ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
1
Solution to HW7
AP8.4 We are given the forward and feedback path transfer functions for a unity negative
feedback system. The forward path transfer function is
G(s) =
s2
1
+ 3.2s + 16
and the feedback path transfer function is
H(s) =
1
.
0.5s + 1
We are asked to sketch the Bode diagram for G(s)H(s).
Solution: First we find the corner frequencies ω. For H(s) we have a pole at s = −2,
hence a corner at ω = 2 and for G(s) we have ω n = 4 and ζ = 3.2/(2ωn ) = 0.4. Thus for
the complex pole pair,
q
and
ωr = ωn 1 − 2ζ 2 = 3.2
1
= 1.36.
2ζ 1 − ζ 2
To determine the magnitude for low frequencies, we must rewrite the transfer function as
Mpω = |G(ωr )| =
G(s)H(s) =
p
1/16
.
(s2 /16 + 3.2s/16 + 1)(s/2 + 1)
Then for frequencies below ω = 2 we have horizontal asymptote |G(ω)H(ω)| = 20 log 10 (1/16) =
−24 dB. From ω = 2 to ω = 4 our asymptote has a slope of −20 dB/decade. Superimposed
on this is a peak of height 20 log 10 (Mpω ) about ωr . Beyond ω = 4, we have asymptotic
slope −60 dB/decade since there are a total of three poles. For the phase plot, we have an
asymptote that decreases from 0 to −90 degrees over the frequency range 2/10 to 2 ∗ 10
rad/s (2 decades centered at ω = 2 rad/s). Another goes from −90 to −180 over the
range 4/10 to 4 ∗ 10 rad/s (2 decades centered at ω = 4 rad/s). We add these to get the
asymptotic phase curve which goes from zero degrees at frequencies below ω = 2/10 to
−270 degrees at ω = 4 ∗ 10 and above.
From this information we could generate the plots by hand. The exact magnitude and
phase curves generated by Matlab are shown in Figure 1. The transcript of the Matlab
session is given below.
>> GH = tf([1/16],conv([1/16 3.2/16 1],[1/2 1]))
Transfer function:
0.0625
-----------------------------------0.03125 s^3 + 0.1625 s^2 + 0.7 s + 1
>> bode(GH); grid;
>> title(’Magnitude and Phase Plots for AP8.4’)
>> print -deps ap8_4.eps
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
2
Figure 1: Definitions of angles for departure angle calculation
Magnitude and Phase Plots for AP8.4
−20
−30
−40
Magnitude (dB)
−50
−60
−70
−80
−90
−100
−110
Phase (deg)
−120
0
−90
−180
−270
−2
10
−1
0
10
1
10
10
2
10
Frequency (rad/sec)
AP8.5 We are given the closed–loop transfer function
T (s) =
10(s + 1)
+ 9s + 10
s2
for a “unity feedback” system and asked to find the open–loop transfer function, generate
a log–magnitude–phase plot, and determine whether the open– and closed–loop systems
are stable.
Solution: The answers to these questions depend upon whether we assume negative
feedback or positive feedback. Probably the authors intended to use negative feedback.
(a) Assuming negative feedback we have
T (s) =
G(s)
1 + G(s)
so
G(s) = (1 + G(s))
10(s + 1)
s2 + 9s + 10
ECE382/ME482 Spring 2004
then
Homework 7 Solution
April 4, 2004
3
s2 + 9s + 10 G(s) = 10(s + 1) (1 + G(s))
and solving for G(s) yields
G(s) =
10(s + 1)
.
s(s − 1)
If we assume positive feedback, we would similarly obtain
G(s) =
s2
10(s + 1)
.
+ 19s + 20
(b) The log–magnitude–phase plots for the three transfer functions are given in Figure 2.
In the plot legend, “NF” stands for negative feedback and “PF” stands for positive
feedback. The Matlab script that generated this figure is given below.
clsys = tf([10 10],[1 9 10]);
olsysn = tf([10 10],[1 -1 0]);
olsysp = tf([10 10],[1 19 20]);
w = logspace(-1,3,1000);
[magc,pc,wc] = bode(clsys,w);
[magon,pon,won] = bode(olsysn,w);
[magop,pop,wop] = bode(olsysp,w);
plot(squeeze(pc),20*log10(squeeze(magc)))
hold on
plot(squeeze(pon),20*log10(squeeze(magon)),’--’)
plot(squeeze(pop),20*log10(squeeze(magop)),’-.’)
legend(’Closed-Loop’,’Open-Loop, NF’,’Open-Loop, PF’)
a = find(wc >= 1);
b = find(wc >= 10);
c = find(wc >= 50);
d = find(wc >= 110);
e = find(wc >= 500);
omega_index = [a(1) b(1) c(1) d(1) e(1)]
wc(omega_index)
for index = omega_index,
plot(pc(1,1,index),20*log10(magc(1,1,index)),’xr’)
plot(pon(1,1,index),20*log10(magon(1,1,index)),’xr’)
plot(pop(1,1,index),20*log10(magop(1,1,index)),’xr’)
end;
title(’Log-Magnitude-Phase Plot for AP8.5b’)
xlabel(’Phase (deg)’)
ylabel(’Magnitude (dB)’)
grid
print -depsc ap8_5.eps
(c) The poles of the given closed–loop system are −1.3 and −7.7 so the closed–loop system
is stable. If we assume negative feedback, the open–loop system obtained has a pole
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
4
Figure 2: Plot generated by Matlab for AP8.5
Log−Magnitude−Phase Plot for AP8.5b
40
Closed−Loop
Open−Loop, NF
Open−Loop, PF
30
20
Magnitude (dB)
10
0
−10
−20
−30
−40
−50
−100
−50
0
50
100
Phase (deg)
150
200
250
300
in the right–half plane, hence is unstable. If we assume positive feedback, the poles
are −1.12 and −17.88, hence stable.
DP8.2 We are given a negative unity feedback system with a controller preceding the plant in
the forward path. The controller transfer function is G c (s) = K(s + 1)/(s + 5) and the
plant transfer function is
1
.
G(s) =
2
s(s + 2s + 10)
We are asked to sketch the Bode diagrams for the open– and closed–loop transfer functions
when K = 20. We are then asked to determine the frequencies for which the gain is 0
dB and for which the phase is −180 degrees in the open–loop case. Next, we are asked to
determine Mpω , ωr and ωB for the closed–loop transfer function for two different values of
the gain, K = 20 and K = 40. Finally, we are asked to choose the one of these gains that
gives the shortest settling time while keeping the overshoot in response to a step input less
than 35%.
Solution:
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
5
(a) With K = 20, the open–loop transfer function is
Gc (s)G(s) =
20(s + 1)
20(s + 1)/50
=
2
s(s + 5)(s + 2s + 10)
s(s/5 + 1)(s2 /10 + 2s/10 + 1)
which yields a corner frequency of ω = 1 for the
√ real zero, another of ω = 5 for
the real pole √
at s = −5 and another at ω n =p 10 due to the complex pole pair.
With ζ = 1/ 10 = 0.32 we find Mpω = (2ζ 1 − ζ 2 )−1 = 1.67. We could now
use asymptotes to draw the Bode diagram. The magnitude plot would have slope
−20dB/decade for ω < 1, value 0+20 log 10 (2/5) at ω = 1 and slope of −60 dB/decade
for large ω. The phase plot would start at −90 degrees (due to the pole at the origin),
increase due to the zero at s = −1, and decrease due to each of the poles to −270
degrees for large ω. From the Bode plot we determine that the gain is 0 dB at
approximately ω = 0.4 rad/s and the phase is −180 degrees at approximately ω = 4
rad/s. (For the exact values, see the output of the margin command in the transcript
below.) The Bode plot is given in Figure 3.
(b) The closed–loop transfer function is
T (s) =
20s + 20
s+1
Gc (s)G(s)
= 4
= 4
1 + Gc (s)G(s)
s + 7s3 + 20s2 + 70s + 20
s /20 + 7s3 /20 + s2 + 7s/2 + 1
which has poles −5.55, −0.57 ± 3.36j, and −0.31. The Bode plots generated for the
closed–loop transfer function using Matlab are shown in Figure 3.
The transcript of the commands that generated this plot is given below.
>> K = 20;
>> olsys = tf(K*[1 1],conv([1 5 0],[1 2 10]))
Transfer function:
20 s + 20
--------------------------s^4 + 7 s^3 + 20 s^2 + 50 s
>> clsys = feedback(olsys,1)
Transfer function:
20 s + 20
-------------------------------s^4 + 7 s^3 + 20 s^2 + 70 s + 20
>> [mag,phase,w] = bode(olsys);
>> [Gm,Pm,Wcg,Wcp] = margin(mag,phase,w)
Gm =
3.1772
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
6
Figure 3: Bode plot generated by Matlab for Problem DP8.2, parts (a) and (b)
Magnitude and Phase Plots for DP8.2
Magnitude (dB)
50
0
−50
Phase (deg)
−100
0
Open−Loop
Closed−Loop
−90
−180
−270
−2
−1
10
10
0
10
Frequency (rad/sec)
Pm =
103.6808
Wcg =
4.0269
Wcp =
0.4433
>> 20*log10(Gm)
ans =
%% gain margin in dB
1
10
2
10
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
7
10.0410
>>
>>
>>
>>
bode(olsys,’-’,clsys,’-.’); grid;
title(’Magnitude and Phase Plots for DP8.2’)
legend(’Open-Loop’,’Closed-Loop’)
print -deps dp8_2.eps
(c) To determine Mpω and ωr , we must factor the denominators of the closed–loop transfer
functions obtained for K = 20 and K = 40. We then identify the complex pole
pairs and use the Matlab command poly to obtain the second order polynomial
corresponding to the complex pole pair. Identifying the polynomial with s 2 +2ζωn s+
ωn2 allows us to easily compute ζ for each case. We then use equations (8.36) and
(8.37) from the text to compute ωr and Mpω , respectively. For the case K = 20,
we find ωr = 3.31 and Mpω = 3.04. For the case K = 40, we find ωr = 3.68 and
Mpω = 7.04.
To determine the bandwidths, we plot the Bode diagrams for both cases, as shown
in Figure 4. We notice that for both cases, the magnitude dips more than 3 dB
below the low frequency value before the peak occurs, so that if we automated the
bandwidth search procedure we would have to exercise caution to identify the −3
dB point after the peak rather than the one before. Clicking on points on the Bode
plots to determine the values leads us to conclude that the bandwidth for the case
K = 20 is approximately 3.5 rad/s whereas the bandwidth for the case K = 40 is
approximately 4.6 rad/s. The transcript of the session that generated the required
values and the Bode plots if given below.
>> K = 40;
>> olsys2 = tf(K*[1 1],conv([1 5 0],[1 2 10]))
Transfer function:
40 s + 40
--------------------------s^4 + 7 s^3 + 20 s^2 + 50 s
>> clsys2 = feedback(olsys2,1)
Transfer function:
40 s + 40
-------------------------------s^4 + 7 s^3 + 20 s^2 + 90 s + 40
>> [num2,den2] = tfdata(clsys2,’v’)
num2 =
0
0
0
40
40
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
8
Figure 4: Bode plots generated by Matlab for Problem DP8.2, part (c)
Magnitude and Phase Plots for DP8.2, part (c)
50
Magnitude (dB)
0
−50
−100
Phase (deg)
−150
0
K=20
K=40
−90
−180
−270
−2
−1
10
0
10
1
10
10
Frequency (rad/sec)
den2 =
1
7
20
90
40
>> poles2 = roots(den2)
poles2 =
-5.9844
-0.2632 + 3.6872i
-0.2632 - 3.6872i
-0.4891
>> [num,den]=tfdata(clsys,’v’)
num =
2
10
3
10
ECE382/ME482 Spring 2004
Homework 7 Solution
0
0
0
20
20
1
7
20
70
20
den =
>> poles = roots(den)
poles =
-5.5518
-0.5689 + 3.3590i
-0.5689 - 3.3590i
-0.3104
>> p1 = poly([poles(2) poles(3)])
p1 =
1.0000
1.1379
11.6066
>> p2 = poly([poles2(2) poles2(3)])
p2 =
1.0000
0.5265
13.6647
>> zeta1 = p1(2)/2/sqrt(p1(3))
zeta1 =
0.1670
>> zeta2 = p2(2)/2/sqrt(p2(3))
zeta2 =
0.0712
>> wn1=sqrt(p1(3))
wn1 =
3.4068
April 4, 2004
9
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
10
>> wn2=sqrt(p2(3))
wn2 =
3.6966
>> wr1 = wn1*sqrt(1-2*zeta1^2)
wr1 =
3.3105
>> wr2 = wn2*sqrt(1-2*zeta2^2)
wr2 =
3.6778
>> Mpw1 = 1/(2*zeta1*sqrt(1-zeta1^2))
Mpw1 =
3.0367
>> Mpw2 = 1/(2*zeta2*sqrt(1-zeta2^2))
Mpw2 =
7.0394
>>
>>
>>
>>
bode(clsys,’-’,clsys2,’-.’); grid
title(’Magnitude and Phase Plots for DP8.2, part (c)’)
legend(’K=20’,’K=40’);
print -depsc dp8_2c.eps
(d) Here we plot the step responses, shown in Figure 5, for the transfer functions corresponding to the two gains. The peak overshoot is less than 35% in both cases. We see
that the settling times appear to be identical, so to check more carefully, we isolate
the segment of the step response between 98% and 102% of the final value. This is
shown in Figure 6, which indicates that the 2% settling times are essentially identical,
whereas the K = 20 system has a shorter 1% settling time than the K = 40 system.
If we are only concerned with the settling time we would choose K = 20. Otherwise,
if the rise time were the most important remaining factor in determining our choice
of K we would choose the faster system with K = 40 but if large oscillations were
undesirable we would tend to choose the slower system with K = 20. The Matlab
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
11
script used is included below.
Figure 5: Step responses generated by Matlab for Problem DP8.2, part (d)
Step responses for DP8.2
1.4
K=20
K=40
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
Time (s)
20
K = 20; olsys = tf(K*[1 1],conv([1 5 0],[1 2 10]))
clsys = feedback(olsys,1)
K = 40; olsys2 = tf(K*[1 1],conv([1 5 0],[1 2 10]))
clsys2 = feedback(olsys2,1)
t = [0:.01:30];
y = step(clsys,t);
y40 = step(clsys2,t);
figure(13);
plot(t,y,t,y40,’-.’);grid;
title(’Step responses for DP8.2’),
xlabel(’Time (s)’);
legend(’K=20’,’K=40’);
print -deps dp8_2d1.eps
25
30
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
12
Figure 6: Close-up of step responses generated by Matlab for Problem DP8.2, part (d)
Close−up Plot, DP8.2 Step Response
1.025
K=20
K=40
1.02
1.015
Truncated Response
1.01
1.005
1
0.995
0.99
0.985
0.98
0
5
10
15
Time (s)
20
25
30
% check to see which has shorter 2% settling time
for index = 1:3001,
test1(index) = min(max(y(index),.98),1.02);
test2(index) = min(max(y40(index),.98),1.02);
end;
figure(15);
plot(t,test1,t,test2);
title(’Close-up Plot, DP8.2 Step Response’);
xlabel(’Time (s)’);
ylabel(’Truncated Response’);
legend(’K=20’,’K=40’);
print -deps dp8_2d2.eps
DP8.4 We are given a negative unity feedback system with a controller preceding the plant in
the forward path. The controller transfer function is G c (s) = K(τ s + 1)/(s/10 + 1) and
ECE382/ME482 Spring 2004
Homework 7 Solution
the plant transfer function is
G(s) =
April 4, 2004
13
1
.
(s/2 + 1)2
We are asked to select a gain K and a controller constant τ so that the bandwidth of the
closed–loop system is maximized subject to the constraining M pω < 1.5, and to determine
the bandwidth achieved using the selected values.
Solution: First, let’s invert the expression for M pω in terms of ζ to find the range of ζ’s
that correspond to Mpω ≤ 1.5.
M pω =
1
≤ 1.5
2ζ 1 − ζ 2
p
holds if
1 ≤ 1.5(2)ζ
q
1 − ζ 2.
Squaring both sides and rearranging terms we determine that the condition is equivalent
to ζ 4 − ζ 2 + 1/9 ≤ 0, so solving the quadratic yields the constraint 0.3464 < ζ < 0.9327.
Next, we will apply a two–step root locus procedure (following the method of section 7.5,
pp. 354–359 of the text) to the closed–loop transfer function
T (s) =
Gc (s)G(s)
K(τ s + 1)
=
.
1 + Gc (s)G(s)
(s/10 + 1)(s/2 + 1)2 + K(τ s + 1)
Expanding the denominator yields the third–order polynomial
0.025s3 + 0.35s2 + (1.1 + Kτ )s + (1 + K)
or, multiplying by 40,
s3 + 14s2 + 40(1.1 + Kτ )s + 40(1 + K).
We will work with this last version.
Dividing both sides of
s3 + 14s2 + 40(1.1 + Kτ )s + 40(1 + K) = 0
by s3 + 14s2 + 44s + 40(1 + K) yields
1+
s3
+
14s2
40Kτ s
= 0.
+ 44s + 40(1 + K)
(1)
We then apply the root locus procedure to the denominator of the second term. Dividing
both sides of
s3 + 14s2 + 44s + 40(1 + K) = 0
by s3 + 14s2 + 44s + 40 yields
1+
s3
40K
= 0.
+ 14s2 + 44s + 40
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
14
Figure 7: Root locus plot generated by Matlab for K in Problem DP8.4,
Root Locus for Gain K in DP8.4
20
15
System: untitled1
Gain: 13.9
Pole: −0.0212 + 6.54i
Damping: 0.00324
Overshoot (%): 99
Frequency (rad/sec): 6.54
10
Imaginary Axis
5
0
−5
−10
−15
−20
−30
−25
−20
−15
−10
−5
0
5
10
Real Axis
The root locus for this equation in shown in Figure 7, and the limiting value for the gain
K is determined by finding the value of K for which the loci cross the imaginary axis.
This yields an upper bound on K for the system to remain stable. We choose K = 13.9,
corresponding to poles at −0.0212 ± 6.54j (slightly to the left of the imaginary axis).
Now we substitute the value K = 13.9 into (1) to obtain
1+
s3
40(13.9)τ s
= 0.
+ 14s2 + 44s + 40(13.9)
The root locus for τ is now shown in Figure 8. We will achieve a larger bandwidth when
we have a larger ωr so we choose ζ = 0.35 near the lower bound on ζ. We choose a point on
the root locus where the damping is approximately 0.35 and find that the corresponding
gain is 0.483, so we select τ = 0.48. The resulting closed–loop transfer function is then
T (s) =
(13.9)(48s/100 + 1)
.
(s/10 + 1)(s/2 + 1)2 + 13.9(48s/100 + 1)
To verify that we have met our design objectives we generate the Bode plot shown in
Figure 9 using the following commands.
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
15
Figure 8: Root locus plot generated by Matlab for τ in Problem DP8.4,
Root Locus for Gain τ in DP8.4
50
40
System: untitled1
Gain: 0.483
Pole: −5.95 + 15.9i
Damping: 0.35
Overshoot (%): 30.9
Frequency (rad/sec): 17
30
20
Imaginary Axis
10
0
−10
−20
−30
−40
−50
−14
−12
−10
−8
−6
−4
−2
0
Real Axis
bode(tf(13.9*[48/100 1],[0.025 0.35 1.1+13.9*48/100 13.9] )); grid;
title(’Bode Diagram for Closed-Loop Transfer Function DP8.4 with ...
\tau=0.48 and K=13.9’);
By clicking on the plot we find that the peak value of the magnitude plot is approximately
2.86 dB, which is within 20 log 10 (1.5) = 3.52 dB of the low frequency value 0 dB at ω = 0.1
rad/s. Similarly, we find that ωB is approximately 23.2 rad/s.
MP8.5 We are given a unity negative feedback system in which the controller with transfer
function Gc (s) precedes the plant, whose transfer function is G(s), in the forward path.
We are asked to obtain Bode plots of both the open and closed–loop transfer functions for
this system.
Solution: The magnitude and phase plots are shown in Figure 11.
A transcript of the Matlab commands that generated the plots is shown below.
>> Gc = tf([1],[1 1]);
>> G = tf([10],[1 3 1]);
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
16
Figure 9: Bode plot generated by Matlab for Problem DP8.4,
Bode Diagram for Closed−Loop Transfer Function in DP8.4 withτ = 0.48 and K = 13.9
20
10
Magnitude (dB)
0
System: untitled1
−10 Frequency (rad/sec): 0.103
Magnitude (dB): −0.0015
−20
System: untitled1
Frequency (rad/sec): 23.2
Magnitude (dB): −3.01
−30
−40
−50
−60
−70
−80
0
Phase (deg)
−45
−90
−135
−180
−1
10
0
10
1
10
Frequency (rad/sec)
>> olsys = series(Gc,G)
Transfer function:
10
--------------------s^3 + 4 s^2 + 4 s + 1
>> clsys = feedback(olsys,1)
Transfer function:
10
---------------------s^3 + 4 s^2 + 4 s + 11
>> bode(olsys,’-’,clsys’-.’); grid;
>> title(’Magnitude and Phase Plots for MP8.5’)
>> legend(’Open-Loop’,’Closed-Loop’)
2
10
3
10
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
17
Figure 10: Magnitude and Phase Plots generated by Matlab for Problem MP8.5
Magnitude and Phase Plots for MP8.5
20
Magnitude (dB)
0
−20
−40
−60
−80
Phase (deg)
−100
0
Open−Loop
Closed−Loop
−90
−180
−270
−2
10
−1
0
10
10
1
10
2
10
Frequency (rad/sec)
>> print -depsc mp8_5.eps
MP8.6 We are given the forward path transfer function for a unity negative feedback system.
The forward path transfer function is
G(s) =
1
.
s(s + 2p)
We are asked to use Matlab to plot the bandwidth vs. the parameter p for 0 < p < 1.
Solution: The closed–loop transfer function is
G(s) =
1
,
s2 + 2ps + 1
which corresponds to ωn = 1 and ζ = p.
The following script Matlab script generates the requested plot, which is given in Figure 11.
ECE382/ME482 Spring 2004
Homework 7 Solution
April 4, 2004
18
p =
w =
num
for
[0:0.001:1]’;
logspace(-1,1,2000);
= 1;
index = 1:length(p),
den = [1 2*p(index) 0];
clsys = feedback(tf(num,den),1);
[mag,phase,w] = bode(clsys,w);
a = find(mag<0.707);
wb(index) = w(a(1));
end;
figure(10)
plot(p,wb);grid;
xlabel(’p’)
ylabel(’Bandwidth \omega_b (rad/s)’)
title(’Plot for MP8.6’)
print -deps mp8_6.eps
Figure 11: Plot of system bandwidth vs. value of p for Problem MP8.6
Plot for MP8.6
1.6
1.5
1.4
1.2
b
Bandwidth ω (rad/s)
1.3
1.1
1
0.9
0.8
0.7
0.6
0
0.1
0.2
0.3
0.4
0.5
p
0.6
0.7
0.8
0.9
1