1. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 43
Bode Plots Filters
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. Outline: Bode Plots A full Bode DIAGRAM consists of two Separate Plots
|H(f)| (in dB) vs. f in Hz (or ω in rads/S)
H(f) in Degrees vs f in Hz (or ω in rads/S)
Recall: ω = 2π∙f
“Poles” & “Zeros” in H(f) with one or the Other
A “Pole” Frequency, fBp, causes |H|→∞
A “Zero” Frequency, fBz, causes |H| →0
Named after Hendrik W. Bode

3. UnGraded HW Assignment
Students are asked to Work Thru, BY HAND, All the Complex Algebra that follows in the H(f) developments
UNDERSTANDING how to Cast Transfer Functions into STANDARD Form is Crucial to Bode Plotting
Some of what Follows May appear on the Next MidTerm Exam, or on the Final Exam

4. ProtoTypical H(f) for Bode
The 1-”Pole” and 1-”Zero” Xfer fcn:
Then The “dB” Magnitude
And the Phase Angle

5. Bode Plot Construction𝑥 𝑦
Consider a Transfer Function with one “pole”, and one “zero”
K ≡ CONSTANT
First, put the Transfer Function into STANDARD FORM
K, fBz, fBp are all constants. “Pole” implies that under some circumstances (fBp – jf) could equal zero producing a vertical Asymptote. “Zero” implies that under some circumstances (fBz – jf) could equal zero producing H(f) = 0

6. Bode Plot Construction
H(f) in POLAR Form
Then
Since in Std Form x=1 it is usually not shown in the aTan calcs 𝑦 𝑥

7. Bode Plot Construction
Bode Magnitudes are typically plotted in deciBels.
So from the last Slide

8. Magnitude Equation Examine the dB Magnitude Equation
20logK0 is a Constant
Positive for K0 > 1
Zero for K0 = 1
Negative for K0 < 1
If K0 = 2, then
f=0:10:10000;
Mag = 20*log10(2)*ones(1,length(f));
semilogx(f,Mag, 'LineWidth',3), grid
axis([ ])
ylabel('Mag_{dB}')
xlabel('f(Hz)')

9. Straight-Line Magnitude Plots
An Approximate plot of ±20log|1+jf/fB| consists of TWO Straight lines
For SMALL f
For LARGE f
On a log Scale for f>fB 20log|1+jf/fB| is a Straight-Line with a slope of 20dB/decade
This Sloped-Line Intersects the 0dB axis at f = fB
Thus fB is often called the CORNER FREQUENCY on a Bode Plot

10. Bode Plot Example: Taking fBz = 100 Hz, approximate |H(f)|
Note the “Corner” at 102 Hz
Note the 40db Increase over 102 to 104 → 20dB/decade • (10/100)2 = 1% of 1 • (1000/100)2 is 100 times of 1
f = logspace(0,4,500);
Mag = 20*log10(abs(1+j*f/100));;
semilogx(f,Mag, 'LineWidth',2), grid
axis([ ])
ylabel('Mag_{dB}')
xlabel('f (Hz)')
Straight-Line Approximations

11. Bode Plot Example: Converting to dB form with fB = 300Hz
Note again the Corner
This time at 300 Hz
1885 rads/sec

12. Near the Corner:
Why the “10”? →

13. Straight-Line Phase-Angle Plots
Plot The phase angle of a 1st order ZERO Xfer fcn of this form 𝑥 𝑦

14. Straight-Line Phase-Angle Plots
The Phase Angle of constant K0 is zero degrees
The phase angle of a 1st order ZERO Xfer fcn has 3 straight lines (jf/fBz in numerator)
1. For values of f = fBz, 45° point (by α = atan[1/1])
2. For values of f>=10 fBz, it is a straight line at 90° (α = atan[10/1] ≈ 90°)
3. For values of f<=0.1fBz, it is a straight line at 0° (α = atan[0.1/1] ≈ 0°)
4. For values of 0.1fBz <= f <= 10fBz, it is a straight line with a slope of 45°/decade
The phase angle of 1st order POLE Xfer fcn has 3 straight lines (jf/fBp in denominator)
1. For values of f = fBp, −45° point (by −β = −atan[1/1])
2. For values of f>=10 fBp, it is a straight line at −90° (−β = −atan[10/1] ≈ − 90°)
3. For values of f<=0.1fBp, it is a straight line 0° ° (−β = −atan[0.1/1] ≈ 0°)
4. For values 0.1fBp <= f <= 10fBp, it is a straight line with a slope of −45°/decade
The approximate phase angle of −20 log(f/fB) is a straight line having a slope of −45°/decade that intersects the 0 degree axis at f=0.1fB, and intersects the −90° line at f=10fB

15. Example 𝝓 𝑯 : ReCall from the dB development In this case
Also Recall that for this transfer fcn form

16. Angle, 𝝓 𝑯 , Approx:

17. Angle, 𝝓 𝑯 , Exact: f = logspace(-1,4,500); a = atand(f/10);
b = atand(f/100);
angj = 90*ones(1,length(f));
q = a - angj -b;
semilogx(f,a, f,-b, f,-angj, f,q, 'LineWidth',2), grid
axis([ ])
ylabel('\phi_{H} (°)')
xlabel('f (Hz)')

18. Angle, 𝝓 𝑯 , Exact by: (31.26 Hz, −35.10°) f = logspace(-1,4,500);
h = 2*(1+j*f/10)./((j*f).*(1+j*f/100))
a=angle(h);
deg=a*180/pi;
[degmax, Nmax] = max(deg);
fmax = f(Nmax);
semilogx(f,deg, fmax, degmax, '*', 'LineWidth',3), grid
axis([ ])
ylabel('\phi_{H} (°)')
xlabel('f (Hz)')
fmax
degmax

19. Series LCR Analysis
With phasor analysis, this circuit is readily analyzed
Use Loop+Ohm

20. Second Order Transfer Function
So we have:
To find the poles/zeros, put H in Standard form:
One zero at DC frequency (f=0)  can’t conduct DC due to capacitor

21. Resonance For the Series Resonant ckt at Right the Series Impedance:
At the “Resonant” Frequency, f0,The imaginary impedances ADD to ZERO EXACTLY
Thus the Frequency, f0, that produces a PURELY Resistive Impedance where VO = VS:

22. Find Transfer Function Poles
At Resonance H(f) Denom is quadratic in f
f0 is the RESONANT FREQUENCY at which the series impedance is purely resistive. Qs is the Quality-Fraction → ZL/ZR = ZC/ZR

23. 𝑸 𝒔 Complex Algebra
For the Last Equation

24. Simplify Transfer Function
Multiply the Transfer Function by 1
0 at 𝑓= 𝑓 0

25. Magnitude Response The response “peakiness” depends on Q
Normalize the Independent Variable to f/f0 to Plot →

26. BandWidth ReCall the Half-Power Frequency fB At fB:
The Series RLC Circuit has TWO half-power Frequencies for any Given value of Q: fL and fH
The difference Between the Lo & Hi ½-Power Frequencies is defined as the BANDWIDTH, B

27. BandWidth Formally Recall the H(f) Then |H(f)|:
At the ½-Power Level Frequency fB:
Thus
Or for fB = fL & fH

28. BandWidth With Enough Algebra; B
Note that |H(f)| may not be symmetrical about f0
Thus for QS >> 1 approximate:

29. Parallel Resonance Consider the Parallel ckt at Right
The Equivalent Parallel Impedance
At Resonance, f0, the imaginary Impedances Cancel
Then f0:
The Quality Factor is the ratio of ZR/ZL

30. Parallel Resonance
Using QP in the Equivalent Impedance expression find
And vout by Ohm in Frequency Domain
With the previous Bandwidth Definition
The H vs f plot

31. WhiteBoard Work
Show that for Series Resonant Ckt

32. WhiteBoard Work
Design a NOTCH Filter to remove 60Hz “Power Line” Hum from an Audio Circuit
The Audio Circuit has an equivalent Load, Rld that connects to the Filter
Find L, R & C for this passive Filter Design
Want |H(f)| = 0 for:
f0 = 60 Hz (ω0 = 377 rads/sec)
Want w0 = 377 rad/s OR want f0 = 60Hz

33. WhiteBoard: Filter Results
Check the Effectiveness of the Filter Design if
C = 100 µF
L = ???
Rld = 100 Ω
Filter out 60Hz noise that is 5X the Audio Signal; to whit:
L = 70.4 mH
Noise
Signal

34. WhiteBoard: Filter Results
Neglecting Phase-Shifts the Output:
The Signal-Amplitude is now 10X the Noise-Amplitude
Plot the Result using MATLAB

35. WhiteBoard: Filter Results

36. MATLAB Code % Bruce Mayer, PE % ENGR43 • 17Jul11
% Lec6b_Filter60_1107.m %
Rld = 100; L = 70e-3;, C = 100e-6;
H Rld*(1-w.^2*L*C)/((Rld*(1-w.^2*L*C) + j*w*L));
H60 = H(60*2*pi); % 60Hz = 377 rad/sec
H1000 = H(1000*2*pi); % 1000 Hz = 6238 rad/sec
Vo1 = H60*1;
Vo2 = H1000*0.2;
Vm1 = abs(Vo1);
Vm2 = abs(Vo2);
% t in mSec
t = linspace(0,25,1500);
% Calc the INput
vin1 = 1*cos(60*2*pi*t/1000);
vin2 = 0.2*cos(1000*2*pi*t/1000);
vintot = vin1 + vin2;
% Calc the OUTput
vo1 = Vm1*cos(60*2*pi*t/1000);
vo2 = Vm2*cos(1000*2*pi*t/1000);
votot = vo1 + vo2;
plot(t, vintot, t, votot, 'LineWidth',3), grid
Filterlegend = legend('vin(t)','vo(t)');
xlabel('t (mS)'), ylabel('vin, vo (V)')
MATLAB Code

37. 796 Hz BandPass Filter All Done for Today

38. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 43
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

45. WhiteBoard Work
Design a NOTCH Filter to remove 60Hz “Power Line” Hum from An audio Circuit
The Audio Circuit has an equivalent Load, Rld to connect to the Filter
Find L & C for this passive Filter Design

50. Check Rejection for Rld = 100Ω

51. Poles of the Transfer Function
The “f” Roots for the DeNominator
Poles are complex conjugate frequencies
The QS parameter is called the
“quality-factor” or Q-factor
QS is an important Parameter Re Im

52. Obsolete – 20Mar12