LINEAR CONTROL
SYSTEMS
Ali Karimpour
Associate Professor
Ferdowsi University of Mashhad
Lecture 19
Lecture 19
Nyquist stability criteria (Continue).
Topics to be covered include:



Nyquist stability criteria (continue).
Minimum phase systems.
Simplified Nyquist stability criterion.
2
Dr. Ali Karimpour Dec 2013
Lecture 19
Nyquist fundamental
kf (s )
Nyquist path
1  kf (s )
Nyquist plot
Z 0  P0  N0
Nyquist path
Z 1  P1  N0
Z 1  P1  N 1
k 0
Z 1  P1  N1
k 0
Nyquist plot
k 0
3
Dr. Ali Karimpour Dec 2013
Example 1: Check the stability of following system
by Nyquist method.
Lecture 19
R(s)+
-
.‫ پایداری سیستم را توسط روش نایکوئیست بررسی کنید‬:1 ‫مثال‬
1
C (s )
375k
0
s( s  5)(s  10)
f (s) 
375
s( s  5)(s  10)
375
375

  0
50 s 500
375
375
f (s) 

   45
50 s 5045
f (s) 
f (s) 
375
375

   90
50 s 5090
375
375

   270
s3
(90)3
375
375
f ( s)  3 
   135
s
(45)3
375
375
f ( s)  3 
 0
s
(0)3
f ( s) 
-1
Very important
Very important
k
375
s( s  5)(s  10)
4
Dr. Ali Karimpour Dec 2013
Example 1: Check the stability of following
system by Nyquist method.
Lecture 19
R(s)+
-
k
375
s( s  5)(s  10)
C (s )
.‫پایداری سیستم را توسط روش نایکوئیست بررسی کنید‬
1
375k
0
s( s  5)(s  10)
f (s) 
375
s( s  5)(s  10)
-1
f ( j ) 
375
j ( j  5)( j  10)
Very important

f ( j )
0
5
7.07
10
20
  90
0.95  162
0.5  180
0.24  198
0.04   230
5
Dr. Ali Karimpour Dec 2013
Example 1: Check the stability of following
system by Nyquist method.
Lecture 19
R(s)+
-
k
375
s( s  5)(s  10)
C (s )
.‫پایداری سیستم را توسط روش نایکوئیست بررسی کنید‬
f (s) 
375
s( s  5)(s  10)
-1
CheckingtheRHP rootsof
375k
1
0
s( s  5)(s  10)
or stabilityof abovesystem
k  0 Z 1  P1  N1
Z 1  0  N1
0 Stable for 0  k  2
Z 1  N1  
2 Unstable for k6  2
TwoDr.RHP
roots
Ali Karimpour
Dec 2013
Example 1: Check the stability of following
system by Nyquist method.
Lecture 19
R(s)+
-
k
375
s( s  5)(s  10)
C (s )
.‫پایداری سیستم را توسط روش نایکوئیست بررسی کنید‬
f (s) 
375
s( s  5)(s  10)
-1
CheckingtheRHP rootsof
375k
1
0
s( s  5)(s  10)
or stabilityof abovesystem
k 0
Z 1  P1  N1
Z 1  0  N1
Z 1  N1  1 0  1 Unstablefor k  0
1 RHProot
7
Dr. Ali Karimpour Dec 2013
Example 2: Check the stability of following system from the givenLecture 19
Nyquist plot.
.‫ پایداری سیستم را با توجه به منحنی نایکوئیست داده شده بررسی کنید‬:2 ‫مثال‬
1
d (s)
d(s) is a polynomial.
R(s)+
-
k
1
d (s)
C (s )
Z 0  P0  N0
0  k  6 Z 1  P1  N1

0  P0  1
Z 1  1  0
 0 .1
 0
P0  1  P1
Z 1  1
6  k  10 Z 1  P1  N1 Z 1 1  1 z 1  0
z 1  2
k  10 Z 1  P1  N1 Z 1  1  1
Z 1  P1  N1 Z 1  1  0 z 1  1
k 0
1
6
System is unstable( 1 RHP zero)
System is stable
System is unstable( 2 RHP zero)
8
System is unstable(
1 RHPDeczero)
Dr. Ali Karimpour
2013
Example 2: Check the stability of following system from the givenLecture 19
Nyquist plot.
.‫ پایداری سیستم را با توجه به منحنی نایکوئیست داده شده بررسی کنید‬:2 ‫مثال‬
1
d (s)
d(s) is a polynomial.
R(s)+
1
d (s)
k
-
C (s )

1
6
 0 .1
 0
P0  1  P1
0  P0  1
 0 1
0  k  6 System is unstable( 1 RHP zero)
 N 1  P1   1 1
6  k  10 System is stable

 1 1
k  10
System is unstable( 2 RHP zero)
Z 0  P0  N0
Z 1  P1  N1
k 0
Z 1  P1  N1
Z 1  1  0
z 1  1
9
System is unstable(
1 RHPDeczero)
Dr. Ali Karimpour
2013
Example 3: Discuss about the RHP roots of following system.
1 k
Lecture 19
2( s  1)
 0 .‫ بحث کنید‬k ‫ در مورد ریشه های سمت راست معادله روبرو بر حسب مقادیر مختلف‬:3 ‫مثال‬
s ( s  1)
Very important part
f ( s) 
2( s  1)
s( s  1)
?
2 2
f (s) 

   180
s
0
2
2
f (s) 

   225
s
45
2
2
f (s) 

   270
s
90
f (s) 
2
2

   90
s  90
f (s) 
2
2

   45
s  45
f ( s) 
2
2

 0
s  0
10
Dr. Ali Karimpour Dec 2013
Lecture 19
Example 3: Discuss about the RHP roots of the following system.
2( s  1)
1 k
 0 .‫ بحث کنید‬k ‫ در مورد ریشه های سمت راست معادله روبرو بر حسب مقادیر مختلف‬:3 ‫مثال‬
s ( s  1)
f ( s) 
2( s  1)
s( s  1)
 1
2( j  1)
f ( j ) 
j ( j  1)

?1
f ( j )
?2
2
Which point is very important?
f ( j )  num den  180 tan1   (90  tan1 )  90  2  tan1 
f ( j)  0

 1
Or equivalently let Im(f) 0
f ( j1) 
2(1 j  1)
 20
11
j ( j  1)
Dr. Ali Karimpour Dec 2013
Lecture 19
Example 3: Discuss about the RHP roots of the following system for different values
of k.
2( s  1)
1 k
 0 .‫ بحث کنید‬k ‫ در مورد ریشه های سمت راست معادله روبرو بر حسب مقادیر مختلف‬:3 ‫مثال‬
s ( s  1)
k  0 Z 1  P1  N1

 1
1
1
Z 1  1
Unstable( one RHP root)
k  0 Z 1  P1  N1
2

Z 1  0  1
0

Z 1  
2

Z 1  N1
 0.5  k  0 Stable
k  0 .5
Unstable
(two RHP root)
More Study:
Z 0  P0  N0
1 0  1
12
Dr. Ali Karimpour Dec 2013
Lecture 19
Example 4: Discuss about the stability of the following system for different values of k.
.‫ بحث کنید‬k ‫ پایداری سیستم را بر حسب مقادیر مختلف‬:4 ‫مثال‬
+
-
+
-
k (s  2)
k (s  2)
10(s  2)
1 k 3
0
2
s  3s  10
+
-
10
s 2 (s  3)
10
s 3  3s 2  10
:‫فرم استانداردمعادله به صورت زیر است‬
13
Dr. Ali Karimpour Dec 2013
Discuss about the RHP roots of:
10 ( s  2)
1 k 3
0
2
s  3s  10
f (s) 
10( s  2)
s 3  3s 2  10
f (s) 
10(0  2)
2
0  0  10
10
10

   180
2
2
s
(90)
10
10
f (s)  2 
   90
s
(45) 2
Lecture 19
f ( s) 
f ( s) 
1
2
10
10

 0
s 2 (0) 2
f ( j ) 
10( j  2) (10  3 2 )  j 3
f ( j ) 
(10  3 2 )  j 3 (10  3 2 )  j 3
10( j 10  2)
 1180
f ( j 10) 
(10  30)  j10 10
10( j  2)
(10  3 2 )  j 3
10 (10  3 2 )  20 3
Im( f ) 
0
(10  3 2 ) 2   6
How?
f ( j1) 
  0 ,  10
20  10 j
10( j1  2)
 2.6  1.8 j

14
7

j
1
(10  3)  j1
Dr. Ali Karimpour Dec 2013
10 ( s  2)
1 k 3
0
2
s  3s  10
Discuss about the RHP roots of:
Lecture 19
P1  P0  ?
Z 0  P0  N0
1
2
k 0
0  P0  2
Z 1  P1  N1
P1  P0  2
Z 1  2  N1
 2  2  0 k  1
Z 1  N1  2  
 0 2  2 0  k 1
k0
Z 1  P1  N1
 02  2
Z 1  N1  2  
 1  2  1
Stable
Unstable
2 RHP roots
Z 1  2  N1
Z 1  N1  2
 0 .5  k  0
Unstable (2 RHP roots)
k  0 .5
Unstable (1Dr.RHP
roots)
Ali Karimpour
15
Dec 2013
Lecture 19
Minimum Phase systems
‫توابع حداقل فاز‬
f(s) is said to be minimum phase if it has no poles and zeros
on the RHP and on the jω axis (origin is an exception) and
there is no delay.
‫ و روی‬RHP ‫ را حداقل فاز گویند اگر این تابع هیچ قطب و صفری در‬f(s) ‫تابع‬
.‫ نداشته (مبدا استثنا است) و دارای تاخیر نباشد‬jω ‫محور‬
nz
f (s) 
 ( s  zi )
i 1
np
s  (s  p j )
Ty
j 1
Important note:
If it was minimumphase
zi , p j  0
Ty is typeof system
If f(s) is minimumphase then
Z0  P1  P0  0
16
Dr. Ali Karimpour Dec 2013
Lecture 19
Nyquist fundamental for minimum phase systems
kf (s )
Nyquist path
Nyquist plot
k 0
Z 1  P1  N1
Z 1  N1
k 0
Z 1  P1  N1
Z 1  N1
17
Dr. Ali Karimpour Dec 2013
Check the stability of following system by Nyquist
method.
f ( s) 
R(s)+
-
k
40
s ( s  25)
Lecture
C (s ) 19
40
s( s  25)
-1
System is minimum phase
k 0
k 0
2( 90Ty φ-1 ) 2(90  90 )
Z 1  N1 

0


360
360
2( 90Ty  φ1 ) 2(90  90 )

1
Z 1  N1 


360
360
Stable for k  0
Unstablefor k  0
18
One
RHP
zero
Dr. Ali Karimpour Dec 2013
Simplified Nyquist
path plot
Check the stability of following system by Nyquist
method.
f ( s) 
R(s)+
-
k
40
s ( s  25)
Lecture
C (s ) 19
40
s( s  25)
-1
System is minimum phase
2( 90Ty φ-1 )

360 
k 0
Z 1  N1
k 0
2( 90Ty  φ1 )
Z 1  N1 
360 
φ-1 is the angle of polar plot around -1
φ1 is the angle of polar plot around
1
19
Dr. Ali Karimpour Dec 2013
Lecture 19
Example 5: Discuss about the RHP roots of the following system for different values
of k.
.‫ بحث کنید‬k ‫ سیستم زیر بر حسب مقادیر مختلف‬RHP ‫ در مورد ریشه های‬:5 ‫مثال‬
1 k
1
0
1 s
 0
Clearly System is minimum phase so we use simplified Nyquist method
f ( s) 
1
1 s
1
-1
Simplified Nyquist
path plot
k 0
k 0
Polar plot
Z 1  N 1 
Z 1  N1 
2(90Ty   1 )

360
2(90Ty  1 )
360

2 ( 0  0)
0

360
1
180/ 180  1



180   0 / 180  0
No RHP root.
k  1
1  k  0
One RHP root.
No RHP
root.
20
Dr. Ali Karimpour Dec 2013
Lecture 19
Example 7: Discuss about the RHP roots of the following system for different values
of k.
.‫ بحث کنید‬k ‫ سیستم زیر بر حسب مقادیر مختلف‬RHP ‫ در مورد ریشه های‬:6 ‫مثال‬
1 k
1
0
s 2 (1   s)
 0
Clearly System is minimum phase so we use simplified Nyquist method
f ( s) 
1
s 2 (1   s)
1
-1
Simplified Nyquist
path plot
k 0
k 0
Polar plot
2(90Ty   1 ) 2(180  180 )

 2 Two RHP roots.
Z 1  N 1 


360
360
2(90Ty  1 )
2(180  0)

 1 One RHP root.
Z 1  N1 


360
360
21
Dr. Ali Karimpour Dec 2013
Lecture 19
Simplified Nyquist method
kf (s )
-1
Simplified Nyquist
path plot
System is minimum phase
2( 90Ty φ-1 )

360 
k 0
Z 1  N1
k 0
2( 90Ty  φ1 )
Z 1  N1 
360 
Polar plot
φ-1 is the angle of polar plot around -1
φ1 is the angle of polar plot around 1
Important remark: If any of φ-1 or φ1 is greater than zero the system is
unstable but if they were less than zero one must check it!
22
Dr. Ali Karimpour Dec 2013
Lecture 19
Simplified Nyquist method
kf (s )
-1
Unstable
Polar plot
kf (s )
Simplified Nyquist
path plot
-1
Polar plot
23 T
Stability depends on
y
Dr. Ali Karimpour Dec 2013
Lecture 19
Exercises
‫تمرینها‬
1- The open loop transfer function of a unity-feedback (negative sign) is:
G p ( s) 
k
( s  5) n
Apply the Nyquist criterion to determine the range of k for stability. Let
n=1,2,3 and 4
2- The characteristic equation of a linear control system is:
s 3  2s 2  20s  10k  0
Apply the Nyquist criterion to determine the range of k for stability.
24
Dr. Ali Karimpour Dec 2013
Lecture 19
Exercises
‫تمرینها‬
3- The open loop transfer function of a unity-feedback (negative sign)
with PD controller is:
G p ( s) 
10( K p  K d s)
s2
Select the value of Kp so that the parabolic error constant be 100. Find
the equivalent open-loop transfer function Geq(s) for stability analysis
with Kd as a gain factor. Sketch the Nyquist plot and check the stability
for different values of Kd.
25
Dr. Ali Karimpour Dec 2013
Lecture 19
Exercises
‫تمرینها‬
4- The polar plot of an open loop transfer function of a minimum phase
system is:
3
Determine the steady state error of the system to a unit step.
ess 
1
4
:‫جواب‬
5- The open loop transfer function of a unity-feedback (negative sign) is:
keTs
G( s) 
s 1
(k  1)
Derive an expression that make the system stable.
[T k 2  1  tan1 k 2  1]   :‫جواب‬
26
Dr. Ali Karimpour Dec 2013
Lecture 19
Example 6: Discuss about the RHP roots of the following system for different values
of k.
.‫ بحث کنید‬k ‫ سیستم زیر بر حسب مقادیر مختلف‬RHP ‫ در مورد ریشه های‬:7 ‫مثال‬
1 k
1
0
s(1   s)
 0
Clearly System is minimum phase so we use simplified Nyquist method
f ( s) 
1
s(1   s)
-1
Simplified Nyquist
path plot
k 0
Z 1  N 1 
k 0
Z 1  N1 
1
Polar plot
2(90Ty   1 )


2(90  90)
0

360
No RHP root.


2(90  90 )
1

360
One RHP root.
360
2(90Ty  1 )
360
27
Dr. Ali Karimpour Dec 2013
Lecture 19
Example 8: Discuss about the RHP roots of following system for different value of k.
.‫ بحث کنید‬k ‫ سیستم زیر بر حسب مقادیر مختلف‬RHP ‫ در مورد ریشه های‬:8 ‫مثال‬
1 k
1
0
s 3 (1   s)
 0
Clearly System is minimum phase so we use simplified Nyquist method
f ( s) 
1
s 3 (1   s)
1
-1
Simplified Nyquist
path plot
k 0
Z 1  N 1 
k 0
Z 1  N1 
Polar plot
2(90Ty   1 )
360
2(90Ty  1 )

360

2( 270  90 )
 2 Two RHP roots.

360

2(270  90)
1

360
One RHP root.
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Dr. Ali Karimpour Dec 2013