5047 (c) The equivalent plant with unity feedback isG0 = 200 s2(s XXXXXXXXXX200s : Thus the system is type 1 with Kv = 1: If the velocity feedback were zero, the system would be type 2 with Ka = 200...

5047
(c) The equivalent plant with unity feedback isG0 =
200
s2(s XXXXXXXXXX200s
:
Thus the system is type 1 with Kv = 1: If the velocity feedback
were zero, the system would be type 2 with Ka =
200
40
= 5:
(d) The transfer function
Y
W1
=
100s2
s2(s2 + 12s XXXXXXXXXXs+ 1)
: The
system is thus type 2 with Ka = 100:
(e) The transfer function
Y
W2
=
100
s2(s2 + 12s XXXXXXXXXXs+ 1)
: The
system here is type 0 with Kp = 1:
(f) To get more damping in the closed-loop response, the controller needs
to have a lead compensation.
33. Consider the plant transfer function
G(s) =
s+ k
s2[mMs2 + (M +m)bs+ (M +m)k]
to be put in the unity feedback loop of Fig XXXXXXXXXXThis is the transfe
function relating the input force u(t) and the position y(t) of mass M in
the non-collocated sensor and actuator problem. In this problem we will
use root-locus techniques to design a controller D(s) so that the closed-
loop step response has a rise time of less than 0.1 sec and an overshoot of
less than 10%. You may use MATLAB for any of the following questions.
(a) Approximate G(s) by assuming that m �= 0, and let M = 1, k = 1,
= 0:1, and D(s) = K. Can K be chosen to satisfy the performance
speci…cations? Why or why not?
(b) Repeat part (a) assuming D(s) = K(s+ z), and show that K and z
can be chosen to meet the speci…cations.
(c) Repeat part (b) but with a practical controller given by the transfe
function
D(s) = K
p(s+ z)
s+ p
;
and pick p so that the values forK and z computed in part (b) remain
more or less valid.
(d) Now suppose that the small mass m is not negligible, but is given by
m = M=10. Check to see if the controller you designed in part (c)
still meets the given speci…cations. If not, adjust the controller pa-
ameters so that the speci…cations are met.
Solution:
(a) The locus in this case is the imaginary axis and cannot meet the
specs for any K:
5048 CHAPTER 5. THE ROOT-LOCUS DESIGN METHOD
(b) The specs require that � > 0:6; !n > 18: Select z = 15 for a
start. The locus will be a circle with radius 15: Because of the zero,
the overshoot will be increased and Figure 3.32 indicates that we’d
etter make the damping greater than 0.7. As a matter of fact,
experimentation shows that we can lower the overshoot of less than
10% only by setting the zero at a low value and putting the poles on
the real axis. The plot shows the result if D = 25(s+ 4):
(c) In this case, we take D(s) = 20
s+ 4
:01s+ 1
:
(d) With the resonance present, the only chance we have is to introduce
a notch as well as a lead. The compensation resulting in the plots
shown is D(s) = 11
s+ 4
(:01s+ 1)
s2=9:25 + s=9:25 + 1
s2=3600 + s=30 + 1
: The design gain
was obtained by a cycle of repeated loci, root location …nding, and
step responses. Refer to the …le ch5p35.m for the design aid.
Root Locus
Real Axis
Im
ag
A
xi
s
XXXXXXXXXX -10 0
-20
-10
0
10
20
XXXXXXXXXX
0
0.5
1
1.5
Root Locus
Real Axis
Im
ag
A
xi
s
XXXXXXXXXX -10 0
-20
-10
0
10
20
XXXXXXXXXX
0
0.5
1
1.5
Root Locus
Real Axis
Im
ag
A
xi
s
XXXXXXXXXX
-4
-2
0
2
4
XXXXXXXXXX
0
0.5
1
Root loci and step responses for Problem 33
34. Consider the type 1 system drawn in Fig XXXXXXXXXXWe would like to design the
compensation D(s) to meet the following requirements: (1) The steady-
state value of y due to a constant unit distu
ance w should be less than
4
5 , and (2) the damping ratio � = 0:7. Using root-locus techniques:
(a) Show that proportional control alone is not adequate.

6156 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
58. For a unity feedback system with
G(s) =
1
s( s20 + 1)(
s2
XXXXXXXXXX:5
s
100 + 1)
(2)
(a) A lead compensator is introduced with � = 1=5 and a zero at 1=T =
20. How must the gain be changed to obtain crossover at !c =
31:6 rad/sec, and what is the resulting value of Kv?
(b) With the lead compensator in place, what is the required value of K
for a lag compensator that will readjust the gain to a Kv value of
100?
(c) Place the pole of the lag compensator at 3.16 rad/sec, and determine
the zero location that will maintain the crossover frequency at !c =
31:6 rad/sec. Plot the compensated frequency response on the same
graph.
(d) Determine the PM of the compensated design.
Solution :
(a) From a sketch of the asymptotes with the lead compensation (with
K1 = 1) :
D1(s) = K1
s
20 + 1
s
100 + 1
in place, we see that the slope is -1 from zero frequency to ! = 100
ad/sec. Therefore, to obtain crossover at !c = 31:6 rad/sec, the
gain K1 = 31:6 is required. Therefore,
Kv = 31:6
6157
(b) To increase Kv to be 100, we need an additional gain of 3.16 from
the lag compensator at very low frequencies to yield Kv = 100:
(c) For a low frequency gain increase of 3.16, and the pole at 3.16 rad/sec,
the zero needs to be at 10 in order to maintain the crossover at
!c = 31:6 rad/sec. So the lag compensator is
D2(s) = 3:16
s
10
+ 1
s
3:16
+ 1
and
D1(s)D2(s) = 100
s
20 + 1
s
100 + 1
s
10
+ 1
s
3:16
+ 1
The Bode plots of the system before and after adding the lag com-
pensation are
XXXXXXXXXX
10-2
100
102
ω (rad/sec)
M
ag
ni
tu
de
Bode Diagrams
XXXXXXXXXX
-400
-300
-200
-100
0
ω (rad/sec)
P
ha
se
(d
eg
)
XXXXXXXXXX
10-2
100
102
ω (rad/sec)
M
ag
ni
tu
de
Bode Diagrams
Lead and Lag
Lead only
XXXXXXXXXX
-400
-300
-200
-100
0
ω (rad/sec)
P
ha
se
(d
eg
)
6158 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
(d) By using the margin routine from Matlab, we see that
PM = 49� (!c = 34:5 deg/sec)
59. Golden Nugget Airlines had great success with their free bar near the tail
of the airplane. (See Problem 5.39) However, when they purchased a much
larger airplane to handle the passenger demand, they discovered that there
was some ‡exibility in the fuselage that caused a lot of unpleasant yawing
motion at the rear of the airplane when in tu
ulence and was causing the
evelers to spill their drinks. The approximate transfer function for the
dutch roll mode (See Section XXXXXXXXXXis
(s)
�r(s)
=
8:75(4s2 + 0:4s+ 1)
(s=0: XXXXXXXXXXs2 + 0:24s+ 1)
where r is the airplane’s yaw rate and �r is the rudder angle. In performing
a Finite Element Analysis (FEA) of the fuselage structure and adding
those dynamics to the dutch roll motion, they found that the transfe
function needed additional terms that re‡ected the fuselage lateral bending
that occu
ed due to excitation from the rudder and tu
ulence. The
evised transfer function is
(s)
�r(s)
=
8:75(4s2 + 0:4s+ 1)
(s=0: XXXXXXXXXXs2 + 0:24s+ 1)
� 1
( s
2
!2
+ 2� s!b + 1)
where !b is the frequency of the bending mode (= 10 rad/sec) and � is the
ending mode damping ratio (= 0:02). Most swept wing airplanes have
a “yaw damper”which essentially feeds back yaw rate measured by a rate
gyro to the rudder with a simple proportional control law. For the new
Golden Nugget airplane, the proportional feedback gain, K = 1; where
�r(s) = �Kr(s): (3)
(a) Make a Bode plot of the open-loop system, determine the PM and
GM for the nominal design, and plot the step response and Bode
magnitude of the closed-loop system. What is the frequency of the
lightly damped mode that is causing the di¢ culty?
(b) Investigate remedies to quiet down the oscillations, but maintain the
same low frequency gain in order not to a¤ect the quality of the
dutch roll damping provided by the yaw rate feedback. Speci…cally,
investigate one at a time:
i. increasing the damping of the bending mode from � = 0:02 to
� = 0:04: (Would require adding energy abso
ing material in
the fuselage structure)
ii. increasing the frequency of the bending mode from !b = 10
ad/sec to !b = 20 rad/sec. (Would require stronger and heavie
structural elements)

6101
(d) The characteristic equation for PM of 45� :
1 +
1:1
s(s+ 1)
h�
s
5
�2
+ 0:4
�
s
5
�
+ 1
i = 0
=) s4 + 3s3 + 27s2 + 25s+ 27:88 = 0
=) s = �1:03� j4:78; �0:47� j0:97
32. For the system depicted in Fig. 6.94(a), the transfer-function blocks are
de…ned by
G(s) =
1
(s+ 2)2(s+ 4)
and H(s) =
1
s+ 1
:
(a) Using rlocus and rloc…nd, determine the value of K at the stability
oundary.
(b) Using rlocus and rloc…nd, determine the value of K that will produce
oots with damping co
esponding to � = 0:707.
(c) What is the gain margin of the system if the gain is set to the value
determined in part (b)? Answer this question without using any
frequency response methods.
(d) Create the Bode plots for the system, and determine the gain margin
that results for PM = 65�. What damping ratio would you expect
for this PM?
6102 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
Figure 6.94: Block diagram for Problem 32: (a) unity feedback; (b) H(s) in
feedback
(e) Sketch a root locus for the system shown in Fig. 6.94(b).. How does
it di¤er from the one in part (a)?
(f) For the systems in Figs. 6.94(a) and (b), how does the transfer func-
tion Y2(s)=R(s) di¤er from Y1(s)=R(s)? Would you expect the step
esponse to r(t) be di¤erent for the two cases?
Solution :
(a) The root locus crosses j! axis at s0 = j2.
K =
1
jH(s0)G(s0)j
js0=j2
= jj2 + 1j jj2 + 4j jj2 + 2j2
=) K = 80
6103
(b)
� = 0:707 =) 0:707 = sin � =) � = 45�
From the root locus given,
s1 = �0:91 + j0:91
K =
1
jH(s1)G(s1)j
js1=�0:91+j0:91
= j0:01 + j0:91j j3:09 + j0:91j j1:09 + j0:91j2
=) K = 5:9
(c)
GM =
Ka
K
=
80
5:9
=