MAE 143B
Linear Control
Prof. M. Krstic
FINAL
June 12, 2012
• One sheet of hand-written notes (two pages).
• Present your reasoning and calculations clearly. Inconsistent etchings will not be graded.
• Write answers only in the blue book.
• Total points: 25. Time: 1 hour 15 minutes.
Problem 1: Nyquist Plots (12 points) For each of the given transfer functions, sketch the Nyquist
plot and find the range of K > 0 that makes the following closed-loop system stable.
r
K
+-
y
G(s)
.01(s − 10)(s + 100)
(s + 1)2
s
1
s
Solution. G(s) = −10 (s+1)
+ 1 100
+ 1 . We first draw the Bode plot.
− 10
2
(a) (3 points) G(s) =
Bode Diagram
20
Magnitude (dB)
10
0
-10
-20
-30
-40
-1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
Phase (deg)
180
90
0
-90
-1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
The phase plot indicates that the Nyquist plot (1) starts at the negative real axis, (2) moves
in a clockwise way, (3) crosses the positive real axis, (4) turns around and moves back in a
counter-clockwise way, and (5) converges to the positive real axis.
The magnitude plot indicates that the Nyquist plot (1) starts at −10 due to 20dB and 180◦ ,
(2) gets closer to the origin as ω increases, and (3) ends at 0.01 due to −40dB and 0◦ .
1
Nyquist Diagram
Imaginary Axis
6
4
2
0
-2
-10
-8
-6
-4
Real Axis
-2
0
Nyquist Diagram magnified around the origin
Imaginary Axis
0.05
0
-0.05
-0.05
-0.04
-0.03
-0.02
-0.01
0
Real Axis
0.01
0.02
0.03
0.04
0.05
Since G(s) is stable,
the closed-loop system is stable if, and only if, there is no encirclement
1
around − K , j0 . Therefore, 0 < K < 0.1
s(s + 100)
(b) (3 points) G(s) = s2 + .1s + 1
Solution. G(s) = 100s
1
s2 +.1s+1
s
100
+ 1 . We first draw the Bode plot.
Bode Diagram
60
Magnitude (dB)
50
40
30
20
10
0
-1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
Phase (deg)
90
0
-90
-1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
The phase plot indicates that the Nyquist plot (1) starts at the positive imaginary axis, (2)
moves in a clockwise way, (3) crosses the positive real axis, (4) turns around and moves back
in a counter-clockwise way, and (5) converges to the positive real axis.
2
The magnitude plot indicates that the Nyquist plot (1) starts at the origin, (2) moves, first,
away from and, then, towards the origin, and (3) ends at 1 due to 0dB and 0◦ .
Nyquist Diagram
600
Imaginary Axis
400
200
0
-200
-400
-600
0
100
200
300
400
500
Real Axis
600
700
800
900
1000
Nyquist Diagram magnified around the origin
10
Imaginary Axis
5
0
-5
-10
-4
-3
-2
-1
0
Real Axis
1
2
3
4
Since G(s) is stable,
the closed-loop system is stable if, and only if, there is no encirclement
1
around − K , j0 . Therefore, K > 0
(c) (3 points) G(s) =
10(s − 1)2
s2 (s2 + 10s + 100)
0.1
(s
s2
Solution. G(s) =
− 1)2
2
( 10s )
1
.
s
+ 10
+1)
We first draw the Bode plot.
Bode Diagram
Magnitude (dB)
20
0
-20
-40
-60
-1
10
0
1
10
10
2
10
Frequency (rad/sec)
Phase (deg)
180
90
0
-90
-180
-1
10
0
1
10
10
Frequency (rad/sec)
3
2
10
The phase plot indicates that the Nyquist plot (1) starts with phase −180◦ , (2) moves in a
clockwise way, and (3) converges to the negative real axis.
The magnitude plot indicates that the Nyquist plot (1) comes from −∞, (2) moves towards
to the origin, and (3) converges to the origin.
Nyquist Diagram
Imaginary Axis
3
2
1
0
-1
-10
-8
-6
-4
Real Axis
-2
0
Nyquist Diagram magnified around the origin
Imaginary Axis
0.2
0.1
0
-0.1
-0.2
-0.4
-0.3
-0.2
-0.1
0
Real Axis
0.1
0.2
0.3
0.4
(1) The number of poles of G(s) on the right half plane is 0.
(2) Around − K1 , j0 , we have #CCW = 0 and #CW = 1 for any K > 0.
Therefore, the Nyquist stability criterion is not satisfied for any K > 0, which means that
the closed-loop system is unstable for any K > 0.
4
Problem 2: Root Locus, Bode Plot, and Nyquist Plot (13 points) We can learn a lot about the
behavior of a system from its Bode plot. Suppose we perform an experiment on a dynamic
system, and we obtain the following Bode plot:
Bode Diagram
10
0
Magnitude (dB)
−10
−20
−30
−40
−50
−60
−70
−80
0
Phase (deg)
−90
−180
−270
−360
−450
−2
10
−1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
4
10
(a) (4 points) Sketch the Nyquist plot.
The phase plot indicates that the Nyquist plot (1) starts at the positive real axis, (2) moves
in a clockwise way until it reaches and crosses the positive real axis, and (3) converges to
the negative imaginary axis.
The magnitude plot indicates that the Nyquist plot (1) starts at 1 due to 0dB and 0◦ , (2)
moves towards the origin except for the peak, and (3) converges to the origin.
Nyquist Diagram
0.5
Imaginary Axis
0
-0.5
-1
-1.5
-2
-2.5
-1.5
-1
-0.5
0
Real Axis
0.5
1
1.5
0.04
0.06
Nyquist Diagram magnified around the origin
0.03
Imaginary Axis
0.02
0.01
0
-0.01
-0.02
-0.03
-0.06
-0.04
-0.02
0
Real Axis
5
0.02
(b) (2 points) Determine the gain and phase margin.
When the phase plot crosses −180◦ , the magnitude plot crosses about −7dB, which means
that the gain margin is about 7dB. When the magnitude plot crosses 0dB, the phase plot
crosses about −156◦ , which means that the phase margin is about −156◦ − (−180◦ ) = 24◦ .
(c) (4 points) Determine the transfer function of the open-loop system. (Use powers of 10 for
your breakpoints) (Hint: Use the magnitude and frequency corresponding to the phase of
−180◦ in order to find ζ for the resonant pair of poles.)
From the Bode plot, the transfer function is determined as
s
2
(s − 10)2
1
1
=
− +1
G(s) = 2
s
+1
s + 2ζs + 1
10
(s2 + 2ζs + 1)(s + 100)
102
And from the phase plot, the frequency corresponding to the phase −180◦ is about 1.8 rad/sec.
At 1.8 rad/sec, we have
1.8
−1.8
3.6ζ
(10 − j1.8)2
◦
= 2 arctan
− arctan
+ 180 −arctan
,
∠ G(j1.8) = ∠
(−2.24 + j3.6ζ)(100 + j1.8)
10
−2.24
100
which should be −180◦ . Thus, we have
−1.8
1.8
2.24
tan 2 arctan
− arctan
ζ=−
= 0.2443
3.6
10
100
Therefore, the transfer function is
(s − 10)2
G(s) = 2
(s + 0.4886s + 1)(s + 100)
(d) (3 points) Sketch the Root Locus to confirm that high gain leads to instability.
(1)
√ The open-loop zeros are at s = 10, 10 (2) The open-loop poles are at s = −100, −0.2443±
j 1 − 0.24432 (3) The relative degree is 1. Thus, there is one asymptote.
Root Locus
Imaginary Axis
5
0
-5
-100
-80
-60
-40
Real Axis
-20
0
20
Root Locus magnified around the origin
Imaginary Axis
5
0
-5
-10
-8
-6
-4
-2
0
Real Axis
6
2
4
6
8
10