MAE143B Final Exam Solutions — Tuesday December 8, 2009
Question 1
Part (i):
Lead compensator portion,
s/
2
..
5+1
s/
25+1
. For a lead compensator, the pole is larger than the zero
and both are real and negative. It is designed to improve stability and bandwidth by adding phase
lead.
Lag compensator portion,
s
+0
.
1
s
+0
.
02
. For a lag compensator, the zero is larger than the pole and both
are real and negative. It is designed to improve steadystate error without affecting too much the
bandwidth and stability, in spite of it adding phase lag.
The design was performed in three parts; analysis of the openloop plant with proportional control,
lead compensator design to achieve closedloop bandwidth, lag compensator design to achieve static
accuracy speciﬁcation.
Figure 2 shows that
G
(
s
)
with unity feedback is marginally stable. It could be stabilized by reducing
the proportional gain. But for 20
◦
phase margin, this would yield a closedloop bandwidth much less
than one radian per second.
In Figure 3 we see that the lead compensator has been included to achieve the 20
◦
phase margin at the
target frequency of 3.9 radians/sec. This is achieved with the compensator
s/
2
..
5+1
s/
25+1
and unity feedback.
The static accuracy of this closedloop system to a unit ramp is given by.
lim
t
→∞
e
t
= lim
s
→
0
sE
(
s
)
,
= lim
s
→
0
s
×
1
1 +
s/
2
..
5+1
s/
25+1

s
+2
s
(
s
+1)
×
1
s
2
,
= 0
.
5
.
In Figure 4 we see that a lag compensator,
s
+0
.
1
s
+0
.
02
, has been included with the lead compensator
to increase the velocity constant by a factor of ﬁve low frequencies – and thereby deal with the
static accuracy speciﬁcation – while leaving the bandwidth and phase margin at 4 radians per second
relatively unaltered.
Part (ii):
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 Fall '09
 CALLAFON
 lim, lead compensator, lag compensator

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