Control system bode plot

Control System
Bode Plot
Nilesh Bhaskarrao Bahadure
nbahadure@gmail.com
https://www.sites.google.com/site/nileshbbahadure/home
June 30, 2021
Nilesh Bhaskarrao Bahadure (Ph.D.) Control System June 30, 2021 1 / 51

Overview I
1 Frequency Response Specifications
Band Width
Cut off Frequency
2 Bode Plot of Transfer Function
Bode gain of factor K1
Poles at origin
Zeros at origin
First Order Poles
First Order Zeros
Second Order Poles
Second Order Zeros
Summary
3 Steps for solving Bode plots
4 Examples on Bode Plot
Example:01
Example:02

Overview II
5 Thank You

Bode Plots

What is Band Width?
What is Band Width?

What is Band Width?
What is Band Width?
It is defined as the range of frequencies over which the system will respond
satisfactorily
Figure : Closed loop system

What is Band Width?
What is Band Width?
It is defined as the range of frequencies over which the system will respond
satisfactorily
Figure : Closed loop system
from Figure 1 the range o to ωb is nothing but band width.

What Cut-off Frequency (wb)?

The frequency at which the magnitude of the closed loop response is 3dB
down from its zero frequency value is called cut-off frequency.
What Cut-off Rate?

What Cut-off Rate?
The slope of resultant magnitude curve near the cut-off frequency is called
cut-off rate.
What Resonant Peak (Mr)?

What Cut-off Rate?
cut-off rate.
It is maximum value of magnitude of closed loop frequency response
What Resonant Frequency (Wr)?

What Cut-off Rate?
cut-off rate.
It is maximum value of magnitude of closed loop frequency response
What Resonant Frequency (Wr)?
The frequency at which resonant peak occur.

Gain Cross Over Frequency ωgc

The frequency at which magnitude of G(jω).H(jω) is unity i.e. 1 is called
gain cross over frequency.
|G(jω).H(jω)| = 1 (1)
in dB,
20log10|G(jω).H(jω)| = 20log101 = 0dB (2)
Phase Cross Over Frequency ωpc

The frequency at which magnitude of G(jω).H(jω) is unity i.e. 1 is called
gain cross over frequency.
|G(jω).H(jω)| = 1 (1)
in dB,
20log10|G(jω).H(jω)| = 20log101 = 0dB (2)
Phase Cross Over Frequency ωpc
The frequency at which phase angle of G(jω).H(jω) is −180o is called
phase cross over frequency ωpc.

Gain Margin (GM)

Gain Margin (GM)
It is defined as the magnitude of the reciprocal of the closed loop transfer
function, evaluate at the phase cross over frequency ωpc which is the
frequency at which phase is −180o. That is,
Gain Margin (GM) =
1
|G(jω).H(jω)|
(3)
where,
∠G(jω).H(jω) = −180o

Figure : Closed loop Gain and Phase Margin

Phase Margin (PM)

Phase Margin (PM)
It is defined as 180o plus the phase angle φ of the open loop transfer
function at unity gain i.e.
φPM = 180 + ∠G(jω).H(jω) (4)

Bode Plot of Transfer Function

The frequency transfer function can be written in a generalized form as
the ratio of polynomials
G(s).H(s) =
N(s)
D(s)
=
a0sm + a1sm−1 + a2sm−2 + ....
b0sn + b1sn−1 + b2sn−2 + ....
(5)

The frequency transfer function can be written in a generalized form as
the ratio of polynomials
G(s).H(s) =
N(s)
D(s)
=
a0sm + a1sm−1 + a2sm−2 + ....
b0sn + b1sn−1 + b2sn−2 + ....
(5)
This must be factorized to obtain a general type,
G(s).H(s) =
N(s)
D(s)
=
K1sg (s + z1)(s + z2)....
sh(s + p1)(s + p2)....
(6)
We convert the above equation in time constant form,
G(s).H(s) =
N(s)
D(s)
=
K1sg (1 + s
z1
)(1 + s
z2
)....
sh(1 + s
p1
)(1 + s
p2
)...
(7)

Note
K1 =
Kz1z2
p1p2

Note
K1 =
Kz1z2
p1p2
we now put s = jω to convert the transfer function to the frequency
domain
G(jω).H(jω) =
K1jωg (1 + jω
z1
)(1 + jω
z2
)....
jωh(1 + jω
p1
)(1 + jω
p2
)...
(8)

Note
K1 =
Kz1z2
p1p2
domain
G(jω).H(jω) =
K1jωg (1 + jω
z1
)(1 + jω
z2
)....
jωh(1 + jω
p1
)(1 + jω
p2
)...
(8)
Here K1 is the gain constant or Bode gain. Equation 8 is the standard
form or Bode form.

Note
K1 =
Kz1z2
p1p2
domain
G(jω).H(jω) =
K1jωg (1 + jω
z1
)(1 + jω
z2
)....
jωh(1 + jω
p1
)(1 + jω
p2
)...
(8)
Here K1 is the gain constant or Bode gain. Equation 8 is the standard
form or Bode form.
The following is a list of possible factors of G(jω).H(jω) from Equation 8

Bode Plot of Transfer Function...
1 Bode gain K1 or constant factor

2 Poles at origin 1
(jω)K of order K

2 Poles at origin 1
(jω)K of order K
3 Zeros at origin (jω)g of order g

2 Poles at origin 1
(jω)K of order K
4 Finite simple poles at 1
1+ jω
p1
or first order poles.

2 Poles at origin 1
(jω)K of order K
1+ jω
p1
5 Finite simple zero at (1 + jω
z1
) or first order zeros.

2 Poles at origin 1
(jω)K of order K
1+ jω
p1
z1
6 Second order poles 1
1+j2ξω−ω2

2 Poles at origin 1
(jω)K of order K
1+ jω
p1
z1
1+j2ξω−ω2
7 Second order zeros (1 + j2ξω − ω2)

2 Poles at origin 1
(jω)K of order K
1+ jω
p1
z1
1+j2ξω−ω2
7 Second order zeros (1 + j2ξω − ω2)
Any transfer function will contain a few of the listed several factors. We
shall now understand how each of these 7 terms can be represented on the
magnitude plot and the phase plot.

Bode Plot of Standard Factors
Magnitude Plot

Magnitude Plot
consider G(s).H(s) = K1, replace s = jω to convert it to frequency
domain, therefore G(jω).K(jω) = K1
For the magnitude in dB we use the standard formula
20log10G(jω).H(jω) = 20log10|K1| = MdB

Magnitude Plot
This is a straight line independent of the input frequency. Hence we
represent the gain factor by a straight line on the Y-axis.

Magnitude Plot
This is a straight line independent of the input frequency. Hence we
represent the gain factor by a straight line on the Y-axis.
Example, Bode magnitude for K1 = 5 will give us a straight line of
20log5 = 13.98dB on the Y-axis.

Bode Plot of Standard Factors...
Phase Plot for Bode gain K1

Since K1 is a constant, it can be written as,
G(jω)H(jω) = K1 + j0

For finding phase angle in degree we have the equation
φ = tan−1 Imaginary Part
Real Part
φ = tan−1 0
K1
= 0
i.e. for positive value of K1, the phase is zero

Real Part
φ = tan−1 0
K1
= 0
Hence a positive value of K1 has no effect on the phase. The phase angle is equal to
−180o
for all ω for negative value of K1

Real Part
φ = tan−1 0
K1
= 0
Hence a positive value of K1 has no effect on the phase. The phase angle is equal to
−180o
for all ω for negative value of K1
Both magnitude and phase are shown below figure (3). They have been drawn
separately for better understanding. In practice magnitude and phase have to be drawn
one below the other.

Figure : Magnitude and Phase plot for gain K1

Poles at Origin or Integral factor ( 1
jω )K
jω
)K
Magnitude Plot

jω )K
jω
)K
Magnitude Plot
Consider G(s).H(s) = 1
sK , replace s = jω to convert it to frequency domain, therefore
G(jω).H(jω) = 1
(jω)K = (jω)−K
Since the magnitude is plotted in decibel dB we have
20log10G(jω).H(jω) = 20log10|(jω)|−K
= −20Klog10(ω)dB

jω )K
jω
)K
Magnitude Plot
G(jω).H(jω) = 1
(jω)K = (jω)−K
= −20Klog10(ω)dB
This is a straight line passing though the point ω = rad/sec and Y = 0dB

jω )K
jω
)K
Magnitude Plot
G(jω).H(jω) = 1
(jω)K = (jω)−K
= −20Klog10(ω)dB
This is because at ω = 1rad/sec, Y = −20Klog10(ω)dB = −20Klog10(0)dB = 0dB
1 If K = 1 straight line passes through (0 dB, ω = 1), at slope −20db/dec

jω )K
jω
)K
Magnitude Plot
G(jω).H(jω) = 1
(jω)K = (jω)−K
= −20Klog10(ω)dB
This is because at ω = 1rad/sec, Y = −20Klog10(ω)dB = −20Klog10(0)dB = 0dB
1 If K = 1 straight line passes through (0 dB, ω = 1), at slope −20db/dec
2 If K = 2 straight line passes through (0 dB, ω = 1), at slope −40db/dec and so
on.

jω
)K
Phase Plot

jω
)K
Phase Plot
G(jω).H(jω) =
1
(jω)K
= (jω)−K

jω
)K
Phase Plot
G(jω).H(jω) =
1
(jω)K
= (jω)−K
For K = 1, ∠G.H(jω) = ∠ 1
jω

jω
)K
Phase Plot
G(jω).H(jω) =
1
(jω)K
= (jω)−K
For K = 1, ∠G.H(jω) = ∠ 1
jω
= ∠
1 + j0
0 + jω
=
tan−1 0
1
tan−1 ω
0
=
0
tan−1∞
=
0
90o
= −90o

jω
)K
Phase Plot
G(jω).H(jω) =
1
(jω)K
= (jω)−K
For K = 1, ∠G.H(jω) = ∠ 1
jω
= ∠
1 + j0
0 + jω
=
tan−1 0
1
tan−1 ω
0
=
0
tan−1∞
=
0
90o
= −90o
∠G.H(jω) = ∠
1
(jω)K
=
1
jω
+
1
jω
+ ... +
1
jω

jω
)K
Phase Plot
G(jω).H(jω) =
1
(jω)K
= (jω)−K
For K = 1, ∠G.H(jω) = ∠ 1
jω
= ∠
1 + j0
0 + jω
=
tan−1 0
1
tan−1 ω
0
=
0
tan−1∞
=
0
90o
= −90o
∠G.H(jω) = ∠
1
(jω)K
=
1
jω
+
1
jω
+ ... +
1
jω
Each term is pole at origin

jω
)K
Phase Plot
G(jω).H(jω) =
1
(jω)K
= (jω)−K
For K = 1, ∠G.H(jω) = ∠ 1
jω
= ∠
1 + j0
0 + jω
=
tan−1 0
1
tan−1 ω
0
=
0
tan−1∞
=
0
90o
= −90o
∠G.H(jω) = ∠
1
(jω)K
=
1
jω
+
1
jω
+ ... +
1
jω
Each term is pole at origin
∠G.H(jω) = ∠
1
(jω)K
=
1 + j0
0 + jω
+
1 + j0
0 + jω
+ ... +
1 + j0
0 + jω

jω
)K
Phase Plot...

jω
)K
Phase Plot...
We now find phase φ for each term
φ =
0
90
+
0
90
+
0
90
+ .... +
0
90
φ = −90o
+ (−90o
) + (−90o
) + .... + (−90o
)

jω
)K
Phase Plot...
φ =
0
90
+
0
90
+
0
90
+ .... +
0
90
φ = −90o
+ (−90o
) + (−90o
) + .... + (−90o
)
Hence each pole at origin contributes a phase of −90o

jω
)K
Phase Plot...
φ =
0
90
+
0
90
+
0
90
+ .... +
0
90
φ = −90o
+ (−90o
) + (−90o
) + .... + (−90o
)
Hence if we have 1 pole at the origin, the phase φ = −90o

jω
)K
Phase Plot...
φ =
0
90
+
0
90
+
0
90
+ .... +
0
90
φ = −90o
+ (−90o
) + (−90o
) + .... + (−90o
)
if we have 2 pole at the origin, the phase φ = −180o

jω
)K
Phase Plot...
φ =
0
90
+
0
90
+
0
90
+ .... +
0
90
φ = −90o
+ (−90o
) + (−90o
) + .... + (−90o
)
and so on

Zeros at Origin or Derivative factor (jω)g
Magnitude Plot

Magnitude Plot
Consider G(s).H(s) = (sg
), replace s = jω to convert it to frequency domain, therefore
G(jω).H(jω) = (jω)g
20log10G(jω).H(jω) = 20log10|(jω)|g
= 20glog10(ω)dB

Magnitude Plot
= 20glog10(ω)dB
This is a plot of slope of 20gdB/dec
1 For g = 1 straight line passes through (0 dB, ω = 1), at slope +20db/dec

Magnitude Plot
= 20glog10(ω)dB
This is a plot of slope of 20gdB/dec
1 For g = 1 straight line passes through (0 dB, ω = 1), at slope +20db/dec
2 For g = 2 straight line passes through (0 dB, ω = 1), at slope +40db/dec and so
on.

Phase Plot

Phase Plot
Consider G(s).H(s) = sg
, replace s = jω to convert it to frequency domain, therefore

Phase Plot

Phase Plot
For g = 1, ∠G.H(jω) = ∠jω

Phase Plot
= ∠0 + jω = tan−1 ω
0
= 90o

Phase Plot
= ∠0 + jω = tan−1 ω
0
= 90o
∠G.H(jω) = ∠(jω)g
= jω + jω + ... + jω

Phase Plot
= ∠0 + jω = tan−1 ω
0
= 90o
= jω + jω + ... + jω
Each term is zero at origin

Phase Plot
= ∠0 + jω = tan−1 ω
0
= 90o
= jω + jω + ... + jω
Each term is zero at origin
∠G.H(jω) = ∠(jω)K
= 0 + jω + 0 + jω + ... + 0 + jω

Phase Plot...

Phase Plot...
φ = 90o
+ 90o
+ 90o
+ .... + 90o

Phase Plot...
φ = 90o
+ 90o
+ 90o
+ .... + 90o
Hence each zero at origin contributes a phase of 90o

Phase Plot...
φ = 90o
+ 90o
+ 90o
+ .... + 90o
Hence if we have 1 zero at the origin, the phase φ = 90o

Phase Plot...
φ = 90o
+ 90o
+ 90o
+ .... + 90o
if we have 2 zero at the origin, the phase φ = 180o

Phase Plot...
φ = 90o
+ 90o
+ 90o
+ .... + 90o
and so on

...
Figure : Magnitude plot

First Order Poles 1
1+jω
p1
First Order Poles 1
1+jω
p1
Magnitude Plot

First Order Poles 1
1+jω
p1
First Order Poles 1
1+jω
p1
Magnitude Plot
1+ s
p1
G(jω).H(jω) = 1
1+ jω
p1

First Order Poles 1
1+jω
p1
First Order Poles 1
1+jω
p1
Magnitude Plot
1+ s
p1
G(jω).H(jω) = 1
1+ jω
p1
20log10G(jω).H(jω) = 20log10|
1
1 + jω
p1
| = 20log10(
1
q
12 + ω2
p2
1
)dB

First Order Poles 1
1+jω
p1
First Order Poles 1
1+jω
p1
Magnitude Plot
1+ s
p1
G(jω).H(jω) = 1
1+ jω
p1
20log10G(jω).H(jω) = 20log10|
1
1 + jω
p1
| = 20log10(
1
q
12 + ω2
p2
1
)dB
20log10G(jω).H(jω) = −20log10
s
12 +
ω2
p2
1
dB

First Order Poles 1
1+jω
p1
Magnitude Plot...

First Order Poles 1
1+jω
p1
Magnitude Plot...
We consider range of ω
1 The graph has a slope of 0 db for ω
p1
<< 1 i.e. ω < p1

First Order Poles 1
1+jω
p1
Magnitude Plot...
p1
<< 1 i.e. ω < p1
2 The graph has a slope of -20 db for ω
p1
>> 1 i.e. ω > p1

First Order Poles 1
1+jω
p1
Magnitude Plot...
p1
<< 1 i.e. ω < p1
2 The graph has a slope of -20 db for ω
p1
>> 1 i.e. ω > p1
3 The change slope occurs at −ω
p1
= 1 i.e. ω = p1

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First Order Poles 1
1+jω
p1
Phase Plot

First Order Poles 1
1+jω
p1
Phase Plot
G(jω).H(jω) =
1
1 + jω
p1

First Order Poles 1
1+jω
p1
Phase Plot
G(jω).H(jω) =
1
1 + jω
p1
G(jω).H(jω) =
1 + j0
1 + jω
p1

First Order Poles 1
1+jω
p1
Phase Plot
G(jω).H(jω) =
1
1 + jω
p1
G(jω).H(jω) =
1 + j0
1 + jω
p1
∠GH(jω) =
tan−1
(0
1
)
tan−1( ω
p1
)

First Order Poles 1
1+jω
p1
Phase Plot
G(jω).H(jω) =
1
1 + jω
p1
G(jω).H(jω) =
1 + j0
1 + jω
p1
∠GH(jω) =
tan−1
(0
1
)
tan−1( ω
p1
)
∠GH(jω) =
0
tan−1( ω
p1
)

First Order Poles 1
1+jω
p1
Phase Plot
G(jω).H(jω) =
1
1 + jω
p1
G(jω).H(jω) =
1 + j0
1 + jω
p1
∠GH(jω) =
tan−1
(0
1
)
tan−1( ω
p1
)
∠GH(jω) =
0
tan−1( ω
p1
)
∠GH(jω) = −tan−1
(
ω
p1
)

First Order Poles 1
1+jω
p1
Phase Plot
G(jω).H(jω) =
1
1 + jω
p1
G(jω).H(jω) =
1 + j0
1 + jω
p1
∠GH(jω) =
tan−1
(0
1
)
tan−1( ω
p1
)
∠GH(jω) =
0
tan−1( ω
p1
)
∠GH(jω) = −tan−1
(
ω
p1
)
As ω becomes larger the phase angle value approaches −90o

First Order Poles 1 + jω
z1
z1
Magnitude Plot

z1
z1
Magnitude Plot
The analysis has the same treatment as first order poles
1 For all ω ≤ z1 we obtain a straight line of -20 dB/dec

z1
z1
Magnitude Plot
2 For all ω > z1 we obtain a straight line of +20 dB/dec

z1
z1
Magnitude Plot
2 For all ω > z1 we obtain a straight line of +20 dB/dec
3 The change of slope occurs at ω = z1

z1
Phase Plot

z1
Phase Plot
G(jω).H(jω) = 1 +
jω
z1

z1
Phase Plot
G(jω).H(jω) = 1 +
jω
z1
∠GH(jω) = tan−1
(
ω
z1
)

z1
Phase Plot
G(jω).H(jω) = 1 +
jω
z1
∠GH(jω) = tan−1
(
ω
z1
)
To put the phase, we take different values of ω starting from 0.1 rad/sec and compute
the angle.

z1
...

Second Order Poles
Second Order Poles
Magnitude Plot

Second Order Poles
Second Order Poles
Magnitude Plot
GH(s) =
ω2
n
s2 + 2ξωns + ω2
n

Second Order Poles
Second Order Poles
Magnitude Plot
GH(s) =
ω2
n
s2 + 2ξωns + ω2
n
Put s = jω

Second Order Poles
Second Order Poles
Magnitude Plot
GH(s) =
ω2
n
s2 + 2ξωns + ω2
n
Put s = jω
GH(jω) =
ω2
n
(jω)2 + 2ξωn(jω) + ω2
n

Second Order Poles
Second Order Poles
Magnitude Plot
GH(s) =
ω2
n
s2 + 2ξωns + ω2
n
Put s = jω
GH(jω) =
ω2
n
(jω)2 + 2ξωn(jω) + ω2
n
GH(jω) =
ω2
n
1 − ( ω
ωn
)2 + 2jξ ω
ωn
M = |GH(jω)| =
1
q
(1 − ω
ωn
)2 + (2ξ ω
ωn
)2

Second Order Poles
Second Order Poles
Magnitude Plot
GH(s) =
ω2
n
s2 + 2ξωns + ω2
n
Put s = jω
GH(jω) =
ω2
n
(jω)2 + 2ξωn(jω) + ω2
n
GH(jω) =
ω2
n
1 − ( ω
ωn
)2 + 2jξ ω
ωn
M = |GH(jω)| =
1
q
(1 − ω
ωn
)2 + (2ξ ω
ωn
)2
As GH(ω) is a second order pole, approximate Bode magnitude plot is shown
1 0 dB line for ω < ωn

Second Order Poles
Second Order Poles
Magnitude Plot
GH(s) =
ω2
n
s2 + 2ξωns + ω2
n
Put s = jω
GH(jω) =
ω2
n
(jω)2 + 2ξωn(jω) + ω2
n
GH(jω) =
ω2
n
1 − ( ω
ωn
)2 + 2jξ ω
ωn
M = |GH(jω)| =
1
q
(1 − ω
ωn
)2 + (2ξ ω
ωn
)2
As GH(ω) is a second order pole, approximate Bode magnitude plot is shown
2 -40 db/dec line for ω > ωn

Second Order Poles
Phase Plot

Second Order Poles
Phase Plot
∠G(jω).H(jω) = φ =
tan−1
(0
1
)
tan−1(
2ξ ω
ωn
1−( ω
ωn
)2 )
φ = −tan−1
"
2ξ ω
ωn
1 − ( ω
ωn
)2
#

Second Order Zeros
Second Order Zeros
Magnitude Plot

Second Order Zeros
Second Order Zeros
Magnitude Plot
GH(jω) = 1 − (
ω
ωn
)2
+ 2jξ
ω
ωn

Second Order Zeros
Second Order Zeros
Magnitude Plot
GH(jω) = 1 − (
ω
ωn
)2
+ 2jξ
ω
ωn
2 +40 db/dec line for ω > ωn

Second Order Zeros
Phase Plot
φ = tan−1
"
2ξ ω
ωn
1 − ( ω
ωn
)2
#

Second Order Zeros...

Summary of Bode Parameters
Type of term G(jω)H(jω) Slope(dB/dec) Magnitude (dB) Phase an-
gle(degrees)

gle(degrees)
Constant
K 0 20logK 0

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90
’n’ zeros at origin
(jω)n 20n 20nlog(ω) 90n

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90
Pole at origin
1
jω
-20 −20log(ω) -90 or 270

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90
Pole at origin
1
jω
-20 −20log(ω) -90 or 270
’n’ poles at origin
1
(jω)n -20n −20nlog(ω) -90n or 270n

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90
Pole at origin
1
jω
-20 −20log(ω) -90 or 270
1
(jω)n -20n −20nlog(ω) -90n or 270n
Simple zero
1 + jωr 20 0 for ω < 1
r
0 for ω < 1
r
20logωr for ω > 1
r
90 for ω > 1
r

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90
Pole at origin
1
jω
-20 −20log(ω) -90 or 270
1
(jω)n -20n −20nlog(ω) -90n or 270n
Simple zero
1 + jωr 20 0 for ω < 1
r
0 for ω < 1
r
20logωr for ω > 1
r
90 for ω > 1
r
Simple pole
1
1+jωr
-20 0 for ω < 1
r
0 for ω < 1
r
−20logωr for ω >
1
r
-90 or 270 for ω > 1
r

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90
Pole at origin
1
jω
-20 −20log(ω) -90 or 270
1
(jω)n -20n −20nlog(ω) -90n or 270n
Simple zero
1 + jωr 20 0 for ω < 1
r
0 for ω < 1
r
20logωr for ω > 1
r
90 for ω > 1
r
Simple pole
1
1+jωr
-20 0 for ω < 1
r
0 for ω < 1
r
1
r
-90 or 270 for ω > 1
r
Second order deriva-
tive term
ω2
n(1 − ω2
ω2
n
+ 2jξω
ωn
) 40 40logωn for ω < ωn 0 for ω < ωn
20log(2ξω2
n) for ω =
ωn
90 for ω = ωn
40logω for ω > ωn 180 for ω > ωn

gle(degrees)
Constant
K 0 20logK 0
Zero at origin
jω 20 20log(ω) 90
Pole at origin
1
jω
-20 −20log(ω) -90 or 270
1
(jω)n -20n −20nlog(ω) -90n or 270n
Simple zero
1 + jωr 20 0 for ω < 1
r
0 for ω < 1
r
20logωr for ω > 1
r
90 for ω > 1
r
Simple pole
1
1+jωr
-20 0 for ω < 1
r
0 for ω < 1
r
1
r
-90 or 270 for ω > 1
r
Second order deriva-
tive term
ω2
n(1 − ω2
ω2
n
+ 2jξω
ωn
) 40 40logωn for ω < ωn 0 for ω < ωn
20log(2ξω2
n) for ω =
ωn
90 for ω = ωn
40logω for ω > ωn 180 for ω > ωn
Second order integral
term
1
ω2
n(1− ω2
ω2
n
+
2jξω
ωn
)
-40 −40logωn for ω <
ωn
0 for ω < ωn
−20log(2ξω2
n) for
ω = ωn
-90 for ω = ωn
−40logω for ω >
ωn
-180 for ω > ωn
Table : Bode Parameters

Steps for solving Bode plots:
1 Bring the given G(s).H(s) transfer function in to standard time
constant form

constant form
2 Replace all s by jω to get the frequency domain transfer function

constant form
3 Prepare a list of standard factor present in the given transfer function
and prepare tables for factors

constant form
4 Prepare the table for magnitude plot

constant form
5 Choose suitable Y and X scales for magnitude plot. Draw the lines of
corresponding factors from starting point to corner frequency (end
point)

constant form
point)
6 Write the phase angle equation corresponding to the phase of each
factor present

constant form
point)
factor present
7 Prepare the phase angle table & obtain the resultant phase angle by
actual calculation

constant form
point)
factor present
7 Prepare the phase angle table & obtain the resultant phase angle by
actual calculation
8 Depending upon largest frequency term present, choose the starting
point of log scale. Plot the points as per phase angle table φ & draw
the smooth curve to obtain the phase angle plot.

Steps for solving Bode plots:...
Note:

Steps for solving Bode plots:...
Note:
Consider the starting frequency of the Bode plot as 1
10th of the minimum
corner frequency or 0.1 rad/sec whichever is smaller value and draw the
Bode plot upto 10 times maximum corner frequency.

Example: 01
Example:
A unity feedback control system has
G(s) =
100
s(s + 0.5)(s + 10)
Draw the Bode plot and determine GM, PM, ωgc, and ωpc

Solution:

Solution:
Step 1: Bring equation in standard form:

Solution:
G(s).H(s) =
100
0.5 × 10 × s(1 + s
0.5
)(1 + s
10
)
=
20
s(1 + s
0.5
)(1 + s
10
)

Solution:
G(s).H(s) =
100
0.5 × 10 × s(1 + s
0.5
)(1 + s
10
)
=
20
s(1 + s
0.5
)(1 + s
10
)
Step 2: Convert to frequency domain by replacing s = jω

Solution:
G(s).H(s) =
100
0.5 × 10 × s(1 + s
0.5
)(1 + s
10
)
=
20
s(1 + s
0.5
)(1 + s
10
)
G(jω).H(jω) =
20
jω(1 + jω
0.5
)(1 + jω
10
)

Solution:
G(s).H(s) =
100
0.5 × 10 × s(1 + s
0.5
)(1 + s
10
)
=
20
s(1 + s
0.5
)(1 + s
10
)
G(jω).H(jω) =
20
jω(1 + jω
0.5
)(1 + jω
10
)
Step 3: The following factors are present:
1 Constant K = 20

Solution:
G(s).H(s) =
100
0.5 × 10 × s(1 + s
0.5
)(1 + s
10
)
=
20
s(1 + s
0.5
)(1 + s
10
)
G(jω).H(jω) =
20
jω(1 + jω
0.5
)(1 + jω
10
)
1 Constant K = 20
2 Poles at origin 1
jω

Solution:
G(s).H(s) =
100
0.5 × 10 × s(1 + s
0.5
)(1 + s
10
)
=
20
s(1 + s
0.5
)(1 + s
10
)
G(jω).H(jω) =
20
jω(1 + jω
0.5
)(1 + jω
10
)
1 Constant K = 20
2 Poles at origin 1
jω
3 First order pole 1
1+ jω
0.5

Solution:
G(s).H(s) =
100
0.5 × 10 × s(1 + s
0.5
)(1 + s
10
)
=
20
s(1 + s
0.5
)(1 + s
10
)
G(jω).H(jω) =
20
jω(1 + jω
0.5
)(1 + jω
10
)
1 Constant K = 20
2 Poles at origin 1
jω
1+ jω
0.5
1+ jω
10

Solution:
Sr. No. Factors Magnitude Curve Phase Curve

Solution:
1 K = 20 straight line φ = 0
20log1020 = 26.02dB

Solution:
20log1020 = 26.02dB
2 1
jω
slope of -20dB/dec passing though φ = −90
0 dB at ω = 1rad/sec

Solution:
20log1020 = 26.02dB
2 1
jω
3 1
1+ jω
0.5
Line slopes are φ = −tan−1 ω
0.5
1. 0 dB for ω < 0.5
2. -20 dB for ω ≥ 0.5

Solution:
20log1020 = 26.02dB
2 1
jω
3 1
1+ jω
0.5
0.5
1. 0 dB for ω < 0.5
2. -20 dB for ω ≥ 0.5
4 1
1+ jω
10
10
1. 0 dB for ω < 10
2. -20 dB for ω ≥ 10

Solution:

Solution:
Step 4: Magnitude plot table:
Sr. No. Factors Resultant slope S.P. (ω) E.P. (ω)

Solution:
1 K = 20 straight line of 26.02 dB 0.1 ∞

Solution:
2 1
jω
-20 dB/dec 0.1 0.5

Solution:
2 1
jω
-20 dB/dec 0.1 0.5
3 1
1+ jω
0.5
-20 + (-20) = -40 dB/dec 0.5 10

Solution:
2 1
jω
-20 dB/dec 0.1 0.5
3 1
1+ jω
0.5
-20 + (-20) = -40 dB/dec 0.5 10
4 1
1+ jω
10
-40 + (-20) = -60 dB/dec 10 ∞

Solution:

Solution:
Step 5: Phase angle:

Solution:
Step 5: Phase angle: φresultant (ω) = −90o
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2
3 −90o
-80.53 -16.6 -187.13

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2
3 −90o
-80.53 -16.6 -187.13
4 −90o
-82.87 -21.8 -194.67

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2
3 −90o
-80.53 -16.6 -187.13
4 −90o
-82.87 -21.8 -194.67
5 −90o
-84.2 -26.2 -200.7

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2
3 −90o
-80.53 -16.6 -187.13
4 −90o
-82.87 -21.8 -194.67
5 −90o
-84.2 -26.2 -200.7
10 −90o
-87.1 -45 -222.1

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2
3 −90o
-80.53 -16.6 -187.13
4 −90o
-82.87 -21.8 -194.67
5 −90o
-84.2 -26.2 -200.7
10 −90o
-87.1 -45 -222.1
100 −90o
-89.7 -84.2 -263.9

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2
3 −90o
-80.53 -16.6 -187.13
4 −90o
-82.87 -21.8 -194.67
5 −90o
-84.2 -26.2 -200.7
10 −90o
-87.1 -45 -222.1
100 −90o
-89.7 -84.2 -263.9
500 −90o
-89.94 -88.8 -268.7

Solution:
− tan−1
( ω
0.5
) − tan−1
( ω
10
)
ω 1
jω
−tan−1
( ω
0.5
) −tan−1
( ω
10
) φresultant (ω)
0.1 −90o
-11.3 -0.5 -101.8
0.5 −90o
-45 -2.8 -137.8
1 −90o
-63.4 -5.7 -159.1
2 −90o
-75.96 -11.3 -177.2
3 −90o
-80.53 -16.6 -187.13
4 −90o
-82.87 -21.8 -194.67
5 −90o
-84.2 -26.2 -200.7
10 −90o
-87.1 -45 -222.1
100 −90o
-89.7 -84.2 -263.9
500 −90o
-89.94 -88.8 -268.7
1000 −90o
-89.97 -89.4 -289.3

Example: 02
Example:
A feedback control system has
G(s)H(s) =
100(s + 3)
s(s + 1)(s + 5)
Draw the Bode plot and comments on stability

Solution:

Solution:
G(s).H(s) =
100 × 3(1 + s
3
)
5 × s(1 + s)(1 + s
5
)
=
60(1 + s
3
)
s(1 + s)(1 + s
5
)

Solution:
G(s).H(s) =
100 × 3(1 + s
3
)
5 × s(1 + s)(1 + s
5
)
=
60(1 + s
3
)
s(1 + s)(1 + s
5
)
G(jω).H(jω) =
60(1 + jω
3
)
jω(1 + jω)(1 + jω
5
)

Solution:
G(s).H(s) =
100 × 3(1 + s
3
)
5 × s(1 + s)(1 + s
5
)
=
60(1 + s
3
)
s(1 + s)(1 + s
5
)
G(jω).H(jω) =
60(1 + jω
3
)
jω(1 + jω)(1 + jω
5
)
1 Constant K = 60

Solution:
G(s).H(s) =
100 × 3(1 + s
3
)
5 × s(1 + s)(1 + s
5
)
=
60(1 + s
3
)
s(1 + s)(1 + s
5
)
G(jω).H(jω) =
60(1 + jω
3
)
jω(1 + jω)(1 + jω
5
)
1 Constant K = 60
2 Poles at origin 1
jω

Solution:
G(s).H(s) =
100 × 3(1 + s
3
)
5 × s(1 + s)(1 + s
5
)
=
60(1 + s
3
)
s(1 + s)(1 + s
5
)
G(jω).H(jω) =
60(1 + jω
3
)
jω(1 + jω)(1 + jω
5
)
1 Constant K = 60
2 Poles at origin 1
jω
1+jω

Solution:
G(s).H(s) =
100 × 3(1 + s
3
)
5 × s(1 + s)(1 + s
5
)
=
60(1 + s
3
)
s(1 + s)(1 + s
5
)
G(jω).H(jω) =
60(1 + jω
3
)
jω(1 + jω)(1 + jω
5
)
1 Constant K = 60
2 Poles at origin 1
jω
1+jω
1+ jω
5

Solution:
G(s).H(s) =
100 × 3(1 + s
3
)
5 × s(1 + s)(1 + s
5
)
=
60(1 + s
3
)
s(1 + s)(1 + s
5
)
G(jω).H(jω) =
60(1 + jω
3
)
jω(1 + jω)(1 + jω
5
)
1 Constant K = 60
2 Poles at origin 1
jω
1+jω
1+ jω
5
5 First order zero 1 + jω
3

Solution:

Solution:
20log1060 = 35.56dB

Solution:
20log1060 = 35.56dB
2 1
jω

Solution:
20log1060 = 35.56dB
2 1
jω
3 1
1+jω
Line slopes are φ = −tan−1
(ω)
1. 0 dB for ω < 1
2. -20 dB for ω ≥ 1

Solution:
20log1060 = 35.56dB
2 1
jω
3 1
1+jω
(ω)
1. 0 dB for ω < 1
2. -20 dB for ω ≥ 1
4 1
1+ jω
5
5
1. 0 dB for ω < 5
2. -20 dB for ω ≥ 5

Solution:
20log1060 = 35.56dB
2 1
jω
3 1
1+jω
(ω)
1. 0 dB for ω < 1
2. -20 dB for ω ≥ 1
4 1
1+ jω
5
5
1. 0 dB for ω < 5
2. -20 dB for ω ≥ 5
5 1 + jω
3
Line slopes are φ = tan−1 ω
3
1. 0 dB for ω < 3
2. +20 dB for ω ≥ 3

Solution:

Solution:
2 1
jω
-20 dB/dec 0.1 1

Solution:
2 1
jω
-20 dB/dec 0.1 1
3 1
1+jω
-20 + (-20) = -40 dB/dec 1 3

Solution:
2 1
jω
-20 dB/dec 0.1 1
3 1
1+jω
-20 + (-20) = -40 dB/dec 1 3
4 1 + jω
3
-40 + (+20) = -20 dB/dec 3 5

Solution:
2 1
jω
-20 dB/dec 0.1 1
3 1
1+jω
-20 + (-20) = -40 dB/dec 1 3
4 1 + jω
3
-40 + (+20) = -20 dB/dec 3 5
5 1
1+ jω
3
-20 + (-20) = -40 dB/dec 5 ∞

Solution:

Solution:
Step 5: Phase angle:

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9
0.5 −90o
-26.5 -5.7 9.46 -112.9

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9
0.5 −90o
-26.5 -5.7 9.46 -112.9
1 −90o
-45 -11.3 18.4 -127.9

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9
0.5 −90o
-26.5 -5.7 9.46 -112.9
1 −90o
-45 -11.3 18.4 -127.9
5 −90o
-78.69 -45 59 -154.79

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9
0.5 −90o
-26.5 -5.7 9.46 -112.9
1 −90o
-45 -11.3 18.4 -127.9
5 −90o
-78.69 -45 59 -154.79
10 −90o
-84.3 -63.4 73.3 -164.4

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9
0.5 −90o
-26.5 -5.7 9.46 -112.9
1 −90o
-45 -11.3 18.4 -127.9
5 −90o
-78.69 -45 59 -154.79
10 −90o
-84.3 -63.4 73.3 -164.4
100 −90o
-89.4 -87.1 88.3 -178.2

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9
0.5 −90o
-26.5 -5.7 9.46 -112.9
1 −90o
-45 -11.3 18.4 -127.9
5 −90o
-78.69 -45 59 -154.79
10 −90o
-84.3 -63.4 73.3 -164.4
100 −90o
-89.4 -87.1 88.3 -178.2
500 −90o
-89.8 -89.4 89.6 -179.6

Solution:
− tan−1
(ω) − tan−1
(ω
5
) + tan−1
(ω
3
)
ω 1
jω
−tan−1
(ω) −tan−1
(ω
5
) +tan−1
(ω
3
) φresultant (ω)
0.1 −90o
-5.7 -1.1 1.9 -94.9
0.5 −90o
-26.5 -5.7 9.46 -112.9
1 −90o
-45 -11.3 18.4 -127.9
5 −90o
-78.69 -45 59 -154.79
10 −90o
-84.3 -63.4 73.3 -164.4
100 −90o
-89.4 -87.1 88.3 -178.2
500 −90o
-89.8 -89.4 89.6 -179.6
1000 −90o
-89.9 -89.71 89.8 -179.8

Thank you
Please send your feedback at nbahadure@gmail.com

Thank you
Please send your feedback at nbahadure@gmail.com
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Control system bode plot

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