VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
1
SYSTEM ANALYSIS USING LOGARITHMIC PLOTS (BODE PLOTS)
Frequency Response:
The term frequency response refers to the steady state response of a system
subjected to a sinusoidal input of fixed amplitude, but with the frequency
varied over some range. This concept is illustrated in the figure shown, in
which a linear system is forced by an input a sin t.
b (sinωt+Φ)
a sinωt
Input
a
b
0
Φ
Output
Frequency Response
The output is b Sin (t + ). The expression for the output is given in general
form in which +  indicates the addition of an angle which may be either
positive or negative.
The input and output wave forms are as shown and of interest is the ratio of
the amplitudes and the phase angle , both of which are functions of
frequency.
This ratio, is commonly called as “Magnitude Ratio” and will be designated
as M().
The phase angle (or phase Shift) will be denoted by ()
The main concern is . . . . ?
“Determining the frequency response information analytically although
such data can be obtained experimentally if the system exists”
VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
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Procedure to obtain frequency response data analytically.
Step.1: Obtain transfer function for the element or combination of elements
o( S )
 F (S )
involved i.e.,
I (S )
Step.2: In the transfer function replace S, by Jw
(Justification . . . . ?)
Step.3: For various values of frequency , determine the magnitude ratio ‘M’
and the phase angle .
Step.4: Plot the results form the above step 3, in polar co-ordinate or
rectangular coordinator form. These plots are not only convenient
means for presenting frequency response data, but are also the
basis for analytical and design methods.
Frequency response data can be presented in rectangular co-ordinate form.
The most useful representation is a “logarithmic” plot which consists of two
graphs one consists of logarithm of | G(j) | and the other phase angle of
G(Jw) both plotted against frequency in “logarithmic scale”.
These plots are called “BODE PLOTS”
The standard representation of the logarithmic magnitude of G(Jw)
20 log10 |G(Jw)| decibels.
is
Here curves are drawn on Semi-log paper, using log scale for frequency and
the linear scale for either magnitude (db) or phase angle (degrees).
The main advantage of using the log-plot is that multiplication of
magnitudes can be converted into additions.
Basic Factors:
Four basic factors,
i) Constant (Gain K)
ii) Integral & derivative factors,  j 1 (Zeros or poles at origin)
iii) First order factor 1  jT 1 (Simple zeros or poles)
iv) Quadratic factors,
2

 jw   jw  
1  2      

 n   n  
1
VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
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We can draw log plots of these basic factions, then it is possible to use these
basic plots to construct a composite log plot of any general factors of G (jw)
H(jw) by sketching curves for each factor and adding individual curves
graphically because adding log of gains corresponds to multiplying them
together
Now, the standard representation of logarithmic magnitude of
G(jw) = 20 log10 | G(j) |
Unit, = decibels (db)
 Mag = 20 log10 | G(jw) | db
1
3 5 7 1
+ 80
+ 60
+ 40
(M) + 20
ZERO
db
- 20
- 40
- 60
0.1
1.0
10
2
 10
103
104
Seme – Log Graph Sheet
Gain K :
K (indb) = 20 log10 K db
i.e A log-mag curve for a constant gain K, is a horizontal straight line at the
magnitude of 20 log10 K db
The phase angle of K is ZERO
 |M| = 20 log10 K
 = ZERO
VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
1
3 5 7 1
+ 80
a. Gain K K 1
(i) (K)-1
+ 60
K = 102
+ 40
(K)+1
(M) + 20
K = 10
ZERO
K = 10
- 20
db
(K)-1
K = 102
- 40
- 60
0.1
1.0
10
103
2
 10
104
(b) Integral or derivative factor
i)
 jw1,
1
= -20 log10
jw
 |M| = 20 log

1
10
100
1000
1
jw
= -20 log10  db
M
Zero
- 20
- 40
- 60
 Slope = -20 db / decade
 = - tan-  = - 90
1
3 5 7 1
+ 80
+ 60
b. Integral or derivative factor
 j 1
(i) (J)-1
+ 40
(M) + 20
ZERO
db
- 20
(J)-1
-20 db/decade,
Slope,
- 40
- 60
0.1
1.0
10
2
 10
103
104
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
ii)
 jw1
 M = 20 log |jw|
= 20 log10 |w| db

1
10
102
103
M
Zero
+ 20
+ 40
+ 60
 Slope = + 20 db / decade
 = + tan  = + 90
1
3 5 7 1
+ 80
+ 60
b. Integral or derivative factor  j 1
(ii) (J)+1
+ 40
(J)+1
(M) + 20
+20 db/decade,
Slope,
ZERO
db
- 20
- 40
- 60
0.1
1.0
10
2
 10
103
104
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
What about
 j  n
1
( j ) n
(iii) (a) (j)-n =
M = 20 log
1
( j ) n
= -20 log (j)n
= -20 x n log (j)
 Slope = - 20 x n db / decade
Similarly  = - 90 x n
(b) (j)+n
 M = 20 log (j)n
= 20 x n log (j)
 Slope = 20 x n db / decade
Similarly  = + 90 x n
1
3 5 7 1
+ 80
+ 60
b (iii)  j  n
+ 40
+ 60 db/de (n=3)
+ 40 db/de (n=2)
 j  n
(M) + 20
ZERO
db
 j  n
- 20
- 40
- 40 db/de (n=2)
- 60
- 60 db/de (n=3)
2
10
103
 10
0.1
1.0
104
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
+ 160
b. (iii)  j  n
+ 120
n=4
+ 80
n=3 j 
+ 40
n=2
n
ZERO
- 40
db
n=2
 j  n
- 80
- 120
0.01
0.1
1.0
(c) First order factors : (1 + jwT)
(i) (1 + jwT)-1 =
M = 20 log10=
1
(1  jwT )
1
(1  jwT )
= - 20 log10 (1 + jwT)
= - 20 log10
12   2T 2 db
For low frequencies , i.e.,  < <
i.e.  T << 1
Then M = -20log10 1= 0, (ZERO)
If  t >>
1
i.e., T >> 1
T
Then M = -20 log10 T db
1
T
1
n=3
n=4
102
 10
103
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
Now

1
T
10
T
10
T
8
M
Zero
- 20
 Slope = - 20 db / decade
- 40
 = - tan-1 T


0
0
1
- tan-1 1 = -45
T

= - tan-1  = - 90
+ 80
C First order Factors 1  jT 
1
+ 60
+ 40
(i) 1  jT 1
T = 0.1
(M) + 20
CF =
ZERO
db
- 20
1
T
= 10
-20 db/de
- 40
1
 
T 
- 60
0.1
1.0
10
 10 
 
T 
2
 10
 102 


T


103
104
VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
(ii)
(1+ J T)
+1
| M | = 20 log | 1 + J T |
= 20 log10
db
12   2T 2
Now, for low frequencies, i.e  <<
1
T
i.e, T << 1
|M| = 20 log10 = Zero
If T >> 1, Then,
|M| = 20 log10 T db

 If
(M)
1
T
10
T
Zero
+ 20 db
10 2
T
+ 40 db
103
T
+ 60 db
 = - tan-1

 Slope = +20 db/decade
(T)

0
0
1
T
tan-1 1 = 45

tan-1  = 90
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
+ 80
C First order Factors 1  jT 
1
+ 60
(ii) 1  jT 1
+ 40
(M) + 20
+20 db/de
T = 0.1
CF = 10
ZERO
- 20
db
1
 
T 
- 40
 10 
 
T 
 102 


 T 
- 60
0.1
+ 80
+ 60
+ 40
1.0
10
2
 10
ZERO
- 20
- 40
- 60
0.1
104
C First order Factors1  jT 1
3
1
(iii)
2 1  jT 
1
(M) + 20
db
103
1.0
10
2
 10
T CF
1 0.2 5.0
2 0.4 2.5
3 0.8 1.25
1
1
2 1  jT 
3
103
104
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
(iii) for (1+jT) n
We can show that in case of (1+jT)-n
db
decade
Slope = - 20 x n
 for n = 2 Slope = -40 db/de
=3
:
10
= -60 db/de
:
= -200 db/de
again,  = - tan-1 T x n
+ 80
n=4
C First order
1
Factors 1  jT 
+ 60
+ 40
n=3
1  jT  n
(iv)
n=2
(M) + 20
ZERO
db
- 20
n=2 1  jT  n
- 40
n=3
- 60
0.1
1.0
10
In case of (1+jT)+n
Slope = +20 x n db/decade
and  = tan-1 T x n
2
 10
103
104
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
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HOW TO GET FIRST ORDER FACTORS:
Consider the system shown in figure the differential equation is
x
C
dy
 ky  kxt 
dt
K
.
c .
 y  k  xt    y  k  x(t )
k
y
C
Taking laplace, and rearranging the terms, we get transfer function.
 TF 
Where  
C
k
Y (S )
1
1


X ( S ) 1  S 1  TS
= Time constant = T
Now making substitution, S=J and assuming some suitable value for
 = 0.1 see, yields
1
Y ( j )
=
1  j
x ( j )
=
1
1  0 .1 j 
Now a table can be constructed giving the magnitude ratio M(w) and the
phase angle  (w)
To illustrate this, let w = 10, as an example,

Y
x
=
=
1
1  j1
1
1.41
= 0.71
=
45
45
1
1  12
2
VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
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Based on the above calculation for different values of , = 0, 2,5,10 . . . . . . .
 can be tabulated as shown below

(rad/sec)
M()
 () (degrees)
Zero
O
1.0
Zero
2
0.98
-11.3
5
0.89
-26.6
10
0.71
-45
20
0.45
-63.4
40
0.24
-76.0

0.00
-90

(rad/sec)
M()
M=20log10M
(decibels)
O
1.0
Zero
2
0.98
-0.175
5
0.89
-1.01
10
0.71
-2.97
20
0.45
-6.93
40
0.24
-12.39

0.00

HOW TO GET QUADRATIC FACTORS:
Consider k-m-c, system as shown the differential equation, is
..
x
and transfer function is
K
Y (s)
k

2
X ( s ) Ms  CS  k
M
C
.
M y  cy  k y  kx(t )
y

1
M 2 C
s  S 1
k
k

1
s
2

S 1
2
n n
2
VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
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To form a tabular column, as done earlier,
let us assume, natural frequency = 10 rad/sec and a damping ratio of 0.5,
and making the substitution s = j,yields,
1
Y ( j )
=
2
0.01( j )  0.1 j  1
x ( j )
=
1
(1  0.01 2 )  j.0.1
Now, a table is made giving the magnitude ratio and phase angle for various
values of , let =5, as an example, then
Y
x
=
1
(0  1.25)  j 0.5
=
1
0 .9
-33.7
= 1.11 -33.7

M()
 () (degrees)
O
1.0
Zero
2
1.02
-11.8
5
1.11
-33.7
8
1.14
-65.8
10
1.0
-90
12
0.78
-110.1
15
0.51
-129.8
20
0.28
-146.3
40
0.06
-165.1
70
0.02
-171.7

0.00
-180.0
(rad/sec)
VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering

(rad/sec)
M()
M()
(decibels)
O
1.0
Zero
2
1.02
0.172
5
1.11
0.906
8
1.14
1.138
10
1.0
Zero
12
0.78
-2.15
15
0.51
-5.84
20
0.28
-11.5
40
0.06
-24.43
70
0.02
-33.97

0.00

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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
2

 j   j  
  
 
1  2 



 n   n  
(d) Quadratic factors:
(i)
2

 j   j  
  
 
1  2 

 n   n  
= M = 20 log10
1
1
2

 j   j  
  
 
1  2 



 n   n  
2


 
 
= -20 log10 1  2  j    j  
   

 n   n  
  2  
 
= -20 log10 1   2   j  2  

n 

n 

1/ 2
2
2
  2  
  
 
= -20 log10 1  2    2
 n   n  


If  <<
If  >>
n , then
(M) = - 20 log10 1 = ZERO db
n , then,
 

(M) = - 20 log10 

 n
2
 
 db
= - 40 log10 
 n 
n ,
If  =
n
= 102  n
= 10
M = -40 log1 = ZERO
= -40 log10 = -40 db
= -40 log102 = -80 db
 Slope = -40 db/decade
1
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
Phase angle:
 2 / n 
2
1   / n  
 = - tan-1 
The phase angle is a function of both  and 
 If  = 0,
 = - tan-1 0 = ZERO
2
0

= -tan-1
2
= n
= -tan-1
=
1
= -tan  = - 90
= - 180
3 5 7 1
+ 80
(d) Quadratic factors
+ 60
+ 40 1
(i) [
2

j  j  
 
1  2
 
n  n  


] -1
3 5 7 1
1
+ 80
(M) + 20
(C) Quadratic factors
+ 60
ZERO
(ii) [
] -1
C.F = 1.0
+ -40
20
C.F = 1.0
db
Slope:+40 db/de
(M) +-20
40
- 60
ZERO
db
Slope:-40 db/de
0.1
1.0
0.1
1.0
10
2
 10
103
104
- 20
- 40
- 60
10
102
103
104
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
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VTU – Edusat Programme – 5th Semester Mechanical - ME 55 – Control Engineering
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