MAE143B Homework Set 3  Solutions
Question 1
FPEN Question 5.30: Assume that the closedloop system has plant transfer function
P
(
s
) =
1
s
(
s
+ 2)
.
Design a lag compensator so that the dominant poles of the closedloop system are located at
s
=

1
±
j
and the steadystate error to a unit ramp input is less than 0.2.
Without compensation, the root locus of the system is as shown in the following figure. By choosing
−
2.5
−
2
−
1.5
−
1
−
0.5
0
0.5
−
1.5
−
1
−
0.5
0
0.5
1
1.5
Root Locus
Real Axis
Imaginary Axis
a constant feedback gain of
K
= 2
, we can place the closed loop poles at
s
=

1
±
j
. However, this
leaves the steadystate error to a ramp as
lim
t
→∞
e
t
= lim
s
→
0
sE
(
s
) = lim
s
→
0
s
s
2
+ 2
s
s
2
+ 2
s
+ 1
1
s
2
= 1
.
We need to increase the lowfrequency gain via a lag compensator without affecting the closedloop
poles very much. Since we need an increase in gain by a factor of five to get our error down to 0.2, let’s
try
C
(
s
) =
s
+0
.
1
s
+0
.
02
. This has the lowfrequency gain of five and has a pole and zero which effectively
cancel out when
s
is as big as

1
±
j
.
With this new choice of compensator, we plot the root locus of
C
(
s
)
P
(
s
) =
s
+ 0
.
1
s
+ 0
.
02
1
s
(
s
+ 2)
=
s
+ 0
.
1
s
3
+ 2
.
02
s
2
+ 0
.
04
s
.
This yields Note that the gain for poles close to
s
=

1 +
j
is given by
K
=

s
3
+ 2
.
02
s
2
+ 0
.
04
s
s
+ 0
.
1
s
=

1+
j
= 1
.
9900

0
.
0101
j
≈
2
.
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−
2.5
−
2
−
1.5
−
1
−
0.5
0
0.5
−
0.8
−
0.6
−
0.4
−
0.2
0
0.2
0.4
0.6
0.8
Root Locus
Real Axis
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 Fall '09
 CALLAFON
 Root Locus

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