1

MEM 640 Lecture 2: Bode Plots

2

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

3

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

4

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 =−=Δt

deg 502 =Δ

=Ttπφ

5

Bode Plot: Step-by-Step InstructionsProblem: Below is data collected by applying sine voltage inputs into a low pass RC filter. Sketch the Bode diagram on semi-log paper

6

Step 1: Label axis on semi-log paper. Choose units and be consistent!

Solution

7

Step 2: Plot your points

8

Step 3: Draw asymptotes, mark -3 dB point, cutoff frequency and label slopes

9

Step 4: On new semi-log paper, plot the phase angle versus frequencies

10

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

11

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

1==τHence

15s1TF2+

=

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

2.0=cωThe cutoff frequency

8.0=cωThe cutoff frequency The time constant is

25.18.0

1==τ

Hence

125.1TF3 += s

12

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

The cutoff frequency is 10=cω The time constant

1.0101==τ

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 ωj

( ) ( ) ( )11.0log22015log

220log2025.11log

220log20)(log20 222222 +−+−−++= ωωωωKsG

(1)

13

From Bode magnitude plot, see that at 1.0=ω have dB 60)(log20 =ωjG

Thus substituting this frequency into the (1)

4103.4969.0200673.0log20dB 60 −×−−++= K

90.40log20 =K

9.110=K

Hence

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

+++

=sss

ssG