Download - Bode Diagram 1
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Bode Diagram (1)
Hany FerdinandoDept. of Electrical Engineering
Petra Christian University
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General Overview
This section discusses the steady-state response of sinusoidal input
The frequency response analysis uses Bode diagram
Students also learn how to plot Bode diagram
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What is frequency response?
G(s)X(s) Y(s)
If x(t) = X sin If x(t) = X sin t then y(t) = Y sin (t then y(t) = Y sin (t + t + ))
The sinusoidal transfer function G(jThe sinusoidal transfer function G(j) is a ) is a complex quantity and can be represented complex quantity and can be represented by the magnitude and phase angle with by the magnitude and phase angle with
frequency as a parameterfrequency as a parameter
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Presenting the freq. resp.
Bode Diagram or logarithmic plot (this section)
Nyquist plot or polar plot Log-magnitude versus phase plot
Matlab can be used to plot both Bode diagram and Nyquist plot
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Preparation
Bode diagram uses the open loop transfer function
The plot is a pair of magnitude and phase plots The representation of logarithmic
magnitude is 20 log |G(j)| in dB
The main advantage: multiplication is converted into addition
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Basic factor of G(j)
Gain K Derivative and integral factors (j)±1
First-order factors (1+j)±1
Quadratic factors [1+2(jn)+(jn)2]±1
One must pay attention to the corner frequency and the form of the equation must be fitted to
above
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Gain K It is real part only no phase angle The log-magnitude is a straight line at
20 log (K) If K > 1, then the magnitude is positive If K < 1, then the magnitude is negative
Varying K only influences the log-magnitude plot, the phase angle remains the same
Slope is 0 at corner frequency 0 rad/s
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Integral (j)-1
It has only imaginary part Log-magnitude = -20 log () Phase angle = 90o (constant) Slope is -20 dB/decade at corner
frequency =1 rad/s
)log(201
log201
log20
j
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Derivative (j)
It has only imaginary part Log-magnitude: 20 log () Phase angle: 90o (constant) Slope is 20 dB/decade at corner
frequency =1 rad/s
)log(20log20log20 j
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First order (1+j)±1 (1) Integral:
Corner frequency is at =1/T Slope is -20 dB/decade Phase angle is -45o at corner
frequency Derivative:
Corner frequency is at =1/T Slope is 20 dB/decade Phase angle is 45o at corner frequency
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First order (1+j)±1 (2)
-30
-25
-20
-15
-10
-5
0
Mag
nitu
de (
dB)
10-2
10-1
100
101
-90
-45
0
Pha
se (
deg)
Bode Diagram
Frequency (rad/sec)
21
1
j
45o at =0,5 rad/s
Slope: 20dB/dec
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Quadratic factors (1) Integral:
Corner frequency is at =n Slope is – 40 dB/decade Phase angle is -90o at corner frequency
Derivative: Corner frequency is at =n
Slope is 40 dB/decade Phase angle is 90o at corner frequency
Resonant freq: 221 nr
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Quadratic factors (2)12
2
)(
21
jj
n = √2
-80
-60
-40
-20
0
Mag
nitu
de (
dB)
10-1
100
101
102
-180
-135
-90
-45
0
Pha
se (
deg)
Bode Diagram
Frequency (rad/sec)
90o at corner freq. n = √2
Slope = 40dB/dec
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Example:
)2)(2(
)3(10)(
2
ssss
ssG
Draw Bode diagram for the following transfer function:
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Answer:
)2))((2)((
)3(10)(
2
jjjj
jsG
Substitute the s with j! We got
Make it to the standard form… Proot it!!!
122
)(1
2)(
13
5.7)(
2
jjjj
j
sG
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Answer:
122
)(1
2)(
13
5.7)(
2
jjjj
j
sG
13
j12
122
)(
jj1
12
j1)( j
from
We got
7.5
With = 0.35
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Answer:
7.5 is gain Log-magnitude = 20 log (7.5) Phase = 0o
The
Slope = 20 dB/decade Phase = 45o at = 3 rad/s
13
j
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Answer
The Slope = -20 dB/decade Phase = -90o (constant)
The
Slope = -20 dB/decade Phase = - 45o at = 2 rad/s
1)( j
1
12
j
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Answer:
The
Slope = - 40 dB/decade Phase = -90o at = √2 rad/s
12
122
)(
jj
The next step is to combine all magnitudes and phases respectively,
then add all of them to from a sketch of Bode diagram
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Next…
The Bode diagram has been discussed here, the next topic is Phase and Gain
Margin. Several important points will be discussed as well
Please prepare yourself by reading the book!