Magnitude Bode Plot for the Low Pass Filter

Magnitude Bode Plot for the Low Pass Filter

* Transfer function is

-V---o---u---t = 1-----+-----j-1-ω----R----C---
Vin

“The magnitude of the ratio is the ratio of the magnitudes:”

-V---o---u---t = 20log -----------1------------ = 20 log  -----1----+-----(---ω1-----⁄---ω----o---)--2-
Vin dB 1 + jωRC 



* ωο = 1 / RC is the “break frequency” or “-3 dB frequency”
ω << ωο results in a magnitude of 20 log (1/1) = 0 dB
ω >> ωο results in a magnitude of

V----o---u---t = 20 log  -----1----+-----(---ω1-----⁄---ω----o---)--2- ≅ 20log  (---ω------⁄1--ω----o----) = –20log  -ω-ω---o-
Vin dB   



* substitute ω = 10 ωο, 100 ωο, 1000 ωο

EE 105 Spring 2000 Page 1 Week 12, Lecture 28

Approximate Magnitude Bode Plot

* Sketch asymtotes above and below the break frequency

Vout
Vin dB 0

-20
-40

-60
-80

ωο ω

At ω = ωο , the exact magnitude is:

V----o---u---t = 20log ----------------1----------------- = 20log -------1--------- = –3dB
Vin dB 1 + j(ωo ⁄ ωo) 1+1

EE 105 Spring 2000 Page 2 Week 12, Lecture 28

Phase Bode Plot for Low Pass Filter

* From the definition of the phase,
∠-V-V--o--i-u-n--t = ∠-1----+-----j-1-ω----R----C--- = ∠1 – ∠(1 + j(ω ⁄ ωo)) = –∠(1 + j(ω ⁄ ωo))

* Substituting the arctangent,
∠V--V--o--i-u-n--t = –atan (ω ⁄ ωo)

* Look at asymtotes, again:
ω << ωο results in a phase of - atan(0) = 0
ω >> ωο results in a phase of - atan(infinity) = - 90o
ω = (1/10)ωο results in a phase of - atan(0.1) = - 6o
ω = (10)ωο results in a phase of - atan(10) = - 84o
ω = ωο results in a phase of - atan(1) = - 45o

Vout 0.1ωο ωο 10ωο ω
Vin

−45

−90

EE 105 Spring 2000 Page 3 Week 12, Lecture 28

Finding the Output Waveform from the Bode Plot

* Suppose that vin(t) = 100 mV cos (ωοt + 0o)

note that the input signal frequency is equal to the break frequency and that the
phase is 0o ... the input signal phase is arbitrary and is generally selected to be 0.

the output phasor is:

Vout = Vin -1----+-----j--(--ω--1---o----⁄--ω-----o---) = Vin -----1-----
1+j

magnitude: … Vout = --V----i--n-- = 1---0---0----m-----V-- = 71mV
V----o---u---t = –3dB 2 2
Vin dB

phase:

∠V--V--o--i-u-n--t = ∠1 – ∠(1 + j) = 0 – 45° ∠Vout = –45°

Vout = (71mV)e–j45°

output waveform vout(t) is given by:

vout(t) = R e  V j ωo t = R e ( 71 mV e –j 45 ° e j ω o t )
 
out e

vout(t) = 71 mVcos(ωot – 45o)

EE 105 Spring 2000 Page 4 Week 12, Lecture 28

Bode Plots of General Transfer Functions

* Procedure is to identify standard forms in the transfer functions, apply
asymptotic techniques to sketch each form, and then combine the sketches
graphically

H(jω) = -A----j--ω----(--1-----+-----j--ω----τ---2---)---(--1-----+----j--ω-----τ---4--.--)--.--.-.--(---1----+-----j--ω-----τ--n----)
(1 + jωτ1)(1 + jωτ3)...(1 + jωτn – 1)

where the τi are time constants -- (1/τi) are the break frequencies, which are
called poles when in the demoninator and zeroes when in the numerator

* From complex algebra, the factors can be dealt with separately in the magnitude
and in the phase and the results added up to find |H(jω) | and phase (H(jω))

Three types of factors:

1. poles (binomial factors in the denominator)
2. zeroes (binomial factors in the numerator)
3. jω in the numerator (or denominator)

(note: we aren’t going to consider complex poles)

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Rapid Sketching of Bode Plots

* Poles: - 3 dB and -45o at break frequency
0 dB below and -20 dB/decade above
0o for low frequencies and -90o for high frequencies; width of transition is
between 10 and (1/10) break frequency

* Zeros: +3 dB and +45o at break frequency
0 dB below and + 20 dB/decade above
0o for low frequencies and +90o for high frequencies; width of transition is
between 10 and (1/10) break frequency

* jω: +20 dB/decade (0 dB at ω = 1 rad/s) and +90o contribution to phase

Example I:

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Magnitude and Phase Bode Plot Sketches

|H(jω)|dB

ω

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Phase Bode Plot Sketches

H(jω)

ω

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