BODE PLOT FOR CLOSE LOOP CONTROL SYSTEMS

Poles & Zeros and Transfer Functions
Transfer Function: A transfer function is defined as the ratio of the Laplace
transform of the output to the input with all initial
conditions equal to zero. Transfer functions are defined
only for linear time invariant systems.
Considerations: Transfer functions can usually be expressed as the ratio
of two polynomials in the complex variable, s.
Factorization: A transfer function can be factored into the following form.
)
(
.
.
.
)
)(
(
)
(
.
.
.
)
)(
(
)
(
2
1
2
1
n
m
p
s
p
s
p
s
z
s
z
s
z
s
K
s
G







The roots of the numerator polynomial are called zeros.
The roots of the denominator polynomial are called poles.

Poles, Zeros and S-Plane
An Example: You are given the following transfer function. Show the
poles and zeros in the s-plane.
)
10
)(
4
(
)
14
)(
8
(
)
(





s
s
s
s
s
s
G
S - plane
x
x
o
x
o
0
-4
-8
-10
-14
origin
 axis
j axis

Bode Plot
• It is graphical representation of transfer function
to find out the stability of control system.
• It consists of two plots
– Magnitude (in dB) Vs frequency plot
– Phase angle Vs frequency plot

Bode Plot…
• Consider following T.F
• Put s=jw
• Arrange in following form
























 2
2
1
2
1
1
2
1
)
(
.
.
.
)
)(
(
)
(
)
(
.
.
.
)
)(
(
)
(
n
n
n
N
m
w
jw
w
jw
p
jw
p
jw
p
jw
jw
z
jw
z
jw
z
jw
K
jw
G

























 2
2
1
2
1
1
2
1
)
(
.
.
.
)
)(
(
)
(
.
.
.
)
)(
(
)
(
n
n
n
N
m
w
s
w
s
p
s
p
s
p
s
s
z
s
z
s
z
s
K
s
G

)
(
|
)
(
|
)
( jw
G
jw
G
jw
G 

























 2
2
1
1
2
1
)
1
(
.
.
.
)
1
)(
1
(
)
(
)
1
(
.
.
.
)
1
)(
1
(
)
(
n
n
n
b
a
N
m
w
jw
w
jw
jwT
jwT
jwT
jw
jwT
jwT
jwT
K
jw
G


Bode Plot…
• So
• Hence Bode Plot consists of two plots
– Magnitude ( dB) Vs frequency plot (w)
– Phase angle ( )Vs frequency plot (w)
|
)
(
|
log
20 10 jw
G
dB
in
Magnitude 
)
(
|
)
(
|
)
( jw
G
jw
G
jw
G 

Magnitude Phase Angle
|
)
(
|
log
20 10 jw
G
)
( jw
G


Bode Plot…
|
1
|
log
20
|
1
|
log
20
|
|
log
20
|
)
(
|
log
20 2
10
1
10
10
10 jwT
jwT
K
jw
G 



 |
1
|
log
20
... 10 n
jwT


etc
jwT
jwT
jwT m
b
a 





 |
1
|
log
20
...
|
1
|
log
20
|
1
|
log
20 10
10
10
Magnitude in dB
Phase Angle
     
n
jwT
jwT
jwT 1
2
1
1
1
0
tan
...
tan
tan
90 






      etc
jwT
jwT
jwT m
b
a 


 

 1
1
1
tan
...
tan
tan
      etc
jwT
jwT
jwT m
b
a 







 1
...
1
1
     
n
jwT
jwT
jwT
K
jw
G 










 1
...
1
1
)
(
)
( 2
1

Bode Plot…
Type of
System
Initial Slope Intersection with
0 dB line
0 0 dB/dec Parallel to 0 axis
1 -20dB/dec =K
2 -40dB/dec =K1/2
3 -60dB/dec =K1/3
. . .
. . .
. . .
N -20NdB/dec =K1/N

Bode Plot Procedure
• Steps to draw Bode Plot
1. Convert the TF in following standard form & put s=jw
2. Find out corner frequencies by using
























 2
2
1
1
2
1
)
1
(
.
.
.
)
1
)(
1
(
)
(
)
1
(
.
.
.
)
1
)(
1
(
)
(
n
n
n
b
a
N
m
w
jw
w
jw
jwT
jwT
jwT
jw
jwT
jwT
jwT
K
jw
G

etc
Rad
T
T
T
T
T
T c
b
a
sec
/
1
,
1
,
1
...
1
,
1
,
1
3
2
1

Bode Plot Procedure …
3. Draw the magnitude plot. The slope will change at each
corner frequency by +20dB/dec for zero and -20dB/dec for
pole.
 For complex conjugate zero and pole the slope will change
by
4. Starting plot
i. For type Zero (N=0) system, draw a line up to first (lowest)
corner frequency having 0dB/dec slope of magnitude (height)
20log10K
ii. For type One (N=1) system, draw a line having slope
-20dB/dec from w=K and mark first (lowest) corner
frequency.
iii. For type One (N=2) system, draw a line having slope
-40dB/dec from w=K1/2
and mark first (lowest) corner
frequency.
dec
dB /
40


Bode Plot Procedure …
5. Draw a line up to second corner frequency by
adding the slope of next pole or zero to the
previous slope and so on….
i. Slope due to a zero = +20dB/dec
ii. Slope due to a pole = -20dB/dec
6. Calculate phase angle for different value of ‘w’
and draw phase angle Vs frequency curve

Bode Plot GM & PM
• Gain Margin: It is the amount of gain in db that can be
added to the system before the system become
unstable
– GM in dB = 20log10(1/|G(jw|) = -20log10|G(jw|
– Gain cross-over frequency: Frequency where magnitude plot
intersect the 0dB line (x-axis) denoted by wg
• Phase Margin: It is the amount of phase lag in degree
that can be added to the system before the system
become unstable
– PM in degree = 1800
+angle[G(jw)]
– Phase cross-over frequency: Frequency where phase plot
intersect the 1800
dB line (x-axis) denoted by wp
– Less PM => More oscillating system

Bode Plot & Stability
Stability by Bode Plot
1. Stable
If wg<wp => GM & PM are +ve
2. Unstable
If wg>wp => GM & PM are –ve
3. Marginally stable
If wg=wp => GM & PM are zero

Bode Plot Examples
Example 1:Sketech the Bode plot for the TF
Determine
(i) GM
(ii) PM
(iii) Stability
)
001
.
0
1
)(
1
.
0
1
(
1000
)
(
s
s
s
G




Bode Plot Examples…
Solution:
1. Convert the TF in following standard form & put s=jw
2. Find out corner frequencies
So corner frequencies are 10, 1000 rad/sec
)
001
.
0
1
)(
1
.
0
1
(
1000
)
(
jw
jw
jw
G



10
1
.
0
1
 1000
001
.
0
1


• How to draw different slopes

• Magnitude Plot

• Phase Plot
S.N
o
W Angle (G(jw)
1 1 ----
2 100 -900
3 200 -980
4 1000 -134.420
5 2000 -153.150
6 3000 -161.360
7 5000 -168.570
8 8000 -172.790
9 Infi -1800

• Phase Plot

• Phase Plot …

• So Complete Bode Plot

References
• Automatic Control System By Hasan Saeed
– Katson Publication

BODE PLOT FOR CLOSE LOOP CONTROL SYSTEMS

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