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EE105 - Fall 2005Microelectronic Devices and Circuits

Lecture 17

Frequency-Domain Analysis

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AnnouncementsHomework 7 due todayHomework 8 due next weekLab 6 this weekReading: Chapter 10 (10.1)

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Lecture Material

Last lectureCommon drain amplifierReview phasors

This lectureFrequency-domain analysisBode plots

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RC Circuit with Sinusoidal Input

R

vc(t)

C

iR

iC vs(t) +

-

vs(t)= Vs cos(ωt) : set phase of source to zero (use as the reference)

vc(t)= Vc cos(ωt + φ) : solution is a sinusoidal signal with the same frequency, but with a different amplitude and phase-shifted with respect to the source

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A Better Technique

It is much more efficient to work with imaginary exponentials as “representing” sinusoids, since these functions are direct solutions of linear differential equations:

ddt----- ejωt( ) jω ejωt( )=

• Note that EEs use j = (-1)1/2 rather than i, since thesymbol i is already taken for current

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Using Imaginary Exponentials

scc vv

dtdvRC =+

Substitute:tj

ss evtv ω=)(

)()( φ+ω= tjcc evtv

Result:

tjs

tjc

tjc evevevj ωφ+ωφ+ω =+ωτ )()()(

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Finding the Amplitude Ratio

tjs

tjjc

jc eveevevj ωωφφ =+ωτ ])([

scjj vveej =+ωτ φφ ][

)1(1

ωτ+=

+ωτ=

φ−

φφ je

eejvv j

jjs

cAmplitudeRatio:

Answer is a real number, so take magnitude

… use to find amplitude and phase

( )21

1

ωτ+=

s

cvv

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Graphical Result for Amplitude Ratio

ω 1/τ 10/τ 1/10τ

1.0

0.5

0.707

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Amplitude: A New Representation

• We are interested in very small ratios (e.g., Vc/Vs = 0.0001)

• Therefore, we use a log plot … but we also define a new function called the deciBel (after Alex. Graham Bell)

(Vc/Vs )dB = 20 log10 (Vc/Vs )

• Examples: Vc/Vs = 0.0001 (Vc/Vs )dB = -80 dB

Vc/Vs = 0.707 (Vc/Vs )dB = -3 dB

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Finding the Phase

Collect real and imaginary parts; latter must be zero:

Use Euler’s formula to convert to rectangular form:

csjj vveej /][ =+ωτ φφ

(a real number)

cs vvjjj /)sin(cos)sin(cos =φ+φ+φ+φωτ

0sincos)Im( =φ+φωτ=⋅

ωτ−=φtan

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Graphical Result for Phase φ

ω

1/τ 10/τ 1/10τ

0

-45

-90

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Finding the “Real” Waveform

How to connect the imaginary exponential solution to the measured waveform v(t)? Conventionally, v(t) is the real part of the of the imaginary exponential

)cos()Re( )( φ+ω=φ+ω tvve tj

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Pushing This Idea Further …There are two parameters needed to define a sinusoidal signal:

* magnitude* phase

Why not work with a complex number as the signal and eliminatethe imaginary exponential from the analysis (it cancels out)?

Define the complex number consisting of the amplitude and phasea sinusoidal signal as a phasor

tjVetvtvtv ω=⇔φ+ω= )()cos()(φ= jveV

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Using Phasors: Capacitor Current

C vc(t)+

-ic(t) dt

dvCti cc =)(

tjcc eIti ω=)( tj

cc eVtv ω=)(

Result:

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Impedance of a Capacitor

Definition: the impedance Z of a two-terminal circuit element is the ratio of the phasor voltage to the phasorcurrent (positive reference convention)

C+

-Ic Vc Zc =

Admittance: Yc = 1 / Zc =

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Using Phasors: Inductor Voltage

+

-

L vL(t)iL(t) dt

diLtv LL =)(

tjLL eIti ω=)(tj

LL eVtv ω=)(

Result:

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Inductor Impedance

ZL =

Admittance: YL = 1 / ZL =

+

-

L VL

IL

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Circuit Analysis with PhasorsAssumption: sources are sinusoidal, steady-state!

+-vs(t)

RC

+

vc(t)

-

==s

cVVH

Ratio of output to input phasor is the transfer function of the circuit:

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Bode Plots

1. Plot magnitude | H | in dB vs. ω (log scale)2. Plot phase H in degrees vs. ω (log scale)

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Bode Plots for Low-Pass Filter

1. Plot magnitude | H | in dB vs. ω (log scale)2. Plot phase H in degrees vs. ω (log scale)

dBdBdB jj

H ⎥⎦

⎤⎢⎣

⎡+

=+

=|1|

|1||1

1|||ωτωτ

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Sketching the Magnitude Plot

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=⎥⎦

⎤⎢⎣

⎡+

=2)(1

1log20|1|

|1|||ωτωτ dB

dB jH

Low-frequency (ωτ <<1) asymptote

High-frequency (ωτ >>1) asymptote

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The Break Frequency ω-3dB = (1/τ)

1/τ 10/τ 100/τ 1/(10τ) ω| H |dB

-20

-40

-60

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Finding the Phase Plot

( ) )arctan(01

1ωτ−=⎥

⎦

⎤⎢⎣

⎡ωτ+

∠=∠j

H

Why?

Low-frequency asymptote

High-frequency asymptote

Approx. linear with ω for 1/(10τ) < ω < 10/ τ

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Rapidly Sketching the Phase Plot

1/τ 10/τ 100/τ 1/(10τ) ωPhase H

-45

-90

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The High-Pass Filter

vs(t) +- R

C +

vr(t)

-

+- R

1/(jωC) +

Vr

-

Vs ==sr

VVH

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Magnitude Bode Plot

dBdBdBdB jj

jjH |

11||||

1|||

ωτωτ

ωτωτ

++=

+=

First term (numerator):

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Graphical Addition of Magnitudes

| H |dB

1/τ 10/τ 100/τ 1/(10τ)

ω

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-40

20

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Phase Bode Plot for HPF

1/τ 10/τ 100/τ 1/(10τ) ω

Phase H

+90

-90

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Get to know your logs!

A “margin” of 6dB is a factor of 2 (power)!Knowing a few logs by memory can help you calculate logs of different ratios by employing properties of log. For instance, knowing that the ratio of 2 is 6 dB, what’s the ratio of 4?

dB ratio dB ratio-20 0.100 20 10.000-10 0.316 10 3.162

-5 0.562 5 1.778-3 0.708 3 1.413-2 0.794 2 1.259-1 0.891 1 1.122

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Bode Plot OverviewTechnique for estimating a complicated transfer function (several poles and zeros) quickly

Break frequencies :

)1()1)(1()1()1)(1()()(

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pmpp

znzzK

jjjjjjjGHωτωτωτωτωτωτωω

++++++

=L

L

ii τ

ω 1=

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Summary of Individual Factors Simple Pole:

Simple Zero:

DC Zero:

DC Pole:

ωτ+ j11

ωτ+ j1

ωτj

ωτj1

τω 1

=dB0

dB0

dB0

dB0

90−

90+

90−

90+

τω 1

=

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ExampleConsider the following transfer function

Break frequencies: invert time constants

)1)(1()1(10)(31

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ωτωτωτωωjj

jjjH++

+=

−

ps100ns10ns100

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2

1

===

τττ

Grad/s10Mrad/s100Mrad/s10 321 === ωωω

)1)(1(

)1(10)(

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ωω

ωω

ωωω

ωjj

jj

jH++

+=

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Breaking Down the MagnitudeRecall log of products is sum of logs

Let’s plot each factor separately and add them graphically

)1)(1(

)1(10log20)(

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dB

ωω

ωω

ωωω

ωjj

jj

jH++

+=

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1log201log20

1log2010

log20

ωω

ωω

ωωω

jj

jj

+−+−

++=

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Breaking Down the PhaseSince

Let’s plot each factor separately and add them graphically

)1)(1()1(10)(31

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ωτωτωτωωjj

jjjH++

+∠=∠

−

baba ∠+∠=⋅∠

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110

)(

ωω

ωω

ωωωω

jj

jjjH

+∠−+∠−

+∠+∠=∠

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Magnitude Bode Plot: DC Zero80

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60

40

-20

-60

-80

-40

104 105 106 107 108 109 1010 1011ω

510ωj

0 dB

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Phase Bode Plot: DC Zero180

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135

90

-45

-135

-180

-90

104 105 106 107 108 109 1010 1011ω

510ωj

∠

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Magnitude Bode Plot: Add First Pole80

20

60

40

-20

-60

-80

-40

104 105 106 107 108 109 1010 1011ω

dB510

ωj

dB710

1

1ωj+

Mrad/s101 =ω

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Phase Bode Plot: Add First Pole180

45

135

90

-45

-135

-180

-90

104 105 106 107 108 109 1010 1011ω

510ωj

∠

7101

1ωj+

∠

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Magnitude Bode Plot: Add 2nd Zero80

20

60

40

-20

-60

-80

-40

104 105 106 107 108 109 1010 1011ω

dB810

1 ωj+Mrad/s1002 =ω

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Phase Bode Plot: Add 2nd Zero180

45

135

90

-45

-135

-180

-90

104 105 106 107 108 109 1010 1011ω

8101 ωj+∠

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Magnitude Bode Plot: Add 2nd Pole80

20

60

40

-20

-60

-80

-40

104 105 106 107 108 109 1010 1011ω

dB1010

1

1ωj+

Grad/s103 =ω

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Phase Bode Plot: Add 2nd Pole180

45

135

90

-45

-135

-180

-90

104 105 106 107 108 109 1010 1011ω

10101 ωj+∠−

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Comparison to Actual Mag Plot

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Comparison to Actual Phase Plot