Introduction to the Nyquist criterion
The Nyquist criterion relates the stability of
a closed system to the open-loop frequency
response and open loop pole location.
Mapping. If we take a complex number on the
s-plane and substitute into a function F (s), another complex number results. e.g. substituting s = 4 + j3 into F (s) = s2 + 2s + 1 yields
16 + j30.
Contour. Consider a collection of points, called
a contour A. Contour A can be mapped into
Contour B, as shown in the next Figure.
Figure above; Mapping contour A through F (s)
to contour B.
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Assuming
(s − z1)(s − z2) · · ·
F (s) =
(s − p1)(s − p2) · · ·
If we assume a clockwise direction for mapping
the points on contour A, the contour B maps
in a clockwise direction if F (s) has just one
zero. If the zero is enclosed by contour A,
then contour B enclose origin.
Alternatively, the mapping is in a counterclockwise direction if F (s) has just one pole, and if
the pole is enclosed by contour A, then contour
B enclose origin.
If there is the one pole and one zero is enclosed
by contour A, then contour B does not enclose
origin.
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Figure above; Examples of contour mapping.
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Consider the system in the Figure below.
Figure above; closed loop control system
Letting
G(s) =
NG
N
, H(s) = H ,
DG
DH
We found
T (s) =
G(s)
NGDH
=
1 + G(s)H(s)
DGDH + NGNH
Note that
DGDH + NGNH
1 + G(s)H(s) =
DGDH
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The poles of 1 + G(s)H(s) are the same as the
poles of G(s)H(s), the open-looped system,
that are known. The zeros of 1 + G(s)H(s)
are the same as the poles of T (s), the closedlooped system, that are unknown.
Because stable systems have T (s) with poles
only in the left half-plane, we apply the concept
of contour to use the entire right half-plane as
contour A, as shown in the Figure below.
Figure above; Contour enclosing right halfplane to determine stability.
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We try to construct contour B via
F (s) = G(s)H(s)
which is the same as that of 1 + G(s)H(s),
except that it is shifted to the right by (1, j0).
The mapping is called the Nyquist diagram of
G(s)H(s).
Assuming that A starts from origin, A is a path
traveling up the jω axis, from j0 to j∞, then a
semicircular arc, with radius → ∞, followed by
a path traveling up the jω axis, from −j∞ to
origin. So substituting s = jω, with ω changing
from 0 to ∞, we obtain part of contour B,
which is exactly the polar plot of G(s)H(s).
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Each zero or pole of 1 + G(s)H(s) that is inside contour A (the right half-plane), yields a
rotation around (−1, j0) (clockwise for zero
and counterclockwise for pole) for the resultant Nyquist diagram. The total number of
counterclockwise revolution, N , around (−1, j0)
is N = P − Z, where P is the number of openloop poles,and Z is the number of closed loop
poles.
Thus we determine that that the number of
closed loop poles, Z, in the right half-plane
equals the number of open-loop poles, P , that
are in the right half-plane minus the number
of counterclockwise revolution, N , around −1
of the mapping, i.e. Z = P − N .
Use Nyquist criterion to determine stability
If P = 0 (open loop stable system), for a closed
systems to be stable (i.e. Z = 0), we should
have N = 0. That is, the contour should not
enclose (−1, j0). This is as shown in next Figure (a).
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On the other hand, another system with P = 0
(open loop stable) has generated two clockwise encirclement of (−1, j0), (N = −2), as
shown in Figure (b) below. Thus Z = P − N =
2.
The system is unstable with two closed-loop
poles in the right hand plane.
Figure above; Mapping examples: (a) contour
does not enclose closed loop poles; (b) contour
does enclose closed loop poles;
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Example: Apply the Nyquist criterion to determine the stability of the following unit-feedback
systems with
(i)
s+3
G(s) =
.
(s + 2)(s2 + 2s + 25)
s + 20
.
(s + 2)(s + 7)(s + 50)
(ii)
G(s) =
(iii)
500(s − 2)
G(s) =
.
(s + 2)(s + 7)(s + 50)
Solution: For (i) and (ii), check polar plots in
the previous lecture.
For both systems we have P = 0 (open loop
stable system).
The two nyquist plots does not enclose (−1, j0),
(N=0)
Thus Z = P −N = 0. Both systems (i) and (ii)
are stable since there are no close-loop poles
in the right half plane.
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For (iii), we run
≫ numg=500* [1 -2];;
≫ deng=conv([1 2],[1 7]);
≫ deng=conv(deng,[1 50]);
≫ G=tf(numg,deng);
≫ nyquist(G);
≫ grid on;
Nyquist Diagram
1.5
2 dB
0 dB
−2 dB
−4 dB
4 dB
1
6 dB
Imaginary Axis
0.5
−6 dB
10 dB
−10 dB
20 dB
−20 dB
0
−0.5
−1
−1.5
−1.5
−1
−0.5
0
Real Axis
0.5
1
1.5
Figure above; The polar plots for
G(s) =
500(s − 2)
.
(s + 2)(s + 7)(s + 50)
We have P = 0 (open loop stable system),
but N = −1, so System (iii) is unstable with
one closed loop pole in the right half plane.
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