Sketching Bode Magnitude and Phase Plots
MEM 639 – Real-time Microcomputer Controls 1
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What is a Bode Plot and why would I care?
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Characterizes if should lead or lag….
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Bode Plots: Why Study Them?
Recall: RC Low-pass filter time response plot
Input: Vi (t )  1.0
Output (dots): Vo (t )
• Output reaches 63.6% of steady-state at 0.47 ms
• Note: 1/0.00047 = 2127 radians/sec
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Low Pass Filter Bode Plot
Definition: A Bode diagram consists of 2 plots. The first plots the output/input ratio
[dB] versus frequency. The second plots the phase angle versus frequency.
Typically a semi-log plot for frequency is used
Low Pass Filter Bode Plot Diagram:
-3 dB
2127 radians/sec
Thus the -3 dB point represents the frequency corresponding to about 1 time constant
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Phase Shift Significance
Bode phase plot on previous slide says 45-degree lag at 2127 radians/sec [338 Hz]
Observe: Apply a sine voltage input (338 Hz) into a low-pass RC filter
Period T  0.0049  0.0019  0.003 sec
0.003 sec = 333.3 Hz
t  0.0019  0.00148  0.00042 sec
t
  2  50 deg
T
Hence see that this is approximately the 45 degree lag shown on Bode plot
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Step 1: Semi-Log Paper
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Step 2: Label Axis
Note units
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Step 3: Calculate magnitude [dB] and phase angle [deg]
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2
3
4
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Step 4: Plot magnitude [dB] versus frequency [Hz]
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2
3
4
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Step 5: Sketch slope and -3 dB point, identify cutoff frequencies
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Step 6: Plot phase angle [deg] versus frequency [Hz]. Label inflection point
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System ID Given the Bode Plot
Problem: Given the following Bode Plot, calculate the transfer function
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Step 1: Note that the magnitude is not zero at start and slope is -20 dB/dec
TF1 
K
s
Step 2: Slope decreases -20 dB/dec, hence have another pole.
The cutoff frequency c  0.2 with time constant
Hence
1
5
0.2
1
TF2 
5s  1

Step 3: Slope increase +20 dB/dec, hence a zero has been added.
The cutoff frequency c  0.8 The time constant is

1
 1.25
0.8
Hence
TF3  1.25s  1
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Step 4: Slope decreases another 20 dB/dec. This means a simple pole was added
The cutoff frequency is c  10

The time constant
1
 0.1
10
Hence
TF4 
1
0.1s  1
Step 5: Take product of all sub transfer functions
TF  G ( s ) 
K(1.25s  1)
s (5s  1)(0.1s  1)
Step 6: Determine the value of K
Take decibel log of each side and replace s with j
20 log G ( s )  20 log K 






20
20
20
log 1  1.252  2  20 log   log 52  2  1  log 0.12  2  1
2
2
2
(1)
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From Bode magnitude plot, see that at   0.1 have 20 log G ( j )  60 dB
Thus substituting this frequency into the (1)
60 dB  20 log K  0.0673  20  0.969  4.3  104
20 log K  40.90
K  110.9
Hence
G (s) 
110.9(1.25s  1)
s (5s  1)(0.1s  1)
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