Bode plot

BODE PLOT
Dr.R.Subasri
Professor, Kongu Engineering College,
Perundurai, Erode, Tamilnadu, INDIA
1Dr.R.Subasri,KEC,INDIA

2
Frequency Response Techniques
• Analysing the system on frequency basis
e.g communication systems
• Developed by Nyquist and Bode in 1930
• Older than root locus method given by
Evan in 1948
Dr.R.Subasri,KEC,INDIA

3
– Radio telescopes used for deep space
communication require precise positioning to
function effectively. To achieve a high degree
of precision ,an accurate mathematical model
of antenna dynamics is to used.
– To construct the model, engineers can test
the antenna by vibrating it with sinusoidal
forces of different frequencies and measuring
the vibrations. These measurements form the
basis for the model.

4
Chapter objectives
• Definition of frequency response
• How to plot frequency response
• How to use frequency response to analyse
stability
• How to use frequency response to design
the gain to meet stability specifications.

5
-Advantages
• modelling transfer functions from physical
data.
• Designing compensators to meet steady
state error and transient response
requirements.
• Finding stability of systems

6
Frequency Response
• In the steady state,
sinusoidal inputs to a
linear system
generate sinusoidal
responses of the
same frequency.
• But output differs in
amplitudes and phase
angle from the input
• Differences are
function of frequency.

7
Frequency Response
)(
)(
)(



i
o
M
M
M =
)()()(  io −=
)()(  M
)()(  iiM  )()(  ooM 
)()(  ooM  = )]()([)()(  + iMiM
Magnitude frequency response =
Phase frequency response =
Combination of magnitude and phase frequency responses is Frequency
response
In general Frequency response of a system with transfer function G(s) is
)()(  M
 jSsGjG →= /)()(Dr.R.Subasri,KEC,INDIA

8
Frequency response plots
❖Bode plot(or)Asymptotic
Approximation Plot (or)Logarithmic
plot
❖Polar plot – Nyquist plot
❖Log Magnitude Vs phase angle

9
Logarithmic Frequency Scales
On a logarithmic scale, the variable is multiplied by a
given factor for equal increments of length along the
axis.
Decade Change –log10
Octave change – log2

10
Bode Plot
• Factors of a transfer function
• Constant term K
• - Pure Differentiator-zero at origin -s
-Pure Integrator – pole at origin-1/s
• First order terms
- zero at real axis 1+s
-pole at real axis 1/1+s
• Second order terms
- complex zero s2+2ns+ n
2
-complex pole 1/ s2+2ns+n
2

11
Constant K
consatnt factor G(s) = K
-20
0
20
40
0.1 1 10 100
log w
magnitudeindb
consatnt factor G(s) = K
-90
-45
0
45
90
0.1 1 10 100
log w
phaseangleindeg
Magnitude plot is constant and independent of frequency
Phase plot is constant at zero and independent of
frequency

12
Pure Integrator 1/s
pure integrator-pole at origin G(s)=1/s
-40
-20
0
20
40
0.1 1 10 100
logw
magindb
Pure Integrator G(s) = 1 / s
-180
-135
-90
-45
0
45
90
0.1 1 10 100
log w
phaseangleindeg
starting point of magnitude plot is - 20 log  , where  is
the starting frequency in the plot. The plot starts from that
point and has the slope of - 20 db / dec.
Phase angle is - 90 - a straight line with no slope

13
Pure Differentiator- s
pure differentiator-zero at origin G(s)=s
-40
-20
0
20
40
0.1 1 10 100
logw
magindb
Pure Differentiator G(s) = s
-90
-45
0
45
90
135
180
0.1 1 10 100
log w
phaseangleindeg
starting point of magnitude plot is + 20 log  , where  is
the starting frequency in the plot. The plot starts from
that point and has the slope of + 20 db / dec.
Phase angle is + 90 - a straight line with no slopeDr.R.Subasri,KEC,INDIA

14
Zero at real axis - 1+s
first order term-zero at real axis
G(s)=1+sT
-40
-20
0
20
40
0.1/T 1/T 10/T 100/T
logwT
magindb
First order term-zero at real axis G(s)=1+sT
-90
-45
0
45
90
135
180
0.1/T 1/T 10/T 100/T
log wT
phaseangleindeg
For a constant zero at real axis, in low frequency region, phase angle
is 0 and in high frequency region, phase angle is + 90,in between the
slope is + 45 / dec.
For a constant zero at real axis, find the corner frequency c = 1/T.Up
to c,the magnitude plot is a straight line at 0 db and beyond c, the
plot has a line of slope + 20 db / dec.Dr.R.Subasri,KEC,INDIA

15
Pole at real axis - 1/s+T
For a constant pole at real axis, in low frequency region, phase angle is
0 and in high frequency region, phase angle is -90,in between the
slope is - 45 / dec.
For a constant zero at real axis, find the corner frequency c = T. Up to
c,the magnitude plot is a straight line at 0 db and beyond c, the plot
has a line of slope - 20 db / dec.

16
Complex zero - s2+2ns+n
2
second order term- complex zero
-40
-20
0
20
40
60
80
100
0.1 1 10 100
log w
magindb
second order term- complex zero
-180
-135
-90
-45
0
45
90
135
180
0.1 1 10 100
log w
phaseangleindeg
For a complex zero, find the corner frequency c =n .Up to c,the
magnitude plot is a straight line at +40 log n and beyond c, the plot
has a line of slope + 40 db / dec.
For a complex zero, in low frequency region, phase angle is 0 and in
high frequency region, phase angle is + 180,in between the slope is
+ 90 / dec. Dr.R.Subasri,KEC,INDIA

17
Complex pole - 1/ s2+2ns+n
2
second order term- complex pole
-100
-80
-60
-40
-20
0
20
40
0.1 1 10 100
log w
magindb
second order term- complex pole
-180
-135
-90
-45
0
45
90
135
180
0.1 1 10 100
log w
phaseangleindegFor a complex pole,find the corner frequency c =n .Up to c,the
magnitude plot is a straight line at -40 log n and beyond c, the plot has
a line of slope - 40 db / dec.
For a complex zero, in low frequency region, phase angle is 0 and in
high frequency region, phase angle is - 180,in between the slope is
- 90 / dec.

18
Bode Magnitude plot
• For a constant K, magnitude in db is 20
log K-a straight line with no slope
• For a zero at origin or a pole at origin,(s or
1/s) starting point of magnitude plot is  20
log  , where  is the starting frequency in
the plot. The plot starts from that point and
has the slope of  20 db / dec.

19
Bode Magnitude plot
• For a constant zero at real axis or a pole at
real axis(s+T or 1 / s=T),find the corner
frequency c = T.
Up to c,the magnitude plot is a straight line
at 0 db and beyond c, the plot has a line of
slope  20 db / dec.
• For a complex zero or pole in the form
s2+2ns+n
2, find the corner frequency c =n.
Up to c,the magnitude plot is a straight line at
40 log n and beyond c, the plot has a line of
slope  40 db / dec.

20
Bode Phase plot
• For a constant K, Phase angle is 0- a
straight line with no slope
• For a zero at origin or a pole at origin,(s or
1/s)- Phase angle is  90-a straight line
with no slope
• For a constant zero at real axis or a pole
at real axis(s+T or 1 / s+T), in low
frequency region, phase angle is 0 and in
high frequency region, phase angle is 
90,in between the slope is  45 / dec.

21
Bode Phase plot
• For a complex zero or pole, in low
frequency region, phase angle is 0 and in
high frequency region, phase angle is 
180,in between the slope is  90 / dec.
• Low frequency region - the frequencies up
to one decade below the corner frequency
• High frequency region -all the frequencies
beyond one decade above the corner
frequency

22
Starting point in Bode Plot
Magnitude plot
• Sum of the contributions of
• 1. Constant term K ------------ 20 log K
• 2. Zero or pole at origin---------  20 log  ,
where  is the starting frequency in the plot
• 3.real zero or pole -------------- 20 log T
• 3. Complex zero or pole-------- 40 log n
Phase plot
Contribution from
• zero at origin or a pole at origin,(s or 1/s) which
is  90. Dr.R.Subasri,KEC,INDIA

23
Bode magnitude Plot-Example
G(s) = 2s/(s+1)(s+10)
Factor Corner frequency
c
Magnitude in
db
slope
2 -- 20 log 2 -
s - 20 log  20 db / dec
1/1+s 1 Up to =1,20
log1=0 db & 0
slope
Beyond =1,-
20 db / dec
+20 – 20 =
0 db / dec.
1 /s+10 10 Up to =10, -
20 log 10 db
& 0 slope
Beyond =10,-
20 db / dec
0 – 20 =
-20 db /dec

24
Bode magnitude Plot-Example
Magnitude plot
-40
-20
0
20
40
0.01 0.1 1 10 100
log w
magindb
Let the starting frequency  = 0.1, then starting point of the
plot is:20 log 2 + 20 log 0.1-20 log1-20 log 10= -34 dbDr.R.Subasri,KEC,INDIA

25
40db/dec
20db/dec
-20db/dec
45/dec
90/dec

27
Phase plot –slope contribution
Start of
1/1+s
Start of 1
/1+s/10
End of
1/1+s
End of 1
/1+s/10
 0.1 1 10 100
1/1+s -45 -45 - -
1
/1+s/10
- -45 -45 -
Total
slope
-45 / dec -90/ dec -45/ dec -

28
Phase plot
The system has one zero at origin, plot starts at +90.
Phase plot
-180
-135
-90
-45
0
45
90
135
180
0.01 0.1 1 10 100 1000
log w
phaseangle

29
Plot to transfer function
Magnitude plot
-12
-8
-4
0
1 2.5 10 25 50
log w
magindb
Change in magnitude in db(y axis) = slope x
difference in frequency (x axis)

30
• Step 1:
Change in magnitude in db = -20 x (log 2.5 – log 1)
= -7.95
= -12 + starting db(y axis)
Starting db = - 4.05db
• Step 2:
From the starting point ,there is a slope of -20 db / dec,
hence there is a pole at origin
Starting db = 20 log K -20 log 
At =1,
-4.05 = 20 log K -20 log 1+20 log 2.5+20 log 10-20 log
25
K=0.627 Dr.R.Subasri,KEC,INDIA

31
• Step 3:
First corner frequency is 2.5
Beyond =2.5 , the slope is changed from -20 db / dec
to zero db / dec, hence there should be a zero at 2.5
Factor is s+ 2.5
• Step 4:
Second corner frequency is 10
Beyond =10 , the slope is changed from 0 db / dec to
+20 db / dec, hence there should be a zero at 10
Factor is s+ 10
• Step 3:
Third corner frequency is 25
Beyond =25, the slope is changed from= 20 db / dec to
0 db / dec, hence there should be a pole at 25
Factor is 1/s+25
0.627(s+2.5) (s+10)/s (s+25)Dr.R.Subasri,KEC,INDIA

Thank you
Dr.R.Subasri,KEC,INDIA 33

Bode plot

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