This document provides information about Nyquist plots and the Nyquist stability criterion for analyzing control systems. It begins with background knowledge on system stability and characteristics equations. It then explains that a Nyquist plot maps the open-loop transfer function to show the number of encirclements of the point -1 in the complex plane, which determines stability. The document gives steps for constructing a Nyquist plot using MATLAB and provides an example of analyzing stability by adjusting the open-loop gain K to move closed-loop poles into the left half plane. It concludes that the Nyquist criterion allows determining stability by finding a range of K values that keep the number of encirclements of -1 equal to the number of open-loop poles

1. Bhagwan Mahavir College Of Engineering & Technology, Surat
Subject – Control System Engineering
(2141004)
Topic :- Nyquist Plots
Guided By
Mrs. Sumita D Chakrabortty
Prepared By
Hitesh Atre (140063110001)

2. Knowledge Before
Studying Nyquist Criterion
)
(
)
(
1
)
(
)
(
s
H
s
G
s
G
s
T


)
(
)
(
)
(
s
D
s
N
s
G
G
G

)
(
)
(
)
(
s
D
s
N
s
H
H
H

unstable if there is any pole on RHP (right half plane)

3. )
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
1
)
(
)
(
s
N
s
N
s
D
s
D
s
D
s
N
s
H
s
G
s
G
s
T
H
G
H
G
H
G




)
(
)
(
)
(
)
(
)
(
)
(
s
D
s
N
s
D
s
N
s
H
s
G
H
H
G
G

H
G
H
G
H
G
H
G
H
G
D
D
N
N
D
D
D
D
N
N
s
H
s
G




 1
)
(
)
(
1
poles of G(s)H(s) and 1+G(s)H(s) are the same
zero of 1+G(s)H(s) is pole of T(s)
Characteristic equation:
Open-loop system:
Closed-loop system:

4. )
8
)(
7
)(
6
)(
5
(
)
4
)(
3
)(
2
)(
1
(
)
(
)
(









s
s
s
s
s
s
s
s
s
H
s
G
)
(
)
(
1
)
(
s
H
s
G
s
G

)
(
)
( s
H
s
G )
(
)
(
1 s
H
s
G

Zero – 1,2,3,4
Poles – 5,6,7,8
Zero – a,b,c,d
Poles – 5,6,7,8
Zero – ?,?,?,?
Poles – a,b,c,d
To know stability, we have to know a,b,c,d

5. Stability from Nyquist
plot
From a Nyquist plot, we can tell a number of closed-
loop poles on the right half plane.
o If there is any closed-loop pole on the right half plane, the system goes
unstable.
o If there is no closed-loop pole on the right half plane, the system is stable.

6. Nyquist Criterion
Nyquist plot is a plot used to verify stability
of the system.
function
)
)(
(
)
)(
(
)
(
2
1
2
1
p
s
p
s
z
s
z
s
s
F





mapping all points (contour) from one plane to another
by function F(s).
mapping contour

7. )
)(
(
)
)(
(
)
(
2
1
2
1
p
s
p
s
z
s
z
s
s
F






8. • Pole/zero inside the
contour has 360 deg.
angular change.
• Pole/zero outside
contour has 0 deg.
angular change.
• Move clockwise
around contour, zero
inside yields rotation
in clockwise, pole
inside yields rotation
in counterclockwise

9. Characteristic equation
N = P-Z
N = # of counterclockwise direction about the origin
P = # of poles of characteristic equation inside contour
= # of poles of open-loop system
z = # of zeros of characteristic equation inside contour
= # of poles of closed-loop system
Z = P-N
)
(
)
(
1
)
( s
H
s
G
s
F 


10. Characteristic equation
• Increase size of the contour to cover the right half plane
• More convenient to consider the open-loop system (with
known pole/zero)

11. ‘Open-loop system’
Mapping from characteristic equ. to open-loop
system by shifting to the left one step
Z = P-N
Z = # of closed-loop poles inside the right half plane
P = # of open-loop poles inside the right half plane
N = # of counterclockwise revolutions around -1
)
(
)
( s
H
s
G
Nyquist diagram of

13. Properties of Nyquist plot
If there is a gain, K, in front of open-loop transfer
function, the Nyquist plot will expand by a factor of K.

14. Nyquist plot example
2
1
)
(


s
s
G
)
1
(
1
)
(
1
)
(


 s
S
G
s
G
• Open loop system has pole at
2
• Closed-loop system has pole
at 1
• If we multiply the open-loop
with a gain, K, then we can
move the closed-loop pole’s
position to the left-half plane

15. Nyquist plot example
(cont.)
• New look of open-loop system:
• Corresponding closed-loop system:
• Evaluate value of K for stability
2
)
(


s
K
s
G
)
2
(
)
(
1
)
(



 K
s
K
s
G
s
G
2

K

16. Adjusting an open-loop gain to guarantee stability
Step I: sketch a Nyquist Diagram
Step II: find a range of K that makes the system stable!

17. How to make a Nyquist plot?
Easy way by Matlab
o Nyquist: ‘nyquist’
o Bode: ‘bode’

18. Step I: make a Nyquist plot
• Starts from an open-loop transfer function (set
K=1)
• Set and find frequency response
o At dc,
o Find at which the imaginary part equals zero

j
s 
0
0 

 s



19. 2
2
2
2
3
2
2
2
2
2
2
2
2
2
2
2
2
2
6
)
8
(
)
14
154
(
48
)
8
)(
15
(
6
)
8
(
6
)
8
(
6
)
8
(
8
)
15
(
6
)
8
(
8
)
15
(
8
6
15
8
)
(
)
(
8
6
15
8
)
4
)(
2
(
)
5
)(
3
(
)
(
)
(

































































j
j
j
j
j
j
j
j
j
j
H
j
G
s
s
s
s
s
s
s
s
s
H
s
G
11
,
0


Need the imaginary term = 0,
31
.
1
412
540
)
11
(
6
)
11
8
(
)
11
(
48
)
11
8
)(
11
15
(
2
2









Substitute back in to the transfer function
And get
11


33
.
1
)
( 

s
G

20. At dc, s=0,
At imaginary part=0

21. Step II: satisfying stability condition
• P = 2, N has to be 2 to guarantee stability
• Marginally stable if the plot intersects -1
• For stability, 1.33K has to be greater than 1
K > 1/1.33
or K > 0.75

22. Example
Evaluate a range of K that makes the system stable
)
2
)(
2
2
(
)
( 2




s
s
s
K
s
G

23. 2
2
2
2
2
2
2
2
)
6
(
)
1
(
16
)
6
(
)
1
(
4
)
2
)(
2
2
)
((
)
(





















j
j
j
j
K
j
G
At 6
,
0

 the imaginary part = 0
Plug back in the transfer function
and get G = -0.05
Step I: find frequency at which imaginary part = 0

j
s 
6


Set

24. Step II: consider stability
condition
• P = 0, N has to be 0 to guarantee stability
• Marginally stable if the plot intersects -1
• For stability, 0.05K has to be less than 1
K < 1/0.05
or K < 20

25. Thank U !