Stability via the Nyquist diagram
Range of gain for stability
Problem: For the unity feedback system below, where
K
G(s) =
,
s(s + 3)(s + 5)
find the range of gain, K, for stability, instability and the value of K for marginal stability.
For marginal stability, also find the frequency
of oscillation. Use the Nyquist criterion.
Figure above; Closed-loop unity feedback system.
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Solution:
G(jω) =
K
|s→jω
s(s + 3)(s + 5)
−8Kω − j · K(15 − ω 2)
=
64ω 3 + ω(15 − ω 2)2
When K = 1,
−8ω − j · (15 − ω 2)
G(jω) =
64ω 3 + ω(15 − ω 2)2
Important points:
Starting point: ω = 0, G(jω) = −0.0356 − j∞
Ending point: ω = ∞, G(jω) = 06 − 270◦
Real axis crossing: found by setting the imaginary part of G(jω) as zero,
ω=
√
15,
K
{−
, j0}
120
2
When K = 1, P = 0, from the Nyquist plot,
N is zero, so the system is stable. The real
K does not encircle [−1, j0)]
axis crossing − 120
until K = 120. At that point, the system is
marginally stable, and the frequency of oscilla√
tion is ω = 15 rad/s.
Nyquist Diagram
Nyquist Diagram
0.05
2
0.04
1.5
0.03
System: G
Real: −0.00824
Imag: 1e−005
Frequency (rad/sec): −3.91
0.01
1
Imaginary Axis
Imaginary Axis
0.02
0
−0.01
0.5
0
−0.5
−0.02
−1
−0.03
−1.5
−0.04
−0.05
−0.1
−0.08 −0.06 −0.04 −0.02
0
0.02
Real Axis
(a)
0.04
0.06
0.08
0.1
−2
−3
−2.5
−2
−1.5
Real Axis
−1
−0.5
0
(b)
K
;
Figure above; Nyquist plots of G(s) = s(s+3)(s+5)
(a) K = 1; (b)K = 120.
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Gain/phase margin via the Nyquist diagram
We use the Nyquist diagram to define two
quantitative measures of how stable a system
is. These are called gain margin and phase
margin. Systems with greater gain margin and
phase margins can withstand greater changes
in system parameters before becoming unstable.
Gain margin, GM , The gain margin is the change
in open-loop gain, expressed in decibels (dB),
required at 180◦ of phase shift to make the
closed-loop system unstable.
Phase margin, ΦM , The phase margin is the
change in open-loop phase shift, required at
unity gain to make the closed-loop system unstable.
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Figure above; Nyquist diagram showing gain
and phase margins
Problem: Find the gain and phase margin for
the unity feedback system with
G(s) =
6
.
2
(s + 2s + 2)(s + 2)
5
Solution: From G(s), we see P = 0. The
Nyquist diagram shows that N is zero, so the
closed-loop system is stable.
G(jω) =
6
|s→jω
2
(s + 2s + 2)(s + 2)
6[4(1 − ω 2) − jω(6 − ω 2)]
=
16(1 − ω 2)2 + ω 2(6 − ω 2)2
The Nyquist diagram
√ crosses the real axis at a
frequency of ω = 6. The real part is found
to be −0.3.
Nyquist Diagram
1.5
Imaginary Axis
1
0.5
0
−0.5
−1
−1.5
−1
−0.5
0
0.5
1
1.5
Real Axis
Figure above; Nyquist diagram for
G(s) =
6
.
2
(s + 2s + 2)(s + 2)
6
The gain margin is
GM = 20 log(1/0.3) = 10.45dB.
To find the phase margin find the frequency
for which G(jω) has a unit gain. Using a computing tool, we can find G(jω) has a unit gain
at a frequency of 1.251, at this frequency the
phase angle is −112.3◦. The difference of this
angle with −180◦ is 67.7◦, which is the phase
margin ΦM .
Gain/phase margin via the Bode plots
Figure above; Gain and phase margins on the
Bode plots.
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Problem: Let a unit feedback system have
K
G(s) =
.
(s + 2)(s + 4)(s + 5)
Use Bode plots to determine the range of gain
within which the system is stable. If K = 200
find the gain margin and the phase margin.
The low frequency gain is found by setting s to
zero. Thus the Bode magnitude plots starts
at K/40. For convenience set K = 40, so
that the log-magnitude plots starts at 0dB. At
each break frequency, 2, 4 and 5, a slope of
-20dB/decade is added.
The phase diagram starts at 0◦ until 0.2rad/s
(a decade below the break frequency of 2),
the curve decreases at a slope of 45◦/decade
at each subsequent frequency at 0.4rad/s and
0.5rad/s (a decade below the break frequency
of 4 and 5 respectively). Finally at 20rad/s,
40rad/s and 50rad/s (a decade above the break
frequencies of 2,4,5), a slope of +20dB/decade
is added, until the curve levels out.
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Figure above; Bode log-magnitude and phase
40
diagram for G(s) = (s+2)(s+4)(s+5)
.
The Nyquist criterion tells us that we want zero
encirclement of {−1, j0}. Thus the Bode logmagnitude plot must be less than unity when
the Bode phase plot is −180◦. Accordingly we
see that at frequency 7 rad/s, when the phase
plot is −180◦. The magnitude is -20dB.
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Thus an increase of 20dB is possible before the
system becomes unstable, which is a gain of
10, so the gain for instability is K > 10 × 40 =
400.
If K = 200 (five times greater than K = 40),
the magnitude plot would be 20 log 5 = 13.98dB
higher, as the Bodes plots was scaled to a gain
of 40.
At −180◦, the gain is −20 + 13.98 = −6.02dB,
so GM = 6.02dB.
To find phase margin, we look on the magnitude plot for the frequency where the gain
is 0dB. As the plot should be 13.98dB higher,
so we look at −13.98dB crossing to find the
frequency is 5.5rad/s. At this frequency, the
phase angle is −165◦. Thus
ΦM = −165◦ − (−180)◦ = 15◦.
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