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Chapter 6

Frequency Response & Systems Concepts

•AC circuit analysis methods to study the frequency response of electrical circuits

•Understanding of frequency response aided by the concepts of phasors and impedance.

•Filtering – a new concept will be explored

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Objectives

• Understand significance of frequency domain analysis

• Introduction of Fourier series as a tool for computation of Fourier spectrum.

• Analyze first and second-order electrical filters by determining their filtering properties.

• Computation of frequency response and its graphical representation as Bode plot.

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Sinusoidal Frequency Response

• Provides a circuit response to a sinusoidal input of arbitrary frequency.

• The frequency response of a circuit is a measure of voltage or current (magnitude and phase) as a function of the frequency of excitation (source) signal.

)(

)()(

jV

jVjH

S

Lv

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Methods to Compute Frequency Response

• Thevenin equivalent source circuit:

~

Zs Z1

Z2VTVL

+

-

~

ZT

VT

21

21

ZZZ

ZZZZ

s

sT

21

2

ZZZ

ZVV

sST

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Load Voltage VT

~

ZT

VTZL

SSSL

L

SS

S

SL

L

TTL

LL

VZZZZZZZ

ZZ

VZZZ

Z

ZZZ

ZZZZ

Z

VZZ

ZV

2121

2

21

2

21

21

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Frequency Response

• From definition:

• VL(j) is a phase-shifted and amplitude-scaled version of VS(j) ⇨

2121

2

ZZZZZZZ

ZZ

jV

jVjH

SSL

L

S

LV

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Frequency Response (cont)

• Phasor form of the load voltage:

SVL

SVL

jHjSV

jL

jS

jHjV

jL

SVL

jH

jVjHjV

ejVjHejV

ejVejHejV

jVjHjV

SVL

SVL

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Example 6.1

• Compute the frequency response Hv(j) of the circuit for R1= 1k, C=10F; and RL= 10k.

VS C

R1

RL VL

110arctan

110

10022

2121

2

2121

2

jH

ZZZZZ

ZZjH

VZZZZZ

ZZ

VZZ

ZV

V

L

LV

SL

L

TTL

LL

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Magnitude & Phaze

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Fourier Analysis

• Let x(t) be a periodic signal with period T.– x(t) = x(t+nT) for n=1,2,3,…

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Fourier Series• A signal x(t) can be expressed as an infinite summation

of sinusoidal components know as Fourier Series:• Sine-cosine (quadrature) representation

• Magnitude and Phase form:

• Fundamental Frequency and Period T:

110

2sin

2cos

nn

nn t

Tnbt

Tnaatx

10

10

2cos

2sin

nnn

nnn

tT

ncctx

tT

ncctx

Tf

22 00

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Fourier Series

• It can be shown

• Or similarly

n

n

n

n

nnn

n

n

nnn

a

bcba

a

bcba

tan

cot

22

22

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Fourier Series Aproximation

• Infinite summation practically not possible

• Replaced by finite summation that leads to approximation.– Higher order coefficients; n, are associated

with higher frequencies; (2/T)n. ⇒– Better approximations require larger

bandwidths.

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Fourier Series

• Odd and Even Functions

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Frequency Spectrum

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Computation of Fourier Series Coefficients

dttT

ntxT

dttT

ntxT

b

dttT

ntxT

dttT

ntxT

a

dttxT

dttxT

a

T

T

T

n

T

T

T

n

T

T

T

2sin

22sin

2

2cos

22cos

2

x(t)of valueaverage11

2

20

2

20

2

20

0

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Example of Fourier Series Approximation

• Square wave and its representation by a Fourier series. (a) Square wave (even function); (b) first three terms; (c) sum of first three terms

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Example 6.3 Computation of Fourier Series Coefficients

• Problem: Compute the complete Fourier Spectrum of the sawtooth function shown in the Figure below for T=1 and A=1:

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Solution

• x(t) is an odd function.

• Evaluate the integral in equation

TtT

tAtx

0 ,

21)(

1,2,3,...n 2

2cos2

4

2cos

/2

2sin

/2

140

2sin

42cos

2

2

2sin

222sin

2

2sin

21

2

2

2

0

222

02

0

00

0

πn

Aπn

πn

T

T

A

tT

πn

Tπn

tt

T

πn

TπnT

A

dttT

πn-t

T

At

T

πn

πn

T-

T

A

dttT

πn

T

t-

T

A dtt

T

πn

T

A

dttT

πn

T

t-A

Tb

T

TT

TT

T

n

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Solution (cont)• Spectrum computation:

00

cotcot 11

22

n

n

n

nnn

b

a

b

bbac

n

n

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Matlab Simulation• Components of the sawtooth wave

function:

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Matlab Simulation• Fourier Series approximation of sawtooth

wave function

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Example 6.4

• Problem: Compute the complete Fourier series expansion of the pulse waveform shown in the Figure for /T=0.2

• Plot the spectrum of the signal

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Solution

• Expression for x(t)

• Evaluate Integral Equations:

Tt

tAtx

0

0)( {

52.0|

10

0

0

AA

T

At

T

AAdt

Ta

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Solution (cont)

5

2cos1

2cos

2

2

2sin

2

5

2sin

2sin

2

2

2cos

2

0

0

0

0

πn

πn

At

T

πn

πn

AT

T

dttT

πn A

Tb

πn

πn

At

T

πn

πn

AT

T

dttT

πn A

Ta

n

n

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Spectrum Computation

• Magnitude:

• Phase:

22

22

5

2cos1

5

2sin

n

n

An

n

Abac

nnn

52sin

52cos1cotcot 11

nnA

nnA

a

b

n

n

n

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Graphical Representation

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Matlab Simulation

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Matlab Simulation

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Linear Systems Response to Periodic Inputs

• Any periodic signal x(t) can be represented as a sum of finite number of pure periodic terms:

10

2sin

nnn t

Tncctx

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General Input-Output Representation of a System

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Linear Systems• For Linear Systems - by definition

Principle of superposition applies:T{ax1(t) + bx2(t)} = aT{x1(t)} + bT{x2(t)}

x1

x2

a

b

T{} y= T{ax1(t) + bx2(t)} ax1[n] + bx2[n]

x1

x2

T{}

T{}

a

bbT{x2[n]}

y= aT{x1(n)}+bT{x2(n)}

aT{x1[n]}

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Linear System View of a Circuit

• Output of a circuit y(t) as a function of the input x(t):

1

sinn

nnnnn jHtcjHty

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Example 6.6 Response of Linear System to Periodic Input

• Problem: – Linear system:

– Input: sawtooth waveform approximated with only first two Fourier components of the input waveform.

2.01

2

jjH

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Solution• Approximation of the sawtooth function

with first two terms of Fourier Series:

• Spectrum Computation:

tttA

tA

tx

16sin2

8sin4

25.0

4sin

25.0

2sin

2)(

1622

814

01222

22

0122

1

2

111

bbac

bbac

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Frequency Response• Magnitude and Phase

• Computation of Frequency Response for two frequency values of 1 = 8 and 2 = 16:

5arctan

2.01

2

2.01

22

jjH

022

222

2

011

221

1

32.8447.116*2.0arctan2.0arctan

1980.016*2.01

2

2.01

2

75.7837.18*2.0arctan2.0arctan

3902.08*2.01

2

2.01

2

radj

jH

radj

jH

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Frequency Response (cont)

• Computation of steady-state periodic output of the system:

47.116sin2

1980.037.18sin4

3902.0

sin2

1

tt

jHtcjHtyn

nnnnn

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Matlab Simulation

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Matlab Simulation

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Filters

• Low-Pass Filters

C

R

VoVi

+ +

--

Simple RC Filter

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Low-Pass Filter

RCjjV

jV

jVRCj

CjR

CjjVjV

jV

jVjH

i

o

iio

i

o

1

1

1

11

1

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Low-Pass Filter

=0– H(j)=1 ⇨ Vo(j)=Vi(j)

>0

RC

RC

j

i

o

eRC

e

e

RC

RCjjV

jVjH

arctan

2

1arctan

0

2

1

1

1

1

1

1

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Low-Pass Filter

RC

RCjH

RCjH

ejHjH

o

o

o

jHj

1

arctanarctan

1

1

1

122

Cutoff Frequency

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Example 6.7

• Compute the response of the RC filter to sinusoidal inputs at the frequencies of 60 and 10,000 Hz.

• R=1k, C=0.47F, vi(t)=5cos(t) V

0=1/RC= 2,128 rad/sec = 120 rad/sec ⇒ /0 = 0.177

= 20,000 rad/sec ⇒ /0 = 29.5

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Solution

537.1000,102cos923.4

175.0602cos923.4

537.150345.0000,102

128,2000,102

1

1000,102

175.05985.0602

128,2602

1

1602

1

1

0

tv

tv

VjVj

jV

VjVj

jV

jVj

jV

o

o

io

io

io

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High-Pass Filters

VoVi

+ +

--

C

R

RCj

RCj

jV

jV

jVRCj

RCj

CjR

RjVjV

jV

jVjH

i

o

iio

i

o

1

11

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High-Pass Filter• The expression in previous slide can be

written in magnitude-and-phase form:

RCjH

RC

RCjH

eRC

RC

eRC

RCe

RCj

RCjjH

RCj

RCj

j

arctan90

1

1

11

0

2

arctan2

2

1arctan

2

2

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High-Pass Filter Response

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Band-Pass Filters

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Frequency Response of Band-Pass Filter

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Analysis of the Second Order Circuit

21

2

111

1

jj

jA

LCjRCj

RCj

LjCj

R

R

jV

jVjH

i

o

1, 2 and are the two frequencies that determine the pass-band (bandwidth) of the filter.

•A – gain of the filter.

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Magnitude and Phase form

21

21

arctanarctan2

2

2

2

1

arctanarctan

2

2

2

2

1

21

11

11

11

j

jj

j

eA

ee

eA

jj

jAjH

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Magnitude and Phase form

21

2

2

2

1

arctanarctan2

11

jH

AjH

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Frequency Response and Bandwidth

2

2

2

11

1

21

2

1

nn

n

nn

n

i

o

jQ

j

Qj

jj

j

LCjRCj

RCj

jV

jVjH

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Frequency Response and Bandwidth

Ratio Damping; 22

1

FactorQuality ; R

11

2

1Q

FrequencyResonant or Natural; 1

L

CR

Q

C

L

RC

LC

n

n

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Normalized Frequency Response & Bandwidth

Bandwidth ; Q

B n

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Frequency Response of the Filter with R=1 k; C=10 F; L=5 mH

rad/s 10,000Q

B n

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Frequency Response of the Filter with R=10 ; C=10 F; L=5 mH

rad/s 100Q

B n

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Bode Plots

• Logarithmic Plots of System’s Frequency Response:

• Both plots are function of frequency also represented in log scale.

Re

Imarctan

(dB) decibelsin log20

jH

A

A

A

AjH

i

o

i

o

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Bode Plot

• Example of Low-Pass Filter:

02

0

0

arctan

1

1

1

1

jV

jV

jjV

jVjH

i

o

i

o

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Bode Plot of Low-Pass Filter

0103.3log202log10log20 1

log20log20log20 1

log20 1

1log10log20

1

log20

0

00

0

2

02

0

KKjV

jV

KjV

jV

KjV

jV

KK

jV

jVjH

dBi

o

dBi

o

dBi

o

dBi

o

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Bode Plot of Low-Pass Filter

n⇨ Cut-off Frequency

-20 dB Slope

3dB

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Bode Plot of Low-Pass Filter

0

0

0

0

2

4

0

arctan

when

when

when

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Bode Plot of Low-Pass Filter

-450 dB Slope

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Correction Factors

/0Magnitude

Response Error in dB

Phase Response Error in dB

0.1 0 -5.7

0.5 -1 4.9

1 -3 0

2 -1 -4.9

10 0 +5/7

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Bode Plots of Higher-Order Filters

• Bode Plots Higher-Order Filters may be obtained by combining Bode-Plots of lower-order functions:– H(j) = H1(j) H2(j) H3(j) … Hn(j)

– |H(j)|dB = |H1(j)|dB + |H2(j)|dB + |H3(j)|dB + … + |Hn(j)|dB

H(j) = H1(j) + H2(j) + H3(j) + … + Hn(j)

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Example of High-Order Filter

1001

101

51005.0

1001log20

101log20

51log20005.0log20

1001

101

51005.0

1001100

10110

515

10010

5

jjjjH

jjjjH

jj

j

jj

jjH

jj

jjH

dB

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Magnitude and Phase in Bode Plots

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Composite Bode Plot

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Bode Plot Approximation Example

50001

101

200120

50001log20

101log20

2001log2020log20

50001

101

200120

1002.0102

1.020235

jjjjH

jjjjH

jjj

jjH

jjj

jjH

dB

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Straight Line Aproximation of Bode Plots

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Actual Magnitude and Phase

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Bode Plot Approximation Example

30001

1001

30111.0

30001log20

1001log20

301log201log201.0log20

30001

301

10011.0

1000,90

030,31091

1.0102

4

23

jjjjH

jjjjjH

jj

jjjH

jj

jjjH

dB

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Bode Plots

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Bode Plots