EEB 311 (2013/2014)Electrical Network Theory

The Fourier Series 1.Introduction2.Trigonometric Fourier Series3.Symmetry Considerations4.Circuit Applications

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• Previously, have considered analysis of circuits with sinusoidal sources.

• The Fourier series provides a means of analyzing circuits with periodic non-sinusoidal excitations.

• Fourier is a technique for expressing any practical periodic function as a sum of sinusoids.

• Fourier representation + superposition theorem, allows to find response of circuits to arbitrary periodic inputs using phasor techniques.

IntroductionIntroduction

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• The Fourier series of a periodic functionperiodic function f(t) is a representation that resolves f(t) into a dc componentdc component and an ac componentac component comprising an infinite series of harmonic sinusoids.

• Given a periodic function f(t)=f(t)=f(t+nT) ) where nn is an integer and TT is the period of the function.

where w0=2π/T is called the fundamental frequency in radians per second.

Trigonometric Fourier Series

ac

nnn

dc

tnbtnaatf

1

000 )sincos()(

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• and Fourier coefficients, an and bn , are:

Trigonometric Fourier SeriesTrigonometric Fourier Series

T

on dttntfT

a0

)cos()(2

)(tan , 1n

22

n

nnnn a

bbaA

T

on dttntfT

b0

)sin()(2

• in alternative form of f(t)

where

ac

nnn

dc

tnAatf

1

00 )cos(()(

T

dttfT

a00 )(

1

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ConditionsConditions ((Dirichlet conditions conditions)) on on f(t)f(t) to yield a to yield a convergent Fourier seriesconvergent Fourier series:

1. f(t) is single-valued everywhere.

2. f(t) has a finite number of finite discontinuities in any one period.

3. f(t) has a finite number of maxima and minima in any one period.

4. The integral

Trigonometric Fourier SeriesTrigonometric Fourier Series

.any for )( 0

0

0

tdttfTt

t

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Example

Determine the Fourier series of the waveform shown below. Obtain the amplitude and phase spectra

Trigonometric Fourier SeriesTrigonometric Fourier Series

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Solution:

Trigonometric Fourier Series

)2()( and 21 ,0

10 ,1)(

tftft

ttf

evenn ,0

oddn,/2)sin()(

2

and 0)cos()(2

0 0

0 0

ndttntf

Tb

dttntfT

a

T

n

T

n

1

12 ),sin(12

2

1)(

k

kntnn

tf

evenn ,0

oddn,90

evenn ,0

oddn,/2

n

n

nA

Truncating the series at N=11

a) Amplitude andb) Phase spectrum

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Three types of symmetry

1. Even Symmetry : a function f(t) if its plot is symmetrical about the vertical axis.

In this case,

Symmetry Considerations

)()( tftf

0

)cos()(4

)(2

2/

0 0

2/

00

n

T

n

T

b

dttntfT

a

dttfT

a

Typical examples of even periodic function

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2. Odd Symmetry : a function f(t) if its plot is anti-symmetrical about the vertical axis.

In this case,

Symmetry Considerations

)()( tftf

2/

0 0

0

)sin()(4

0

T

n dttntfT

b

a

Typical examples of odd periodic function

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3. Half-wave Symmetry : a function f(t) if

Symmetry Considerations

)()2

( tfT

tf

even n for ,

odd n for ,

even n for ,

odd n for ,

0

)sin()(4

0

)cos()(4

0

2/

0 0

2/

0 0

0

T

n

T

n

dttntfTb

dttntfTa

a

Typical examples of half-wave odd periodic functions

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Example

Find the Fourier series expansion of f(t) given below.

Symmetry Considerations

1 2sin

2cos1

12)(

n

tnn

ntf

Ans:

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Example

Determine the Fourier series for the half-wave cosine function as shown below.

Symmetry Considerations

1

2212 ,cos

14

2

1)(

k

knntn

tf

Ans:

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Circuit ApplicationsCircuit Applications

Steps for Applying Fourier Series

1. Express the excitation as a Fourier series.Example, for periodic voltage source:

2. Transform the circuit from the time domain to the frequency domain.

3. Find the response of the dc and ac components in the Fourier series.

4. Add the individual dc and ac response using the superposition principle.

ac

1nn0n

dc

0 )tncos(V(V)t(v

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Example

Find the response v0(t) of the circuit below when the voltage source vs(t) is given by

Circuit Applications

12 ,sin12

2

1)(

1

kntnn

tvn

s

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Solution

Phasor of the circuit

For dc component, (n=0 or n=0), Vs = ½ => Vo = 0

For nth harmonic,

In time domain,

Circuit Applications

s0 V25

2V

nj

nj

)5

2tan(c

425

4)(

1

1

220

k

ntnos

ntv

s22

1

0 V425

5/2tan4V ,90

2V

n

n

nS

Amplitude spectrum of the output voltage

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Given:

Useful Formula