MEM 640 Lecture 2: Bode Plots
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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
Δt = 0.0019 − 0.00148 = 0.00042 sec
Δt
φ = 2π = 50 deg
T
Hence see that this is approximately the 45 degree lag shown on Bode plot
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Bode Plot: Step-by-Step Instructions
Problem: Below is data collected by applying sine voltage inputs into a
low pass RC filter. Sketch the Bode diagram on semi-log paper
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Solution
Step 1: Label axis on semi-log paper. Choose units and be consistent!
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Step 2: Plot your points
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Step 3: Draw asymptotes, mark -3 dB point, cutoff frequency and label slopes
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Step 4: On new semi-log paper, plot the phase angle versus frequencies
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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 ωc = 0.2 with time constant
Hence
1
=5
0 .2
1
TF2 =
5s + 1
τ=
Step 3: Slope increase +20 dB/dec, hence a zero has been added.
The cutoff frequency ωc = 0.8 The time constant is
τ=
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 ωc = 10
τ=
The time constant
1
= 0 .1
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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 jω
20 log G ( s ) = 20 log K +
(
)
(
)
(
)
20
20
20
log 1 + 1.252 ω 2 − 20 log ω − log 52 ω 2 + 1 − log 0.12 ω 2 + 1
2
2
2
(1)
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From Bode magnitude plot, see that at ω = 0.1 have 20 log G ( jω ) = 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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