ch11_7ed_Nise

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E L E V E N 
 
 Design via 
Frequency Response 
 
SOLUTIONS TO CASE STUDIES CHALLENGES 
Antenna Control: Gain Design 
a. The required phase margin for 25% overshoot ( = 0.404), found from Eq. (10.73), is 43.49o. 
From the solution to the Case Study Challenge problem of Chapter 10, G(s) = 
50.88K
s(s 1.32)(s 100) 
 
Using the Bode plots for K = 1 from the solution to the Case Study Challenge problem of 
Chapter 10, we find the required phase margin at  = 1.35 rad/s, where the magnitude response is 
-14 dB. Hence, K = 5.01 (14 dB). 
b. 
Program: 
%Input system 
numg=50.88; 
deng=poly([0 -1.32 -100]); 
G=tf(numg,deng); 
%Percent Overshoot to Damping Ratio to Phase Margin 
Po=input('Type %OS '); 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); 
fprintf('\nPercent Overshoot = %g',Po) 
fprintf(', Damping Ratio = %g',z) 
fprintf(', Phase Margin = %g',Pm) 
%Get Bode data 
bode(G) 
pause 
w=0.01:0.05:1000;%Step size can be increased if memory low. 
[M,P]=bode(G,w); 
M=M(:,:); 
P=P(:,:); 
Ph=-180+Pm; 
for i=1:1:length(P); 
if P(i)-Ph<=0; 
M=M(i); 
K=1/M; 
fprintf(', Frequency = %g',w(i)) 
fprintf(', Phase = %g',P(i)) 
fprintf(', Magnitude = %g',M) 
11-2 Chapter 11: Design via Frequency Response 
fprintf(', Magnitude (dB) = %g',20*log10(M)) 
fprintf(', K = %g',K) 
break 
end 
end 
T=feedback(K*G,1); 
step(T) 
 
Computer response: 
Type %OS 25 
 
Percent Overshoot = 25, Damping Ratio = 0.403713, Phase Margin = 43.463, 
Frequency = 1.36, Phase = -136.634, Magnitude = 0.197379, Magnitude (dB) 
= -14.094, K = 5.06641 
 
 
 
 
Solutions to Case Studies Challenges 11-3 
 
 
Antenna Control: Cascade Compensation Design 
a. From the solution to the previous Case Study Challenge in this chapter, G(s) = 
50.88K
s(s 1.32)(s 100) 
. 
For Kv = 20, K = 51.89. Hence, the gain compensated system is 
G(s) = 
2640.16
s(s 1.32)(s 100) 
 
Using Eq. (10.73), 15% overshoot (i.e.  = 0.517) requires a phase margin of 53.18o. Using the Bode 
plots for K = 1 from the solution to the Case Study Challenge problem of Chapter 10, we find the 
required phase margin at  = 0.97 rad/s where the phase is -126.82o. 
To speed up the system, we choose the compensated phase margin frequency to be 4.6 * 0.97 = 4.46 
rad/s. Choose the lag compensator break a decade below this frequency, or  = 0.446 rad/s. 
At the phase margin frequency, the phase angle is -166.067o, or a phase margin of 13.93o. Using 5o 
leeway, we need to add 53.18o - 13.93o + 5o = 44.25o. From Figure 11.8,  = 0.15, or  = 
1
²
 = 
6.667. Using Eq. (11.15), the lag portion of the compensator is 
 
GLag (s) = 
(s 0.446)
0.446
(s )
6.667


 = 
s 0.446
s 0.0669


. 
11-4 Chapter 11: Design via Frequency Response 
Using Eqs. (11.9) and (11.15), T2 = 
max
1
É ²
 = 0.579. From Eq. (11.15), the lead portion of the 
compensator is 
GLead (s) = 
s 1.727
s 11.51


 
The final forward path transfer function is 
G(s)GLag (s)GLead(s) = 
2640.16(s 0.446)(s 1.727)
s(s 1.32)(s 100)(s 0.0669)(s 11.51)
 
   
 
b. 
Program: 
%Input system ***************************** 
K=51.89; 
numg=50.88*K; 
deng=poly([0 -1.32 -100]); 
G=tf(numg,deng); 
Po=15; 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
%Determine required phase margin************** 
Pmreq=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi) 
phreq=Pmreq-(180)%required phase 
w=0.1:0.01:10; 
[M,P]=bode(G,w); 
for i=1:1:length(P);%search for phase angle 
if P(i)-phreq<=0; 
ph=P(i) 
w(i) 
break 
end 
end 
wpm=4.6*w(i) 
[M,P]=bode(G,wpm);%Find phase at wpm 
Pmreqc=Pmreq-(180+P)+5%Find contribution required from compensator+5 
beta=(1-sin(Pmreqc*pi/180))/(1+sin(Pmreqc*pi/180)) 
%Design lag compensator*************** 
zclag=wpm/10; 
pclag=zclag*beta; 
Kclag=beta; 
%Design lead compensator********** 
zclead=wpm*sqrt(beta); 
pclead=zclead/beta; 
Kclead=1/beta; 
%Create compensated forward path********* 
numgclag=Kclag*[1 zclag]; 
dengclag=[1 pclag]; 
'Gclag(s)' 
Gclag=tf(numgclag,dengclag); 
Gclagzpk=zpk(Gclag) 
numgclead=Kclead*[1 zclead]; 
dengclead=[1 pclead]; 
'Gclead(s)' 
Gclead=tf(numgclead,dengclead); 
Gcleadzpk=zpk(Gclead) 
Gc=Gclag*Gclead; 
'Ge(s)=G(s)*Gclag(s)*Gclead(s)' 
Solutions to Case Studies Challenges 11-5 
 
Ge=Gc*G; 
 
11-6 Chapter 11: Design via Frequency Response 
Gezpk=zpk(Ge) 
%Test lag-lead compensator**************** 
T=feedback(Ge,1); 
bode(Ge) 
title('Lag-lead Compensated') 
[Gm,Pm,wcp,wcg]=margin(Ge); 
'Compensated System Results' 
fprintf('\nResulting Phase Margin = %g',Pm) 
fprintf(', Resulting Phase Margin Frequency = %g',wcg) 
pause 
step(T) 
title('Lag-lead Compensated') 
 
Computer response: 
Pmreq = 
 
 53.1718 
 
phreq = 
 
 -126.8282 
 
ph = 
 
 -126.8660 
 
ans = 
 
 0.9700 
 
wpm = 
 4.4620 
Pmreqc = 
 
 44.2468 
 
beta = 
 
 0.1780 
 
ans = 
 
Gclag(s) 
 
Zero/pole/gain: 
0.17803 (s+0.4462) 
------------------ 
 (s+0.07944) 
 
ans = 
 
Gclead(s) 
 
Zero/pole/gain: 
 
Solutions to Case Studies Challenges 11-7 
 
5.617 (s+1.883) 
--------------- 
 (s+10.58) 
 
ans = 
 
Ge(s)=G(s)*Gclag(s)*Gclead(s) 
 
 
Zero/pole/gain: 
 2640.1632 (s+1.883) (s+0.4462) 
---------------------------------------- 
s (s+100) (s+10.58) (s+1.32) (s+0.07944) 
 
ans = 
 
Compensated System Results 
 
Resulting Phase Margin = 57.6157, Resulting Phase Margin Frequency = 
2.68618» 
 
 
 
 
 
11-8 Chapter 11: Design via Frequency Response 
 
 
Answers to Review Questions 
1. Steady-state error requirements can be designed simultaneously with transient response requirements. 
2. Via the phase margin 
3. The lag compensator is a low pass filter. Thus, while the low frequency gain is increased, the high-
frequency gain at 180 o is decreased to make the system stable. 
4. The lag network affects the phase angle at low frequencies, but not at high frequencies. For the 
compensated system, the phase plot is about the same as that of the uncompensated system around and 
above the phase margin frequency yielding the same transient response. 
5. To compensate for the slight negative angle that the lag compensator has near the phase margin 
frequency 
6. Compensated system has higher low-frequency gain than the uncompensated system designed to yield 
the same transient response; compensated and uncompensated system have the same phase margin 
frequency; the compensated system has lower gain around the phase margin frequency; the compensated 
and uncompensated system's have approximately the same phase values around the phase margin 
frequency. 
7. The lead network is a high pass filter. It raises the gain at high frequencies. The phase margin frequency 
is increased.
 
Solutions to Problems 11-9 
 
8. Not only is the magnitude curve increased at higher frequencies, but so is the phase curve. Thus the 180o 
point moves up in frequency with the increase in gain. 
9. To correct for the negative phase angle of the uncompensated system 
10. When designing the lag portion of a lag-lead compensator, we do not worry about the transient design. 
The transient response will be considered when designing the lead portion of a lag-lead compensator. 
 
SOLUTIONS TO PROBLEMS 
 
1. 
a. Plot Bode plots for K = 1; angle is 180o at  = 15.8 rad/s where the magnitude is –76.5 dB. 
Therefore a 66.5 dB ( or K = 2113) increase will yield a 10 dB gain margin. 
 
 
b. Plot Bode plots for K = 1; angle is 180o at  = 6.32 rad/s where the magnitude is –55 dB. 
Therefore a 45 dB ( or K = 177.8) increase will yield a 10 dB gain margin. 
11-10 Chapter 11: Design via Frequency Response 
 
 
c. Plot Bode plots for K = 1; angle is 180o at  = 9.45 rad/s where the magnitude is –63.8 dB. 
Therefore a 53.8 dB ( or K = 489.8) increase will yielda 10 dB gain margin. 
 
 
 
 
 
 
 
Solutions to Problems 11-11 
 
 
2. 
a. For a 40o phase margin, the phase must be -140o when the magnitude plot is zero dB. The phase is 
-140o at  = 9.12 rad/s. At this frequency, the magnitude curve is –67.48 dB. Thus a 67.48 dB 
increase (K = 2365) will yield a 40o phase margin. 
b. For a 40o phase margin, the phase must be -140o when the magnitude plot is zero dB. The phase is 
-140o at  = 2.76 rad/s. At this frequency, the magnitude curve is – 42.86 dB. Thus a 42.86 dB 
increase (K = 139) will yield a 40o phase margin. 
c. For a 40o phase margin, the phase must be -140o when the magnitude plot is zero dB. The phase is 
-140o at  = 5.04 rad/s. At this frequency, the magnitude curve is – 54.4 dB. Thus a 54.4 dB increase 
(K = 525) will yield a 40o phase margin. 
3. 
20% overshoot =>  = 0.456 => M = 48.15
o
. 
a. Looking at the phase diagram, where M = 48.15
o
 (i.e. -131.85
o
), the phase margin frequency = 
4.11 rad/s. At this frequency, the magnitude curve is -55.2 dB. Thus the magnitude curve has to be 
raised by 55.2 dB (K = 575). 
b. Looking at the phase diagram, where M = 48.15
o
 (i.e.  = -131.85
o
), the phase margin frequency = 
7.14 rad/s. At this frequency, the magnitude curve is – 65.6 dB. Thus the magnitude curve has to be 
raised by 65.6 dB (K = 1905). 
c. Looking at the phase diagram, where M = 48.15
o
 (i.e.  = -131.85
o
), the phase margin frequency = 
8.2 rad/s. At this frequency, the magnitude curve is – 67.3 dB. Thus the magnitude curve has to be 
raised by 67.3 dB (K = 2317). 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
11-12 Chapter 11: Design via Frequency Response 
4. 
a. A 20% overshoot corresponds to a damping factor of 0.456  and from equation 10.73 to a 
48.15
PM
  . The bode plot of ( )G s with 1K  is 
 
We search on the Bode diagram for the frequency at which the phase is 
180 48.15 131.85    . This occurs at a frequency of 1.37 rad/sec. At this frequency 
( 1.37) 7.11G j   dB. Thus the magnitude characteristic must be raised by this quantity; 
7.11
2010 2.267K   . 
b. 
>> s=tf('s'); 
>> G=2.267*(s+10)*(s+15)/s/(s+2)/(s+5)/(s+20); 
>> T=feedback(G,1); 
>> step(T) 
Solutions to Problems 11-13 
 
 
 
5. 
For Kv = 50, K = 350. Plot the Bode plots for this gain. 
 
 
 Also, since %OS = 15%,  = 0.517. Using Eq. (10.73), M = 53.17
o. Increasing M by 10
o we will 
design for a phase margin of 63.17o. The phase margin frequency is where the phase angle is 
 
 
 
11-14 Chapter 11: Design via Frequency Response 
 63.17 - 180o = -116.83o, or M = 3.54 rad/s. At this frequency, the magnitude is 22 dB. Start the 
magnitude of the compensator at - 22 dB and draw it to 1 decade below M. 
 
 
 Then begin +20 dB/dec until zero dB is reached. Read the break frequencies as 0.028 rad/s and 0.354 
rad/s from the Bode plot and form a lag transfer function that has unity dc gain: 
Gc(s) =0.0791 
s 0.354
s 0.028


 
 The compensated forward path is 
 
G(s) = 
350*0.0791(s 0.354) 27.69(s 0.354)
s(s 7)(s 0.028) s(s 7)(s 0.028)
 

   
 
6. 
a. For Kv = 1000, K = 1473. Plotting the Bode for this value of K: 
 
 
 
Solutions to Problems 11-15 
 
 
Using Eqs. (4.39) and (10.73) a percent overshoot = 15 is equivalent to a  = 0.517 and M = 53.17. 
Using an extra 10o, the phase margin is 63.17o. The phase-margin frequency = 1.21 rad/s. At this 
frequency, the magnitude = 57.55 dB = 754.2. Hence the lag compensator K = 1/754.2 = 0.001326. 
Following Steps 3 and 4 of the lag compensator design procedure in Section 11.3, 
Glag(s) = 0.001326 


s 0.121
s 0.0001604
 
b. 
Program: 
%Input system 
numg=1473*poly([-10 -11]); 
deng=poly([0 -3 -6 -9]); 
G=tf(numg,deng); 
numc=0.001326*[1 0.121]; 
denc=[1 0.0001604]; 
Gc=tf(numc,denc); 
Ge=G*Gc; 
T=feedback(Ge,1); 
step(T) 
 
Computer response: 
 
 
 
 
 
 
 
 
 
11-16 Chapter 11: Design via Frequency Response 
 
 
 
7. 
The MATLAB M-file given in Appendix B for Example 11.2 was modified as follows to assist in 
solving this problem: 
 
'(ch11p7) Problem 11.7' % Display label. 
numg = 1; % Set numerator of G(s)to unity. 
deng = poly([-1 -5 -10]); 
Gu = tf(numg, deng); % G(s)with numerator set to unity. 
rlocus(Gp); % Plot the root locus of Gu(s). 
title('Full Root Locus') 
pause 
axis ([- 4, 0, 0, 8]); % Zoom-in the root locus of Gu(s). 
pos=15; % Desired percent overshoot. 
z=(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2)); 
 % Calculate required damping ratio,z. 
sgrid(z,0) 
title('Root Locus Zoomed-in for Dominant Poles at 15% O.S.') 
[K1,p]=rlocfind(Gu); 
pause 
Kpo = K1/50; % Find uncompensated error constant. 
ess = 1/(1+Kpo); % Find uncompensated step error. 
ec = ess/5; % Required compensated step error. 
Kpn = (1/ec)-1; % Required compensated error constant. 
K = 50*Kpn; % Calculate required open-loop gain 
numg=[K]; % Define numerator of G(s). 
'G(s)' % Display label. 
G=tf(numg,deng) % Create and display G(s). 
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi)+10; 
 % Calculate required phase margin and 
 % add 10 degrees to compensate for the 
 % effect of the lag compensator. 
w=0.01:0.01:100; % Set range of frequency from 0.01 to 
 % 100 in steps of 0.01. 
bode(G) % Bode plot for G(s). 
grid 
pause 
[M,P]=bode(G,w); % Get Bode data. 
Ph=-180+Pm; % Calculate required phase angle. 
for k=1:1:length(P); % Search Bode data for required phase 
 % angle. 
if P(k)-Ph<=0; % If required phase angle is found, 
 % find the value of 
M=M(k); % magnitude at the same frequency. 
wf=w(k); % At this frequency the magnitude 
 % plot must go through 0 dB. 
break % Stop the loop. 
end % End if. 
end % End for. 
wh=wf/10; % Calculate the high-frequency break 
 % of the lag compensator. 
 
 
 
wl=(wh/M); % Calculate the low-frequency break 
Solutions to Problems 11-17 
 
 % of the lag compensator; found from 
 % lag compensator, Gc(s)=Kc(s+wh)/(s+wl), 
 % high & low frequency gain requirements. 
 % At low w, gain=1. Thus, Kc*wh/wl=1. 
 % At high w, gain=1/M. Thus Kc=1/M. Hence 
 % Kc=wl/wh=1/M, or wl=wh/M. 
numc=[1 wh]; % Generate numerator of lag 
 % compensator, Gc(s). 
denc=[1 wl]; % Generate denominator of lag 
 % compensator, Gc(s). 
Kc=wl/wh; % Generate K for Gc(s). 
'Lag compensator' % Display label. 
Kc % Display lag compensator K. 
'Gc(s)' % Display label. 
Gc=tf(Kc*numc,denc) % Create and display Gc(s). 
'Gc(s)G(s)' % Display label. 
GcG=Gc*G % Create and display Gc(s)G(s). 
T=feedback(GcG,1); % Create T(s). 
step(T) % Generate a closed-loop, lag- 
 % compensated step response. 
title('Closed-Loop Step Response for Lag-Compensated System') 
 % Add title to step response. 
pause 
 
 
 
Uncompensated system:The full root locus for the uncompensated system, Gu(s), with the gain set to 
unity, is shown below. 
 
 
Searching the zoomed-in root locus (shown below) along the 121.1
o
 line (15% overshoot; e.g., 
=0.517), the dominant poles were found to be at – 2.0711 ± j3.4161 with K = 139.1. Therefore, the 
11-18 Chapter 11: Design via Frequency Response 
uncompensated static error constant is 
139.1
2.782
50
po
K   and the steady-state error is 
1 1
( ) 0.26441, 
1 1 2.782
step
po
e
K
   
 
 which may be also obtained in the command window 
as: ess = 0.2644. 
 
Compensated system: 
For a 5 times improvement in steady-state error: 
1
( ) 0.053
1
step
pn
e
K
  

, yielding, 17.9086
pn
K  and K 
= 895.4. The Bode plot obtained for the uncompensated G(s) at that gain is shown below. Its data along with the 
desired phase margin (calculated as 63.17
o
 & adding 10
o
 to compensate for the effect of the lag compensator) were 
used to generate the transfer function of the compensator as: Gc(s) = 0.2044 s + 0.05888 / (s + 0.05888); e.g.; 
 
 
 
 
 
Solutions to Problems 11-19 
 
0.2044( 0.28806)
( )
( 0.05888)
C
s
G s
s



 
 
 
Thus, the compensated forward path was found to be: 
 
 
       
( 0.28806) 183.01( 0.28806)
( ) ( ) 0.2044*895.4
( 1)( 5)( 10)( 0.05888) ( 1)( 5)( 10)( 0.05888)
C
s s
G s G s
s s s s s s s s
 
From the step response, shown below, we see that the closed-loop lag-compensated system has a final value of 
0.947); e.g. a steady-state error of 0.053, and an overshoot of less than 8%. 
11-20 Chapter 11: Design via Frequency Response 
 
 
 
8. 
For 
(4)
100
(2)(6)(8)
p
K
K   , K = 2400. Plotting the Bode plot for this gain, 
 
 
 
We will design the system for a phase margin 10
0
 larger than the specification. Thus m = 55
0
. The 
phase margin frequency is where the phase angle is –180
0
 + 55
0
 = -125
0
. From the Bode plot this 
frequency is 11
m
  rad/s. At this frequency the magnitude is 23.37 dB. Start the magnitude of the 
Solutions to Problems 11-21 
 
lag compensator at –23.37 dB and draw it to 1 decade below 11
m
  , or 1.1 rad/s. Then begin a 
+20 dB/dec climb until 0 dB is reached. Read the break frequencies as 0.0746 rad/s and 1.1 rad/s from 
the Bode plot as shown below. 
 
 
Ensuring unity dc gain, the transfer function of the lag is 
( 1.1)
( ) 0.06782
( 0.0746)
lag
s
G s
s



. The 
compensated forward-path transfer function is thus the product of the plant and the compensator, or 
 
162.8( 4)( 1.1)
( )
( 2)( 6)( 8)( 0.0746)
e
s s
G s
s s s s
 

   
 
9. 
From Example 11.1, K = 58251 yields 9.48% overshoot or a phase margin of 59.19o. Also, 
G(s) = 
 
58251
s(s 36)(s 100)
 
Allowing for a 10o contribution from the PI controller, we want a phase margin of 69.19o, or a phase 
angle of -180o + 69.19o = -110.81o. This phase angle occurs at  = 9.8 rad/s where the magnitude is 4 
dB. Thus, the PI controller should contribute - 4 dB at  = 9.8 rad/s. Selecting a break frequency a 
decade below the phase margin frequency, 
Gc(s) = 
s 0.98
s
 
11-22 Chapter 11: Design via Frequency Response 
This function has a high-frequency gain of zero dB. Since we want a high-frequency gain of 
-4 dB (a gain of 0.631), 
Gc(s) = 0.631 
s 0.98
s

 
The compensated forward path is 
G(s) = 
58251*0.631(s 0.98)
s(s+36)(s+100)

 = 
36756.38(s 0.98)
s(s+36)(s+100)

 
10. 
 Bode plots for K = 1: 
 
Using Eqs. (4.39) and (10.73) a percent overshoot = 15 is equivalent to a  = 0.517 and M = 53.17
o. 
The phase-margin frequency = 1.66 rad/s. The magnitude = -9.174 dB = 0.3478. Hence K = 1/ 
0.3478 = 2.876. 
b. 
Bode plots for K = 2.876. 
Solutions to Problems 11-23 
 
 
Adding 10o to the phase margin yields 63.17. Thus, the required phase is –180
0
 + 63.17
0
 = -116.83
0
, 
which occurs at a frequency of 1.21 rad/s. The magnitude = 3.366 dB = 1.473. Hence, the lag 
compensator K = 1/ 1.473 = =0.6787. Selecting the break a decade below the phase-margin 
frequency, 
 
Gc(s) = 0.6787 
s 0.121
s

 
c. 
Program: 
%Input system 
numg=2.876*poly([-10 -11]); 
deng=poly([0 -3 -6 -9]); 
G=tf(numg,deng); 
numc=0.6787*[1 0.121]; 
denc=[1 0]; 
Gc=tf(numc,denc); 
Ge=G*Gc; 
T=feedback(Ge,1); 
step(T) 
 
11-24 Chapter 11: Design via Frequency Response 
Computer response: 
 
 
 
 
 
11. 
Program: 
%PI Compensator Design via Frequency Response 
%Input system 
G=zpk([],[-5 -10],1); 
G=tf(G); 
%Percent Overshoot to Damping Ratio to Phase Margin 
Po=input('Type %OS '); 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi)+10; 
fprintf('\nPercent Overshoot = %g',Po) 
fprintf(', Damping Ratio = %g',z) 
fprintf(', Phase Margin = %g',Pm) 
%Get Bode data 
bode(G) 
title('Uncompensated') 
pause 
%Find frequency at desired phase margin and the gain at this frequency 
w=logspace(-1,2,10000); 
%w=.1:0.1:100; 
[M,P,w]=bode(G,w); 
Ph=-180+Pm 
for i=1:1:length(P); 
if P(i)-Ph<=0 
Mag=M(i) 
wf=w(i); 
 
Solutions to Problems 11-25 
 
fprintf(', Frequency = %g',wf) 
fprintf(', Phase = %g',P(i)) 
fprintf(', Magnitude = %g',Mag) 
fprintf(', Magnitude (dB) = %g',20*log10(Mag)) 
break 
end 
end 
%Design PI compensator 
%Break frequency is a decade below phase margin frequency 
wh=wf/10; 
%Magnitude is reciprocal of magnitude of G at the phase margin frequency 
%so net magnitude is 0 dB at the phase margin frequency 
Kc=1/Mag 
'PI Compensator' 
Gpi=tf(Kc*[1 wh],[1 0]) 
bode(Gpi) 
title(['PI compensator']) 
pause 
'G(s)Gpi(s)' 
Ge=series(G,Gpi); 
Ge=zpk(Ge) 
bode(Ge) 
title('PI Compensated') 
[Gm,Pm,Wcp,Wcg]=margin(Ge); 
'Gain margin(dB); Phase margin(deg.); 0 dB freq. (r/s);' 
'180 deg. freq. (r/s)' 
margins=[20*log10(Gm),Pm,Wcg,Wcp] 
pause 
T=feedback(Ge,1); 
step(T) 
title('PI Compensated') 
 
Computer response: 
Type %OS 25 
 
Percent Overshoot = 25, Damping Ratio = 0.403713, Phase Margin = 53.463 
Ph = 
 
 -126.5370 
 
 
Mag = 
 
 0.0037 
 
, Frequency = 14.5518, Phase = -126.54, Magnitude = 0.00368082, Magnitude 
(dB) = -48.6811 
Kc = 
 
 271.6786 
 
 
ans = 
 
11-26 Chapter 11: Design via Frequency Response 
 
PI Compensator 
 
 
Transfer function: 
271.7 s + 395.3 
--------------- 
 s 
 
 
ans = 
 
G(s)Gpi(s) 
 
 
Zero/pole/gain: 
271.6786 (s+1.455) 
------------------ 
 s (s+10) (s+5) 
 
 
ans = 
 
Gain margin(dB); Phase margin(deg.); 0 dB freq. (r/s); 
 
 
ans = 
 
180 deg. freq. (r/s) 
 
 
margins = 
 
 Inf 47.6277 14.5975 Inf 
 
 
 
Solutions to Problems 11-27 
 
 
 
 
 
 
11-28 Chapter 11: Design via Frequency Response 
 
 edegr 
 
 
 
 
 
 
 
 
 
 
 
 
 
12. 
Solutions to Problems 11-29 
 
4
500
v
K
K   or 2000K  . We obtain the Bode diagram: 
 
 
 
11-30 Chapter 11: Design via Frequency Response 
The uncompensated system shows a 19.9
M
  at a frequency of 2.45 rad/sec. To achieve the 
phase margin specification we need 45 19.9 25.1  and we add an arbitrary 10 to 
compensate for the fact that the phase margin will increase. So we will specify 36 of phase 
lead. Using equations (11.11) and (11.12) we get that 0.26  and max( ) 1.96 5.85G j   
dB. Searching on the Bode diagram we find that ( 3.57) 5.85G j   dB, so we chose 
3.57  rad/sec as the frequency for the compensator. Using equations (11.6) and (11.9) we 
get that the compensator is 
 
3.846( 1.82)
( )
7
c
s
G s
s



 
Applying this compensator it is found that the resulting 37.4
M
  . Adjusting slightly the 
poles and zeros of the compensator one gets that 
 
4.611( 1.8)
( )
( 8.3)
c
s
G s
s



 
for a 40
M
  . The resulting open loop transfer functionis: 
 
2000*4.611( 1.8) 9222( 1.8)
( )
( 2)( 8.3)( 10)( 20) ( 2)( 8.3)( 10)( 20)
s s
G s
s s s s s s s s s s
 
 
       
 
 
13. 
a. Gain-compensated time response: 
 
 
Solutions to Problems 11-31 
 
 
Bode plots for K = 1000 (Kv = 10): 
 
 
 
The specifications for the gain compensated system are: K = 1000, percent overshoot = 10,  = 
0.591155, peak time = 0.5 s, current phase margin = 22.5362o. 
To meet the requirements: required phase margin (Eq. 10.73) = 58.5931o, required phase margin with 
correction factor of 20o = 78.5931, required bandwidth (Eq. 10.56) = 9.03591, required phase 
contribution from compensator = 78.5931o - 22.5362o = 56.0569o, compensator beta (Eq. 11.11) = 
0.0931398, new phase margin frequency (Eq. 11.12) = 11.51. 
Now design the compensator: Compensator gain Kc = 1/ = 10.7366, compensator zero (Eq. 11.12) 
= -3.51272, compensator pole = zc/ = -37.7144. 
Lead-compensated Bode plots: 
 
11-32 Chapter 11: Design via Frequency Response 
 
 
Lead-compensated phase margin = 50.2352. 
 
b. 
Program: 
numg=1000; 
deng=poly([0 -5 -20]); 
G=tf(numg,deng); 
numc=[1 3.51272]; 
denc=[1 37.7144]; 
Gc=tf(numc,denc); 
Ge=G*10.7366*Gc; 
T=feedback(Ge,1); 
step(T) 
 
 
 
Computer response: 
 
Solutions to Problems 11-33 
 
 
 
 
 
14. 
The following MATLAB M-file was written to assist in solving this problem: 
 
'G(s)' % Display label. 
G=zpk(K*Gp) % Define G(s), put K into it, 
 % convert 
 % to factored form, and display. 
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); 
 % Calculate phase margin. 
Tsd= input('Type Desired Settling Time, Tsd=(Ts/2) = '); 
 % Input Desired Ts. 
wn=4/(z*Tsd); % Calculate required natural 
 % frequency. 
wBW=wn*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2)); 
 % Determine required bandwidth. 
w=0.01:0.5:1000; % Set range of frequency from 0.01 
 % to 1000 in steps of 0.5. 
bode(G) % Display the Bode plots. 
pause 
[M,P]=bode(G,w); % Get Bode data. 
[Gm,Pm,Wcg,Wcp]=margin(G); % Find current phase margin. 
Pmreq=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); 
 % Calculate required phase margin. 
Pmreqc=Pmreq+10; % Add a correction factor of 10 
 % degrees. 
Pc=Pmreqc-Pm; % Calculate phase contribution 
 % required from lead compensator. 
% Design lead compensator 
beta=(1-sin(Pc*pi/180))/(1+sin(Pc*pi/180)); 
 % Find compensator beta. 
magpc=1/sqrt(beta); % Find compensator peak 
 % magnitude. 
11-34 Chapter 11: Design via Frequency Response 
for k=1:1:length(M); % Find frequency at which 
 % uncompensated system has a 
 % magnitude of 1/magpc. 
 % This frequency will be the new 
 % phase margin frequency. 
if M(k)-(1/magpc)<=0; % Look for peak magnitude. 
wmax=w(k); % This is the frequency at the 
 % peak magnitude. 
break % Stop the loop. 
end % End if. 
end % End for. 
% Calculate lead compensator zero, pole, and gain. 
zc=wmax*sqrt(beta); % Calculate the lead compensator’s 
 % low break frequency. 
pc=zc/beta; % Calculate the lead compensator’s 
 % high break frequency. 
Kc=1/beta; % Calculate the lead compensator’s 
 % gain. 
'Gc(s)' % Display label. 
Gc=tf(Kc*[1 zc],[1 pc]); % Create Gc(s). 
Gc=zpk(Gc) % Convert Gc(s) to factored form 
 % and display. 
'Ge(s)=G(s)Gc(s)' % Display label. 
Ge=G*Gc % Form Ge(s)=Gc(s)G(s). 
Ge=minreal(Ge); % Cancel common factors. 
Kp=dcgain(Ge) % Calculate Kp. 
T=feedback(Ge,1); % Find T(s). 
step(T) % Generate closed-loop, lead- 
 % compensated step response. 
title('Lead-Compensated Step Response') 
 % Add title to lead-compensated 
 % step response. 
pause 
 
Uncompensated System: The full root locus for the uncompensated system, Gp(s), with the gain set 
to unity, is shown below. Searching the zoomed-in locus (shown below the full locus) along the  = 
0.456 line (20% O.S.), find the dominant pole Q = –7.54 + j 14.7 with a gain of 239. 
 
Hence: 
a. Ts = 4 /  n = 4 / (0.456×16.5) = 0.532 sec; 
b. 
239 4
Kp 7.967
2 5 12

 
 
. 
 
Solutions to Problems 11-35 
 
 
 
 
 
 
c. The Bode plot for the uncompensated system is shown below. From that plot we see that the 
phase margin and the phase-margin frequency are 30.8 degrees and 26.4 rad/sec, respectively. 
11-36 Chapter 11: Design via Frequency Response 
 
 
d. As could be seen from the above M-file, frequency response techniques were used to design a 
compensator that would yield a threefold improvement in KP and a twofold reduction in settling 
time while keeping the overshoot at 20%. That lead compensator was found to have a transfer 
function: 
 
GC (s) = 


2.6961(s 21.32)
s 57( .49)
 
Compensated system: 
 Thus, the compensated forward path was found to be: 
Ge (s) = G (s) GC (s) = 
 
   
2090.1(s 4)(s 21.32)
(s 2)(s 5)(s 12)(s 57.49)
 
 
The step response of the lead-compensated system, shown below, indicates that all requirements 
are met: We obtained more than threefold improvement in the error constant (KP was increased 
more than 3.2 times to 25.84) and the settling time was reduced to 0.186 seconds (almost 2.9 
fold reduction) while the overshoot was kept almost at 20%. 
Solutions to Problems 11-37 
 
 
 
 
15. 
If G(s) = 
144000
s(s 36)(s 100) 
, Kv = 40. Also, for a 0.1 second peak time, and = 0.456 (20% 
overshoot), Eq. (10.56) yields a required bandwidth of 46.59 rad/s. Using Eq. (10.73), the required 
phase margin is 48.15o. Let us assume that we raise the phase margin frequency to 39 rad/s. The 
phase angle of the uncompensated system at this frequency is -158.6o. To obtain the required phase 
margin, the compensator must contribute 26.75o more at 39 rad/s. Assume the following form for the 
compensator: Gc(s) = K'KD(s+
D
1
K
). The angle contributed by the compensator is 
c = tan
-1 
D
1 / K

 = 26.75o. Letting  = 39 rad/s, KD = 0.0129. Hence, the compensator is 
Gc(s) = 0.0129 (s+77.37). The compensated forward path is 
 
144000*0.0129(s 77.37) 1857.6(s 77.37)
G(s)
s(s 36)(s 100) s(s 36)(s 100)
 
 
   
 
 
The closed-loop bandwidth is approximately 50 rad/s, which meets the requirements. 
The lag compensated forward path is 
11-38 Chapter 11: Design via Frequency Response 
G(s) = 7.759 
2
(s 0.058)
s(s 2s+5)(s 3)(s 0.0015)

  
 
16. 
a. Bode plots and specifications for gain compensated system are the same as Problem 13. Required 
phase margin and required bandwidth is the same as Problem 13. Select a phase margin frequency 7 
rad/s higher than the bandwidth = 9 + 7 = 16 rad/s. The phase angle at the new phase-margin 
frequency is -201.6
0
. Thephase contribution required from the compensator is –180
0
 + 201.6
0
 + 
58.59
0
 = 80.2
0
 at the phase-margin frequency. Using the geometry below: 
 
 
tan (80.2) = 
C
16
z
. Therefore, zc = 2.76. Thus, Gc (s) = 
1
(s 2.76)
2.76
 . 
The PD compensated Bode plots: 
 
 
Solutions to Problems 11-39 
 
Compensated phase margin is 62.942o. 
 
 
b. 
Program: 
numg=1000; 
deng=poly([0 -5 -20]); 
G=tf(numg,deng); 
numc=(1/2.76)*[1 2.76]; 
denc=1; 
Gc=tf(numc,denc); 
Ge=G*Gc; 
T=feedback(Ge,1); 
step(T) 
title('PD Compensated') 
 
Computer response: 
 
 
 
 
17. 
Program: 
%Lead Compensator Design via Frequency Response 
 %Input system 
K=input('Type K to meet steady-state error '); 
numg=K*[1 1]; 
deng=poly([0 -2 -6]); 
'Open-loop system' 
'G(s)' 
G=tf(numg,deng) 
 %Generate uncompensated step response 
T=feedback(G,1); 
step(T) 
title('Gain Compensated') 
 
 %Input transient response specifications 
Po=input('Type %OS '); 
%Ts=input('Type settling time '); 
Tp=input('Type peak time '); 
 
11-40 Chapter 11: Design via Frequency Response 
 %Determine required bandwidth 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
%wn=4/(z*Ts); 
wn=pi/(Tp*sqrt(1-z^2)); 
wBW=wn*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2)); 
 
 %Make a Bode plot and get Bode data 
%Get Bode data 
bode(G) 
title('Gain Compensated') 
 
w=0.01:0.1:100; 
[M,P]=bode(numg,deng,w); 
 
 %Find current phase margin 
[Gm,Pm,wcp,wcg]=margin(M,P,w); 
 
 %Calculate required phase margin 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
Pmreq=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); 
 
 %Add a correction factor of 10 degrees 
Pmreqc=Pmreq+10; 
 
 %Calculate phase required from compensator 
Pc=Pmreqc-Pm; 
 
 %Design lead compensator 
%Find compensator beta and peak compensator magnitude 
beta=(1-sin(Pc*pi/180))/(1+sin(Pc*pi/180)); 
magpc=1/sqrt(beta); 
%Find frequency at which uncompensated system has a magnitude of 1/magpc 
%This frequency will be the new phase margin frequency 
for i=1:1:length(M); 
if M(i)-(1/magpc)<=0; 
wmax=w(i); 
break 
end 
end 
%Calculate the lead compensator's break frequencies 
zc=wmax*sqrt(beta); 
pc=zc/beta; 
Kc=1/beta; 
numc=[1 zc]; 
denc=[1 pc]; 
'Gc(s)' 
Gc=tf(numc,denc) 
 %Display data 
fprintf('\nK = %g',K) 
fprintf(' Percent Overshoot = %g',Po) 
fprintf(', Damping Ratio = %g',z) 
%fprintf(', Settling Time = %g',Ts) 
fprintf(', Peak Time = %g',Tp) 
 
Solutions to Problems 11-41 
 
fprintf(', Current Phase Margin = %g',Pm) 
fprintf(', Required Phase Margin = %g',Pmreq) 
fprintf(', Required Phase Margin with Correction Factor = %g',Pmreqc) 
fprintf(', Required Bandwidth = %g',wBW) 
fprintf(', Required Phase Contribution from Compensator = %g',Pc) 
fprintf(', Compensator Beta = %g',beta) 
fprintf(', New phase margin frequency = %g',wmax) 
fprintf(' Compensator gain, Kc = %g',Kc) 
fprintf(' Compensator zero,= %g',-zc) 
fprintf(' Compensator pole,= %g',-pc) 
'G(s)Gc(s)' 
Ge=G*Kc*Gc 
pause 
 %Generate compensated Bode plots 
%Make a Bode plot and get Bode data 
%Get Bode data 
bode(Ge) 
title('Lead Compensated') 
 
w=0.01:0.1:1000; 
[M,P]=bode(Ge,w); 
%Find compensated phase margin 
[Gm,Pm,wcp,wcg]=margin(M,P,w); 
fprintf('\nCompensated Phase Margin,= %g',Pm) 
pause 
 %Generate step response 
T=feedback(Ge,1); 
step(T) 
title('Lead Compensated') 
 
Computer response: 
Type K to meet steady-state error 360 
 
ans = 
 
Open-loop system 
 
ans = 
 
G(s) 
 
Transfer function: 
 360 s + 360 
------------------ 
s^3 + 8 s^2 + 12 s 
 
Type %OS 10 
Type peak time 0.1 
 
ans = 
 
Gc(s) 
 
Transfer function: 
s + 11.71 
--------- 
11-42 Chapter 11: Design via Frequency Response 
s + 77.44 
 
K = 360 Percent Overshoot = 10, Damping Ratio = 0.591155, Peak Time = 0.1, 
Current Phase Margin = 21.0851, Required Phase Margin = 58.5931, Required 
Phase Margin with Correction Factor = 68.5931, Required Bandwidth = 
45.1795, Required Phase Contribution from Compensator = 47.508, Compensator 
Beta = 0.151164, New phase margin frequency = 30.11 Compensator gain, Kc = 
6.61532 Compensator zero,= -11.7067 Compensator pole,= -77.4437 
ans = 
 
G(s)Gc(s) 
 
Transfer function: 
 2382 s^2 + 3.026e004 s + 2.788e004 
------------------------------------- 
s^4 + 85.44 s^3 + 631.5 s^2 + 929.3 s 
 
Compensated Phase Margin,= 60.676» 
 
 
 
Solutions to Problems 11-43 
 
 
 
 
 
11-44 Chapter 11: Design via Frequency Response 
 
 
 
 
 
 
18. 
Program: 
%PD Compensator Design via Frequency Response 
 %Input system 
%Uncompensated system 
K=input('Type K to meet steady-state error '); 
numg=K*[1 1]; 
deng=poly([0 -2 -6]); 
G=tf(numg,deng); 
T=feedback(G,1); 
step(T) 
title('Gain Compensated') 
'Open-loop system' 
'G(s)' 
Gzpk=zpk(G) 
 
 %Input transient response specifications 
Po=input('Type %OS '); 
%Ts=input('Type settling time '); 
Tp=input('Type peak time '); 
 
 %Determine required bandwidth 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
%wn=4/(z*Ts); 
wn=pi/(Tp*sqrt(1-z^2)); 
wBW=wn*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2)); 
 
Solutions to Problems 11-45 
 
 
 %Make a Bode plot and get Bode data 
%Get Bode data 
bode(G) 
title('Gain Compensated') 
w=0.01:0.1:100; 
[M,P]=bode(G,w); 
 
 %Find current phase margin 
[Gm,Pm,wcp,wcg]=margin(M,P,w); 
 
 %Calculate required phase margin 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
Pmreq=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi)+20; 
 
 %Determine a phase margin frequency 
wpm=wBW+7; 
 %Find phase angle at new phase margin frequency and 
 
 %calculate phase required from the compensator 
for i=1:1:length(w); 
if w(i)-wpm>=0; 
wpm=w(i); 
Pwpm=P(i); 
break 
end 
end 
 %Design PD compensator 
Pc=Pmreq-(180+Pwpm); 
zc=wpm/tan(Pc*pi/180); 
Kc=1/zc; 
numc=Kc*[1 zc]; 
denc=1; 
'Gc(s)' 
Gc=tf(numc,denc); 
Gczpk=zpk(Gc) 
 %Display data 
fprintf('\nK = %g',K) 
fprintf(' Percent Overshoot = %g',Po) 
fprintf(', Damping Ratio = %g',z) 
%fprintf(', Settling Time = %g',Ts) 
fprintf(', Peak Time = %g',Tp) 
fprintf(', Current Phase Margin = %g',Pm) 
fprintf(', Required Phase Margin = %g',Pmreq) 
fprintf(', Required Bandwidth = %g',wBW) 
fprintf(', New phase margin frequency = %g',wpm) 
fprintf(', Phase angle at new phase margin frequency = %g',Pwpm) 
fprintf(', Required Phase Contribution from Compensator = %g',Pc) 
fprintf(' Compensator gain, Kc = %g',Kc) 
fprintf(' Compensator zero,= %g',-zc) 
 
pause 
 %Generate compensated Bode plots 
 
11-46 Chapter 11: Design via Frequency Response 
%Make a Bode plot and get Bode data 
%Get Bode data 
'G(s)Gc(s)' 
Ge=G*Gc; 
Gezpk=zpk(Ge) 
bode(Ge) 
title('PD Compensated') 
w=0.01:0.1:100; 
[M,P]=bode(Ge,w); 
%Find compensated phase margin 
[Gm,Pm,wcp,wcg]=margin(M,P,w); 
fprintf('\nCompensated Phase Margin,= %g',Pm) 
pause 
 %Generate step response 
T=feedback(Ge,1); 
step(T) 
title('PD Compensated') 
 
Computer response: 
Type K to meet steady-state error 360 
 
ans = 
 
Open-loop system 
 
 
ans = 
 
G(s) 
 
 
Zero/pole/gain: 
 360 (s+1) 
------------- 
s (s+6) (s+2) 
 
Type %OS 10 
Type peak time 0.1 
 
ans = 
 
Gc(s) 
 
 
Zero/pole/gain: 
0.05544 (s+18.04) 
 
 
K = 360 Percent Overshoot = 10, Damping Ratio = 0.591155, Peak Time = 0.1, 
Current Phase Margin = 21.0851, Required Phase Margin = 78.5931, Required 
Bandwidth = 45.1795, New phase margin frequency = 52.21, Phase angle at new 
phase margin frequency = -172.348, Required Phase Contribution from 
Compensator = 70.9409 Compensator gain, Kc = 0.0554397 Compensator zero,=-18.0376 
 
Solutions to Problems 11-47 
 
ans = 
 
G(s)Gc(s) 
 
 
Zero/pole/gain: 
19.9583 (s+18.04) (s+1) 
----------------------- 
 s (s+6) (s+2) 
 
 
Compensated Phase Margin,= 69.546 
 
 
 
 
 
 
 
 
 
11-48 Chapter 11: Design via Frequency Response 
 
 
 
 
 
Solutions to Problems 11-49 
 
 
 
19. 
5
1000
20
v
K
K   , so 4000K  . A 10% overshoot corresponds to a 0.591  damping factor. 
Using equation (4.42) 33.841
n
  . The required bandwidth is obtained using equation (10.54) 
39.26
BW
  rad/sec. The phase margin is obtained using equation (10.73) with an additional 5 ; 
58.6 5 63.6
PM
    . A new phase margin frequency is chosen close to the bandwidth 
frequency 0.8 31.4
PM BW
   rad/sec. The Bode plot for 4000K  is 
 
 
 
 
 
 
11-50 Chapter 11: Design via Frequency Response 
 
At the new phase margin frequency the phase angle is 168 . The contribution from the lead 
compensator must be 63.6 (180 168 ) 51.6   . Using equation (11.11) 0.13  . 
Lag compensator design: 3.14
10
PM
clag
z

  ; 0.41
clag clag
p z   ; 0.131
clag
clag
clag
p
K
z
  ; 
3.14
0.131
0.41
lag
s
G
s



 
Lead compensator design: Using equations (11.6), (11.9) and (11.12) 
1
0.089
PM
T
 
  ; 
1
11.23
clead
z
T
  ; 86.37clead
clead
z
p

  ; 7.691cleadlead
clead
p
K
z
  ; 
11.23
7.691
86.37
lead
s
G
s



 
A unit step input simulation is performed: 
 
 
Note that the design does not satisfy specs. Redesign if required. 
 
20. 
Program: 
%Lag-Lead Compensator Design via Frequency Response 
%Input system 
K=input('Type K to meet steady-state error '); 
numg=K*[1 5]; 
deng=poly([0 -2 -10]); 
G=tf(numg,deng); 
Solutions to Problems 11-51 
 
'G(s)' 
Gzpk=zpk(G) 
%Input transient response specifications 
Po=input('Type %OS '); 
Ts=input('Type settling time '); 
%Tp=input('Type peak time '); 
T=feedback(G,1); 
step(T) 
title('Gain Compensated') 
pause 
%Determine required bandwidth 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
wn=4/(z*Ts); 
%wn=pi/(Tp*sqrt(1-z^2)); 
wBW=wn*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2)); 
%wBW=(4/(Ts*z))*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2)); 
%wBW=(pi/(Tp*sqrt(1-z^2)))*sqrt((1-2*z^2)+sqrt(4*z^4-4*z^2+2)); 
%Determine required phase margin 
Pmreq=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi)+5; 
%Choose new phase margin frequency 
wpm=0.8*wBW; 
%Determine additional phase lead required at the 
%new phase margin frequency from the lead compensator 
[M,P]=bode(G,wpm); 
Pmreqc=Pmreq-(180+P); 
beta=(1-sin(Pmreqc*pi/180))/(1+sin(Pmreqc*pi/180)); 
%Display data 
fprintf('\nPercent Overshoot = %g',Po) 
fprintf(', Settling Time = %g',Ts) 
%fprintf(', Peak Time = %g',Tp) 
fprintf(', Damping Ratio = %g',z) 
fprintf(', Required Phase Margin = %g',Pmreq) 
fprintf(', Required Bandwidth = %g',wBW) 
fprintf(', New Phase Margin Frequency = %g',wpm) 
fprintf(', Required Phase from Lead Compensator = %g',Pmreqc) 
fprintf(', Beta = %g',beta) 
bode(numg,deng) 
title('Gain compensated') 
pause 
%Design lag compensator 
zclag=wpm/10; 
pclag=zclag*beta; 
Kclag=beta; 
'Lag compensator' 
'Gclag' 
Gclag=tf(Kclag*[1 zclag],[1 pclag]); 
Gclagzpk=zpk(Gclag) 
%Design lead compensator 
zclead=wpm*sqrt(beta); 
pclead=zclead/beta; 
Kclead=1/beta; 
'Lead compensator' 
'Gclead' 
Gclead=tf(Kclead*[1 zclead],[1 pclead]); 
Gcleadzpk=zpk(Gclead) 
%Create compensated forward path 
'Gclag(s)Gclead(s)G(s)' 
Ge=G*Gclag*Gclead; 
Gezpk=zpk(Ge) 
%Test lag-lead compensator 
11-52 Chapter 11: Design via Frequency Response 
T=feedback(Ge,1); 
bode(Ge) 
title('Lag-lead Compensated') 
[M,P,w]=bode(Ge); 
[Gm,Pm,wcp,wcg]=margin(M,P,w); 
'Compensated System Results' 
fprintf('\nResulting Phase Margin = %g',Pm) 
fprintf(', Resulting Phase Margin Frequency = %g',wcg) 
pause 
step(T) 
title('Lag-lead Compensated') 
 
Computer Response: 
Type K to meet steady-state error 4000 
 
ans = 
 
G(s) 
 
 
Gzpk = 
 
 4000 (s+5) 
 -------------- 
 s (s+10) (s+2) 
 
Continuous-time zero/pole/gain model. 
 
Type %OS 10 
Type settling time 0.2 
 
Percent Overshoot = 10, Settling Time = 0.2, Damping Ratio = 0.591155, Required Phase Margin = 
63.5931, Required Bandwidth = 39.2424, New Phase Margin Frequency = 31.394, Required Phase 
from Lead Compensator = 51.3288, Beta = 0.123126 
ans = 
 
Lag compensator 
 
 
ans = 
 
Gclag 
 
 
Gclagzpk = 
 
 0.12313 (s+3.139) 
 ----------------- 
 (s+0.3865) 
 
Continuous-time zero/pole/gain model. 
 
 
 
ans = 
 
Solutions to Problems 11-53 
 
Lead compensator 
 
 
ans = 
 
Gclead 
 
 
Gcleadzpk = 
 
 8.1218 (s+11.02) 
 ---------------- 
 (s+89.47) 
 
Continuous-time zero/pole/gain model. 
 
 
ans = 
 
Gclag(s)Gclead(s)G(s) 
 
 
Gezpk = 
 
 4000 (s+11.02) (s+5) (s+3.139) 
 ----------------------------------- 
 s (s+89.47) (s+10) (s+2) (s+0.3865) 
 
Continuous-time zero/pole/gain model. 
 
 
ans = 
 
Compensated System Results 
 
 
Resulting Phase Margin = 55.8196, Resulting Phase Margin Frequency = 41.1584 
 
11-54 Chapter 11: Design via Frequency Response 
 
 
 
Note that the design does not satisfy specs. Redesign if required. 
 
 
21. 
The required bandwidth for a peak time of 1.8 seconds and  = 0.456 (i.e. 20% overshoot) is 2.588 
rad/s. From the Bode diagram plotted for K = 1 and 20% overshoot, we get a M = 48.15
o
, or a phase 
angle of −180
o
 + 48.15
o
 = −131.85
o
. We also find that this occurs at 1.08 rad/s and that at K = 13.5, 
the magnitude curve will intersect zero dB at 1.08 rad/s. Thus, the following function yields 20% 
overshoot: 
 
  
13.5
1.75 6
G s
s s s

 
 
 
The Bode diagram plotted for K = 13.5 is shown below. 
Solutions to Problems 11-55 
 
 
 
PI controller design: Allowing for a 5
o
 margin, we want M = 48.1
o
 + 5
o
 = 53.1
o 
or a phase angle of 
−180
o
 + 53.1
o
 = −126.9
o
. This angle occurs at  = 0.92 rad/s where the magnitude is at 1.75 dB. The 
controller should contribute − 1.75 dB so that the magnitude curve passes through 0 dB at  = 0.92 
rad/s. Choosing the break frequency one decade below the phase margin frequency of 0.92 rad/s, and 
adjusting the controller's gain to yield − 1.75 dB at high frequencies, the ideal integral controller is 
GcPI(s)
 1.73 0.092s
s

 
and the PI compensated forward path is 
GPI(s) = G(s)GcPI(s) 
 
  2
23.355 0.092
1.75 6
s
s s s


 
 
 
Plotting the Bode diagram for the PI compensated system yields, 
11-56 Chapter 11: Design via Frequency Response 
 
 
The magnitude is zero dB at  = 1.6 rad/s. The phase at this frequency is −150.6
o
. Thus, we have a 
phase margin of −29.4
o
. 
 
PID controller design: Let us increase the phase margin frequency to 4 rad/s, at which the phase is 
−191
o
. To obtain the required phase margin of 48.1
o
, the phase curve must be raised an additional 
59.1
o
. Assume the following form for the proportional derivative part of the controller: 
 
GcPD(s)  ( )
1
' D
D
K K s
K
 
 
The angle contributed by GcPD is: c 
1tan 59.1o
1/
D
K
  
Letting  = 4 rad/s  KD = 0.418. Hence, the PD controller is GcPD(s)  0.418 '( 2.392)K s 
The final PID compensated forward path is: 
 
GPID(s) = GPI(s) GcPD(s) 

 
 2
23.355( 0.092)
*0.418 '( 2.392)
( 1.75)( 6)
s
K s
s s s
 
 
 

 

2
9.762 ( 0.092)( 2.392)
( 1.75)( 6)
K s s
s s s
 
 
Letting 1K   , the magnitude of this function at 4 rad/s is -8.841 dB. Thus, K' must be adjusted to 
bring the magnitude to zero dB. Hence, K' = 2.767 (8.841 dB). 
 
Thus, 
Solutions to Problems 11-57 
 
GPID(s) 
 

 2
27.01( 0.092)( 2.392)
( 1.75)( 6)
s s
s s s
 
 
 
The PID compensated Bode plot follows: 
 
 
 
 
 
The PID compensated time responseis shown below: 
11-58 Chapter 11: Design via Frequency Response 
 
 
 
22. 
Program: 
%Input system 
numg1=1; 
deng1=poly([0 -3 -6]); 
G1=tf(numg1,deng1); 
[numg2,deng2]=pade(0.5,5); 
G2=tf(numg2,deng2); 
'G(s)=G1(s)G2(s)' 
G=G1*G2; 
Gzpk=zpk(G) 
Tu=feedback(G,1); 
step(Tu) 
title('K = 1') 
%Percent Overshoot to Damping Ratio to Phase Margin 
Po=input('Type %OS '); 
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2)); 
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); 
fprintf('\nPercent Overshoot = %g',Po) 
fprintf(', Damping Ratio = %g',z) 
fprintf(', Phase Margin = %g',Pm) 
%Get Bode data 
bode(G) 
title('K = 1') 
pause 
w=0.1:0.01:100; 
[M,P]=bode(G,w); 
 
Solutions to Problems 11-59 
 
Ph=-180+Pm; 
for i=1:1:length(P); 
if P(i)-Ph<=0; 
M=M(i); 
K=1/M; 
fprintf(', Frequency = %g',w(i)) 
fprintf(', Phase = %g',P(i)) 
fprintf(', Magnitude = %g',M) 
fprintf(', Magnitude (dB) = %g',20*log10(M)) 
fprintf(', K = %g',K) 
break 
end 
end 
T=feedback(K*G,1); 
step(T) 
title('Gain Compensated') 
 
Computer response: 
ans = 
 
G(s)=G1(s)G2(s) 
 
Zero/pole/gain: 
 - (s-14.59) (s^2 - 26.82s + 228.4) (s^2 - 18.6s + 290.5) 
-------------------------------------------------------------------- 
s (s+14.59) (s+6) (s+3) (s^2 + 26.82s + 228.4) (s^2 + 18.6s + 290.5) 
 
Type %OS 20 
 
Percent Overshoot = 20, Damping Ratio = 0.45595, Phase Margin = 48.1477, 
Frequency = 0.74, Phase = -132.087, Magnitude = 0.0723422, Magnitude (dB) 
= -22.8122, K = 13.8232» 
 
 
 
 
 
11-60 Chapter 11: Design via Frequency Response 
 
 
 
 
 
 
 
Second-order approximation not valid. 
 
 
 
 
 
 
 
Solutions to Design Problems 11-61 
 
 
SOLUTIONS TO DESIGN PROBLEMS 
 
23. 
a. The bode plot for the open loop transmission is obtained and shown next: 
>> syms s 
>> s=tf('s'); 
>> M=0.005/(s+0.005); 
>> P=140625/(s+2.67)/(s+10); 
>> set(P,'inputdelay',0.1) 
>> L=M*P 
Transfer function: 
 703.1 
exp(-0.1*s) * ---------------------------------- 
 s^3 + 12.68 s^2 + 26.76 s + 0.1335 
 
>> bode(L) 
 
11-62 Chapter 11: Design via Frequency Response 
 
It can be seen that with 1L  , the system is closed loop unstable. The desired %OS=15% 
corresponds to a damping factor of 
2 2
ln(% /100)
0.517
ln (% /100)
OS
OS



 

. In turn this 
corresponds to a phase margin of 
1
2 4
2
tan 53.2
2 1 4
M

 
  
  
. For this to occur 
the open loop transmission must have a phase of 180 53.2 126.8    at the point where the 
open loop transmission has a magnitude of 0db. The open loop transmission attains this phase when 
1.17  rad/sec, and at that frequency the open loop transmission has a magnitude of 26.2db. So 
the open loop transmission must be decreased by this amount resulting in 0.049L  . 
 
 
 
 
 
Solutions to Design Problems 11-63 
 
b. 
 
 
 
 
 
 
 
 
 
 
 
The figure shows a %OS slightly smaller than 15%. 
 
11-64 Chapter 11: Design via Frequency Response 
 
 
24. 
a. We calculate 
ss
e for step inputs in the uncompensated system. 
0
( ) 19.21
p
s
K Lim G s

  , so 
1
0.05
1
ss
p
e
K
 

. The uncompensated phase margin can also be obtained through a bode plot 
and is of 77.3. 
A tenfold improvement in steady state error requires multiplying the open loop gain by 10. The 
resulting open loop transfer function frequency response is shown in the following ode plot. The 
desired phase margin for the design will be 77.3+10=87.3. It can be seen there that this phase 
value is achieved when 10.1
sec
rad
  . At this point the magnitude of the open loop transmission is 
25.8db. So the lag compensator must provide -25.8db at 10.1
sec
rad
  . 
 
 
The compensator is designed by postulating -25.8db at high frequencies with a higher break 
frequency of 10.1/10=1.01 rad/sec. The phase lag asymptote predicts -5.8 db one decade earlier at 
Solutions to Design Problems 11-65 
 
0.101 rad/sec, and approximately 0db one octave before that at 0.101/2=0.0505 rad/sec. Maintaining 
unity dc gain for the compensator we get: 
0.05( 1.01)
( ) 10*
( 0.0505)
s
Gc s
s



 
b. 
>> syms s 
>> s=tf('s'); 
>> P=1361/(s^2+69*s+70.85); 
>> Gc=10*0.05*(s+1.01)/(s+0.0505); 
>> T=P/(1+P); 
>> Tc=Gc*P/(1+Gc*P); 
>> step(T,Tc,0.7) 
 
 
 
 
25. 
11-66 Chapter 11: Design via Frequency Response 
a. For a phase margin of 30 the gain is 0.00135K  as shown in the following bode plot 
 
 
 
 
b. The gain margin is 12.1db 
c. From figure 10.48 a phase margin of 30 corresponds to a 0.3  damping factor. This in turn 
corresponds to a %OS=37%. The bode plot above shows three crossovers of -7db. We use the largest 
one as the estimate for 
max
0.328
sec
rad
  ; from equation 10.55 we get 59secsT  
d. 
>>syms s 
>>s=tf(‘s’); 
>>P=(-34.16*s^3-144.4*s^2+7047*s+557.2)/(s^5+13.18*s^4+95.93*s^3+14.61*s^2+31.94*s); 
Solutions to Design Problems 11-67 
 
 >> L=0.00135*P; 
>> T=L/(1+L); 
>> t=linspace(0,350,5000); 
 >> step(T,t) 
 
 
 
The estimate for the %OS is very inaccurate, the settling time estimate is reasonable. 
 
e. The reason the estimate of %OS is very inaccurate is due to the multiple crossovers of the 
magnitude response. The hypothesis of a second order approximation is invalid. 
 
26. 
a. The %OS spec required a damping factor of 
2 2
ln(% /100)
0.5
ln (% /100)
OS
OS



 

, which in 
turn requires a phase margin of 
1
2 4
2
tan 52
2 1 4
M

 
  
  
. The bandwidth 
11-68 Chapter 11: Design via Frequency Response 
requirement is obtained from 
2 4 24 1 2 4 4 2 5088.1
sec
BW
s
rad
T
   

      . To 
obtained the compensator gain requirement to achieve this bandwidth obtain 
 
4
2
3.3333 10
( 5088.1) 0.0013
5088.1
G j

    . The compensator’s gain can be obtained from 
1
0.0013
2
K  or 549.1891K  . 
The required open loop crossover frequency is obtained by solving 
4
2
3.3333 10
549.2 1
c


 , 
giving 4280
sec
c
rad
  . This is the frequency at which the lead compensator should provide 
maximum lead phase. So for the design of the compensator from Figure 11.8 let 0.1  with 
3.5T  giving 4
3.5
8.1776 10
c
T

   . So the designed compensator is 
1
10( 1222.9)
( ) 549.2
1 ( 12229.9)
c
s
K sTG s
s
s
T




 


. The gain of the compensator is now adjusted so 
that the maximum phase lead is provided at the crossover frequency giving
10( 1222.9)
( ) 173.7
( 12229.9)
c
s
G s
s



. However a time domain simulation shows that although the 
settling time spec is satisfied the resulting %OS=22%. The parameters of the phase lead 
compensator are slightly adjusted to provide more phase lead giving 
15( 1000)
( ) 173.7
( 15000)
c
s
G s
s



 
b. 
>> syms s 
>> s=tf('s'); 
>> P=3.3333e4/s^2; 
>> G=173.67*15*(s+1000)/(s+15000); 
>> L=G*P; 
>> T=L/(1+L); 
 >> step(T,4e-3) 
 
Solutions to Design Problems 11-69 
 
 
 
 
27. 
 
We follow the step outlined in the chapter. 
a. 
1. We calculate the required 
2 2
ln(% /100)
0.6
ln (% /100)
OS
OS



 

. The required bandwidth is 
2 4 24 1 2 4 4 2 0.128
sec
BW
s
rad
T
   

      
2. The uncompensated system is 
45.8333 10
( )
( 0.02)( 0.0833)( 0.25)
K
G s
s s s


  
 so 
(0) 1.4
p
K G K  . The steady state requirement requires 
1
0.05
1 1.4K


 or 13.6K  
3. The bode plot for the uncompensated system is 
 
 
 
 
 
 
11-70 Chapter 11: Design via Frequency Response 
 
 
 
 
4. The required phase margin for the design is 
1
2 4
2
tan 59.2
2 1 4
M

 
  
  
 
5. We arbitrarily select 0.1
sec
rad
  
6. It can be seen from the uncompensated bode plot that at 0.1
sec
rad
  the phase margin is29. The 
requirement in the phase lead network is going to be 59.2-29+5 (for the lag compensator 
contribution)=35.2 
7. The lag compensator design is done by schooseng the higher break frequency as 
0.1
0.01
10 sec
rad
 . 
From the lead compensator graphs, figure 11.8 in the text let 0.25  or 
1
4

  . So the lag 
compensator is given by 
2
_
2
1
( )
1 1 ( 0.01)
1 4 ( 0.0025)
( )
C lag
s
T s
G
s
s
T




 


 
Solutions to Design Problems 11-71 
 
8. For the lead compensator design 
max
0.1  and 0.25  . 
max
1
1
0.05
T
   and 
1
1
0.2
T
 Resulting in 1_
1
1
( )
( 0.05)
4
( 0.2)
( )
c lead
s
T s
G
s
s
T




 


 
9. The resulting lag-lead compensator is 
( 0.01)( 0.05)
( ) 13.6
( 0.0025)( 0.2)
c
s s
G s
s s
 

 
. The resulting 
compensated bode plot is shown above. 
b. The step response simulation is 
>> syms s 
>> s=tf('s'); 
>> G=13.6*5.8333e-4/(s+0.02)/(s+0.25)/(s+0.0833); 
>> Gc = (s+0.01)*(s+0.05)/(s+0.0025)/(s+0.2); 
>> L=G*Gc; 
 >> T=L/(1+L); 
>> step(T) 
 
 
 
11-72 Chapter 11: Design via Frequency Response 
The system exhibits a “long tail” response because compensation adds a pole-zero pair very close to 
the origin. However it can be seen that after 60sec the response is very close to steady state. 
28. 
a. Using equation (4.39), x=0.7. Then from equation (10.52) Mp=1=0db. For zero 
steady state error we add an integrator, and then we raise the system’s gain. 
 
Then we design an appropriate lead compensator. So far, 



3.038( 4.6)
( 13.7)
lead
s
G
s s
 as shown in 
the figure with a slight violation of the 1db requirement. 
The resulting Kv=8.306 so a factor of 2.4 is needed to reach the required 20. The lag 
compensator is designed by arbitrarily adding a pole zero pair at 0.1 and 0.24 respectively, 
giving 
0.24
0.1
lag
s
G
s



. The total compensation is 
 

 
3.038( 0.24)( 4.6)
( 0.1)( 13.7)
c
s s
G
s s s
 
b. For this design, we start by adding an integrator to the compensator. The resulting Kv=8.14 
so a K=2.46 is needed to obtain Kv=20. Then enough lead is provided to satisfy the 
Mp<1db requirement. The resulting 



35( 2.18)
( 31)
s
G
s s
. The resulting 32B  rad/sec as shown 
in the next figure. 
Solutions to Design Problems 11-73 
 
 
c. The following figure shows the step response of both designs. The faster design 
corresponds to part b with the larger bandwidth. 
 
11-74 Chapter 11: Design via Frequency Response 
 
 
29. 
a. 
0.1416 30
v
K K  from which we obtain 211.864K  . Also 10% overshoot corresponds 
to a damping factor of 0.591  , which in turn requires (equation(10.73)) a phase margin of 
58.6
PM
  , but since the lag compensator will contribute some phase lag the requirement is 
augmented by 10 to 68.6
PM
  . 
A bode plot with 211.864K  shows that the uncompensated phase margin is 30.3 at a 
frequency of 31.2 rad/sec, thus the need for compensation. 
Solutions to Design Problems 11-75 
 
 
We look on the uncompensated Bode plot for the frequency at which the open loop transfer 
function has 180 68.6 111.14    of phase. This occurs when 11.6  rad/sec. This 
will be the new phase margin frequency. Thus at this frequency we want the gain to be 0 dB. 
The current gain is 8.29 dB. 
To design the compensator we select the high break frequency as 11.6 /10 1.16 rad/sec. We 
use 8.29 dB as the high frequency asymptote of the compensator and we draw a -20 db/sec 
asymptote from the intersection of these points towards 0dB. We find that the 0dB line is 
intersected when 0.45  rad/sec. Thus the lag compensator is 
 
 
1 /1.16
1 / 0.45
lag
s
G
s



 
b. 
The step response is shown below: 
 
11-76 Chapter 11: Design via Frequency Response 
 
30. 
Including the additional integrator and gain the open loop transfer function becomes 
27
1 4 3 2
(0.0169 0.03558)
( )
3.368 3.762 32.19
sK sG s e
s s s s

  
 
Let 2K  to achieve the steady state requirement on ramp inputs. A 13% overshoot corresponds 
to a 0.54  damping factor. Using equation (4.42) 0.0148n  . The required bandwidth is 
obtained using equation (10.54) 0.01815
BW
  rad/sec. The phase margin is obtained using 
equation (10.73) with an additional 5 ; 55 5 60
PM
    . A new phase margin frequency is 
chosen close to the bandwidth frequency 0.8 0.0145
PM BW
   rad/sec. The Bode plot for 
2K  and the integrator is 
Solutions to Design Problems 11-77 
 
 
At the new phase margin frequency the phase angle is 137 . The contribution from the lead 
compensator must be 60 (180 137 ) 17   . Using equation (11.11) 0.54  . 
Lag compensator design: 0.00145
10
PM
clag
z

  ; 0.000783
clag clag
p z   ; 
0.54
clag
clag
clag
p
K
z
  ; 
0.00145
0.54
0.000783
lag
s
G
s



 
Lead compensator design: Using equations (11.6), (11.9) and (11.12) 
1
93.85
PM
T
 
  ; 
1
0.0107
clead
z
T
  ; 0.0197clead
clead
z
p

  ; 1.841cleadlead
clead
p
K
z
  ; 
0.0107
1.841
0.0197
lead
s
G
s



 
A unit step input simulation is performed: 
11-78 Chapter 11: Design via Frequency Response 
 
 
 
The overshoot obtained is certainly unacceptable; it can be corrected by iterating on the pole of the 
lag compensator: 
(1 / 0.00145)
(1 / 0.00045)
lag
s
G
s



 . In this case the overshot is corrected at detriment of 
the settling time: 
Solutions to Design Problems 11-79 
 
 
 
 
31. 
The controlled plant of Problem 10.52 was given by: 
 

   
2
4 3 2
0.1111(4 5 1)
( )
3.1 0.85 0.87 0.1111
P
s s
G s
s s s s
. 
 
The following MATLAB M-file was written: 
 
clc % Clear the Command window. 
clf % Clear the Figure. 
numg = 0.1111*[4 5 1]; 
deng = [1 3.1 0.85 0.87 0.1111]; 
Gp = tf(numg, deng); 
pos = 10; % Desired percent overshoot. 
z=(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2)); 
 % Calculate required damping ratio,z. 
Pm=atan(2*z/(sqrt(-2*z^2+sqrt(1+4*z^4))))*(180/pi); 
 % Calculate required phase margin. 
w=0.01:0.01:100; % Set range of frequency from 0.01 to 
 % 100 in steps of 0.01. 
 
11-80 Chapter 11: Design via Frequency Response 
bode(Gp) % Bode plot for Gp(s). 
grid 
pause 
[M,P]=bode(Gp,w); % Get Bode data. 
Ph=-180+Pm; % Calculate required phase angle. 
for k=1:1:length(P); % Search Bode data for required phase 
 % angle. 
if P(k)-Ph<=0; % If required phase angle is found, 
 % find the value of the magnitude at 
M=M(k); % the same frequency. 
wf=w(k); % At this frequency the magnitude 
 % plot must go through 0 dB. 
break % Stop the loop. 
end % End if. 
end % End for. 
'G(s)' % Display label for G(s). 
G=zpk(Gp/M) % 
bode(G) % Bode plot for G(s). 
grid 
pause 
numgcpi = 0.9613*poly([-0.0557]); 
 % Numerator of PI controller's TF. 
dengcpi = poly([0]); % Denominator of PI controller's TF. 
Gcpi=tf(numgcpi,dengcpi); 
Gpi=zpk(G*Gcpi) 
bode(Gpi) % Bode plot for PI-controlled system. 
grid 
pause 
numgcpd = 12.8463*poly([-3.593]); 
 % Numerator of PD controller's TF. 
dengcpd = 1; % Denominator of PD controller's TF. 
Gcpd=tf(numgcpd,dengcpd); 
Gpid=zpk(Gpi*Gcpd) 
bode(Gpid) % Bode plot for PID-controlled system. 
grid 
pause 
T=feedback(Gpid,1); % FindT(s). 
step(T) % Generate closed-loop, PID-Controlled 
 % step response. 
title('PID-Controlled Step Response') 
 % Add title 
pause 
 
 The required bandwidth for a settling time of 50 sec. and  = 0.456 (i.e. 10% overshoot) is BW = 
0.157 rad/s. The Bode plot (for K = 1 and 10% overshoot) gives a M = 24.1
o
, or a phase angle = 
−180
o
 + 24.1
o
 = −155.9
o
. This occurs at 0.68 rad/s at which K = 0.2588, the magnitude curve will 
intersect zero dB at 0.68 rad/s. Thus, the following transfer function (in pole-zero form) yields 10% 
overshoot: 
 
Solutions to Design Problems 11-81 
 
 

   2
0.11501( 1)( 0.25)
( )
( 2.906)( 0.1373)( 0.0567 0.2784)
s s
G s
s s s s
 
 
 The Bode diagram plotted for G(s) is shown below. 
 
PI controller design: Allowing for a 5
o
 margin, we want M = 50.1
o
 + 5
o
 = 55.1
o 
or a phase angle of 
−180
o
 + 55.1
o
 = −124.9
o
. This angle occurs at  = 0.557 rad/s where the magnitude is at 0.343 dB. The 
controller should contribute − 0.343 dB so that the magnitude curve passes through 0 dB at  = 0.557 
rad/s. Choosing the break frequency one decade below the phase margin frequency and adjusting the 
controller's gain to yield − 0.343 dB at high frequencies, the ideal integral controller is: 
 
 
 
 
GcPI(s) 


0.9613( 0.0557)s
s
 
 
11-82 Chapter 11: Design via Frequency Response 
Thus, the PI compensated forward path is: 
 
 
   2
0.11056( 1)( 0.25)(s+0.0557)
( 2.906)( 0.1373)( 0.0567 0.2784)
PI cPI
s s
G (s) G(s)G (s)
s s s s s
 
 
From the Bode diagram for the PI compensated system (shown below), we see that the phase margin is 
46
o
at a frequency  = 0.558 rad/s. 
 
 
 
 
 
 
PID controller design: Let us increase the phase margin frequency to 2 rad/s, at which the phase is 
−154
o
. To obtain the required phase margin of 55.1
o
, the phase curve must be raised an additional 
29.1
o
. Assume the following form for the proportional derivative part of the controller: 
 
Solutions to Design Problems 11-83 
 
GcPD(s)  ( )
1
' D
D
K K s
K
 
 
The angle contributed by GcPD is: 
1 29.1
1 /
o
C
D
tan
K

   
 
Letting  = 2 rad/s  KD = 0.2783. Hence, the PD controller is GcPD(s)  0.2783 '( 3.593)K s 
 
The final PID compensated forward path is: 
 
GPID(s) = GPI(s) GcPD(s) 
   

   

2
0.03077 (s 1)(s 0.25)(s 0.0557)(s 3.593)
( 2.906)( 0.1373)( 0.0567 0.2784)
K
s s s s s
 
 
Letting 1K   , the magnitude of this function at 2 rad/s is -33.28 dB. Thus, K' must be adjusted to 
bring the magnitude to zero dB. Hence, K' = 46.16 (33.2846 dB). 
 
Thus, 
GPID(s) 
   

   2
1.4203(s 1)(s 0.25)(s 0.0557)(s 3.593)
( 2.906)( 0.1373)( 0.0567 0.2784)s s s s s
 
 
 
The PID compensated Bode plot follows: 
11-84 Chapter 11: Design via Frequency Response 
 
 
 
The PID compensated time response is shown below. Since it satisfies all 
requirements, the design is acceptable. 
. 
 
 
Solutions to Design Problems 11-85 
 
 
32. 
a. For an overdamped system 
4
s
BW
T

 . So for this system 0.04
sec
BW
rad
  
b. The bode plot is: 
 
 
11-86 Chapter 11: Design via Frequency Response 
At -90 of open loop phase lag the crossover frequency of the open loop transmission equals the 
bandwidth of the closed loop system, so it can be seen that we have to drop the magnitude 73.8db. 
So 2 4K e  . Note that in the low range of frequencies the phase of the loop transmission is 
approximately -90, lowering the gain will maintain closed loop stability. 
c. 
>> syms s 
>> s=tf(‘s’); 
>> G=-2e-4*(s^2+0.04*s+0.0048)/(s+0.02)/s; 
>> L=G*P; 
>> T=L/(1+L); 
>> step(T) 
 
 
 
No further gain adjustments are necessary. 
 
Solutions to Design Problems 11-87 
 
33. 
a. For a zero steady-state error for step inputs and a steady-state error for ramp 
inputs ≤2 %, the required value of K may be found from: 
1
( ) 0.02
ramp
v
e
K
   , where 
 

  
0 0
( 0.6)
( ) 50lim lim
( 0.5858)
v
s s
K s
K sG s
s
. 
Hence:    0.6
0.5858
50 48.82K K 
The following MATLAB file was used to plot the Bode magnitude and phase plots for that system 
and to obtain the response of the system, c(t), to a step input, r(t) = 4 u(t). 
 
numg = [1 0.6]; 
deng = poly ([0 -0.5858]); 
G = 48.82*tf(numg, deng); 
bode (G); 
grid 
pause 
T = feedback(G,1); %T is the closed-loop TF of the PI-controlled system 
T = minreal(T); 
step(4*T); 
axis 
grid 
xlabel ('Time') 
ylabel ('Speed Sensor Output, c(t) in volts') 
title ('PI-controlled Systems Response to a 4 volt Input Step') 
 
The Bode magnitude and phase plots obtained are shown below with the minimum stability 
margins displayed on the phase plot. 
11-88 Chapter 11: Design via Frequency Response 
 
For a %OS ≤ 4.32 %, the damping ratio is 
2 2 2 2
ln(% /100) ln(4.32 /100)
0.707
ln (% /100) ln (4.32 /100)
OS
OS

 
 
  
 
. Using Eq. (10.73) we find 
the phase margin needed to meet the damping ratio requirement: 
1 o
2 4
2
tan 65.52
2 1 4
M

 
  
  
 
 
 
 
The phase margin found from the Bode plot obtained in step (1) is greater than the required value. 
Therefore, the response of the system, c(t), to a step input, r(t) = 4 u(t), has been plotted and is 
shown below. The settling time, Ts, and the final value of the output are noted. 
Solutions to Design Problems 11-89 
 
 
 
As could be seen from the graph and the analysis presented above all requirements are met. 
Therefore, the design has been completed. 
b. 1) When the PI-controller zero, ZI, moves to – 0.01304: 
 
 
    
 0 0
( 0.6)( 0.01304 )
( ) 50 61.02lim lim
( 0.0163)( 0.5858)
v
s s
K s s
K sG s K
s s
. 
The phase margin found from the Bode plot obtained is still greater than the required value. 
Therefore, c(t) was plotted and is shown below with the settling time, Ts, and the final value of the 
output noted on the graph. As could be seen from the graph, the settling and rise times are less by 
~ 20% than the respective values obtained for ZI at – 0.0163. 
11-90 Chapter 11: Design via Frequency Response 
 
2) When the PI-controller zero, ZI, moves to – 0.01956: 
 
 
    
 0 0
( 0.6)( 0.01956 )
( ) 50 40.68lim lim
( 0.0163)( 0.5858)
v
s s
K s s
K sG s K
s s
. 
The phase margin found from the Bode plot obtained is still greater than the required value. 
Therefore, c(t) was plotted and is shown below with the settling time, Ts, and the final value of the 
output noted on the graph. In this case, however, the settling and rise times are higher by ~ 20% 
than the respective values obtained for ZI at – 0.0163. Nevertheless, all requirements are still met. 
Solutions to Design Problems 11-91 
 
 
 
The responses obtained in all cases here are closer to the response of a first-order system rather 
than a second order system. Note that when ZI is at – 0.01956, for example, two of the closed-loop 
poles (at – 0.6002 and – 0.01956) are very close, respectively, to the closed-loop zeros located at – 
0.5999 and – 0.0196. Therefore, the system behaves as if it has only one closed-loop pole, which 
is at – 40.66. 
Since the PI-controller designed meets all requirements (even when pole-zero 
cancellation is not achieved and the controller’s zero changes by ± 20%), there no need to 
add a derivative mode. 
 
 
 
 
 
 
 
 
 
 
 
 
 
11-92 Chapter 11: Design via Frequency Response 
 
34. 
a. The resulting Nyquist diagram is 
 
 
 
 
b. Using the gain margin, the new range of closed loop stability for K is 0<K<2.86. 
c. We use equation (10.73) to find that a 0.5  damping factor corresponds to a 
phase margin of 
51.8
M
  , which corresponds to arg( ( )) 180 51.8 128.2
c
G j      . 
Examination ofthe frequency response shows that this phase is achieved when 
0.0089  rad/sec with a magnitude of -0.973db (when K=1). Thus to achieve this 
damping factor K=1.12 . 
d. The corresponding unit step response for the closed system is shown below: 
Solutions to Design Problems 11-93 
 
 
 
 
 
ONLINEFFIRS 11/25/2014 13:29:37 Page 1
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Library of Congress Cataloging-in-Publication Data
Nise, Norman S.
Control systems engineering / Norman S. Nise, California State Polytechnic University, Pomona. — Seventh edition.
1 online resource.
Includes bibliographical references and index.
Description based on print version record and CIP data provided by publisher; resource not viewed.
ISBN 978-1-118-80082-9 (pdf) — ISBN 978-1-118-17051-9 (cloth : alk. paper)
1. Automatic control–Textbooks. 2. Systems engineering–Textbooks. I. Title.
TJ213
629.8–dc23
2014037468
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