Bode Plot Notes Step by Step

Part of Lecture Series on Automatic Control Systems delivered by me to Final year Diploma in Engg. Students. Equally useful for higher level. Easy language and step by step procedure for drawing Bode Plots. Three illustrative examples are included.

1. BEE-502
Automati
Unit-4, B
 
 
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an
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c Control Syst
Bode Plot Supp
mination
y low freq
er terms h
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tude in dB
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aring this e
he slope is
we consider
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al part of t
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System: n
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for variou
n = 0
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the Bode p
is 20 log K
n = 1
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Diploma in E
ectrical/ Instr.
5
th
Seme
s
f Bode Pl
rm
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ribution to
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stem has a
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tem
uctor: Mohd. U
Date
Pag
all other te
nitude, he
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a slope 0 d
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Umar Rehman
e: 7. 11. 2017
ge 1 of 8
erms as
nce the
quencies
can say
dB/dec,

2. BEE-502
Automati
Unit-4, B
 
 
T
in
eq
(iii) T
T
in
eq
c Control Syst
Bode Plot Supp
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quation to
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part of the
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he Bode p
0 dB axis
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ω
0
20
n = 2
()gGjHω
e Bode plot
0 dB axis
logK
ω
ω
2
0
20
Diploma in E
ectrical/ Instr.
5
th
Seme
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plot has a
s can be f
log
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=
=
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Engg.
& Control)
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Date
Pag
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Umar Rehman
e: 7. 11. 2017
ge 2 of 8
point of
f above
point of
f above

3. BEE-502
Automatic Control Systems
Unit-4, Bode Plot Supplementary Notes
Diploma in Engg.
(Electrical/ Instr. & Control)
5
th
Semester
Instructor: Mohd. Umar Rehman
Date: 7. 11. 2017  
Page 3 of 8
 

Let us see some examples on Bode plot.
Example 1: Draw the Bode plot of the unity feedback system with forward gain as
(
(
)
)
)(ss s
Gs=
++
200
220

Also determine the gain margin, phase margin and comment on the stability of the
system.
Solution: Step-1: First convert the given transfer function into time constant form
( )( )
() (
)( )
.
)
(
.
ss
GsH
s
s
s
ss
s
ss
=
⎛⎞⎛ ⎞++
⎟⎟⎜⎜
⎟⎟×× + +⎜⎜⎟⎟⎜⎜⎟⎟⎜⎜⎝⎠⎝ ⎠
=
=
++
200 200
220
2201 1
220
5
1051005

Step-2: Determine the sinusoidal transfer function
( )( )
()()
..
Gj Hj
jj j
ωω
ωω ω
=
++
5
105 1005
 
Step-3: Identify different parts of the Bode plot
(i) Constant term: K = 5
(ii) Type of system: 1, this means initial slope is −20 dB/dec, and intersection of
the initial part of the plot with 0 dB axis occurs at
rad/sKω== 5 .
(iii) Corner frequencies:
c1 c2
rad/s, rad/s
..
ωω== = =
11
220
05 005
Step-4: Draw reference slopes on top left corner of the semi-log graph paper. For
magnitude plot draw a 0 dB axis line, with positive values above it like 20 dB, 40 dB
etc. and negative values below it like −20 dB, 40 dB etc. For phase plot, draw a
−180° axis line. Above it the values will be like −150°, −120°, −90° etc. and the
values below it will be like −210°, −240°, −270° etc.

4. BEE-502
Automatic Control Systems
Unit-4, Bode Plot Supplementary Notes
Diploma in Engg.
(Electrical/ Instr. & Control)
5
th
Semester
Instructor: Mohd. Umar Rehman
Date: 7. 11. 2017  
Page 4 of 8
 
Step-5: Start plotting the magnitude plot by drawing a line with initial slope of
−20 dB/dec that intersects the 0 dB axis at 5 rad/s. This slope continues till first
corner frequency of 2 rad/s. Since this corner frequency is due to a pole, an
additional slope of −20 dB/dec will be added, hence the slope now will be
−40 dB/dec. This slope will continue till second corner frequency of 20 rad/s. Since
the second corner frequency is also due to a pole, an additional slope of −20 dB/dec
will be added, and hence the slope now will be −60 dB/dec. This slope will continue
for all the further values of ω.
Step-6: Phase calculation: Only denominator terms will contribute to phase angle.
The expression for phase angle is given by:
tan()() (.) ( tan . )Gj Hj ωωφωω
−−
∠=−−=−
11
90 0 5 0 05
D

Using this expression, we tabulate some values of ϕ w. r. t. ω as follows:
ω 0.1 1 2 10 20 40 ∞
φ −93.15° −119.42° −140.7° −195.29° −219.28° −240.58° −270°

Step-7: Determine the value of frequency at which the magnitude plot crosses the 0
dB axis. This value is the gain crossover frequency
gc
ω. Drop a perpendicular line at
this value of frequency on the phase plot and read the phase angle value. Here,
gc
. rad/sω=285 and phase angle at this value is ()
gc
φω =− °153 .
Next, determine the value of frequency at which the magnitude plot crosses the
−180° axis. This frequency is the phase crossover frequency
pc
ω. Drop a
perpendicular line at
pc
ωon the magnitude plot and read the corresponding value of
magnitude. Here
pc
. rad/sω=63 and magnitude is ()
pc
Mω=−13 dB.
Step-8: Calculate the stability margins
Gain Margin,
()
pc
()GM M ω=− =−− =+0 0 13 13

dB.
Phase Margin,
()
gc
()PM φω=+ =+−°=°180 180 153 27
DD

Since both the margins are positive, the system is stable.
A Bode plot drawn by software is shown here and all values given here are for
reference only. The closer your values are to these values, the more accurate your
Bode plot will be.

5. BEE-502
Automatic Control Systems
Unit-4, Bode Plot Supplementary Notes
Diploma in Engg.
(Electrical/ Instr. & Control)
5
th
Semester
Instructor: Mohd. Umar Rehman
Date: 7. 11. 2017  
Page 5 of 8
 

Fig. 1. Bode Plot of Example 1
Example 2: Draw the Bode plot of the unity feedback system with forward gain as
(
(
)
)
)(ss
Gs
=
++
40
25

Also determine the gain margin, phase margin and comment on the stability of the
system.
Solution: Repeat the steps as in previous example
Step-1: Convert transfer function to time constant form
( )( )
() (
() ..
)
)(ss
GsHs
ss
=
++ +
=
+
40 4
25 105102

Step-2: Convert to sinusoidal form
( )( )
()()
..
Gj Hj
jj
ωω
ωω++
=
05 012
4
1

6. BEE-502
Automatic Control Systems
Unit-4, Bode Plot Supplementary Notes
Diploma in Engg.
(Electrical/ Instr. & Control)
5
th
Semester
Instructor: Mohd. Umar Rehman
Date: 7. 11. 2017  
Page 6 of 8
 
Step-3: Different parts of the plot
(i) Constant term: K = 4
(ii) Type of system: 0, this means initial slope is 0 dB/dec, and initial magnitude
is 20 log K = 20 log 4 = 12.04 dB
(iii) Corner frequencies:
c1 c2
rad/s, rad/s
..
ωω== ==
11
25
05 02
Step-4: Draw reference slopes and axes.
Step-5: Draw magnitude plot.
Step-6: Draw phase plot,
tan tan()() (.) (.)Gj Hjφωω ω ω
−−
∠=− =−
11
05 02
Using this expression, tabulate some values of
φw. r. t. ω as follows:
ω 0.1 1 2 5 10 20 ∞
φ −40.08° −37.87° −66.80° − 113.2° −142.1° −160.25° −180°

Step-7: Gain cross-over and phase cross-over frequencies and corresponding values
of phase and magnitude.
()
()
gc gc
pc pc
. rad/s,
rad/s, dBM
ωφω
ωω
==
=∞ =−∞
5 18 115
D

Step-8: Stability margins
Gain Margin,
()
pc
() GM M ω=− =−−∞=+∞00 dB.
Phase Margin,
()
gc
PM φω=+ =−=180 180 115 65
DDDD

The phase plot curve is asymptotic to −180° axis at high frequencies and the gain at
high crossover is negative such that the GM is +
∞. Since, both PM and GM are
positive, therefore, the system is stable. Further the GM is infinite, hence, the system
is inherently stable.

7. BEE-502
Automatic Control Systems
Unit-4, Bode Plot Supplementary Notes
Diploma in Engg.
(Electrical/ Instr. & Control)
5
th
Semester
Instructor: Mohd. Umar Rehman
Date: 7. 11. 2017  
Page 7 of 8
 

Fig. 2. Bode Plot of Example 2

Example 3: Determine the open loop transfer function of a system whose
magnitude Bode plot is shown below.
 
Fig. 2. Bode Plot of Example 3 
Solution: We see that the initial slope of the magnitude Bode plot is 0 dB/dec, so it
is a type - 0 system. Also, the initial magnitude is 20 dB.
Hence,
log logKKK=⇒ =⇒=20 120 10

20

8. BEE-502
Automatic Control Systems
Unit-4, Bode Plot Supplementary Notes
Diploma in Engg.
(Electrical/ Instr. & Control)
5
th
Semester
Instructor: Mohd. Umar Rehman
Date: 7. 11. 2017  
Page 8 of 8
 
There is one corner frequency due to pole at ω
1 = 10 rad/s and another corner
frequency due to zero at ω
2 = 1000 rad/s. Hence, we can write the TF as:
()
()
()
()
()
()
()
s
s
s
KsT
Gs
sT s
s
s
ω
ω
−
−
⎛⎞
⎟⎜
⎟+⎜ ⎟⎜ ⎟⎟⎜+ ⎝⎠
==
⎛⎞+
⎟⎜
⎟+⎜ ⎟⎜ ⎟⎟⎜⎝⎠
+
+
==
++
22
3
1
1
1
10 1
1
1
1
10 1 10
1000
10 10110

 
 
 
 
 
 
 
*** End of Unit - 4 ***