Chapter 6 Frequency Response & Systems Concepts

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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.

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Methods to Compute Frequency Response

• Thevenin equivalent source circuit:

Z2VTVL

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Load Voltage VT

VZZZZZZZ

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Frequency Response

• From definition:

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

ZZZZZZZ

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Frequency Response (cont)

• Phasor form of the load voltage:

jVjHjV

ejVjHejV

ejVejHejV

jVjHjV

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Example 6.1

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

110arctan

VZZZZZ

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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:

Tnaatx

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Fourier Series

• It can be shown

• Or similarly

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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

x(t)of valueaverage11

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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

1,2,3,...n 2

222sin

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Solution (cont)• Spectrum computation:

cotcot 11

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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:

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Solution (cont)

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Spectrum Computation

• Magnitude:

• Phase:

52cos1cotcot 11

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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:

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)}

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

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):

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.

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Solution• Approximation of the sawtooth function

with first two terms of Fourier Series:

• Spectrum Computation:

16sin2

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Frequency Response• Magnitude and Phase

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

5arctan

32.8447.116*2.0arctan2.0arctan

1980.016*2.01

75.7837.18*2.0arctan2.0arctan

3902.08*2.01

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Frequency Response (cont)

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

47.116sin2

1980.037.18sin4

3902.0

jHtcjHtyn

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Matlab Simulation

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Matlab Simulation

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Filters

• Low-Pass Filters

Simple RC Filter

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Low-Pass Filter

CjjVjV

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Low-Pass Filter

=0– H(j)=1 ⇨ Vo(j)=Vi(j)

arctan

1arctan

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Low-Pass Filter

arctanarctan

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

1000,102

175.05985.0602

128,2602

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High-Pass Filters

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

written in magnitude-and-phase form:

arctan90

arctan2

1arctan

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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

LCjRCj

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

arctanarctan2

arctanarctan

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Magnitude and Phase form

arctanarctan2

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Frequency Response and Bandwidth

LCjRCj

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Frequency Response and Bandwidth

Ratio Damping; 22

FactorQuality ; R

FrequencyResonant or Natural; 1

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Normalized Frequency Response & Bandwidth

Bandwidth ; Q

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

rad/s 10,000Q

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

rad/s 100Q

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Bode Plots

• Logarithmic Plots of System’s Frequency Response:

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

Imarctan

(dB) decibelsin log20

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Bode Plot

• Example of Low-Pass Filter:

arctan

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Bode Plot of Low-Pass Filter

0103.3log202log10log20 1

log20log20log20 1

log20 1

1log10log20

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n⇨ Cut-off Frequency

-20 dB Slope

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arctan

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-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

51005.0

1001log20

101log20

51log20005.0log20

51005.0

1001100

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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

200120

50001log20

101log20

2001log2020log20

200120

1002.0102

1.020235

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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

30111.0

30001log20

1001log20

301log201log201.0log20

10011.0

1000,90

030,31091

1.0102

jjjjjH

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Bode Plots

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Bode Plots

Chapter 6 Frequency Response & Systems Concepts

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