Nyquesy Stability

7/25/2019 Nyquesy Stability

1/20

NYQUIST STABILITY CRITERION


2/20

Nyquist Stability Criterion(NSC)

Relates the location of the roots of characteristics equation to

open loop frequency response of a system Unlike root-locus technique, computation of closed loop poles is

not necessary in NSC.

( ) ( )G j H j

a y s u y can e carr e ou grap ca y rom open oop

frequency response.

Experimentally determined open loop frequency response can

be used directly for the study of stability when the feedback

path is closed(i.e. closed system stability)


3/20

Mathematical Preliminaries Consider a function

S=complex variable

= also complex variable

-

( )q s

1 2

1 2

( )( ) ( )( )

( )( ) ( )m

n

s s sq s

s s s

=

L

L

( )q s

(1)

.

q(s) is analytic (function and all its derivatives exist) we canfind a corresponding point q(s) in the q(s)-plane.

Since any number of points in s-plane can be mapped into

q(s)-plane, for a contour in the s-plane which does not go

through any singular points (points in the s-plane where the

function or its derivatives does not exist), there corresponds a

contour in the q(s)-plane.


4/20

Mathematical Preliminaries

The region to the right of a closed-contour is considered enclosed by the

contour when the contour is traversed in clockwise direction. Thus, the shaded

area in Fig.1 is enclosed by the closed contour.

We are not interested in exact shape of the q(s)-plane contour. But, it isimportant to know the encirclement of the origin by q(s)-plane contour for NSC

To investigate the encirclement:

Consider an s-plane contour which encloses only one of the zeros of q(s), say , while

all the poles and remaining zeros are distributed in the s-plane outside the contour.1

s =

.


5/20

Mathematical Preliminaries For any non-singular point s on the s-plane contour, there corresponds a

point q(s) on q(s)-plane contour.

From (1), the point q(s) is given by 1 2

1 2

( ) ( ) ( )( )( ) ( ) ( )

m

n

s s sq ss s s

=

L

L

1 2 1 2( ) ( ) ( ) ( ) ( )q s s s s s = + + L L

Fig2

From Fig.2(a) it is found that as the point s follows the prescribed path ( i.e.

clockwise direction on the s-plane contour, eventually returning to the

starting point, the phasor generates a net angle of ,while all

other phasors generate net angles.1

( )s 2


6/20


This implies tip of q(s)-phasor must describe a closed contour

about the origin of the q(s)-plane in clockwise direction. As said before, exact shape of the closed contour in q(s)-plane

is not important for us rather it is sufficient to observe that this

.

If the contour in the s-plane is so chosen that it does not enclose

any zero or pole, the corresponding contour in q(s)-plane then

it will not encircle the origin.


7/20

Mathematical Preliminaries If s-plane contour encloses 2 zeros (say , ) , q(s)-plane contour

encircles origin twice in clockwise direction.

Fig.3

1s =

2s =

We can say, that for each zero of q(s), enclosed by s-plane

contour, the corresponding q(s)-plane contour encircles origin once

in CW direction.


8/20


-=

Phasor generates an angle as s traverses the prescribed path.

Since is in denominator of Eq.(1), the q(s)-plane contour experiences

an angle change of , which means one contour CW encirclement of the

origin.

1

1( )s 2

1( )s

2


9/20

Mathematical Preliminaries This argument holds for other poles of q(s). Thus, if there are P poles and Z zeros of

q(s) enclosed by s-plane contour, then the corresponding q(s)-plane contour must

encircle the origin Z times in CW direction and P times in CCW direction resulting in

a net-encirclement of the origin (p-z) times in CCW direction.

This relation between enclosure of poles and zeros of q(s) by the s-plane contour

- .

Ex: 1 zero and 3 poles enclosed by the s-plane contour, the net encirclement of the

origin by the q(s)-plane contour is rad i.e. two contour CCW

revolutions.

2 (3 1) 4 =

Fig.4


10/20

Nyquist Stability Criterion Consider Single-loop feedback system

(3)

Ch. Eq:

Open loop transfer function (pole zero form):

From (3) and (4)

( ) 1 ( ) ( ) 0q s G s H s= + =

1 2

1 2

( )( ) ( )( ) ( )

( )( ) ( )

m

n

K s z s z s zG s H s

s p s p s p

+ + +=

+ + +

L

L

(4)

1 2

1 2

( )( ) ( )( ) 1

( )( ) ( )

m

n

s z s z s zq s K

s p s p s p

+ + += +

+ + +

L

L


11/20

Nyquist Stability Criterion1 2

1 2

1 2 1 2

1 2

1 1 1

1 2

1 2

( )( ) ( )( ) 1

( )( ) ( )

( )( ) ( ) ( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

m

n

n m

n

n

n

s z s z s zq s K

s p s p s p

s p s p s p K s z s z s z

s p s p s p

s z s z s z

s p s p s p

+ + += +

+ + +

+ + + + + + +=

+ + +

+ + +=

+ + +

L

L

L L

L

L

L

(5)

(6)

From (6) Zeros of q(s) at are the roots of Ch.Eq. Closed looppoles of q(s) at are the same as the open loop poles of the

system.

For a system to be stable, the roots of Ch.Eq and hence zeros of q(s) must

lie in LHSP

Note: Even if some of OL poles lie in RHSP, all the zeros of q(s) i.e. CL poles

may lie in LHSP meaning there by that an OL unstable system may lead to

a CL stable operation.

1 1 11 2, , nz z z

L

1 2, ,

np p p L


12/20

Nyquist Stability Criterion In order to investigate the presence of any zero of q(s)=1+G(s)H(s) in the

RHSP, let us choose a contour which completely encloses the RHSP. Such a

contour C is called Nyquist Contour (Fig.6)

Fig.6

It is directed CW and comprises of an infinie segment C1 along axis

and an arc C2 of infinite radius. Along C1: with s varying from to

Along C2: with varying from to 0 to

j

s j= j j

Rej

s

= 2

2


13/20

Nyquist Stability Criterion The Nyquist contour so defined encloses all the right half s-plane zeros and

poles of

Let there be Z zeros and P poles of q(s) in the RHSP.

As S moves along the Nyquist contour in the s-plane, a closed contour is

traversed in the q(s)-plane which encloses the origin N=P-Z times in CCW

( ) 1 ( ) ( )q s G s H s= +

irection.

In order the system to be stable, there should be no zeros of

q(s)=1+G(s)H(s) in the RHSP i.e. Z=0.

This condition is met if N=P, i.e. for a CL system to be stable, the no. of

CCW encirclements of the origin of the q(s)-plane by the contour should be

equal the no. of right half s-plane poles of q(s) which are the poles of the

OL TF of G(s)H(s).


14/20

Nyquist Stability Criterion In special case of p=0 (OL stable system), the CL system is stable if N=P=0

which means that net encirclement of the origin of the q(s)-plane by contour

should be zero.

It is easily observer that

From 9: Contour of G(s)H(s) corresponding to Nyquist contour is s-plane

[ ]( ) ( ) 1 ( ) ( ) 1G s H s G s H s= + (9)

GH

is t e same as contour o + s s rawn rom t e point

Thus, the encirclement of the origin by the contour is equivalent to the

encirclement of the point by the contour (Fig.7)

Fig7

q ( 1 0)j +

q

( 1 0)j +GH


15/20

Nyquist Stability Criterion If the contour of the OL TF G(s)H(s) corresponding to the Nyquist contour

in the s-plane encircles the point in the CCW direction as many times

as the no. of RHSP poles of G(s)H(s), the CL system is stable.

The CL system is stable if the contour of G(s)H(s) does not encircle

point, i.e. the net encirclement is zero.

GH

( 1 0)j +

GH ( 1 0)j +

apping o yquist contour into contour is carrie out as o ows :

Mapping of Imaginary axis: Substitute axis in G(s)H(s). This converts mapping

function into a frequency function of

In physical systems

real constant

Thus the infinite arc of Nyquist contour maps into a point on the real-axis.

Complete contour is thus the polar plot of with varying from

Nyquist plot is symmetrical about the real axis since

GH

j

( ) ( )G j H j

( )m n

Relim ( ) ( )

js

G s H s

=

=

GH ( ) ( )G j H j to

* *( ) ( ) ( ) ( )G j H j G j H j =


16/20

Example-1: Consider a feedback system whose OL TF is

1 2

( ) ( )( 1)( 1)

KG s H sT s T s

=+ +

( ) ( ) K

G j H j = (1)

2 2 2 2

1 2

( ) ( )(1 ) (1 )

KG j H jT T

=

+ +

1 1

1 2( ) ( ) tan tanG j H j T T =

lim ( ) ( ) , lim 0G j H j K GH

= =

0lim 0, lim 180GH GH

= =

Rationalize (1) & separate the real & imaginary parts2

1 2 1 2

2 2 2 2 2 2 2 21 2 1 2 1 2

(1 ) ( )

(1 )(1 ) (1 )(1 ) (1 )(1 )

K T T T T K Kj

j T j T T T T T

+=

+ + + + + +

Fig.7


17/20

Example-1:

Equating real parts to zero, we get1 2

1

T T=

1 2

1 21

1 2T T

KT TGH

T T==

+

The plot of GH is shown in Figure. The infinite semicircular arc of the Nyquist contour

maps into the origin.

As the point is not encircled by the plot, N=0, P=0, Z=0 which implies stable

system.

1 0j +

does not encircle point for any +ve values of

Therefore system is stable for all +ve values of

( ) ( )G j H j ( 1 0)j + 1 2, &K T T

1 2, &K T T


18/20

Example-2:

Determine whether the system is stable when the feedback path is closed.

2( ) ( )

( 1)( 1)

sG s H s

s s

+=

+

From OL TF, one pole in RHSP therefore P=1,

locus encircles point once on CCW direction therefore N=1=P

Thus Z=0, i.e. there is no zeros of in RHSP and hence the CL system is stable

( ) ( )G j H j ( 1 0)j +

1 ( ) ( )G s H s+

Fig.8


19/20

OL Poles on the imaginary axis:

If G(s)H(s) and therefore 1+G(s)H(s) has any poles on the imaginary axis, the Nyquist

Contour is defined as

Fig.9


20/20

To study the stability in such cases, the Nyquist contour must ne modified so as to bypass

any imaginary axis poles. This is accompanied by indenting the Nyquist contour around

The imaginary axis poles along a semicircle of radius , where . Fig.10 0

Fig.10

Nyquesy Stability

Documents

Transcript of Nyquesy Stability