Module 13
Nyquist Stability Criterion
(4 hours)
13.1 Conformal Mapping,Cauchy’s Theorem
( 保角映射,柯西定理 )
Recall Stability Problem:
To determine the relative stability of a closed-loop system,
we must investigate the characteristic equation of the
system:
0)(1 sGH
Where GH(s) or 1+ GH(s) is a complex function of s,and the
difference between GH(s) and 1+ GH(s) is only 1,So
(1) We can investigate 1+ GH(s) through GH(s) ;
(2) How to investigate GH(s)? ---- If s has a variation,then GH(s)
has a variation certainly,We can suppose the variation of s,to see
the change of GH(s).
M
ps
zs
sF
i
j
)(
)(
)(
)( sFs M a p p i n g
We are concern with the mapping of contours in the s – plane by a
function F(s),A contour map is a contour or trajectory in one
plane mapped or translated another plane by a relation F(s),
Since s is a complex variable,s = σ +jω,the function F(s) is itself
complex; it can be defined as F(s) = u + jv and can be
represented on a complex F(s) – plane with coordinates u and v,
jω
σ
[s]
u
[F(s)]
jv
S1=-1+j1 F1=-2+j2
Mapping
F(S)=2s
i
j
ps
zs
M
ij pszs?
s-plane
F(s)-plane
As an example,let us consider a function F(s) = 2s + 1 and a
contour in the s – plane,The mapping of the s – plane unit square
contour to the F(s) – plane is accomplished through the relation
F(s),and so
1)(212)( jssFjvu
2
12
v
u
Example 2,2)( s ssF
1211 11)(11,jj jsFjsD D
Example 3.
21
)(
s
ssF
M
ps
zs
sF
i
j
)(
)(
)(
i
j
ps
zs
M
)()( ij pszs?
i
j
ps
zs
M )()(
ij pszs?
The encirclement of the poles and zeros of F(s) can be
related to the encirclement of the origin in the F(s)-plane
by Cauchy’s theorem,commonly known as the
principle of the argument,which state:
Cauchy’s Theorem
If a contour Γs in the s-plane encircles Z zeros and P
poles of F(s) and does not pass through any poles or
zeros of F(s) and the traversal is in the clockwise
direction along the contour,the corresponding contour
ΓF in the F(s)-plane encircles the origin of the F(s)-
plane N = Z – P times in the clockwise direction.
Ex,4
Ex,5
13.2 The application of Cauchy’s Theorem:
Nyquist Stability Criterion
13.2.1 F(s) — plane
)(1)( sGHsFSuppose:
)(
)()(
i
j
ps
zsKsGH?
)(
)(
)(
)()(
)(
)(
1)(
i
i
i
ji
i
j
ps
ss
ps
zsKps
ps
zsK
sF
F(s) GH(s)
( Investigate F(s) by GH(s) )
ip
— open–loop pole,and the pole of GH(s) and F(s).
jz
— open–loop zero.
is
— the zero of F(s),that is the closed–loop pole.
For s system F(s) to be stable,all the zeros of F(s) must lie
in the left-hand s-plane,Thus the roots of a stable system
[ the zeros of F(s) ] must lie to the left of the jω-axis in the
s-plane,Therefore we choose a contour Γs in the s-plane
that encloses the entire right-hand s-plane.
We determine whether any zeros of F(s) lie within Γs by
utilizing Cauchy’s theorem.
That is,we plot ΓF in the F(s)-
plane and determine the
number of encirclements of the
origin N,Then the number of
zeros of F(s) within the Γs
contour (and therefore unstable
zeros of F(s)) is,Z = N + P.
For a system to be stable:
P poles
jv
u
P-times
For a system to be unstable:
P poles
jv
u
N-times
N≠P,N=Z-P; Z=N+P
Z zeros
When a system is stable,Z will be 0,Therefore for the
contour ΓF (F(s)),the number of clockwise
encirclements of the origin is Ncw = – P,( the number of
counterclockwise encirclements of the origin is Nccw =
P).
That is,a feedback control system is stable if and only if,
for the contour ΓF (F(s)),the number of counter-
clockwise encirclements of the origin is equal to the
number of poles of F(s) (or GH(s),that is open-loop
poles) with positive parts.
When all the poles of F(s) is located in the left-hand s-
plane,P = 0,a system is stable if and only if the contour
ΓF in the F(s)-plane does not encircle the origin,
Nyquist Stability Criterion in F(s) plane
13.2.2 Nyquist Stability Criterion in GH(s) plane
1)()(
)(1)(
sFsGH
sGHsF
GH(s)F(s)
1)(
0)(
sGH
sF
A feedback control system is stable if and only if,for
the contour ΓGH (GH(s)),the number of counter-
clockwise encirclements of the (-1,j0) point is equal to
the number of poles of GH(s) with positive parts.
When all the poles of GH(s) is located in the left-hand s-
plane,P = 0,a system is stable if and only if the contour
ΓGH in the GH(s)-plane does not encircle the (-1,j0)
point,
13.2.3 Nyquist Stability Criterion in GH(jω) plane
(1) s=jω
jeGH
jGHsGHjs
)()(:)1(
is frequency response.
(2) s= -jω
jeGH
jGHsGHjs
)()(:)2(
is negative frequency
response.
)22(Re)3( js
mntc o n s
mn
ps
zsK
sGHs
n
i
i
m
j
j
R
j
,t a n
,0
)(
)(
)(:Re)3(
1
1
22
is a point (origin or
a point in real-axis
of GH(s)-plane.
Ex,6
Ex,7
)1)(1()( 21 sTsT
KsGH
R
r
If GH(s) has the pole in origin (integral element),Γs must have the
fourth part to avoid pass through the pole of GH(s),Otherwise
mapping will not be conformal.
This fourth part is:
)22,0( a n drres j
)
22
,0(
,
'
l i m)(
)1(
)1('
)(
)(
)(
rresw h e n
e
s
K
sGH
sTs
sK
pss
zsK
sGH
j
jN
N
res
i
N
j
i
N
j
j
The mapping curve of the
fourth part of Γs is N half circle
with infinity radius in
clockwise direction.
Ex,8
)1()( Tss
KsGH
(1)
(1)
(2)
(2)
(3)
(3)
(4)
(4)
Ex,9
)1)(1()( 21 sTsTs
KsGH
Ex,10
)1()( 2 Tss
KsGH
-- unstable.
P = 0;
Ncw = 2
Or Nccw = -2
So
Z = P + Ncw
= P - Nccw
= 0 +2 = 2
Ex,11
)1()( ss
KsGH
-- unstable.
P = 1;
Ncw = 1
or Nccw = -1
So
Z = P + Ncw
= P - Nccw
= 1 +1 = 2
-- stable.
P = 1;
Ncw = -1
or Nccw = 1
So
Z = P + Ncw
= P - Nccw
= 1 -1 = 0
Ex,12
)1(
)1()( 21
ss
sKKsGH
Other Example
Instructional objectives:
At the end of this lecture students
should be able to
Determine the stability of a feedback
control system by the Nyquist Criterion
using Nyquist diagram GH(jω)
Determine the number of closed-loop
poles on the RHP using Nyquist Criterion
Nyquist Stability Criterion
(4 hours)
13.1 Conformal Mapping,Cauchy’s Theorem
( 保角映射,柯西定理 )
Recall Stability Problem:
To determine the relative stability of a closed-loop system,
we must investigate the characteristic equation of the
system:
0)(1 sGH
Where GH(s) or 1+ GH(s) is a complex function of s,and the
difference between GH(s) and 1+ GH(s) is only 1,So
(1) We can investigate 1+ GH(s) through GH(s) ;
(2) How to investigate GH(s)? ---- If s has a variation,then GH(s)
has a variation certainly,We can suppose the variation of s,to see
the change of GH(s).
M
ps
zs
sF
i
j
)(
)(
)(
)( sFs M a p p i n g
We are concern with the mapping of contours in the s – plane by a
function F(s),A contour map is a contour or trajectory in one
plane mapped or translated another plane by a relation F(s),
Since s is a complex variable,s = σ +jω,the function F(s) is itself
complex; it can be defined as F(s) = u + jv and can be
represented on a complex F(s) – plane with coordinates u and v,
jω
σ
[s]
u
[F(s)]
jv
S1=-1+j1 F1=-2+j2
Mapping
F(S)=2s
i
j
ps
zs
M
ij pszs?
s-plane
F(s)-plane
As an example,let us consider a function F(s) = 2s + 1 and a
contour in the s – plane,The mapping of the s – plane unit square
contour to the F(s) – plane is accomplished through the relation
F(s),and so
1)(212)( jssFjvu
2
12
v
u
Example 2,2)( s ssF
1211 11)(11,jj jsFjsD D
Example 3.
21
)(
s
ssF
M
ps
zs
sF
i
j
)(
)(
)(
i
j
ps
zs
M
)()( ij pszs?
i
j
ps
zs
M )()(
ij pszs?
The encirclement of the poles and zeros of F(s) can be
related to the encirclement of the origin in the F(s)-plane
by Cauchy’s theorem,commonly known as the
principle of the argument,which state:
Cauchy’s Theorem
If a contour Γs in the s-plane encircles Z zeros and P
poles of F(s) and does not pass through any poles or
zeros of F(s) and the traversal is in the clockwise
direction along the contour,the corresponding contour
ΓF in the F(s)-plane encircles the origin of the F(s)-
plane N = Z – P times in the clockwise direction.
Ex,4
Ex,5
13.2 The application of Cauchy’s Theorem:
Nyquist Stability Criterion
13.2.1 F(s) — plane
)(1)( sGHsFSuppose:
)(
)()(
i
j
ps
zsKsGH?
)(
)(
)(
)()(
)(
)(
1)(
i
i
i
ji
i
j
ps
ss
ps
zsKps
ps
zsK
sF
F(s) GH(s)
( Investigate F(s) by GH(s) )
ip
— open–loop pole,and the pole of GH(s) and F(s).
jz
— open–loop zero.
is
— the zero of F(s),that is the closed–loop pole.
For s system F(s) to be stable,all the zeros of F(s) must lie
in the left-hand s-plane,Thus the roots of a stable system
[ the zeros of F(s) ] must lie to the left of the jω-axis in the
s-plane,Therefore we choose a contour Γs in the s-plane
that encloses the entire right-hand s-plane.
We determine whether any zeros of F(s) lie within Γs by
utilizing Cauchy’s theorem.
That is,we plot ΓF in the F(s)-
plane and determine the
number of encirclements of the
origin N,Then the number of
zeros of F(s) within the Γs
contour (and therefore unstable
zeros of F(s)) is,Z = N + P.
For a system to be stable:
P poles
jv
u
P-times
For a system to be unstable:
P poles
jv
u
N-times
N≠P,N=Z-P; Z=N+P
Z zeros
When a system is stable,Z will be 0,Therefore for the
contour ΓF (F(s)),the number of clockwise
encirclements of the origin is Ncw = – P,( the number of
counterclockwise encirclements of the origin is Nccw =
P).
That is,a feedback control system is stable if and only if,
for the contour ΓF (F(s)),the number of counter-
clockwise encirclements of the origin is equal to the
number of poles of F(s) (or GH(s),that is open-loop
poles) with positive parts.
When all the poles of F(s) is located in the left-hand s-
plane,P = 0,a system is stable if and only if the contour
ΓF in the F(s)-plane does not encircle the origin,
Nyquist Stability Criterion in F(s) plane
13.2.2 Nyquist Stability Criterion in GH(s) plane
1)()(
)(1)(
sFsGH
sGHsF
GH(s)F(s)
1)(
0)(
sGH
sF
A feedback control system is stable if and only if,for
the contour ΓGH (GH(s)),the number of counter-
clockwise encirclements of the (-1,j0) point is equal to
the number of poles of GH(s) with positive parts.
When all the poles of GH(s) is located in the left-hand s-
plane,P = 0,a system is stable if and only if the contour
ΓGH in the GH(s)-plane does not encircle the (-1,j0)
point,
13.2.3 Nyquist Stability Criterion in GH(jω) plane
(1) s=jω
jeGH
jGHsGHjs
)()(:)1(
is frequency response.
(2) s= -jω
jeGH
jGHsGHjs
)()(:)2(
is negative frequency
response.
)22(Re)3( js
mntc o n s
mn
ps
zsK
sGHs
n
i
i
m
j
j
R
j
,t a n
,0
)(
)(
)(:Re)3(
1
1
22
is a point (origin or
a point in real-axis
of GH(s)-plane.
Ex,6
Ex,7
)1)(1()( 21 sTsT
KsGH
R
r
If GH(s) has the pole in origin (integral element),Γs must have the
fourth part to avoid pass through the pole of GH(s),Otherwise
mapping will not be conformal.
This fourth part is:
)22,0( a n drres j
)
22
,0(
,
'
l i m)(
)1(
)1('
)(
)(
)(
rresw h e n
e
s
K
sGH
sTs
sK
pss
zsK
sGH
j
jN
N
res
i
N
j
i
N
j
j
The mapping curve of the
fourth part of Γs is N half circle
with infinity radius in
clockwise direction.
Ex,8
)1()( Tss
KsGH
(1)
(1)
(2)
(2)
(3)
(3)
(4)
(4)
Ex,9
)1)(1()( 21 sTsTs
KsGH
Ex,10
)1()( 2 Tss
KsGH
-- unstable.
P = 0;
Ncw = 2
Or Nccw = -2
So
Z = P + Ncw
= P - Nccw
= 0 +2 = 2
Ex,11
)1()( ss
KsGH
-- unstable.
P = 1;
Ncw = 1
or Nccw = -1
So
Z = P + Ncw
= P - Nccw
= 1 +1 = 2
-- stable.
P = 1;
Ncw = -1
or Nccw = 1
So
Z = P + Ncw
= P - Nccw
= 1 -1 = 0
Ex,12
)1(
)1()( 21
ss
sKKsGH
Other Example
Instructional objectives:
At the end of this lecture students
should be able to
Determine the stability of a feedback
control system by the Nyquist Criterion
using Nyquist diagram GH(jω)
Determine the number of closed-loop
poles on the RHP using Nyquist Criterion
