7 1 Fourier Series

8/12/2019 7 1 Fourier Series

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FourierAnalysis

S. Awad, Ph.D.

M. Corless, M.S.E.E.D. Cinpinski

E.C.E. Department

University of Michigan-Dearborn

Math Review with Matlab:

Fourier Series


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U of M-Dearborn ECE Depar tmentMath Review with Matlab

Fourier Analysis: Fourier Series

2

Periodic Signal Definition

Parsevals Theorem

Fourier Series

Complex Exponential Representation

Magnitude and Phase Spectra of Fourier Series

Fourier Series Representation of Periodic Signals

Fourier Series Coefficients Orthogonal Signals

Example: Full Wave Rectifier

Example: Finding Complex Coefficients

Example: Orthogonal Signals


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3

For

example, the normal U.S. AC from wall outlet has asine wave with a peak voltage of 170 V (110 Vrms)

The Periodof a signal is the amount of time it takes for agiven signal to complete one cycle.

What is a Periodic Signal ? A Periodic Signalis a signal that repeats itself every

period


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4

General Sinusoid A general cosine wave, v(t), has the form:

)cos()( q tMtv

= Phase Shift, angular offset in radians

F = Frequencyin Hz

T = Periodin seconds (T=1/F)

t = Timein seconds

M= Magnitude, amplitude, maximum value

= Angular Frequencyin radians/sec (=2pF)


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5

General Sinusoid

Plot in Blue:

))60(2sin(5 tp

Plot in Red:

)2

)60(2sin(5 p

p t

1 Period = 1/60 sec.

= 16.67 ms.

/2 Phase Shift

Amplitude = 5


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6

AC Wall

Voltage Sine Wave

1 Period

1 Period


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7

Represent Periodic Signals

For a general periodic signal x(t)

shown to the right:

x(t+nT) = x(t) for all

t

where n is any integer, i.e. n = 0, 1, 2,

T

x(t)

-T/2 T/2

t......


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8

Frequency of Periodic Signals

The frequency of a signal is defined as the

inverse of the period and has the unit

number of cycles/sec. Tfo

1

is the fundamental frequency.of

The frequency of a US standard outlet is 1/T = 60 Hz

T is the period and


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9

What is Fourier Series ? Fourier Series is a technique developed by J. Fourier.

This technique (studied by Fourier) allows us to representperiodic signals as a summation of sine functions ofdifferent frequency, amplitude, and phase shift.


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10

Represent a Square Wave

Represent the

Square Wave at

the right using

Fourier Series

Notice that as

more and more

terms are summed,the approximation

becomes better


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11

Fourier Series Representation

of Periodic Signals Any periodic function can be represented in terms of sine

and cosine functions:

...2sinsin

...2coscos)(

21

210

tbtb

tataatx

oo

oo

This can also be written as:

1

0 )sincos()(

n

onon tnbtnaatx


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12

Fourier Series Coefficients

The above

a0, an, and bnare known as the Fourier Series

Coefficients.

These coefficients are calculated as follows.

1

0 )sincos()(n

onon tnbtnaatx


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Calculating the a0 Coefficient ao, the coefficient outside the summation, is known

as the average value or the dc component

ao is calculated as follows:

2

2)(

1 T

T

o dttxTa


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14

Calculating the anand bn

Coefficients

2

2

,)cos()(2

T

T

on dttntxT

a n = 1, 2,

2

2

,)sin()(2

T

T

on dttntxT

b n = 1, 2,

The an and bn coefficients are calculated as follows:


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15

Orthogonal Signals

Two periodic signals g1(t) and g2(t) are said to be

Orthogonal if the the integral of their product over one

period is equal to zero.

2/

2/

0)(2)(1

T

T

dttgtg


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16

Example: Orthogonal Signals

2/

2/

)()( 21

T

T

dttgtg

2/

2/

2/

2/

)cos()sin(22

1

)cos()sin(

T

T

T

T

dttt

dttt

Show that the

following signals are

orthogonal:

cos(t)(t)g

sin(t)(t)g

2

1


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Orthogonal Signals

0

)cos()cos(41

)2cos(4

1)2sin(

2

1

)cos()sin(22

1

2/

2/

2/

2/

2/

2/

TT

tdtt

dttt

T

T

T

T

T

T


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18

Note that the rectified wave has a period equal to

one-half of the source wave period.

Example: Full Wave Rectifier

y(t)=|sin(ot)|

t

y=|x|

x

y

x(t)

tT/2

one period

one period

T

Consider the output of a full-wave rectifier:


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Function Characteristics The period of y(t) = T/2 and the fundamental

frequency of y(t) is 2o(rad/sec).

1

0 )2cos()(n

on tnaaty

Thus,

Now bn=0 since y(t) is an even function.


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20

Finding ao

2/

0

2

0

2

2

)sin(2

)sin(2

1

)(1

T

oo

T

oo

T

T

o

dttT

a

dttT

a

dttyT

a


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21

Finding ao

1)2

)2

(cos(

)2

(

2

)0*cos()2

cos(2

)cos(22/

0

TT

T

T

a

TT

a

tT

a

o

oo

o

o

T

o

o

o

p

p

* Use o= 2pi/T


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22

Finding ao

p

p

p

p

2

111

1)cos(1

o

o

o

a

a

a


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23

Finding an, n = 1, 2, .

4

0

4

4

)2cos()sin()2(4

)2cos()sin(2

2

T

oo

T

T

oon

dttntT

dttntT

a

2

2

,)cos()(2

T

T

on dttntyT

a


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24

Solution for an

, n = 1, 2, .

4

0

4

0

4

0

)12()12(cos

)12()12(cos4

)12(sin)12(sin2

18

T

o

oT

o

on

T

oon

ntn

ntn

Ta

dttntnT

a

)12(

1

)12(

1

2

4

)12(

1

)12(

14

nn

nnT oo

p


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Solution for an

, n = 1, 2, .

So:

144

14

2222

nnan

pp

Thus:

...6cos35

14cos

15

12cos

3

142)( tttny ooo

pp

Note:We can only obtain an output signal with a

nonzero average value by using a nonlinear system

with our zero average value input signal


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26

Eulers Identity

)sin()cos( qqq jMMMe j

We could also say:

)sin()cos( qqq je j

)sin()cos(

)sin()cos()(

qq

qq

q

q

je

je

j

j

and ...

i i i S i


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Representing Sin and Cos

with Complex Exponentials

)sin()cos( qqq je j )sin()cos( qqq je j

2

)cos(

)cos(2qq

qq

q

qjj

jj

eeee

j

eejee

jj

jj

2)sin(

)sin(2qq

qq

q

q

Add the equations: Subtract the

equations:

F i A l i F i S i


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Complex Exponential

Representation

The Sine and Cosine functions can be written

in terms of complex exponentials.

tjntjno oo eetn 2

1cos

tjntjno oo eej

tn

2

1sin

F i A l i F i S i


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Complex Exponential Fourier Series

From previous slides

1

02

1

2

1)(

n

tjntjn

n

tjntjn

noooo ee

jbeeaatx

tjntjn

ooo

eetn

2

1cos

tjntjn

ooo

eejtn

2

1sin

Using the Complex Exponential representation of Sine and

Cosine, the Fourier series can be written as:

1

0 )sincos()(n

onon tnbtnaatx

F i A l i F i S i


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Fourier Series with

Complex Exponentials

1

0

10

2

1

2

1)(

22

1

)(

n

tjn

nn

tjn

nn

n

tjntjn

n

tjntjn

n

oo

oooo

ejbaejbaatx

ee

j

beeaatx

Noting that 1/j = -j, we can write:

1

0

2

1

2

1)(

n

tjntjn

n

tjntjn

noooo ee

j

beeaatx

F i A l i F i S i


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Fourier Series with

Complex Exponentials

11

0

1

0

1

0

)(

)(

2

1

2

1)(

n

tjn

n

n

tjn

n

n

tjn

n

tjn

n

n

tjn

nn

tjn

nn

oo

oo

oo

ececctx

ececctx

ejbaejbaatx

Make the following substitutions:

n

tjn

noectx

)(

)(

2

1),(

2

1,00 nnnnnn jbacjbacac

F i A l i F i S i


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Fourier Series with

Complex Exponentials The Complex Fourier series can be written as:

n

tjnn

oectx )(

where:

2

2

0)(1

T

T

tjnn dtetx

Tc

Complex cn

*Complex conjugate

Note: if x(t) is real, c-n= cn*

F i A l i F i S i


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Line Spectra

Line Spectra refers the plotting of discrete coefficients

corresponding to their frequencies

For a periodic signal x(t), cn, n = 0, 1, 2, are

uniquely determined from x(t). The set cnuniquely determines x(t)

Because cnappears only at discrete frequencies,

n(o, n = 0, 1, 2, the set cnis called the discrete

frequency spectrum or line spectrumof x(t).

F i A l i F i S i


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The Cn coefficients are in general complex.

Line Spectra

The standard practice is to make 2 2D plots.

Plot 1: Magnitude of Coefficient vs. frequency

The standard practice is to make 2 2D plots.

Plot 1: Magnitude of Coefficient vs. frequency

Plot 2: Phase of Coefficient vs. frequency



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Magnitude of Cn

Recall that the magnitude for a complex number a+jb iscalculated as follows:

22 bajba



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Phase of Cn Recall that the phase for a complex number a+jb dependson the quadrant that the angle lies in.

a

bTan 1

Quadrant 1: Quadrant 2:

Quadrant 3: Quadrant 4:

a

bTan 1p

a

bTan 1p

a

bTan 1

Angle(a+jb) =



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Amplitude Spectrum of Cn Note: If x(t) is real then |Cn| is

of even symmetry. nn cc

nc

sec)(radoo o2o2



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Phase Spectrum of Cn

nn cc Note: If x(t) is real then the

Phase of Cn is odd

nc

sec)(radoo o2o2



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Example: Finding Complex

Coefficients

Consider the periodic signal x(t) with period T = 2 sec.

Thus:

secsec2

2

sec

2 radradrad

To p

pp

x(t)

t-2.5 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.50

1



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Finding Co(avg)

Co(avg) = 0.5

1

1

5.0

5.0

1

5.0

5.0

1

)( )()()(21)(

21 dttxdttxdttxdttxC avgo

0 0

5.0)5.0(5.0212121 5.0

5.0

5.0

5.0

tdt

The area under x(t) from -1 to -.5 and from .5 to 1 is zero



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Calculating Cn

0,2

sin

n

n

n

Cnp

p

22

0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

1

1

2/

2/

00

2

1

02

10

2

1

2

1

2

1

)(21

)(1

nj

nj

tjn

tjntjntjn

tjn

T

T

tjn

n

ee

nj

dte

dtedtedte

dtetx

dtetxT

C

o

ooo

o

o



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Now it can be shown that:

sin(np/2) = 0 for n = 2, 4, Cn= 0

sin(2p/2) = sin(p) = 0

sin(-4p/2) = sin(-2p) = 0

etc .

It can be also be shown that:

sin(np/2) = -1 for n = 3, 7, 11,

sin(np/2) = 1 for n = 1, 5, 9, sin(3p/2) = -1

sin(-7p/2) = 1

etc .

Factor Evaluation



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,...11,7,3,n

n)signum(

,...9,5,1n,

n

signum(n)

etc...4,2,,0

nC

C

nC

n

n

n

0,2

sin

nn

n

Cnp

pRecall:

Factor Evaluation

Co(avg) = 0.5



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Note: Cn=pif Cnis negative

Therefore:

0n&evenn,0

oddn,

05.

||

1

n

n

Cn

p

and

otherwise,

...117,3,n,

0

p

nC

Summary of Results



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Plot the Magnitude Response



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Plot the Phase Response



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What is Parsevals Theorem ?

Parsevals Theorem states that the average power of a

periodic signal x(t) is equal to the sum of the squared

amplitudes of all the harmonic components of the signal

x(t).

This theorem is excellent for determining the power

contribution of each harmonic in terms of its coefficients



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Parsevals Theorem Average powerof x(t) is calculated from the time

or frequency domain by:

)(2

1)(

1

1

222

2

2

2

n

nno

T

T

avg baadttxT

P

n n

nonavg cccP1

2222

Time Domain:

Frequency Domain:

7 1 Fourier Series

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