MEM 640 Lecture 2: Bode Plots

1 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 2 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

3 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 = 0.003sec 0.003 sec = 333.3 Hz Δt = 0.0019 − 0.00148 = 0.00042 sec φ Δt = 2π = 50 deg T

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

4 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

5 Solution

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

6 Step 2: Plot your points

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

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

9 System ID Given the Bode Plot

Problem: Given the following Bode Plot, calculate the transfer function

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

K TF1 = s Step 2: Slope decreases -20 dB/dec, hence have another pole.

The cutoff frequency ωc = 0.2 with time constant 1 τ = = 5 Hence 0.2 1 TF2 = 5s +1

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

The cutoff frequency ωc = 0.8 The time constant is 1 τ = =1.25 0.8 Hence

TF3 =1.25s +1

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

The cutoff frequency is ωc =10 The time constant 1 τ = = 0.1 10 Hence 1 TF4 = 0.1s +1

Step 5: Take product of all sub transfer functions K(1.25s +1) TF = G(s) = s(5s +1)(0.1s +1) ω Step 6: Determine the value of K ω Take decibel log of each side and replace s with jω ω

20 20 20 20logG(s) = 20log K + log(1+1.252 2 )− 20log − log(52 2 +1)− log(0.12ω2 +1) 2 2 2 (1)

12 From Bode magnitude plot, see that at ω = 0.1 have 20logG( jω) = 60 dB Thus substituting this frequency into the (1)

60 dB = 20log K + 0.0673 + 20 − 0.969 − 4.3×10−4 20log K = 40.90

K =110.9

Hence 110.9(1.25s +1) G(s) = s(5s +1)(0.1s +1)