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Control Systems
EE
-
475
Lec Sara
Bode plots
Frequency Response
Analysis
ں د � ودى و ں رے دں ى ں
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Frequency Response Analysis
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ں د ے ا ا ا ر ذ ں ا ر ا د
ر ہ ر ا و س ہ
ر و ر ہ ا ر ل ں ا ا ل د
ا
ں ك د ا
ج ا د ا و
ن
ب
ك ے د ا ت ر ف
م
ا � م ى ا ز 3
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1
. Introduction
Steady State Response of System to a Sinusoidalinput. (Effect of input frequency variation onoutput).
Three common frequency response analysismethods are◦ Bode Diagrams
◦ Nyquist Plots
◦ Nicholas Charts
If input is sin wave than output will also be
wave but with different amplitude and phaseangle.
By replacing = we can shift from Laplace tofrequency domain (Fourier)
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Example
Let a System
For sin input put
As it’s a complex number so it s
magnitude and angle can be determined
as
( )1
K G j
jT ω
ω =
+
2 2( )
1
K G j
T ω
ω =
+( )1( ) tanG j T φ ω ω −= ∠ = −
s jω =
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Bode Diagrams/Logarithmic Plots
Consists of two plots magnitude and
phase angle plotted against frequency on
logarithmic scale.
Standard Logarithmic magnitude for base10 is represented as 20 log
Curves are drawn on semi graph
paper using log scale for frequency andlinear scale for magnitude in dB and phase
angle in degrees.
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Cont…
Followings are the basic factors that
frequently occur in arbitrary transfer
functions.
◦ Gain K
◦ Integral and Derivative Factors
◦ First order Factors
◦ Quadratic Factors
( ) 1
jω ±
( ) 1
1 j T ω ±
+ 12
1 2n n
j jω ω ζ
ω ω
± + +
7
ل � s2
s jw
in second order Transfer function
real part imaginary part
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Cont…
Once we become familiar with plots of
these basic factors we will be able to
construct plot of any general transfer
function by adding individual curvesgraphically.
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Gain Factor K
Gain is some constant value so log-
magnitude curve for gain is horizontal
straight line curve at
its phase angle will be zero.
Effect of variation in Gain is that it raise
or lower the log-magnitude curve of
transfer function and have no effect onthe phase curve.
( )1020log K
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Cont…
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Integral and Derivative Factors
For integral part Logarithmic magnitude is
Phase angle is
So log-magnitude plot will be straight linewith a slope of -20db per decade and phaseangle will be straight horizontal line at
( ) 1 1 j jω
ω
−
=
( )10 101
20log 20log dB j
ω ω
= −
90degφ = −
90o−
11
if (jw)-2 then -40log(w),phi=-180...
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Cont…
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Cont…
For derivative part Logarithmic magnitude is
Phase angle is
So log-magnitude plot will be straight linewith a slope of +20db per decade and phaseangle will be straight horizontal line at
+90deg
( ) jω
( )10 1020log 20log j dBω ω =
90degφ =
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Cont…
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First Order Factors
For
Logarithmic magnitude is
For lower frequencies
magnitude is zero
For higher frequencies
magnitude is . = − 20
( )1
1 j T ω +
( )2 2
10 10
120log 20log 1
1T dB
j T ω
ω = − +
+
1
T ω
1
T ω
( )1020log T dbω −
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Cont…
So magnitude plot for this factor canbe approximated by two straight lines
asymptotes.
Horizontal line of 0db for Straight line of slope for
Frequency at which two asymptotes
meets is called Corner/Break frequencythat in this case will be
10 T ω < <
1T
ω < < ∞
1
T ω =
20 dBdecade
−
log
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Cont…
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Cont…
Phase angle will be
At zero frequency
At corner frequency
At infinity
( )1Tan T φ ω −= −
1(0) 0degTanφ −= − =
1
45deg
T
Tan T φ
− = − = −
1
T ω =
1( ) 90degTanφ −= − ∞ = −
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Cont…
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Cont…
Error due to use of asymptotes can becalculated as
At corner frequency
At frequency one octave below
At frequency one octave above
1
T ω =
( ) ( ) ( )10 10 1020 log 1 20 log 1 1 10 log 2 3.03db− + = − −=1
2T ω =
2T
ω =
( )10 10 101 5
20 log 1 20 log 1 20 log 0.97
4 2
db
− + = − = −
( ) ( )10 10 105
20 log 2 20 log 1 4 20 9log 0. 72
db
− + = − =
−
20
octave=2*w and decade = 10 * w
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Cont…
So asymptotic plot can be corrected toexact by making correction of 3 at
corner frequency and 1 at one octave low
and one octave high. Variation in T only shifts the corner freq
to the left or right but no effect on the
shape of plot. So .( )
1
1 j T ω +
21
ر ۔
و س ہ ز omegavaluefactor
and this what a low pass filter does, it just allows lower frequencies
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Cont…
To Draw the Phase curve Accuratelylocate different points on curve
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Cont…
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These are octaves, NOT decades
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Cont…
For
Logarithmic magnitude is
For lower frequencies
magnitude is zero
For higher frequencies
magnitude is
( )1 j T ω +
( )2 2
10 10
20 log 1 20 log 1 j T T dBω ω + = +
1
T ω
1
T ω
( )1020log T dbω
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Cont…
Horizontal line of 0db for
Straight line of slope +20db for
10T
ω < <1
T ω < < ∞
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Cont…
Phase angle will be
At zero frequency
At corner frequency
At infinity
( )1Tan T φ ω −=
1(0) 0degTanφ −= =
1
45deg
T
Tan T φ
− = =
1
T ω =
1( ) 90 degTanφ −= ∞ =
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Cont…
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Quadratic Factor
For quadratic factor
if it will be product of two first order
factors.(Because it’ll be over-damped system will two –real poles) If it will be product of two complex
conjugate factors. (Because it’ll be under-damped system willtwo – complex conjugate poles)
Magnitude and phase of quadratic factordepends upon corner freq. and dampingratio
12
1 2n n
j jω ω ζ ω ω
−
+ +
1ζ >
0 1ζ < <
( )n and ω ζ
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Cont…
Magnitude will be
For
Magnitude will be
For
Magnitude will be
22 2
10 102
120 log 20 log 1 2
1 2 n n
n n
j j
ω ω ς
ω ω ω ω ς ω ω
= − − +
+ +
nω ω >
2
10 10220log 40log
n n
dbω ω
ω ω
− = −
?
29
� ہ
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Cont…
So log-magnitude plot can be estimated bytwo straight line asymptotes.
◦ Horizontal line of 0db for
◦ Straight line with − 40
slope for
The frequency Is corner frequency in thiscase.
Near the corner freq.() a resonant peak occurs.
Damping ratio determines the magnitude of that peak . Asymptotes are independent of the damping ratio.
n
ω ω < < ∞
0 nω ω < <
nω
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Cont…
Phase Angle will be
At zero frequency
At corner frequency
At infinity
1
2
2
1
n
n
Tan
ω ξ
ω φ
ω
ω
−
= − −
0degφ =
90degφ = −
180degφ = −
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Cont…
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Contd.
For quadratic factor
Magnitude will be
2
1 2n n
j jω ω ζ ω ω
+ +
2 2 2
10 1020 log 1 2 20 log 1 2n n n n
j jω ω ω ω ξ ξ
ω ω ω ω
+ + = − +
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Cont…
For
Magnitude will be
For
Magnitude will be
nω ω >
2
10 10220 log 40 log
n n
dbω ω
ω ω
=
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Cont…
So log-magnitude plot can be estimatedby two straight line asymptotes.
◦ Horizontal line of 0db for
◦ Straight line with 40db/decade slope forn
ω ω < < ∞
0n
ω ω < <
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Cont…
Phase Angle will be
At zero frequency
At corner frequency
At infinity
1
2
2
1
n
n
Tan
ω ξ
ω φ
ω
ω
−
= −
0degφ =
90degφ =
180degφ =
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Example#1 Bode Plots
Let a System
Put =
Split into Sub Systems
( )( )( )
50000
5 500
sG s
s s=
+ +
( )( )( )
50000 20
5 5001 1
5 500
j jG j
j j j j
ω ω ω
ω ω ω ω = =
+ + + +
1 ( ) 20G jω =
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Cont…
Draw bode plot for each factor
for
( )
( )
( )
2
1
3
1
4
15
1 500
G j j
jG j
jG j
ω ω
ω ω
ω ω
−
−
=
= +
= +
1( ) 20G jω =
101( ) 20 log (20) 26.02G jω = =
0degφ =
40
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Cont…
25
25.5
26
26.5
27
27.5
M
a g n i t u d e ( d B )
100
101
-1
-0.5
0
0.5
1
P h a s e ( d e g )
Bode Diagram
Frequency (rad/sec)
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Cont…
for ( )2G j jω ω =
2 10( ) 20log ( )G jω ω =
90degφ =
-5
0
5
10
15
20
M a g n i t u d e ( d B )
100
101
89
89.5
90
90.5
91
P h a s e ( d e g )
Bode Diagram
Frequency (rad/sec)
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Cont…
For ( )1
3 15
jG j ω ω
−
= +
3 10
( ) 20 log5
G j for higher freqω
ω
= −
1
5Tan
ω φ −
= −
3 ( ) 0G j for lower freqω =
0 deg 0
45 deg 5
90deg
for
for
for
φ ω
φ ω
φ ω
= =
= − =
= − = ∞
( )1 15
1
51
5
j j T
T
cutt off freqT
ω ω
+ = +
=
= =
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Cont…
-30
-20
-10
0
10
M a g n
i t u d e ( d B )
Bode Diagram
Fre uenc rad/sec
10-1
100
101
102
-90
-45
0
P
h a s e ( d e g )
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Cont…
For ( )1
4 1500 jG j ω ω
−
= +
4 10
( ) 20log500
G j for higher freqω
ω
= −
1
500Tan
ω φ −
= −
4 ( ) 0G j for lower freqω =
0 deg 0
45 deg 500
90deg
for
for
for
φ ω
φ ω
φ ω
= =
= − =
= − = ∞
( )1 1500
1
5001
500
j j T
T
cutt off freqT
ω ω
+ = +
=
= =
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Cont…
Bode Diagram
Fre uenc rad/sec
-30
-20
-10
0
10
M
a g n i t u d e ( d B )
101
102
103
104
-90
-45
0
P h
a s e ( d e g )
46
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Cont…
Now Plot All of them on Single GraphPaper.
-50
0
50
M a g n i t u d e
( d B )
10-1
100
101
102
103
104
-90
-45
0
45
90
P h a s e ( d e g )
Bode Diagram
Fre uenc rad/sec 47
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Cont…
Now add the graphs to get final plot
10-1
100
101
102
103
104
-90
-45
0
45
90
P h a s e ( d
e g )
Bode Diagram
Fre uenc rad/sec
-50
0
50
M a g n i t u
d e ( d B )
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Cont…
0
20
40
60
M a g n i t u d e ( d B )
10-1
100
101
102
103
104-90
-45
0
45
90
P h a s e ( d e g )
Bode Diagram
Fre uenc rad/sec
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Cont…
Draw bode plot for each factor
for
( ) ( )
( ) ( )
1
2
12 6
3
1 .01
100 10
G j j
G j j
ω ω
ω ω ω
−
−
= +
= − + +
71( ) 10G jω =
7
1 10( ) 20 log (10 ) 140G jω = =
0degφ =
51
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Cont…
139
139.5
140
140.5
141
M a g n i t u d e ( d B )
100
101
-1
-0.5
0
0.5
1
P
h a s e ( d e g )
Bode Diagram
Fre uenc rad/sec
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Cont…
For ( ) ( ) 12 1 0.01G j jω ω −= +
( )2 10( ) 20log .01G j for higher freqω ω = −
( )1 .01Tanφ ω −= −
2 ( ) 0G j for lower freqω =
0 deg 0
45deg 100
90deg
for
for
for
φ ω
φ ω
φ ω
= =
= − =
= = ∞
( ) ( )1 1 .01
.01
1 100
j T j
T
cutt off freqT
ω ω + = +
=
= =
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Cont…
-40
-30
-20
-10
0
10
M a g n i t u d e ( d B )
100 101 102 103 104-90
-45
0
P h a s e ( d e g )
Bode Diagram
Frequency (rad/sec)
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Cont…
For ( ) ( ) 1
2 63 100 10G j jω ω ω
−
= − + +
( )3 10
( ) 40log .001G j for higher freqω ω = −
3 ( ) 0G j for lower freqω =
0deg 0
90deg 1000
180deg
for
for
for
φ ω
φ ω
φ ω
= =
= − =
= − = ∞
2 610
1000
2 100
0.05
n
n
n
ω
ω
ξω
ξ
=
=
=
=
1
2
2
1
n
n
Tan
ω ξ
ω φ
ω
ω
−
= − −
21 2 997r n
ω ω ξ = − =
2
110
2 1r M
ξ ξ = =
−
( ) 12
90 sin 87.13
1
r G j
ξ ω
ξ
−
∠ = − + = −
−
Do apply 20 log to get
magnitude of in dB
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Cont…Do apply 20 log to get
magnitude of in dB.20 log 10 = 20
This has been verified in
Matlab
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Cont…
-60
-40-20
0
20
40
M a g n i t u d e ( d B )
100
101
102
103
104
-270
-225
-180
-135
-90
-45
0
45
P
h a s e ( d e g )
Bode Diagram
Fre uenc rad/sec
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