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), an-

other complex number results. e.g. substitut-

ing 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

F(s) =(s − z1)(s − z2) · · ·

(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 counterclock-

wise 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

DG

, H(s) =NH

DH

,

We found

T(s) =G(s)

1 + G(s)H(s)=

NGDH

DGDH + NGNH

Note that

1 + G(s)H(s) =DGDH + NGNH

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 closed-

looped 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 half-

plane 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 in-

side contour A (the right half-plane), yields a

rotation around (−1, j0) (clockwise for zero

and counterclockwise for pole) for the resul-

tant Nyquist diagram. The total number of

counterclockwise revolution, N , around (−1, j0)

is N = P −Z, where P is the number of open-

loop 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 Fig-

ure (a).

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On the other hand, another system with P = 0

(open loop stable) has generated two clock-

wise 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 deter-

mine the stability of the following unit-feedback

systems with

(i) G(s) =s + 3

(s + 2)(s2 + 2s + 25).

(ii) G(s) =s + 20

(s + 2)(s + 7)(s + 50).

(iii) G(s) =500(s − 2)

(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;

−1.5 −1 −0.5 0 0.5 1 1.5−1.5

−1

−0.5

0

0.5

1

1.50 dB

−20 dB

−10 dB

−6 dB

−4 dB

−2 dB

20 dB

10 dB

6 dB

4 dB

2 dB

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

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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