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Sketching Bode Magnitude and Phase Plots © Copyright 2012, Paul Oh MEM 639 – Real-time Microcomputer Controls 1

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Sketching Bode Magnitude and Phase Plots

© Copyright 2012, Paul Oh

MEM 639 – Real-time Microcomputer Controls 1

What is a Bode Plot and why would I care?

© Copyright 2012, Paul Oh

Characterizes if should lead or lag….

© Copyright 2012, Paul Oh

© Copyright 2012, Paul Oh

5

Bode Plots: Why Study Them?

Input: 0.1)( tVi

Output (dots): )(tVo

• Output reaches 63.6% of steady-state at 0.47 ms

• Note: 1/0.00047 = 2127 radians/sec

Recall: RC Low-pass filter time response plot

© Copyright 2012, Paul Oh

6

Low Pass Filter Bode PlotDefinition: 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

© Copyright 2012, Paul Oh

7

Phase Shift Significance

Observe: Apply a sine voltage input (338 Hz) into a low-pass RC filter

Hence see that this is approximately the 45 degree lag shown on Bode plot

Bode phase plot on previous slide says 45-degree lag at 2127 radians/sec [338 Hz]

Period sec003.00019.00049.0 T0.003 sec = 333.3 Hz

sec00042.000148.00019.0 t

deg 502

Tt

© Copyright 2012, Paul Oh

Step 1: Semi-Log Paper

© Copyright 2012, Paul Oh

Step 2: Label Axis

Note units

© Copyright 2012, Paul Oh

1

2

4

3

Step 3: Calculate magnitude [dB] and phase angle [deg]

© Copyright 2012, Paul Oh

1

2

3

4

Step 4: Plot magnitude [dB] versus frequency [Hz]

© Copyright 2012, Paul Oh

Step 5: Sketch slope and -3 dB point, identify cutoff frequencies

© Copyright 2012, Paul Oh

Step 6: Plot phase angle [deg] versus frequency [Hz]. Label inflection point

© Copyright 2012, Paul Oh

14

System ID Given the Bode PlotProblem: Given the following Bode Plot, calculate the transfer function

© Copyright 2012, Paul Oh

15

Step 1: Note that the magnitude is not zero at start and slope is -20 dB/dec

sKTF1

Step 2: Slope decreases -20 dB/dec, hence have another pole. with time constant

52.0

1Hence

15s1TF2

Step 3: Slope increase +20 dB/dec, hence a zero has been added.

2.0cThe cutoff frequency

8.0cThe cutoff frequency The time constant is

25.18.0

1

Hence

125.1TF3 s

16

Step 4: Slope decreases another 20 dB/dec. This means a simple pole was added

The cutoff frequency is 10c The time constant

1.0101

Hence

10.1s1TF4

Step 5: Take product of all sub transfer functions

)11.0)(15(1)K(1.25s)(TF

ssssG

Step 6: Determine the value of K

Take decibel log of each side and replace s with j

11.0log2

2015log2

20log2025.11log2

20log20)(log20 222222 KsG

(1)

© Copyright 2012, Paul Oh

17

From Bode magnitude plot, see that at 1.0 have dB 60)(log20 jG

Thus substituting this frequency into the (1)

4103.4969.0200673.0log20dB 60 K

90.40log20 K

9.110K

Hence

)11.0)(15()125.1(9.110)(

sss

ssG

© Copyright 2012, Paul Oh