MAE170 Homework 8 Solution
Fall 2009
P9.1 The Nyquist plots shown in the following ﬁgures are obtained by the use of
Matlab’s
nyquist
command. For sketching them by hand, ﬁrst sketch the
polar plot
L
(
jω
)
,
ω
= 0
+
→
+
∞
guided by the Bode plots of
L
(
jω
). Then,
the plot of
L
(
jω
)
,
ω
= 0

→ ∞
is the symmetric graph with respect to the
real axis. Finally, if there
N
poles at
s
= 0, (as in case (d)), complete the plot
(to a closed curve) with an arc of
N
×
180
0
, running clockwise from
L
(
j
0

) to
L
(
j
0
+
).
(a) The ﬁrst openloop transfer function is
L
(
s
) =
1
(1 + 0
.
5
s
)(1 + 2
s
)
In this case, the number of openloop unstable poles is
P
= 0, the number
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
Nyquist Diagram
Real Axis
Imaginary Axis
Figure 1: Nyquist Plot for problem P9.1a
of clockwise encirclements of the (1,0) point by the Nyquist plot is
N
= 0
(see Figure 1), therefore the number of unstable closedloop poles is
Z
=
N
+
P
= 0. Hence, the closedloop system is stable.
(b) The second openloop transfer function is
L
(
s
) =
10(
s
2
+ 1
.
4 + 1)
(
s

1)
2
1
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6
4
2
0
2
4
6
8
10
10
8
6
4
2
0
2
4
6
8
10
Nyquist Diagram
Real Axis
Imaginary Axis
Figure 2: Nyquist Plot for problem P9.1b
In this case,
P
= 2 because there are 2 openloop unstable poles at
s
=

1,
N
=

2 because there are 2 counterclockwise encirclements of the (1,0)
point by the Nyquist plot (note that in Figure 2, the Nyquist plot consists
of going twice around the indicated plot—check the Bode plot!), therefore
the number of unstable closedloop poles is
Z
=
N
+
P
= 0. So, the
closedloop system is stable. You can verify that this system has a Gain
Margin
GM
=

16
.
9
dB
, which indicates that multiplying the openloop
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 Fall '08
 staff
 Nyquist plot, Nichols plot

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