INC341
Frequency Response Method
Lecture 11
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System has 20%
overshoot
Design controller to decrease peak time to
2/3 and steady-state error to 0
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3 expressions of sinusoidal signal
Starts from a sinusoidal signal, A cos(t )  B sin(t ) , which can be
rewritten as


A2  B 2 cos t  tan1 (B / A)
• Polar form (showing magnitude and phase shift): M i i
M i  A2  B 2
i   tan1 ( B / A)
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2 expressions of sinusoidal signal (cont.)
• Rectangular form (complex number): A  jB
cos(t   )  cos(t ) cos( )  sin(t ) sin( )
M i cos(t  i )  M i cos(i ) cos(t )  M i sin(i ) sin(t )


A
B
j
• Euler’s formula (exponential): M i e i
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Frequency response of system
• Magnitude response: M ( )
– ratio of output mag. To input mag.
• Phase response:  ( )
– difference in output phase angle and input phase
angle
• Frequency response: M ( ) ( )
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Question
What is the output from a known system
fed by a sinusoidal command?
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Basic property of frequency Response
‘mechanical system’
input = force
output = distance
Answer:
sinusoidal input
gives sinusoidal
output with same
damped frequency
shifted by  (,)
mag. expanded by
M ( )
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The HP 35670A
Dynamic Signal
Analyzer obtains
frequency response
data from a physical
system.
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Finding frequency response from
differential equation
• Get transfer function T (s)
• Set s  j
• Write M ( )  T ( s)
 ( )  T ( s)
• Then the output is composed of
M o ( )  M ( ) M i ( )
o ( )   ( )  i ( )
M o ( )o ( )  M i ( ) M ( )[i ( )   ( )]
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Finding frequency response from
transfer function
1
Substitute
( s  2)
1
1
G ( j ) 

( j  2) (2  j )
G ( s) 
ω = 0, G = 0.5
ω = 2, G = 0.25 – j0.25
ω = 5, G = 0.07 - 0.17i
ω = 10, G =0.019 - j0.096
ω = ∞, G = 0
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s with j
0.5∟0
0.35 ∟-45
0.19 ∟-68.2
0.01∟-78.7
0 ∟-90
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What’s next?
After getting magnitude and phase of the
system, we need to plot them but how???
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Types of frequency response plots
• Polar plot (Nyquist plot): real and
imaginary part of open-loop system.
• Bode plot: magnitude and phase of openloop system (begin with this one!!).
• Nichols chart: magnitude and phase of
open-loop system in a different manner
(not covered in the class).
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Polar plot of
1
G ( s) 
( s  2)
so called ‘Nyquist plot’
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Bode plot
Magnitude
Phase
Note: log frequency and log magnitude
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Bode plot
• 1st order or higher terms that can be written as a
product of 1st order terms
– 4 cases:
1
1
( s  a ),
, s,
(s  a)
s
• 2nd order terms
– 2 cases:
s
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2
2
 2 n s   n ,
1
s 2  2 n s   n2
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First order terms
G( s)  ( s  a)
Case I: one zero at -a
G ( j )  ( j  a )  a ( j
ω= 0
ω >> a
ω= a
INC 341
G ( j )  a
20 log M  20 log a
G ( j )  a ( j


a
 1)
phase = 0
)  j  90 phase = 90
a
20 log M  20 log
G( j )  ( ja  a)
phase = 45
20log M  20log 2a  20log a  3
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Asymptotes (approximation)
Break frequency
 2  101
= freq. at which mag. has changed by 3 db
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3 dB at break frequency
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First order terms
G (s) 
Case II: one pole at -a
1
( s  a)
1
1
G( j ) 

( j  a) a( j  1)
a
G( j )  1 / a
ω= 0
20 log M  20 log(1 / a)
ω >> a
G ( j ) 
1
a( j


)
1
1
   90
j 
phase = 0
phase = -90
a
20 log M  20 log
phase = -45
ω = a 20log M  20log(1/ 2a)  20log(1/ a)  3
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First order terms
Case III: one zero at 0
G(s) = s
Magnitude depends directly on jω
(straight line up passing through zero dB at ω=1)
Phase = 90 (constant)
Case IV: one pole at 0
G(s) = 1/s
Magnitude depends directly on jω
(straight line down passing through zero dB at ω=1)
Phase = - 90 (constant)
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G(s) = s
G(s) = s+a
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G(s) = 1/s
G(s) = 1/(s+a)
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What about
G( s) 
1
( s  2)(s  3)
???
plot each term separately and sum
them up
• log magnitude (s+2) added with log
magnitude (s+3)
It’s convenient for calculation to plot
magnitude in log scale!!!
• phase (s+2) added with phase (s+3)
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Bode Plots
Find magnitude and phase of each term
and sum them up!!!
G ( s) 
G( s) 
K ( s  z1 )(s  z 2 )
s m ( s  p1 )(s  p2 )
K ( s  z1 ) ( s  z 2 )
mag(num)-mag(den)
phase(num)-phase(den)
s m ( s  p1 ) ( s  p2 )
20 logG ( s)  20 log K  20 log ( s  z1 )  20 log ( s  z 2 )
 20 log s m  20 log ( s  p1 )  20 log ( s  p2 )
G ( s)  K  ( s  z1 )  ( s  z 2 )  s m  ( s  p1 )  ( s  p2 )
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Example
( s  3)
sketch bode plot of G ( s ) 
s( s  1)(s  2)
break frequency at 1,2,3
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Slope at each break frequency for magnitude plot
Frequency
small
1
2
3
s
-20
-20
-20
-20
1/(s+1)
0
-20
-20
-20
1/(s+2)
0
0
-20
-20
(s+3)
0
0
0
20
Total Slope
-20
-40
-60
-40
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Magnitude Plot
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Slope at each point for phase plot
Frequency
small
0.1
0.2
0.3
10
20
30
s
0
0
0
0
0
0
0
1/(s+1)
0
-45
-45
-45
0
0
0
1/(s+2)
0
0
-45
-45
-45
0
0
(s+3)
0
0
0
45
45
45
0
Total Slope
0
-45
-90
-45
0
45
0
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Phase Plot
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nd
2
order terms
2
2
Case I: 2 zeros G(s)  s  2n s  n
set s = jω
G( j)  ( j) 2  2 n ( j)  n2  (n2   2 )  j (2 n)
G(s)     0
2
n
Small ω = 0
large ω = ∞

G(s)  ( j)2   2   2180
log magnitude:
INC 341
2
n
20log 2  40log
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Second-order bode plot
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Magnitude plot of
G( s )  s
2
2
 2 n s  n
G( j)  (n2   2 )  j (2 n)
  n
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G( j)  2 n2
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Phase plot of
G( s )  s
2
2
 2 n s  n
G( j)  (n2   2 )  j (2 n)
  n
INC 341
2
(
2

n)
G ( j )  tan 1
 90
0
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Magnitude plot of
Case II: 2 poles
  n
INC 341
G ( j ) 
G( s) 
1
s 2  2 n s  n2
1
( n2   2 )  j ( 2 n )
G ( j ) 
1
2 n2
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Phase plot of
G ( j ) 
  n
INC 341
G( s) 
1
1
s 2  2 n s   n2
( n2   2 )  j (2 n )
2
(
2

n)
G ( j )   tan 1
 90
0
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Example
sketch bode plot of G ( s) 
( s  3)
( s  2)(s 2  2s  25)
• Set s  j then
G ( j ) 
( j  3)
(( j )  2)(( j ) 2  2( j )  25)
3
• At DC, set s=0, G(0) 
50
• Break frequency at 2, 3, 25 (or 5)
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Magnitude Plot
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Phase plot
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Phase plot
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Conclusions
•
•
•
•
Drawing Bode plot
Get transfer function T (s)
Set s  j
Evaluate the break frequency
Approximate mag. and phase at low and high
frequencies, and also at the break frequency
– Mag. plot: slope changes 20 dB / dec for 1st order,
40 dB / dec for 2nd order (at break frequency)

– Phase plot: slope changes  90 / dec for 1st order,
 180 / dec for 2nd order
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