pc = 11.7.
pc − 4
3 21.29
4 Adding the compensator pole and using – 4 + j3 as the test point, K = 172.92.
e. Compensated: Searching the real axis segments for K = 172.92, we find higherorder poles at 14.19, and approximately at –5.26 ± j0.553. Since there is no pole/zero cancellation with the zeros at
6 and –4.5, the system should be simulated to check the settling time. Solutions to Problems 337 f.
Step Response 0.45
0.4 0.35 Amplitude 0.3
0.25 0.2
0.15 0.1
0.05 0 0 0.5 1
Time (sec.) The graph shows about 2% overshoot and a 0.8 second settling time compared to a desired 1.52%
overshoot and a settling time of 1 second.
12.
Program:
clf
numg=[1 6];
deng=poly([2 4 7 8]);
'G(s)'
G=tf(numg,deng);
Gzpk=zpk(G)
rlocus(G)
z=0.8;
pos=exp(pi*z/sqrt(1z^2))*100;
sgrid(z,0)
title(['Uncompensated Root Locus with ' , num2str(z), ' Damping Ratio Line'])
[K,p]=rlocfind(G);
'Closedloop poles = '
p
i=input('Give pole number that is operating point
');
'Summary of estimated specifications'
operatingpoint=p(i)
gain=K
estimated_settling_time=4/abs(real(p(i)))
estimated_peak_time=pi/abs(imag(p(i)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(i))^2+imag(p(i))^2)
Kp=K*numg(max(size(numg)))/deng(max(size(deng)))
'T(s)'
T=feedback(K*G,1)
'Press any key to continue and obtain the step response'
pause
step(T)
title(['Step Response for Uncompensated System with ' , num2str(z),...
' Damping Ratio'])
'Press any key to go to Lead compensation'
pause 338 Chapter 9: Design Via Root Locus 'Compensated system'
b=4.5;
'Lead Zero at 4.5 '
done=1;
while done>0
a=input('Enter a Test Lead Compensator Pole, (s+a). a =
');
'Gc(s)'
Gc=tf([1 b],[1 a])
GGc=G*Gc;
[numggc,denggc]=tfdata(GGc,'v');
'G(s)Gc(s)'
GGczpk=zpk(GGc)
wn=4/((1)*z);
rlocus(GGc);
sgrid(z,wn)
title(['Lead Compensated Root Locus with ' , num2str(z),...
' Damping Ratio Line, Lead Pole at ', num2str(a), ', and Required Wn'])
done=input('Are you done? (y=0,n=1) ');
end
[K,p]=rlocfind(GGc); %Allows input by selecting point on graphic
'Closedloop poles = '
p
i=input('Give pole number that is operating point
');
'Summary of estimated specifications'
operatingpoint=p(i)
gain=K
estimated_settling_time=4/abs(real(p(i)))
estimated_peak_time=pi/abs(imag(p(i)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(i))^2+imag(p(i))^2)
Kp=dcgain(K*GGc)
'T(s)'
T=feedback(K*GGc,1)
'Press any key to continue and obtain the step response'
pause
step(T)
title(['Step Response for Compensated System with ' , num2str(z),...
' Damping Ratio']) Computer response:
ans =
G(s)
Zero/pole/gain:
(s+6)
(s+8) (s+7) (s+4) (s+2)
Select a point in the graphics window
selected_point =
2.7062 + 2.0053i
ans =
Closedloop poles =
p=
9.3056
6.3230 Solutions to Problems 339 2.6857 + 2.0000i
2.6857  2.0000i
Give pole number that is operating point 3 ans =
Summary of estimated specifications
operatingpoint =
2.6857 + 2.0000i
gain =
35.2956
estimated_settling_time =
1.4894
estimated_peak_time =
1.5708
estimated_percent_overshoot =
1.5165
estimated_damping_ratio =
0.8000
estimated_natural_frequency =
3.3486
Kp =
0.4727
ans =
T(s)
Transfer function:
35.3 s + 211.8
s^4 + 21 s^3 + 154 s^2 + 491.3 s + 659.8
ans =
Press any key to continue and obtain the step response
ans = 340 Chapter 9: Design Via Root Locus Press any key to go to Lead compensation
ans =
Compensated system
ans =
Lead Zero at 4.5
Enter a Test Lead Compensator Pole, (s+a). a = 10 ans =
Gc(s)
Transfer function:
s + 4.5
s + 10
ans =
G(s)Gc(s)
Zero/pole/gain:
(s+6) (s+4.5)
(s+10) (s+8) (s+7) (s+4) (s+2)
Are you done? (y=0,n=1) 1
Enter a Test Lead Compensator Pole, (s+a). a =
ans =
Gc(s)
Transfer function:
s + 4.5
s + 11.7
ans =
G(s)Gc(s)
Zero/pole/gain:
(s+6) (s+4.5)
(s+11.7) (s+8) (s+7) (s+4) (s+2)
Are you done? (y=0,n=1) 0
Select a point in the graphics window
selected_point =
3.9885 + 3.0882i 11.7 Solutions to Problems 341 ans =
Closedloop poles =
p=
14.2326
3.9797
3.9797
5.2540
5.2540 +
+
 3.0860i
3.0860i
0.5076i
0.5076i Give pole number that is operating point
ans =
Summary of estimated specifications
operatingpoint =
3.9797 + 3.0860i
gain =
178.3530
estimated_settling_time =
1.0051
estimated_peak_time =
1.0180
estimated_percent_overshoot =
1.5165
estimated_damping_ratio =
0.8000
estimated_natural_frequency =
5.0360
Kp =
0.9187
ans =
T(s)
Transfer function: 2 342 Chapter 9: Design Via Root Locus 178.4 s^2 + 1873 s + 4816
s^5 + 32.7 s^4 + 399.7 s^3 + 2436 s^2
+ 7656 s + 1.006e004 ans =
Press any key to continue and obtain the step
response Solutions to Problems 343 344 Chapter 9: Design Via Root Locus 13. a. Searching along the 117.13o line (%OS = 20%; ζ = 0.456), find the operating point at Solutions to Problems 345 6.39 + j12.47 with K = 9273. Searching along the real axis for K = 9273, we find a higherorder pole
at –47.22. Thus, Ts = 4 ζω n = 4
= 0.626 second.
6.39 b. For the settling time to decrease by a factor of 2, Re = ζωn = 6.39 x 2 = 12.78. The imaginary part is Im = 12.78 tan 117.13o = 24.94. Hence, the compensated closedloop poles are
12.78 ± j24.94. A settling time of 0.313 second would result.
c. Assume a compensator zero at 20. Using the uncompensated system's poles along with the compensator zero, the summation of angles to the design point, 12.78 ± j24.94 is –159.63o. Thus,
the contribution of the compensator pole must be 159.63o180o = 20.37o. Using the following
geometry, 24.94
= tan 20.37o, or pc = 79.95.
pc − 12.78
24.94 20.37
12.78
Adding the compensator pole and using 12.78 ± j24.94 as the test point, K = 74130.
d.
Step Response
Uncompensated & LeadCompensated System 1 Amplitude 0.8 0.6 0.4 0.2 0 0 0.15 0.3 0.45 Time (sec.) 0.6 0.75 346 Chapter 9: Design Via Root Locus 14.
a. Searching along the 110.97o line (%OS = 30%; ζ= 0.358), find the operating point at 2.065 + j5.388 with K = 366.8. Searching along the real axis for K = 366.8, we find a higherorder
pole at –16.87. Thus, Ts = 4 ζω n = 4
= 1.937 seconds. For the settling time to decrease by a
2.065 factor of 2, Re = ζωn = 2.065 x 2 =  4.13. The imaginary part is – 4.13 tan 110.970 = 10.77. Hence,
the compensated dominant poles are – 4.13 ± j10.77. The compensator zero is at 7. Using the
uncompensated system's poles along with the compensator zero, the summation of angles to the
design point, – 4.13 ± j10.77 is –162.06o. Thus, the contribution of the compensator pole must be –
162.06o  180o = 17.94o. Using the following geometry, 10.77
= tan 17.94o, or pc = 37.4.
pc − 4.13 10.77 17.94
4.13
Adding the compensator pole and using – 4.13 ± j10.77 as the test point, K = 5443.
b. Searching the real axis segments for K = 5443 yields higherorder poles at approximately –8.12 and –42.02. The pole at –42.02 can be neglected since it is more than five times further from the
imaginary axis than the dominant pair. The pole at –8.12 may not be canceling the zero at 7. Hence,
simulate to be sure the requirements are met.
c.
Program:
'Uncompensated System G1(s)'
numg1=1;
deng1=poly([15 (3+2*j) (32*j)]);
G1=tf(numg1,deng1)
G1zpk=zpk(G1)
K1=366.8
'T1(s)'
T1=feedback(K1*G1,1);
T1zpk=zpk(T1)
'Compensator Gc(s)'
numc=[1 7];
denc=[1 37.4];
Gc=tf(numc,denc)
'Compensated System G2(s) = G1(s)Gc(s)'
K2=5443
G2=G1*Gc;
G2zpk=zpk(G2)
'T2(s)'
T2=feedback(K2*G2,1);
T2zpk=zpk(T2)
step(T1,T2)
title(['Uncompensated and Lead Compensated Systems']) Solutions to Problems 347 Computer response:
ans =
Uncompensated System G1(s)
Transfer function:
1
s^3 + 21 s^2 + 103 s + 195
Zero/pole/gain:
1
(s+15) (s^2 + 6s + 13)
K1 =
366.8000
ans =
T1(s)
Zero/pole/gain:
366.8
(s+16.87) (s^2 + 4.132s + 33.31)
ans =
Compensator Gc(s)
Transfer function:
s+7
s + 37.4
ans =
Compensated System G2(s) = G1(s)Gc(s)
K2 =
5443
Zero/pole/gain:
(s+7)
(s+37.4) (s+15) (s^2 + 6s + 13)
ans =
T2(s)
Zero/pole/gain: 348 Chapter 9: Design Via Root Locus 5443 (s+7)
(s+42.02) (s+8.118) (s^2 + 8.261s + 133.1) 15.
a. Searching the 15% overshoot line (121.127o) for 180o yields 0.372 + j0.615. Hence, Ts = 4
=
σd 4
0.372 = 10.75 seconds.
4
4
b. For 7 seconds settling time, σd = T = 7 = 0.571. ωd = 0.571 tan (180o  121.127o) = 0.946.
s
Therefore, the design point is 0.571 + j0.946. Summing the angles of the uncompensated system's
poles as well as the compensator pole at 15 yields 213.493o. Therefore, the compensator zero must
contribute (213.493o  180o) = 33.493o. Using the geometry below, jω
splane
j0.946
33.493o
zc 0.571 σ 0.946
o
zc  0.571 = tan (33.493 ) . Hence, zc = 2. The compensated openloop transfer function is Solutions to Problems 349 K(s+2)
. Evaluating the gain for this function at the point, 0.571 + j0.946 yields K
s(s+1)(s2+10s+26)(s+15)
= 207.512.
c.
Program:
numg= 207.512*[1 2];
r=roots([1,10,26]);
deng=poly([0 ,1, r(1),r(2),15]);
'G(s)'
G=tf(numg,deng);
Gzpk=zpk(G)
T=feedback(G,1);
step(T)
title(['Step Response for Design of Ts = 7, %OS = 15']) Computer response:
ans =
G(s)
Zero/pole/gain:
207.512 (s+2)
s (s+15) (s+1) (s^2 + 10s + 26) 16.
a. From 20.5% overshoot evaluate ζ = 0.45 . Also, since ζωn = 44
= , ωn = 2.963 . The
Ts 3 compensated dominant poles are located at ζωn ± jωn 1ζ2 =  1.3333 ± j2.6432. Assuming
the compensator zero at 0.02, the contribution of openloop poles and the compensator zero to the
design point,  1.3333 ± j2.6432 is 175.78o. Hence, the compensator pole must contribute 350 Chapter 9: Design Via Root Locus 175.78o  180o = 4.22o. Using the following geometry, 2.6432
= tan 4.22o , or pc = 37.16
pc − 1.3333 Adding the pole to the system, K = 4401.52 at the design point.. b. Searching along the real axis segments of the root locus for K = 4401.52, we find higherorder poles at 0.0202, 13.46, and 37.02. There is pole/zero cancellation at 0.02. Also, the poles at ,
13.46, and 37.02 are at least 5 times the design point’s real part. Thus, the secondorder
approximation is valid.
c. Solutions to Problems 351 From the plot, Ts = 2.81 seconds, and %OS = 20.8%. Thus, the requirements are met.
17. 4
4
a. ζωn = T = 0.5 = 8. Since ζ = 0.4, ωn = 20. Therefore the compensated closedloop poles are
s
located at  ζωn ± jωn 1ζ2 = 8 ± j18.33.
b. Using the system's poles along with the compensator's pole at 15, the sum of angles to the test point –8 ± j18.33 is 293.4o . Therefore, the compensator's zero must contribute 293.4o  180o =
18.33
113.4o . Using the following geometry, 8  z = tan 66.6o, or zc = 0.0679.
c c. Adding the compensator zero and using –8 ± j18.33 as the test point, K = 7297.
d. Making a secondorder assumption, the predicted performance is as follows:
Uncompensated: Searching along the 133.58o line (ζ = 0.4), find the uncompensated closedloop pole at 5.43 + j12.45 with K = 3353. Hence, Ts =
25.38%; Kp = 4
= 0.74 seconds;
ζωn %OS = e −ζπ / 1−ζ 2 x100 = 3353
= 1.66. Checking the secondorder assumption by searching the real axis
101x20 segments of the root locus for K = 3353, we find a higherorder pole at 29.13. Since this pole is
more than five times further from the imaginary axis than the dominant pair, the second order
assumption is reasonable.
Compensated: Using the compensated dominant pole location,  8 ± j18.33, Ts = seconds; %OS = e −ζπ / 1−ζ 2 x100 = 25.38%; Kp = 4
= 0.5
ζωn 7297x 0.0679
= 0.016. Checking the second101x 20x15 order assumption by searching the real axis segments of the root locus for K = 7297, we find higherorder poles at 2.086 and 36.91. The poles are not five times further from the imaginary axis nor do
they yield pole/zero cancellation. The secondorder assumption is not valid. 352 Chapter 9: Design Via Root Locus e. The uncompensated system exhibits a steadystate error of 0.38, a percent overshoot of 22.5%, and a
settling time of 0.78 seconds. Since there is no pole/zero cancellation the closedloop zero near the origin produces a large steadystate error. The student should be asked to find a viable design solution to this problem by choosing
the compensator zero further from the origin. For example, placing the compensator zero at 20 yields
a compensator pole at 90.75 and a gain of 28730. This design yields a valid secondorder
approximation.
18.
a. Since %OS = 1.5%, ζ = %OS
 ln ( 100 )
4
2
= 0.8. Since Ts =
= 3 second,
ζωn
%OS
π2 + ln2 ( 100 ) ωn = 7.49 rad/s. Hence, the location of the closedloop poles must be 6±j4.49. The summation of angles from openloop poles to 6±j4.49 is 226.3o. Therefore, the design point is not on the root
locus. Solutions to Problems 353
b. A compensator whose angular contribution is 226.3o180o = 46.3o is required. Assume a compensator zero at 5 canceling the pole. Thus, the breakaway from the real axis will be at the
required 6 if the compensator pole is at 9 as shown below. Adding the compensator pole and zero to the system poles, the gain at the design point is found to be
s+5
29.16. Summarizing the results: Gc(s) = s+9 with K = 29.16.
19.
Lead compensator design: Searching along the 120o line (ζ = 0.5), find the operating point at 1.531 + j2.652 with K = 354.5. Thus, Ts = 4
4
= 1.531 = 2.61 seconds. For the settling time to
ζωn 4
decrease by 0.5 second, Ts = 2.11 seconds, or Re = ζωn =  2.11 = 1.9. The imaginary part is
1.9 tan 60o = 3.29. Hence, the compensated dominant poles are 1.9 ± j3.29. The compensator zero
is at 5. Using the uncompensated system's poles along with the compensator zero, the summation of
angles to the design point, 1.9 ± j3.29 is 166.09o. Thus, the contribution of the compensator pole
3.29
must be 166.09o  180o = 13.91o. Using the following geometry, p  1.9 = tan 13.91o, or pc =
c
15.18. Adding the compensator pole and using 1.9 ± j3.29 as the test point, K = 1417.
Computer simulations yield the following: Uncompensated: Ts = 3 seconds, %OS = 14.6%. 354 Chapter 9: Design Via Root Locus Compensated: Ts = 2.3 seconds, %OS = 15.3%.
Lag compensator design: The lead compensated openloop transfer function is GLC(s) = 1417(s + 5)
. The uncompensated
(s + 2)(s + 4)(s + 6)(s + 8)(s + 15.18) 1
Kp = 354.5/(2 x 4 x 6 x 8) = 0.923. Hence, the uncompensated steadystate error is 1+K p = 0.52. Since we want 30 times improvement, the laglead compensated system must have a steadystate
error of 0.52/30 = 0.017. The lead compensated Kp = 1417*5/(2*4*6*8*15.18) = 1.215. Hence, the
1
leadcompensated error is 1+K = 0.451. Thus, the lag compensator must improve the leadp
1
compensated error by 0.451/0.017 = 26.529 times. Thus 0.451/ ( 1+K pllc ) = 26.529, where Kpllc = 57.823 is the leadlag compensated system's position constant. Thus, the improvement in Kp from the
lead to the laglead compensated system is 57.823/1.215 = 47.59. Use a lag compensator, whose zero
is 47.59 times farther than its pole, or Glag =
loop transfer function is GLLC(s) = (s + 0.04759)
. Thus, the leadlag compensated open(s + 0.001) 1417(s + 5)(s + 0.04759)
.
(s + 2)(s + 4)(s + 6)(s + 8)(s + 15.18)(s + 0.001) 20.
Program:
numg=1;
deng=poly([2 4 6 8]);
'G(s)'
G=tf(numg,deng);
Gzpk=zpk(G)
rlocus(G,0:5:500)
z=0.5;
pos=exp(pi*z/sqrt(1z^2))*100;
sgrid(z,0)
title(['Uncompensated Root Locus with ' , num2str(z), ' Damping Ratio
Line'])
[K,p]=rlocfind(G); %Allows input by selecting point on graphic
'Closedloop poles = '
p
i=input('Give pole number that is operating point
');
'Summary of estimated specifications for uncompensated system'
operatingpoint=p(i)
gain=K
estimated_settling_time=4/abs(real(p(i)))
estimated_peak_time=pi/abs(imag(p(i)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(i))^2+imag(p(i))^2)
Kpo=dcgain(K*G)
T=feedback(K*G,1);
'Press any key to continue and obtain the step response'
pause
step(T)
whitebg('w')
title(['Step Response for Uncompensated System with
' Damping Ratio'],'color','black')
'Press any key to go to Lead compensation' ' , num2str(z),... Solutions to Problems 355 pause
'Compensated system'
b=5;
'Lead Zero at b '
done=1;
while done>0
a=input('Enter a Test Lead Compensator Pole, (s+a). a =
');
numgglead=[1 b];
dengglead=conv([1 a],poly([2 4 6 8]));
'G(s)Glead(s)'
GGlead=tf(numgglead,dengglead);
GGleadzpk=zpk(GGlead)
wn=4/((estimated_settling_time0.5)*z);
clf
rlocus(GGlead,0:10:2000)
sgrid(z,wn)
axis([10 0 5 5])
title(['Lead Compensated Root Locus with ' , num2str(z),...
' Damping Ratio Line, Lead Pole at ', num2str(a), ', and Required Wn'])
done=input('Are you done? (y=0,n=1) ');
end
[K,p]=rlocfind(GGlead); %Allows input by selecting point on graphic
'Closedloop poles = '
p
i=input('Give pole number that is operating point
');
'Summary of estimated specifications for leadcompensated system'
operatingpoint=p(i)
gain=K
estimated_settling_time=4/abs(real(p(i)))
estimated_peak_time=pi/abs(imag(p(i)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(i))^2+imag(p(i))^2)
Kplead=dcgain(K*GGlead)
T=feedback(K*GGlead,1);
'Press any key to continue and obtain the step response'
pause
step(T)
whitebg('w')
title(['Step Response for Lead Compensated System with ' , num2str(z),...
' Damping Ratio'],'color','black')
'Press any key to continue and design lag compensation'
pause
'Improvement in steadystate error with lead compensator is'
error_improvement=(1+Kplead)/(1+Kpo)
additional_error_improvement=30/error_improvement
Kpnn=additional_error_improvement*(1+Kplead)1
pc=0.001
zc=pc*(Kpnn/Kplead)
numggleadlag=conv(numgglead,[1 zc]);
denggleadlag=conv(dengglead,[1 pc]);
'G(s)Glead(s)Glag(s)'
GGleadGlag=tf(numggleadlag,denggleadlag);
GGleadGlagzpk=zpk(GGleadGlag)
rlocus(GGleadGlag,0:10:2000)
z=0.5;
pos=exp(pi*z/sqrt(1z^2))*100;
sgrid(z,0)
title(['LagLead Compensated Root Locus with ' , num2str(z), ...
' Damping Ratio Line and Lag Pole at ',num2str(pc)])
[K,p]=rlocfind(GGleadGlag); %Allows input by selecting point on graphic
'Closedloop poles = '
p
i=input('Give pole number that is operating point
');
'Summary of estimated specifications for laglead compensated system'
operatingpoint=p(i) 356 Chapter 9: Design Via Root Locus gain=K
estimated_settling_time=4/abs(real(p(i)))
estimated_peak_time=pi/abs(imag(p(i)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(i))^2+imag(p(i))^2)
Kpleadlag=dcgain(K*GGleadGlag)
T=feedback(K*GGleadGlag,1);
'Press any key to continue and obtain the step response'
pause
step(T)
whitebg('w')
title(['Step Response for LagLead Compensated System with ',
num2str(z),...
' Damping Ratio and Lag Pole at ',num2str(pc)],'color','black') Computer response:
ans =
G(s) Zero/pole/gain:
1
(s+8) (s+6) (s+4) (s+2)
Select a point in the graphics window
selected_point =
1.5036 + 2.6553i ans =
Closedloop poles = p=
8.4807 + 2.6674i
8.4807  2.6674i
1.5193 + 2.6674i
1.5193  2.6674i
Give pole number that is operating point
ans = 3 Solutions to Problems 357 Summary of estimated specifications for uncompensated system operatingpoint =
1.5193 + 2.6674i gain =
360.8014 estimated_settling_time =
2.6328 estimated_peak_time =
1.1778 estimated_percent_overshoot =
16.3034 estimated_damping_ratio =
0.5000 estimated_natural_frequency =
3.0698 Kpo =
0.9396 ans = 358 Chapter 9: Design Via Root Locus Press any key to continue and obtain the step response ans =
Press any key to go to Lead compensation ans =
Compensated system ans =
Lead Zero at b
Enter a Test Lead Compensator Pole, (s+a). a = 10 ans =
G(s)Glead(s) Zero/pole/gain:
(s+5)
(s+10) (s+8) (s+6) (s+4) (s+2)
Are you done? (y=0,n=1) 1 Enter a Test Lead Compensator Pole, (s+a). a =
ans =
G(s)Glead(s) Zero/pole/gain:
(s+5)
(s+15) (s+8) (s+6) (s+4) (s+2)
Are you done? (y=0,n=1) 0 Select a point in the graphics window 15 Solutions to Problems 359 selected_point =
1.9076 + 3.2453i ans =
Closedloop poles = p=
13.0497 + 1.9313i
13.0497  1.9313i
5.0654
1.9176 + 3.2514i
1.9176  3.2514i
Give pole number that is operating point 4 ans =
Summary of estimated specifications for leadcompensated system operatingpoint =
1.9176 + 3.2514i gain =
1.3601e+003 estimated_settling_time =
2.0860 estimated_peak_time =
0.9662 360 Chapter 9: Design Via Root Locus estimated_percent_overshoot =
16.3034 estimated_damping_ratio =
0.5000 estimated_natural_frequency =
3.7747 Kplead =
1.1806 ans =
Press any key to continue and obtain the step response ans =
Press any key to continue and design lag compensation ans =
Improvement in steadystate error with lead compensator is error_improvement =
1.1243 additional_error_improvement =
26.6842 Solutions to Problems 361 Kpnn =
57.1876 pc =
0.0010 zc =
0.0484 ans =
G(s)Glead(s)Glag(s) Zero/pole/gain:
(s+5) (s+0.04844)
(s+15) (s+8) (s+6) (s+4) (s+2) (s+0.001)
Select a point in the graphics window
selected_point =
1.8306 + 3.2919i ans =
Closedloop poles = p=
13.0938 + 2.0650i
13.0938  2.0650i
5.0623
1.8617 + 3.3112i
1.8617  3.3112i
0.0277 362 Chapter 9: Design Via Root Locus Give pole number that is operating point 4 ans =
Summary of estimated specifications for laglead compensated system operatingpoint =
1.8617 + 3.3112i gain =
1.4428e+003 estimated_settling_time =
2.1486 estimated_peak_time =
0.9488 estimated_percent_overshoot =
16.3034 estimated_damping_ratio =
0.5000 estimated_natural_frequency =
3.7987 Kpleadlag = Solutions to Problems 363 60.6673 ans =
Press any key to continue and obtain the step response 364 Chapter 9: Design Via Root Locus Solutions to Problems 365 366 Chapter 9: Design Via Root Locus 21.
a. For the settling time to be 2.86 seconds with 4.32% overshoot, the real part of the compensated 4
4
dominant poles must be T = 2.86 = 1.4. Hence the compensated dominant poles are 1.4 ± j1.4.
s
Assume the compensator zero to be at 1 canceling the system pole at 1. The summation of angles to
the design point at 1.4 ± j1.4 is 176.19o. Thus the contribution of the compensator pole is
1.4
176.19o  180o = 3.81o. Using the geometry below, p  1.4 = tan 3.81o, or pc = 22.42.
c Adding the compensator pole and using 1.4 ± j1.4 as the test point, K = 88.68.
b. Uncompensated: Search the 135o line (4.32% overshoot) and find the uncompensated dominant 1.11
4
4
pole at  0.419 + j0.419 with K = 1.11. Thus Kv = 3 = 0.37. Hence, Ts =
= 0.419 = 9.55
ζωn Solutions to Problems 367 88.68
seconds and %OS = 4.32%. Compensated: Kv = 22.42 x 3 = 1.32 (Note: steadystate error
improvement is greater than 2). Ts = 4
4
= 1.4 = 2.86 seconds and %OS = 4.32%.
ζωn c. Uncompensated: K = 1.11; Compensated: K = 88.68.
d. Uncompensated: Searching the real axis segments for K = 1.11 yields a higherorder pole at 3.16 which is more than five times the real part of the uncompensated dominant poles. Thus the secondorder approximation for the uncompensated system is valid.
Compensated: Searching the real axis segments for K = 88.68 yields a higherorder pole at 22.62 which is more than five times the real part of the compensated dominant poles' real part. Thus the
second order approximation is valid.
e.
Step Response
Uncompensated
1
0.9
0.8 Amplitude 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 0 5 10
Time (sec.) 368 Chapter 9: Design Via Root Locus Ramp Response
Uncompensated 80 70 60 Amplitude 50 40 30 20 10 0 0 10 20 30 40 50 60 70 3 3.5 Time (sec.) Step Response
Compensated
1
0.9
0.8 Amplitude 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 0 0.5 1 1.5 2
Time (sec.) 2.5 80 Solutions to Problems 369 Ramp Response
Compensated 80 70 60 Amplitude 50 40 30 20 10 0 0 10 20 30 40 50 60 70 80 Time (sec.) 22.
a. Searching the 30% overshoot line (ζ = 0.358; 110.97o) for 180o yields 1.464 + j3.818 with a gain, K = 218.6.
b . Tp = π
π
=
= 0.823 second. Kv =
ωd
3.818 218.6
= 3.975.
(5)(11) c. Lead design: From the requirements, the percent overshoot is 15% and the peak time is 0.4115 second. Thus, ζ = ln(%/100)
π2+ln2(%/100) π
= 0.517; ωd = T = 7.634 = ωn 1ζ2 . Hence, ωn = 8.919. The
p design point is located at ζωn + jωn 1ζ2 = 4.61 + j7.634. Assume a lead compensator zero at 5.
Summing the angles of the uncompensated system's poles as well as the compensator zero at 5 yields
–171.2o. Therefore, the compensator pole must contribute (171.2o  180o) = 8.8o. Using the
geometry below, jω
splane X
pc 8.8o
4.61 j7.634
σ 370 Chapter 9: Design Via Root Locus 7.634
= tan (8.8o) . Hence, pc = 53.92. The compensated openloop transfer function is
pc − 4.61
K
. Evaluating the gain for this function at the point, 4.61 + j7.634 yields
s(s + 11)( s + 53.92)
K = 4430.
Lag design: The uncompensated The lead compensated Kv = Kv = 218.6
= 3.975 . The required Kv is 30*3.975 = 119.25.
(5)(11) 4430
= 7.469. Thus, we need an improvement over the lead
(11)(53.92) compensated system of 119.25/7.469 = 15.97. Thus, use a lag compensator
Glag(s) =
23. 4430(s + 0.01597)
s + 0.01597
. The final openloop function is
.
s + 0.001
s(s + 11)( s + 53.92)(s + 0.001) a. Searching along the 10% overshoot line (ζ = 0.591) the operating point is found to be –1.85 + j2.53 with K = 21.27. A third pole is at –10.29. Thus, the estimated performance before
compensation is: 10% overshoot, Ts = 21.27
4
= 0.266 .
= 2.16 seconds, and K p =
(8)(10)
1.85 b. Lead design: Place compensator zero at –3. The desired operating point is found from the desired specifications. ζω n = 4
4
44
= 6.768 . Thus,
= = 4 and ω n = =
Ts 1
ζ 0.591 Im = ω n 1 − ζ = 6.768 1 − 0.591 = 5.46 . Hence the design point is –4 +j5.46. The angular
2 2 contribution of the system poles and compensator zero at the design point is –166.960. Thus, the
compensator pole must contribute –1800 + 166.960 = 13.040. Using the geometry below, jω
splane X
pc j5.46 13.04o
4 σ 5.46
= tan (13.04o) . Hence, pc = 27.57. The compensated openloop transfer function is
pc − 4
K (s + 3)
. Evaluating the gain for this function at the point
2
(s + 4s + 8)( s + 10)( s + 27.57)
4 + j5.46 yields K = 1092 with higherorder poles at –4.055 and –29.52. Solutions to Problems 371 Lag design: For the leadcompensated system, Kp = 1.485. Thus, we need an improvement of (s + 0.06734)
10
. Finally, the equivalent forwardpath
= 6.734 times. Hence, Glag ( s) =
1.485
( s + 0.01)
1092( s + 3)(s + 0.06734 )
.
transfer function is Ge (s ) = 2
(s + 4 s + 8)(s + 10 )(s + 27.57)( s + 0.01)
c.
Step Response
LagLead Compensated
0.9 0.8 0.7 Amplitude 0.6 0.5 0.4 0.3 0.2 0.1 0 0 20 40 60 80 100 120 Time (sec.) Step Response
LagLead Compensated
0.7 0.6 Amplitude 0.5 0.4 0.3 0.2 0.1 0
0 0.2 0.4 0.6 0.8 1 Time (sec.) 1.2 1.4 1.6 1.8 2 372 Chapter 9: Design Via Root Locus 24.
a. Uncompensated: Search the 135o line (4.32% overshoot) for 180o and find the dominant pole at –3 + j3 with K = 10.
Lag Compensated: Search the 135o line (4.32% overshoot) for 180o and find the dominant pole at 2.88 + j2.88 with K = 9.95.
10
b. Uncompensated: Kp = 2 x 4 = 1.25
9.95 x 0.5
Lag compensated: Kp = 2 x 4 x 0.1 = 6.22
c. %OS = 4.32% both cases; Uncompensated Ts = 4
4
4
= 3 1.33 seconds; Compensated Ts = 2.88 = 1.39 seconds
ζωn d. Uncompensated: Exact secondorder system; approximation OK Compensated: Search real axis segments of the root locus and find a higherorder pole at 0.3. System
should be simulated to see if there is effective pole/zero cancellation with zero at  0.5.
e. Solutions to Problems 373 The compensated system's response takes a while to approach the final value.
f. We will design a lead compensator to speed up the system by a factor of 5. The leadcompensated dominant poles will thus be placed at –15 ± j15. Assume a compensator zero at  4 that cancels the
openloop pole at  4. Using the system's poles and the compensator's zero, the sum of angles to the
design point, 15±j15 is 131.69o. Thus, the angular contribution of the compensator pole must be
131.69o  180o =  48.31o. Using the geometry below, pc = 28.36. K(s+0.5)(s+4)
Using the compensated openloop transfer function, Ge(s) = (s+2)(s+4)(s+0.1)(s+28.36) and using
the design point –15 ± j15, K = 404.1.The time response of the laglead compensated system is
shown below. 25. π Since Tp = 1.047, the imaginary part of the compensated closedloop poles will be 1.047 = 3.
Im
Since Re = tan (cos1ζ), the magnitude of the real part will be
point is – 4 + j3. Assume an PI controller, Gc(s) = Im
= 4. Hence, the design
tan(cos1ζ) s+0.1
s , to reduce the steadystate error to zero. 374 Chapter 9: Design Via Root Locus Using the system's poles and the pole and zero of the ideal integral compensator, the summation of
angles to the design point is 225.7o. Hence, the ideal derivative compensator must contribute 225.7o180o = 45.7o . Using the geometry below, zc = 6.93. The PID controller is thus (s+6.93)(s+0.1)
. Using all poles and zeros of the system and PID
s controller, the gain at the design point is K = 3.08. Searching the real axis segment, a higherorder
pole is found at  0.085. A simulation of the system shows the requirements are met.
26.
a. The desired operating point is found from the desired specifications. ζω n = ωn = 2 ζ = 44
= = 2 and
Ts 2 2
= 4.954 . Thus, Im = ωn 1 − ζ 2 = 4.954 1 − 0.4037 2 = 4.5324 . Hence
0.4037 the design point is –2 +j4.5324. Now, add a pole at the origin to increase system type and drive error
to zero for step inputs.
Now design a PD controller. The angular contribution to the design point of the system poles and
pole at the origin is 101.90. Thus, the compensator zero must contribute 1800 – 101.90 =78.10. Using
the geometry below, 4.5324
= tan(78.10 ) . Hence, zc = 2.955. The compensated openloop transfer function with PD
zc − 2
compensation is K ( s + 2.955)
. Adding the compensator zero to the system and
s ( s + 4)( s + 6)( s + 10) Solutions to Problems 375 evaluating the gain for this at the point –2 + j4.5324 yields K = 294.51 with a higherorder pole at
2.66 and 13.34.
PI design: Use G PI (s ) = Ge ( s ) = ( s + 0.01)
. Hence, the equivalent openloop transfer function is
s K ( s + 2.955)( s + 0.01)
with K = 294.75.
s 2 ( s + 4)( s + 6)( s + 10) b.
Program (Step Response):
numg=[2.995 0.01];
deng=[0 0 4 6 10];
K=294.75;
G=zpk(numg,deng,K)
T=feedback(G,1);
step(T) Computer response:
Zero/pole/gain:
294.75 (s+2.995) (s+0.01)
s^2 (s+4) (s+6) (s+10) Program (Ramp Response):
numg=[2.995 0.01];
deng=[0 0 4 6 10];
K=294.75;
G=zpk(numg,deng,K)
T=feedback(G,1);
Ta=tf([1],[1 0]);
step(T*Ta) 376 Chapter 9: Design Via Root Locus Computer response:
Zero/pole/gain:
294.75 (s+2.995) (s+0.01)
s^2 (s+4) (s+6) (s+10) 27.
Program:
numg=
deng=[4 6 10]
'G(s)'
G=zpk(numg,deng,1)
pos=input('Type desired percent overshoot ');
z=log(pos/100)/sqrt(pi^2+[log(pos/100)]^2);
Ts=input('Type desired settling time ');
zci=input(...
'Type desired position of integral controller zero (absolute value) ');
wn=4/(Ts*z);
desired_pole=(z*wn)+(wn*sqrt(1z^2)*i)
angle_at_desired_pole=(180/pi)*angle(evalfr(G,desired_pole))
PD_angle=180angle_at_desired_pole;
zcpd=((imag(desired_pole)/tan(PD_angle*pi/180))real(desired_pole));
'PD Compensator'
numcpd=[1 zcpd];
dencpd=[0 1];
'Gcpd(s)'
Gcpd=tf(numcpd,dencpd)
Gcpi=zpk([zci],[0],1)
Ge=G*Gcpd*Gcpi
rlocus(Ge)
sgrid(z,0)
title(['PID Compensated Root Locus with ' ,...
num2str(pos), '% Damping Ratio Line'])
[K,p]=rlocfind(Ge);
'Closedloop poles = '
p
f=input('Give pole number that is operating point
');
'Summary of estimated specifications for selected point' Solutions to Problems 377 'on PID compensated root locus'
operatingpoint=p(f)
gain=K
estimated_settling_time=4/abs(real(p(f)))
estimated_peak_time=pi/abs(imag(p(f)))
estimated_percent_overshoot=pos
estimated_damping_ratio=z
estimated_natural_frequency=sqrt(real(p(f))^2+imag(p(f))^2)
T=feedback(K*Ge,1);
step(T)
title(['Step Response for PID Compensated System with ' ,...
num2str(pos),'% Damping Ratio Line'])
pause
one_over_s=tf(1,[1 0]);
Tr=T*one_over_s;
t=0:0.01:10;
step(one_over_s,Tr)
title('Ramp Response for PID Compensated System') Computer response:
numg = deng =
0 4 6 10 ans =
G(s)
Zero/pole/gain:
1
s (s+4) (s+6) (s+10)
Type desired percent overshoot 25
Type desired settling time 2
Type desired position of integral controller zero (absolute value) 0.01
desired_pole =
2.0000 + 4.5324i
angle_at_desired_pole =
101.8963
ans =
PD Compensator
ans = 378 Chapter 9: Design Via Root Locus Gcpd(s)
Transfer function:
s + 2.955
Zero/pole/gain:
(s+0.01)
s
Zero/pole/gain:
(s+2.955) (s+0.01)
s^2 (s+4) (s+6) (s+10)
Select a point in the graphics window
selected_point =
1.9931 + 4.5383i
ans =
Closedloop poles =
p=
13.3485
1.9920 + 4.5377i
1.9920  4.5377i
2.6575
0.0100
Give pole number that is operating point 2 ans =
Summary of estimated specifications for selected point
ans =
on PID compensated root locus
operatingpoint =
1.9920 + 4.5377i
gain =
295.6542
estimated_settling_time =
2.0081
estimated_peak_time = Solutions to Problems 379 0.6923
estimated_percent_overshoot =
25
estimated_damping_ratio =
0.4037
estimated_natural_frequency =
4.9557 380 Chapter 9: Design Via Root Locus 28. Openloop poles are at 2,  0.134, and 1.87. An openloop zero is at 3. Searching the 121.13o line
(ζ = 0.517), find the closedloop dominant poles at 0.747 + j1.237 with K = 1.58. Searching the real
axis segments locates a higherorder pole at 2.51. Since the openloop zero is a zero of H(s), it is not
a closedloop zero. Thus, there are no closedloop zeros.
29.
a. The damping ratio for 15% overshoot is 0.517. The desired operating point is found from the desired specifications. ζω n = 1.333 1.333
44
=
= 2.578 . Thus,
= = 1.333 and ω n =
Ts 3
ζ
0.517 Solutions to Problems 381 Im = ω n 1 − ζ = 2.578 1 − 0.517 = 2.207 . Hence the design point is –1.333 + j2.207. The
2 2 angular contribution of the system poles and compensator zero at the design point is 100.80. Thus, the
compensator zero must contribute 1800 – 100.80 = 79.20. Using the geometry below, jω
splane
j2.207
X
zc 79.2o
1.333 σ 2.207
= tan (79.2o) . Hence, zc = 1.754. The compensated openloop transfer function with PD
zc − 1.333
compensation is K ( s + 1.754)
. Evaluating the gain for this function at the point
s (s + 2)(s + 4)( s + 6) –1.333 + j2.207 yields K = 47.28 with higherorder poles at –1.617 and –7.718. Following
Figure 9.49(c) in the text, 1
= 1.754 . Therefore, K f = 0.5701 . Also, using the notation of
Kf Figure 9.49(c), K1 K f = 47.28 , from which K1 = 82.93 . b.
Program:
K1=82.93;
numg=K1;
deng=poly([0 2 4 6]);
'G(s)'
G=tf(numg,deng);
Gzpk=zpk(G)
Kf=0.5701
numh=Kf*[1 1.754];
denh=1
'H(s)'
H=tf(numh,denh);
Hzpk=zpk(H)
'T(s)'
T=feedback(G,H);
T=minreal(T)
step(T)
title('Step Response for Feedback Compensated System') Computer response:
ans =
G(s)
Zero/pole/gain:
82.93
s (s+6) (s+4) (s+2) 382 Chapter 9: Design Via Root Locus Kf =
0.5701
denh =
1
ans =
H(s)
Zero/pole/gain:
0.5701 (s+1.754)
ans =
T(s)
Transfer function:
82.93
s^4 + 12 s^3 + 44 s^2 + 95.28 s + 82.93 30. a. σd = ζωn = 4/Ts = 4/1 = 4. 5% overshoot > ζ = 0.69. Since ζωn = 4, ωn = 5.8.
ωd = ωn 1ζ2 = 4.195. Thus, the design point is –1 + j4.195. The sum of angles from the minor loop's openloop poles to the design point is 263.634o. Thus, the minorloop's openloop zero must Solutions to Problems 383 4.195
contribute 83.634o to yield 180o at the design point. Hence, z  4 = tan 83.634o, or zc = a = 4.468
c from the geometry below. jω
j4.195
83.634o o
83.634
zc splane
σ 4 Adding the zero and calculating the gain at the design point yields K1 = 38.33. Therefore, the minorloop openloop transfer function is K1G(s)H(s) = 38.33(s+4.468)
s(s+4)(s+9) . The equivalent minorloop K1G(s)
38.33
closedloop transfer function is Gml(s) = 1+K G(s)H(s) = 3
. A simulation
1
s +13s2+74.33s+171.258
of the step response of the minor loop is shown below.
Computer response: Minorloop ClosedLoop Response
0.25 Amplitude 0.2
0.15
0.1
0.05
0 0 0.5 Time (secs) b. The majorloop openloop transfer function is G e (s) = 1 1.5 38.33K
.
s + 13s + 74.33s + 171.258
3 2 Drawing the root locus using Ge(s) and searching along the 10% overshoot line (ζ = 0.591) for 180o
yields the point 3.349 + j4.572 with a gain 38.33K = 31.131, or K = 0.812. 384 Chapter 9: Design Via Root Locus c.
Program:
numg=31.131;
deng=[1 13 74.33 171.258];
'G(s)'
G=tf(numg,deng)
T=feedback(G,1);
step(T)
title('Majorloop ClosedLoop Response') Computer response:
G(s)
Transfer function:
31.13
s^3 + 13 s^2 + 74.33 s + 171.3 d. Adding the PI compensator, Ge(s) = 31.131(s+0.1)
.
s(s3+13s2+74.33s+171.258) Program:
numge=31.131*[1 0.1];
denge=[1 13 74.33 171.258 0];
'Ge(s)'
Ge=tf(numge,denge)
T=feedback(Ge,1);
t=0:0.1:10;
step(T,t)
title('Majorloop ClosedLoop Response with PI Compensator')
pause
step(T) Solutions to Problems 385 title('Majorloop ClosedLoop Response with PI Compensator') Computer response:
ans =
Ge(s)
Transfer function:
31.13 s + 3.113
s^4 + 13 s^3 + 74.33 s^2 + 171.3 s 386 Chapter 9: Design Via Root Locus 31. R2
a. PI controller: Using Table 9.10, R
1 1
s+R C
2
s+0.01
=s
, R2C = 100. Let C = 25 µF. Therefore,
s R2 = 4 MΩ. For unity gain, R1 = 4 MΩ. Compensate elsewhere in the loop for the compensator
negative sign.
1
b. PD controller: Using Table 9.10, R2C(s+R C ) = s+2. Hence, R1C = 0.5. Let C = 1 µF.
1
Therefore, R1 = 500 KΩ. For unity gain, R2C = 1, or R2 = 1 MΩ. Compensate elsewhere in the loop
for the compensator negative sign.
32. 1
R2 C
s + 0.1
. Thus, R2C = 10, and
a. Lag compensator: See Table 9.11.
=
1
s + 0.01
s+
(R1 + R 2 )C
s+ (R1 + R2)C = 100. Letting C = 10 µ F, we find R2 = 1 MΩ. Also R1C = 100  R2C = 90, which
yields R1 = 9 MΩ. The loop gain also must be multiplied by R1 + R 2
.
R2 1
s+R C
1
s+2
b. Lead compensator: See Table 9.11.
1
1 = s+5 . Thus, R1C = 0.5, and
s+R C+R C
1
2
1
1
R1C + R2C = 5. Letting C = 1 µF, R2 = 333 KΩ, and R1 = 500 KΩ.
c. Laglead compensation: See Table 9.11. 1
1
(s + R C )(s + R C )
11
22
(s+0.1)(s+1)
=2
. Thus, R1C1 = 1, and
1
1
1
1
s + 10.01s + 0.1
s2 + (R C + R C + R C )s + R R C C
11
22
21
1212
1
1
1
1
R2C2 = 10. Also, R C + R C + R C = 1 + 0.1 + R C = 10.01, or R2C1 = 0.112. Letting C1 =
11
22
21
21
10 µF, we find R1 = 10 MΩ , R2 = 1.12 MΩ, and C2 = 8.9 µF.
33. C1
s+0.1
a. Lag compensator: See Table 9.10 and Figure 9.58. s+0.01 = C
2 1
(s+R C )
11
. Therefore,
1
(s+R C )
22 R1C1 = 10; R2C2 = 100. Letting C1 = C2 = 20 µF, we find R1 = 500 KΩ and R2 = 5 MΩ.
Compensate elsewhere in the loop for the compensator negative sign. Solutions to Design Problems 387 C1
s+2
b. Lead compensator: See Table 9.10 and Figure 9.58. s+5 = C
2 1
(s+R C )
11
. Therefore,
1
(s+R C )
22 R1C1 = 0.5 and R2C2 = 0.2. Letting C1 = C2 = 20 µF, we find R1 = 25 KΩ and R2 = 10 MΩ.
Compensate elsewhere in the loop for the compensator negative sign.
c. Laglead compensator: See Table 9.10 and Figure 9.58. For lag portion, use (a). For lead: C1
s+1
s+10 = C2 1
(s+R C )
11
. Therefore, R1C1 = 1 and R2C2 = 0.1. Letting C1 = C2 = 10 µF, we find
1
(s+R C )
22 R1 = 100 KΩ and R2 = 10 KΩ. The following circuit can be used to implement the design. SOLUTIONS TO DESIGN PROBLEMS
34.
θm(s)
a. E (s) =
a Kb = Kt
RaJ
KtKb
1
s(s+ J (D + R ))
a Ea
5
4
1
1
= 60000 1
= 0.005; Jeq = 5 (10 x 4 )2 = 0.05; Deq = 1 (10 )2 = 0.01;
ω
x 60 x 2π
2π θm(s)
Ts
Kt
0.5
2
Ra = Ea = 5 = 0.1. Therefore, Ea(s) = s(s+0.21) .
b. The block diagram of the system is shown below. 388 Chapter 9: Design Via Root Locus Forming an equivalent unity feedback system, 1000
Now, T(s) = 2
. Thus, ωn = 1000 ; 2ζωn = 0.21 + 0.2Kt. Since ζ = 0.5,
s + (0.21 + 0.2Kt)s + 1000
Kt = 157.06.
1000
c. Uncompensated: Kt = 0; T(s) = 2
; ωn = 31.62 rad/s; ζ = 3.32 x 103;
s + 0.21s + 1000 %OS = e −ζπ /
Tp = π
ωn 1ζ2 1−ζ 2 4 x100 = 98.96%; Ts = ζω = 38.09 seconds;
n 1000
= 9.93 x 102 second; Kv = 0.21 = 4761.9. 1000
Compensated: Kt = 157.06; T(s) = 2
; ωn = 31.62 rad/s; ζ = 0.5;
s + 31.62s + 1000 %OS = e −ζπ / 1−ζ 2 4 x100 = 16.3%; Ts = ζω = 0.253 second; Tp =
n 1000
Kv = 31.62 = 31.63.
35. 25
a. T(s) = 2
; Therefore, ωn = 5; 2ζωn = 1; ζ = 0.1;
s + s + 25 %OS = e −ζπ / 1−ζ 2 4
x100 = 73%; Ts = ζω = 8 seconds.
n π
ωn 1ζ2 = 0.115 second; Solutions to Design Problems 389 b. From Figure P9.6(b), T(s) =
ωn = 25K 1
. Thus,
s + (1 + 25K f )s + 25K 1
2 25K1 ; 2ζωn = 1 + 25Kf. For 25% overshoot, ζ = 0.404. For Ts = 0.2 = Therefore 1 + 25Kf = 2ζωn = 40, or Kf = 1.56. Also, ωn = 4
, ζωn = 20.
ζωn 20
= 49.5.
ζ ωn2
49.52
Hence K1 = 25 = 25 = 98.01. 25
1
c. Uncompensated: G(s) = s(s+1) ; Therefore, Kv = 25, and e(∞) = K v Compensated: G(s) = 1
e(∞) = K v = 0.04. 25K 1
25 x 98.01
; Therefore, Kv = 1+25 x 1.56 = 61.26, and
s(s + 1 + 25K f ) = 0.0163. 36.
a. The transfer functions of the subsystems are as follows: Pot: Gp(s) = K1
5π
1
= 2 ; Amplifier: Ga(s) = s+20 ; Motor and load: Since the time to rise to 63% of
10π the final value is 0.5 second, the pole is at 2. Thus, the motor transfer function is of the form, Gm(s)
K
K
100
= s(s+2) . But, from the problem statement, 2 = 10 , or K = 20. The block diagram of the system
is shown below. 390 Chapter 9: Design Via Root Locus Using the equivalent system, search along the 117.126o line (20% overshoot) and find the dominant
secondorder pole at  0.89 + j1.74 with K = 10K1 = 77.4. Hence, K1 = 7.74.
77.4
1
b. Kv = 2 x 20 = 1.935. Therefore, e(∞) = K v c. %OS = 20%; ζ = Ts = %OS
 ln ( 100 )
%OS
π2 + ln2 ( 100 ) 4
= 4.49 seconds; Tp =
ζωn
ω π 2
n 1ζ = 0.517. = 0.456; ωn = 0.892 + 1.742 = 1.95 rad/s; = 1.81 seconds. d. The block diagram of the minor loop is shown below. 20
The transfer function of the minor loop is GML(s) = s(s+2+20K ) . Hence, the block diagram of the
f equivalent system is where a = 2 + 20Kf. The design point is now found. Since %OS = 20%, ζ = 0.456. Also, since Ts = %OS
 ln ( 100 )
=
%OS
π2 + ln2 ( 100 ) 4
= 2 seconds, ωn = 4.386 rad/s. Hence, the design point is –2 + j3.9.
ζωn Solutions to Design Problems 391 Using just the openloop poles at the origin and at 20, the summation of angles to the design point is
129.37o. The pole at a must then be contributing 129.37o  180o = 50.63o. Using the geometry
below, a = 5.2, or Kf = 0.16. Adding the pole at 5.2 and using the design point, we find 10K1 = 407.23, or K1 = 40.723.
Summarizing the compensated transient characteristics: ζ = 0.456; ωn = 4.386; %OS = 20%; Ts =
4
= 2 seconds; Tp =
ζωn
ω π
1ζ2 n 407.23
= 0.81 seconds; Kv = 20 x 5.2 = 3.92. 37. Block diagram
K1
20π volts
= 2.
Preamplifier/Power amplifier: (s+40) ; Pots:
5(2π) rad.
Torquespeed curve:
T (Nm) 50 v 75
25
50 150 ω (rad/sec) rad
rev
1 min
rev
1 min
rad
where 1432.35 min x 60 sec x 2π rev = 150 rad/sec; 477.45 min x 60 sec x 2π rev = 50 rad/sec.
50
The slope of the line is  100 =  0.5. Thus, its equation is y = 0.5x + b. Substituting one of the
Kt Tstall 100
ea
points, find b = 100. Thus Tstall = 100, and ωno load = 200. R = e
= 50 = 2; Kb =
=
ωno load
a
a
50
200 = 0.25. 392 Chapter 9: Design Via Root Locus
θm(s)
Motor: E (s) =
a Kt/(RaJ)
0.02
KtKb = s(s+0.505) , where J = 100, D = 50.
1
s(s+ J (D+ R ))
a Gears: 0.1 Drawing block diagram: θc (s) + 0.02 K
40
s +140
(s + 40 ) 2 s( s + 0. 505) θL (s) 0.1  2 θc(s) + 0. 004 K1 θL(s) s( s + 0. 505)(s + 40 )  b. Compensator design  Lead 10% overshoot and Ts = 1 sec yield a design point of  4 + j5.458. Sum of angles of uncompensated
system poles to this point is 257.491o. If we place the lead compensator zero over the
uncompensated system pole at 0.505, the angle at the design point is 134.858o. Thus, the lead
compensator pole must contribute 134.858o  180o = 45.142o. Using the geometry below
5.458
o
pc  4 = tan(45.142 ), or pc = 9.431.
X j5.458 45.142o
p
c 4 Using the uncompensated poles and the lead compensator, the gain at the design point is
0.004K1 = 1897.125. Solutions to Design Problems 393 Compensator design  Lag zlag 1000
1897.125
With lead compensation, Kv = (40)(9.431) = 5.0295.029. Since we want Kv = 1000, p
=
=
lag 5.029
198.85. Use plag = 0.001. Hence zlag = 0.1988. The lag compensated 1897.125(s+0.1988)
Ge(s) = s(s+40)(s+9.431)(s+0.001) .
c. Compensator schematic 1
1
lag: R C = 0.1988. Let C = 100 µF. Then R2 = 50.3 kΩ. Now, (R +R )C = 0.001.
2
12
R2
Thus, R1 = 9.95 MΩ. Buffer gain = reciprocal of lag compensator's R + R . Hence buffer
1
2
gain = R1 + R2
= 198.8.
R2 1
1
1
lead: R C = 0.505. Let C = 10 µF. Then R1 = 198 kΩ. Now, R C + R C = 9.431.
1
1
2
Thus, R2 = 11.2 k Ω. d.
Program:
numg= 1897.125*[1 0.1988];
deng=poly([0 40 9.431 .001]);
'G(s)'
G=tf(numg,deng);
Gzpk=zpk(G)
rlocus(G)
pos=10
z=log(pos/100)/sqrt(pi^2+[log(pos/100)]^2)
sgrid(z,0)
title(['Root Locus with ' , num2str(pos), ' Percent Overshoot Line'])
[K,p]=rlocfind(G) %Allows input by selecting point on graphic
pause
T=feedback(K*G,1);
step(T)
title(['Step Response for Design of ' , num2str(pos), ' Percent']) 394 Chapter 9: Design Via Root Locus Computer response:
ans =
G(s)
Zero/pole/gain:
1897.125 (s+0.1988)
s (s+40) (s+9.431) (s+0.001)
pos =
10
z=
0.5912
Select a point in the graphics window
selected_point =
3.3649 + 4.8447i
K=
0.9090
p=
41.3037
3.9602 + 4.9225i
3.9602  4.9225i
0.2080 Solutions to Design Problems 395 396 Chapter 9: Design Via Root Locus 38. Consider only the minor loop. Searching along the 143.13o line (ζ = 0.8), locate the minorloop
dominant poles at 3.36 ± j2.52 with Kf = 8.53. Searching the real axis segments for Kf = 8.53 locates
a higherorder pole at  0.28. Using the minorloop poles as the openloop poles for the entire system,
search along the 120o line (ζ = 0.5) and find the dominant secondorder poles at 1.39 + j2.41 with K
= 27.79. Searching the real axis segment locates a higherorder pole at  4.2.
39. Consider only the minor loop. Searching along the 143.13o line (ζ = 0.8), locate the minorloop
dominant poles at 7.74 ± j5.8 with Kf = 36.71. Searching the real axis segments for Kf = 36.71
locates a higherorder pole at  0.535. Using the minorloop poles at 7.74 ± j5.8 and  0.535 as the
openloop poles (the openloop zero at the origin is not a closedloop zero) for the entire system,
search along the 135o line (ζ = 0.707; 4.32% overshoot) and find the dominant secondorder poles at
 4.38 + j4 .38 with K = 227.91. Searching the real axis segment locates a higherorder pole at 7.26.
Uncompensated system performance: Setting Kf = 0 and searching along the 135o line (4.32%
overshoot) yields 2.39 + j2.39 as the point on the root locus with K = 78.05. Searching the real axis
segments of the root locus for K = 78.05 locates a higherorder pole at 11.2. The following table
compares the predicted uncompensated characteristics with the predicted compensated
characteristics.
Uncompensated
78.05
G(s) = (s+1)(s+5)(s+10)
Dominant poles: 2.39 + j2.39 Compensated
227.91
G(s) = 3
2+101.71s+50
s +16s
Dominant poles:  4.38 + j4 .38 ζ = 0.707 ζ = 0.707 %OS = e −ζπ /
ωn = Ts =
Tp = 1−ζ 2 x100 = 4.32% 2.392+2.392 = 3.38 rad/s
4
= 1.67 seconds
ζωn π
ωn 1ζ2 = 1.31 seconds %OS = e −ζπ /
ωn = Ts =
Tp = 1−ζ 2 x100 = 4.32% 4.382+4.382 = 6.19 rad/s
4
= 0.91 second
ζωn π
ωn 1ζ2 = 0.72 second 78.05
Kp = 1 x 5 x 10 = 1.56 Kp = 227.91
= 4.56
50 Higherorder pole: 11.22 Higherorder pole: 7.26 Secondorder approximation OK Higherorder pole not 5x further from
imaginary axis than dominant poles.
Simulate to be sure of the performance. 40. In Problem 46, Chapter 8 , the dominant poles,  40 ± j57.25, yielded Ts = 0.1 second and 11.14%
overshoot. The unity feedback system consisted of a gain adjusted forward transfer function of Solutions to Design Problems 397 20000K
G(s) = s(s+100)(s+500)(s+800) , where K = 102,300. To reduce the settling time by a factor of 2 to
0.05 seconds and keep the percent overshoot the same, we double the coordinates of the dominant
poles to –80 ± j114.5. Assume a lead compensator with a zero at 100 that cancels the plant's pole at
100. The summation of angles of the remaining plant poles to the design point is 149.23o. Thus, the
angular contribution of the compensator pole must be 149.23o  180o = 30.77o. Using the
114.5
geometry below, p  80 = tan 30.77o, or pc = 272.3.
c Adding this pole to the poles at the origin, 500, and 800 yields K = 9.92 x 109 at the design point,
80 ± j114.5. Any higherorder poles will have a real part greater than 5 times that of the dominant
pair. Thus, the secondorder approximation is OK.
41. 0.35K
Uncompensated: G(s)H(s) = (s+0.4)(s+0.5)(s+0.163)(s+1.537) . Searching the 133.639o line
(%OS = 5%), find the dominant poles at  0.187 ± j0.196 with gain, 0.35K = 2.88 x 102. Hence, the
estimated values are: %OS = 5%; Ts = 4
4
= 0.187 = 21.39 seconds; Tp =
ζωn
ω n π
1ζ2 π = 0.196 = 16.03 seconds; Kp = 0.575.
PD compensated: Design for 8 seconds peak time and 5% overshoot.
ζ= %OS
 ln ( 100 )
π
= 0.69. Since Tp =
= 8 seconds and ωn 1ζ2 = 0.393,
%OS
ωn 1ζ2
π2 + ln2 ( 100 ) ωn = 0.5426. Hence, ζωn = 0.374. Thus, the design point is  0.374 + j0.393. The summation of angles from the system's poles to the design point is 295.34o. Thus, the angular contribution of the
controller zero must be 295.34o180o = 115.34o. Using the geometry below, 398 Chapter 9: Design Via Root Locus 0.393
o
o
0.374  zc = tan (180  115.34 ), from which zc = 0.19. Adding this zero to the system's poles and
using the design point,  0.374 + j0.393, the gain, 0.35K = 0.205.
PID compensated: Assume the integral controller, Gc(s) = s+0.01
. The total openloop transfer
s 0.35K(s+0.19)(s+0.01)
function is GPID(s)H(s) = s(s+0.4)(s+0.5)(s+0.163)(s+1.537) .
Check: The PID compensated system yields a very slow rise time due to the lag zero at 0.01. The rise
time can be sped up by moving the zero further from the imaginary axis with resultant changes in the
transient response. The plots below show the step response with the PI zero at  0.24. The response compares favorably with a twopole system step response that yields 5% overshoot and
a peak time of 8 seconds as shown below. Solutions to Design Problems 399 42.
a. PD compensator design: Pushing the gain, 10, to the right past the summing junction, the system can be represented as an equivalent unity feedback system with G e (s) = 10 6
.
2
(s − 4551)(s + 286) This system is unstable at any gain. For 1% overshoot and Ts = 0.1, the design point is –40 + j27.29.
The summation of angles from the poles of Ge(s) to this point is 216.903o. Therefore, the
compensator zero must contribute 216.903o  180o = 36.903o. Using the following geometry: X j27.29 36.903 o zc 40 27.29
zc  40 = tan (36.903). Thus, zc = 76.34. Adding this zero to the poles of Ge(s), the gain at the design
point is 106K = 23377. The PD compensated response is shown below. 400 Chapter 9: Design Via Root Locus Step Response for zero at 76.34 and 1% overshoot
4 3.5 3 Amplitude 2.5 2 1.5 1 0.5 0 0 0.02 0.04 0.06
0.08
Time (secs) 0.1 0.12 0.14 b. PI compensator design: To reduce the steadystate error to zero, we add a PI controller of the s+1
form s . The PID compensated step response is shown below.
PID Compensated Step Response for zero at 76.34 and 1% overshoot 4 3.5 3 Amplitude 2.5 2 1.5 1 0.5 0
0 0.2 0.4 0.6 0.8 1
1.2
Time (secs) 1.4 1.6 1.8 2 Solutions to Design Problems 401 We can see the 1% overshoot at about 0.1 second as in the PD compensated system above. But the
system now corrects to zero error.
43.
a. Root locus sketch yields;
25 20 15 10 Imag Axis 5 0 5 10 15 20 25 50 40 30 20 10 0 Real Axis Root locus sketch near imaginary axis yields;
CloseUp Root Locus to Determine Stability 0.25 0.2 0.15 0.1 Imag Axis 0.05 0 0.05 0.1 0.15 0.2 0.25
0.25 0.2 0.15 0.1 0.05 0
Real Axis 0.05 0.1 0.15 0.2 0.25 402 Chapter 9: Design Via Root Locus Searching imaginary axis for 180o yields: j0.083 at a gain of 0.072K = 0.0528 and j0.188 at a gain of
0.072K = 0.081. Also, the gain at the origin is 0.0517. Thus, the system is stable for 0.0517 < 0.072K
< 0.0528; 0.072K > 0.081. Equivalently, for 0.7181 < K < 0.7333; 0.072K > 1.125.
b. See (a)
c. Uncompensated system: Searching the 20% overshoot line, we find the operating point at 8.987 + j17.4542 = 19.71∠117.126o at 0.072K = 16.94 for the uncompensated system. Simulating
the response at this gain yields,
Step Response for Uncompensated System 1.4 1.2 Amplitude 1 0.8 0.6 0.4 0.2 0
0 0.05 0.1 0.15 0.2 0.25
0.3
Time (secs) 0.35 0.4 0.45 0.5 For 20% overshoot and Ts = 0.05 s, a design point of –80 + j156.159 is required. The sum of angles
to the design point is 123.897o. To meet the requirements at the design point, a zero would have to
contribute +303.897o, which is too high for a single zero. Let us first add the pole at the origin to
drive the steadystate error to zero to reduce the angle required from the zero. Summing angles with
this pole at the origin yields 241.023. Thus a zero contributing 61.023o is required. Using the
156.159
geometry below with z  80 = tan (61.023), zc = 166.478.
c
X 61.023 zc o 80 The gain at the design point is 0.072K = 181.55. j156.159 Solutions to Design Problems 403 d.
Step Response for Ccompensated System 1.4 1.2 1 0.8 0.6 0.4 0.2 0
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Time (secs) The settling time requirement has been met, but the percent overshoot has not. Repeating the design
for 1% overshoot and a Ts = 0.05 s yields a design point of –80 + j54.575. The compensator zero is
found to be at 47.855 at a gain 0.072K = 180.107. Step Response for Redesigned System 1.4 1.2 1 Amplitude 0.8 0.6 0.4 0.2 0 0 0.01 0.02 0.03 0.04 0.05
Time (secs) 0.06 0.07 0.08 0.09 0.1 404 Chapter 9: Design Via Root Locus 44. 4
ζωn = T = 2.667; ζ =
s %OS
 ln ( 100 )
= 0.591. Thus, ωn = 4.512 rad/s.
2 + ln2 (%OS)
π
100 Im = ω n 1 − ζ = 4.512 1 − 0.591 = 3.64 . Thus, and the operating point is 2.667 ± j3.64.
2 2 Summation of angles, assuming the compensating zero is at –5 (to cancel the openloop pole at –5, is
–170.88o. Therefore, the compensator pole must contribute 180o – 170.88o = 9.12o. Using the
geometry shown below, 3.64 9.12o
2.667 3.64
= tan 9.12o. Thus, pc = 25.34. Adding the compensator pole and using 2.667 ± j3.64 as
pc − 2.667
the test point, 50K = 2504, or K = 50.08. Thus the compensated openloop transfer function is G e (s) = 2504(s + 5)
. Higherorder pole are at –25.12, 5, and4.898. The
s(s + 5)(s + 10s + 50)(s + 25.34)
2 pole at –5 is cancelled by the closedloop zero at 5. The pole at –4.898 is not far enough away from
the dominant secondorder pair. Thus, the system should be simulated to determine if the response
meets the requirements.
Program:
syms s
numg=2504;
deng=expand(s*(s^2+10*s+50)*(s+25.34));
deng=sym2poly(deng);
G=tf(numg,deng);
Gzpk=zpk(G)
T=feedback(G,1);
step(T) Computer response:
Zero/pole/gain:
2504
s (s+25.34) (s^2 + 10s + 50) Solutions to Design Problems 405 45.
a. From Chapter 8, 0.6488K (s+53.85)
Ge(s) = 2 ______ 2 (s + 8.119s + 376.3) (s + 15.47s + 9283) Cascading the notch filter,
0.6488K (s+53.85)(s2 + 16s + 9200)
Get(s) = (s2 + 8.119s + 376.3) (s2 + 15.47s + 9283)(s+60)2 Arbitrarily design for %OS = 30% (ζ = 0.358) and Ts = 0.3 s. This places desired poles at
–13.33 ± j34.79. At the design point, the sum of the angles without the PD controller is 107.190.
Thus, 34.79
= tan 72.81
zc − 13.33 406 Chapter 9: Design Via Root Locus j34.79 72.810
13.33 zc From which, zc = 24.09. Putting this into the forward path,
0.6488K (s+53.85)(s2 + 16s + 9200)(s+24.09)
Get(s) = (s2 + 8.119s + 376.3) (s^2 + 15.47s + 9283)(s+60)2 Using root locus, the gain 0.6488K = 1637, or K = 2523.
b. Add a PI controller G PI (s ) =
Thus, ( s + 0.1)
s 0.6488K (s+53.85)(s2 + 16s + 9200)(s+24.09)(s+0.1)
Get(s) =
s (s2 + 8.119s + 376.3) (s^2 + 15.47s + 9283)(s+60)2 Using root locus, the gain 0.6488K = 1740, or K = 2682.
c.
Program:
syms s
numg=1637*(s+53.85)*(s^2+16*s+9200)*(s+24.09)*(s+0.1);
deng=s*(s^2+15.47*s+9283)*(s^2+8.119*s+376.3)*(s+60)^2;
numg=sym2poly(numg);
deng=sym2poly(deng);
G=tf(numg,deng);
Gzpk=zpk(G)
T=feedback(G,1);
step(T,0:0.01:1)
title(['With PD, Notch, and PI'])
pause
step(T)
title(['With PD, Notch, and PI']) Computer response:
Zero/pole/gain:
1637 (s+53.85) (s+24.09) (s+0.1) (s^2 + 16s + 9200)
s (s+60)^2 (s^2 + 8.119s + 376.3) (s^2 + 15.47s + 9283) _____ Solutions to Design Problems 407 ...
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This note was uploaded on 04/04/2008 for the course MECH COntrol Sy taught by Professor Khurshid during the Spring '08 term at Michigan State University.
 Spring '08
 khurshid

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