Ee602 Fourier Series

7/27/2019 Ee602 Fourier Series

1/110

EE602 CIRCUIT ANALYSIS

The Fourier Series


2/110

A Fourier series is an expansion of a

periodicfunctionf(t) in terms of an infinite sum

ofcosines and sines

Introduction

1

0 )sincos()(

n

nn tnbtnaatf


3/110

In other words, any periodicfunction can be

resolved as a summation of

constant value and cosine and sine functions:

10 )sincos()(

nnn tnbtnaatf

)sincos( 11 tbta 0a

)2sin2cos( 22 tbta

)3sin3cos( 33 tbta

dc ac


4/110

The computation and study of Fourierseries is known as harmonic analysis and

is extremely useful as a way to break up

an arbitrary periodic function into a set of

simple terms that can be plugged in,

solved individually, and then recombined

to obtain the solution to the original

problem or an approximation to it towhatever accuracy is desired or practical.


5/110

=

+ +

+ + +

Periodic Function0a

ta cos1

ta 2cos2

tb sin1

tb 2sin2

f(t)

t


6/110

1000 )sincos()(

nnn tnbtnaatf

where frequenclFundementa2

0 T

Tt

tn dttntfTa

0

00cos)(

2

Tt

tdttf

Ta

0

0

)(1

0

Tt

t

n dttntf

T

b0

0

0sin)(2


7/110

Notice that

componentdc

valueAverage

periodaovergraphbelowArea

)(1 0

00

T

dttfT

aTt

t

[*The right sign of the area below graph must be

obeyed. +ve sign for area above x-axis & ve sign

for area below x-axis]


8/110

Symmetry Considerations

Symmetry functions:

(i) even symmetry

(ii) odd symmetry


9/110

Even symmetry

Any functionf(t) is even if its plot is

symmetrical about the vertical axis, i.e.

)()( tftf


10/110

Even symmetry (cont.)

The examples ofeven functions are:

2)( ttf

t t

t

||)( ttf

ttf cos)(


11/110

Even symmetry (cont.)

The integral of an even function from A to

+A is twice the integral from 0 to +A

t

AA

Adttfdttf

0

eveneven )(2)(A +A

)(even tf


12/110

Odd symmetry

Any functionf(t) is odd if its plot is

antisymmetrical about the vertical axis, i.e.

)()( tftf


13/110

Odd symmetry (cont.)

The examples ofodd functions are:

3)( ttf

t t

t

ttf )(

ttf sin)(


14/110

Odd symmetry (cont.)

The integral of an odd function from A to

+A is zero

t 0)(odd

A

A

dttfA +A

)(odd

tf


15/110

Even and odd functions

(even)(even) = (even)

(odd)(odd) = (even)

(even)(odd) = (odd) (odd)(even) = (odd)

The product properties ofeven and odd

functions are:


16/110

Symmetry consideration

From the properties ofeven and odd

functions, we can show that:

foreven periodic function;

2/

0

cos)(4

T

n tdtntfT

a 0nb

forodd periodic function;

2/

0

sin)(4

T

n tdtntfT

b 00 naa


17/110

How?? [Even function]

2

T

2

T

2/

0

2/

2/

cos)(4

cos)(2

TT

T

n tdtntfT

tdtntfT

a

(even)

(even)

| |

(even)

0sin)(2

2/

2/

T

T

n tdtntfT

b

(even) (odd)

| |

(odd)

)(tf

t


18/110

How?? [Odd function]

2

T

2

T

2/

0

2/

2/

sin)(4

sin)(2

TT

T

n tdtntfT

tdtntfT

b

(odd)

(odd)

| |

(even)

0cos)(22/

2/

T

T

n tdtntfT

a

(odd) (even)

| |

(odd)

)(tf

t

0)(2

2/

2/

0

T

T

dttfT

a

(odd)


19/110

The amplitude-phase form

is known as the sine-cosine form

We can also express the Fourier series in

the cosine form only, that is

This form is called as the amplitude-phase

form

1

000 )sincos()(n

nn tnbtnaatf

100 )cos()(

nnn tnAatf


20/110

From this form, we can plot the amplitude

spectrum, vs. n and the phasespectrum, vs. n.

It can be shown that the combination of

cosine and sine function can be expressedas a cosine function only:

Comparing both sides of eqn:

n

A

n

xAxA

xxA

xAxbxa

sin)sin(cos)cos(

)sinsincos(cos

)cos(sincos

.....(1cos aA .....(2)sin bA


21/110

22222

22222

222222

)sin(cos

sincos

baAbaA

baA

baAA

a

b

a

ba

b

A

A

1tantan

cos

sin

:)2()1( 22

:

)1(

)2(


22/110

Hence

)cos(sincos 000 nnnn tnAtnbtna

where

22

amplitude, nnn baA

n

n

na

b1tanphase,

Or in phasor/complex form:

n

nnnnnnn

a

bbajbaA 122 tan

00 aA


23/110

Example 1

Determine the Fourier series of the following

waveform. Obtain the amplitude and phase

spectra.


24/110

SolutionFirst, determine the period & describe the one period

of the function:

T= 2

21,0

10,1)(

t

ttf )()2( tftf

T

2

0We find that


25/110

Then, obtain the coefficients a0, an and bn:

2

1)01(2

1012

1)(2

1

)(1

2

1

1

0

2

0

0

0

dtdtdttf

dttfT

aT

Or

21

211graphbelowArea

0 T

a


26/110

n

n

n

tndttdtn

tdtntf

T

an

sinsin0cos1

cos)(2

1

0

2

1

1

0

2

0

0

Notice that n is integer which leads ,

since

0sin n03sin2sinsin

Therefore, .0na


27/110

n

n

n

tndttdtn

tdtntfT

bn

cos1cos0sin1

sin)(2

1

0

2

1

1

0

2

00

15cos3coscos 16cos4cos2cos

Notice that

Therefore,

even,0

odd,/2)1(1

n

nn

nb

n

n

orn

n )1(cos


28/110

ttt

tnn

tn

n

tnbtnaatf

nn

n

n

n

nn

5sin5

23sin

3

2sin

2

2

1

sin2

2

1

sin)1(1

2

1

)sincos()(

odd1

1

1

000

Finally,


29/110

)905cos(5

2

)903cos(3

2)90cos(

2

2

1

)90cos(221

sin2

2

1)(

odd1

odd1

t

tt

tnn

tnn

tf

nn

nn

In amplitude-phase form,


30/110

Amplitude spectrum: Phase spectrum:


31/110

Some helpful identities

Forn integers,n

n )1(cos 0sin n02sin n 12cos n

xx sin)sin(

xx cos)cos(

)90cos(sin xx

)90cos(sin xx

)180cos(cos xx


32/110

The sum of the Fourier series terms can

evolve (progress) into the original

waveform

From Example 1, we obtain

ttttf

5sin

5

23sin

3

2sin

2

2

1)(

It can be demonstrated that the sum will

lead to the square wave:

Notes:


33/110

tttt

7sin7

25sin

5

23sin

3

2sin

2ttt

5sin

5

23sin

3

2sin

2

tt

3sin3

2sin

2t

sin

2

(a) (b)

(c) (d)


34/110

ttttt

9sin9

27sin

7

25sin

5

23sin

3

2sin

2

ttt

23sin23

23sin

3

2sin

2

2

1

(e)

(f)


35/110


36/110

Example 2

Given ,)( ttf 11 t

)()2( tftf

Sketch the graph off(t) such that .33 t

Then compute the Fourier series expansion off(t).

Plot the amplitude and phase spectra until the forthharmonic.


37/110

Solution

The function is described by the following graph:

T= 2

T

20We find that


38/110

nnn

n

n

n

ntn

nn

dtn

tn

n

tnt

tdtnttdtntfT

b

nn

n

1

22

1

0

22

1

0

1

0

1

0

1

0

0

)1(20

)1(2sin2cos2

sin2cos2

cos2

cos2

sin2

4sin)(

4

Then we compute the coefficients:

00 naa sincef(t) is an odd function.


39/110

)904cos(2

1)903cos(

3

2

)902cos(1

)90cos(2

4sin2

13sin

3

22sin

1sin

2

sin)1(2

)sincos()(

1

1

1

000

tt

tt

tttt

tn

n

tnbtnaatf

n

n

n

nn

Hence,


40/110


0.64

0.32

0.210.16


41/110

Example 3

Given

42,0

20,2)(

t

tttv

)()4( tvtv

(i) Sketch the graph ofv (t) such that .120 t

(ii) Compute the trigonometric Fourier series ofv (t).

(iii) Express the Fourier series ofv (t) in the

amplitude-phase form. Then plot the amplitude

and phase spectra until the forth harmonic.


42/110

Solution

(i) The function is described by the following graph:

T= 4

2

20

T

We find that

0 2 4 6 8 10 12t

v (t)

2


43/110

(ii) Then we compute the coefficients:

21

22

41)2(

41

0)2(4

1)(

1

2

0

22

0

4

2

2

0

4

0

0

ttdtt

dtdttdttvT

a

Or

2

1

4

2221

0

a


44/110

odd,/4

even,0])1(1[2)cos1(2

2

2cos1cos

2

10

sin

2

1sin)2(

2

1

0cos)2(21cos)(2

222222

2

0

2

0

2

0

2

0

2

0

2

0 0

0

2

00

0

4

2

2

0

0

4

0

0

nn

n

nn

n

n

n

n

tn

dt

n

tn

n

tnt

tdtnttdtntvT

a

n

n


45/110

nnnn

n

n

tn

n

dtn

tn

n

tnt

tdtnttdtntvT

bn

212

2sin1

sin

2

11

cos

2

1cos)2(

2

1

0sin)2(2

1sin)(

2

0

2

0

2

0

0

2

0

2

0

2

0

0

2

0 0

0

2

00

0

4

2

2

00

4

00

since 0sin2sin 0 nn


46/110

1

00

122

1

000

)cos(

2sin

2

2cos

])1(1[2

2

1

)sincos()(

n

nn

n

n

n

nn

tnAa

tn

n

tn

n

tnbtnaatv

Hence,


47/110

odd,2

tan142

even,902

odd,22

even,2

1

22n

n

nn

n

n

njnn

nn

j

jbaA nnnn

(iii) Where


48/110

5.5775.011 A

0.7822.033 A

9032.022 A

9016.044 A

5.000 aA

We obtained that

902cos16.00.782

3cos22.0

90cos32.05.572cos75.05.0

)cos()(1

00

tt

t

t

tnAatvn

nn


49/110


0.5

0.75

0.32

0.22

0.16

0 /2 3/2 2

/2 3/2 2

57.5

78.090 90


50/110

Example 4

Given

21,1

11,

12,1

)(

t

tt

t

tf

)()4( tftf

Sketch the graph off(t) such that .66 t

Then compute the Fourier series expansion off(t).


51/110

Solution

The function is described by the following graph:

T= 4

2

2

T

We find that

046 2 4 6t

f(t)

2

1

1


52/110

Then we compute the coefficients. Sincef(t) is

an odd function, then

0)(2

2

2

0

dttfT

a

0cos)(2

2

2

tdtntf

T

an

and


53/110

2222

1

0

22

2

1

1

0

1

0

2

1

1

0

2

0

2

2

)2/sin(4cos2sin2cos

cos2cossincos

coscoscos

sin1sin4

4

sin)(4

sin)(2

n

n

n

n

n

n

n

nn

nn

n

tn

n

n

ntndt

ntn

ntnt

tdtntdtnt

tdtntfT

tdtntfT

bn

since 0sin2sin

nn


54/110

1

1

1

1

0

2sin)1(2

2sin

cos2

)sincos(2

)(

n

n

n

n

nn

tnn

tn

n

n

tnbtnaa

tf

Finally,


55/110

Example 5

Compute the Fourier series expansion off(t).


56/110

Solution

The function is described by

T= 3

3

220

Tand

32,1

21,2

10,1

)(

t

t

t

tf

)()3( tftf

T= 3


57/110

Then we compute the coefficients.

3

81

2

32)01(

3

421

3

4)(

4)(

22/3

1

1

0

2/3

0

3

0

0

dtdtdttfT

dttfT

a

3

8)23()12(2)01(

3

2121

3

2)(

2 3

2

2

1

1

0

3

0

0

dtdtdtdttfTa

Or, sincef(t) is an even function, then

Or, simply

3

84

3

2

periodain

graphbelowareaTotal2)(

23

0

0

T

dttfT

a


58/110

3

2sin

2

3

2sinsin2

2

sin2

3sin2

3

4

sin2

3sin2sin

3

4

sin2

3

4sin

3

4

cos2cos13

4

cos)(4

cos)(2

2/3

1

1

0

2/3

1

1

0

2/3

0

3

0

n

n

nn

n

nn

n

nn

nn

n

tn

n

tn

tdtntdtn

tdtntfT

tdtntfT

an

;3

2


59/110

1

1

1

0

3

2cos

3

2sin

12

3

4

3

2cos

3

2sin

2

3

4

)sincos(2

)(

n

n

n

nn

tnn

n

tnn

n

tnbtnaa

tf

Finally,

and 0nb sincef(t) is an even function.


60/110

Parsevals Theorem

Parservals theorem states that the

average power in a periodic signal is equal

to the sum of the average power in its DCcomponent and the average powers in its

harmonics


61/110

=

+ +

+ + +

ta cos1

ta 2cos2

tb sin1

tb 2sin2

f(t)

t

PavgPdc

Pa1 Pb1

Pa2 Pb2

0a


62/110

For sinusoidal (cosine or sine) signal,

R

V

R

V

R

VP

2

peak

2

peak2

rms

2

12

For simplicity, we often assumeR= 1,

which yields2

peak

2

rms21VVP


63/110

And since ,

Therefore, the total power of all harmonics

is

2

2

2

2

2

1

2

1

2

0

dcavg

2

1

2

1

2

1

2

1

2211

babaa

PPPPPP baba

1

22

0

1

222

0avg2

1)(

2

1

n

n

n

nn AabaaP

2rm sVP

1

22

0

1

222

0rms

2

1)(

2

1

n

n

n

nn AabaaV


64/110

Circuit applications

Steps for applying Fourier series:

1. Express the excitation as a Fourier

series

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 responses

using the superposition principle


65/110

The signal in Figure 6.1 is applied to the circuit in

Figure 6.2.

(i) Find the Fourier series of is(t).(ii) Plot the amplitude spectrum for the dc

component and the first three non-zero

harmonics ofis(t).

(iii) Find the load current iL(t).(iv) Find the average power dissipated in the

resistorRL.

Example 6


66/110

)(tis

)(tis

H1

1 1

LR

)(tiL

Figure 6.2

Figure 6.1


67/110

Solution

(i) The current source is described by

and

0,

0,

,||)(

tt

tt

tttis

1,2 0 T


68/110

Compute the Fourier series. Since is(t) is an even

function,

or

22

1

2

2)(

2

0

2

0

2/

0

0

ttdtdtti

Ta

T

s

22

11

0over

graphbelowArea10

t

a


69/110

even,0

odd,/4]1)1[(2

)1(cos2

cos2sin2

sin2sin2cos

2

4

cos)(4

2

2

2

0

2

000

2/

0

0

n

nn

n

n

n

nnt

nn

dtn

nt

n

nttntdtt

tdtntiT

a

n

T

sn


70/110

Therefore, the Fourier series ofis

(t) is

0nb

)1805cos(

25

4

)1803cos(94)180cos(4

2

)180cos(4

2

cos4

2)(

odd1

2

odd1

2

t

tt

ntn

ntn

ti

nn

nn

s


71/110

nsI nsI

n n

1.5711.273

0.141

0.051

(ii) Hence,

,5,3,1,18042 nn

Ins 20

sI


72/110

(iii) To compute iL(t), separate to dc and ac analysis:

dc analysis (n = 0)

S/c the inductor, hence

0sI

11

42

2/

2

0

0

s

L

II


73/110

1802

tantan4

14

180

4

2

1

,)(

11

2

2

2

2

0

nn

n

n

n

njn

jn

nnILjRR

LjRI nsnL

nL nn

s

L

L ILjRR

LjRI)(

ac analysis (n 1)

or


74/110

)49.1905cos(048.0)26.1953cos(124.0

)43.198cos(805.0785.0)(

tt

ttiL

Hence,

49.190048.0

,26.195124.0,43.198805.0

,785.0

5

31

0

L

LL

L

I

II

I


75/110

W949.0)048.0124.0805.0(2

1785.0

2

1

2222

1

22

2)(

0

L

n

LL

LrmsL

RII

RIP

n

(iv) Average power absorbed byRl is


76/110

The signal in Figure 7.1 is applied to the circuit in

Figure 7.2.

(i) Find the Fourier series of vs(t).

(ii) Find the output voltage vo(t).(iii) Plot the amplitude spectrum for the dc

component and the first three non-zero harmonics

of the output voltage vo(t).

(iv) Find the r.m.s value of vo(t) .

Example 7

)(tv


77/110

)(tvsF1

1

)(tvo

Figure 7.2

Figure 7.1

1

)(tvs


78/110

Solution

(i) The voltage source is described by

and

01,1

10,1)(

t

ttv

s

0,2T

Compute the Fourier series. Since vs(t) is an odd

function,

00 naa


79/110

even,0

odd,/4])1(1[2

)cos1(2

cos2sin

2

4

sin)(4

1

0

1

0

2/

00

n

nn

n

n

n

n

tntdtn

tdtntvT

b

n

T

sn


80/110

Therefore, the Fourier series ofvs(t) is

)905cos(5

4

)903cos(3

4)90cos(

4

)90cos(4sin4)(

odd1

odd1

t

tt

tnn

tnn

tv

nn

nn

s

Hence,

,5,3,1,904

nn

Vn

s

0

0

sV


81/110

(ii) To compute vo(t), separate to dc and ac analysis:

dc analysis (n = 0)

O/c the capacitor, hence

00 sV

1

00 oV

1

0I


82/110

902tantan41

14

90

4

21

1

,21

1

11

22

22

0

nnn

n

n

nnj

jn

nnVRCj

RCjV nsn

no nn

sso VRCjRCjV

CjRRCjRV

211

)]/1([)/1(

ac analysis (n 1)

or


83/110

)82.915cos(128.0

)02.933cos(213.0)61.98cos(660.0)(

t

tttvo

Hence,

82.91128.0,02.93213.0,61.98660.0

,0

5

31

0

o

oo

o

VVV

V


84/110

noV

noV

n

n

0.660

0.213

0.128

98.61

(iii)

93.02

91.82

(iv)

V705.0)128.0213.0660.0(2

1

2

1

222

1

22

)rms( 0

nooo nVVV


85/110

Exponential Fourier series

Recall that, from the Eulers identity,

xjxe jx sincos

yields

2

cosjxjx ee

x

2

sin

j

eex

jxjx and


86/110

Then the Fourier series representation becomes

11

0

1

0

1

0

1

0

1

0

222

222

222

222

)sincos(2

)(

n

tjnnn

n

tjnnn

n

tjnnntjnnn

n

tjntjn

n

tjntjn

n

n

tjntjn

n

tjntjn

n

n

nn

ejba

ejbaa

ejbaejbaa

eejb

eea

a

j

eeb

eea

a

tnbtnaatf

jb jb


87/110

Here, let we name

11

0

222)(

n

tjnnn

n

tjnnn ejba

ejbaa

tf

2

nnn

jbac

,

2

nnn

jbac

Hence,

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

ececcec

ececc

ececc

1

0

1

11

0

11

0

and .20

0

a

c

c0 cncn


88/110

Then, the coefficient cn can be derived from

T

tjn

T

TT

TT

nnn

dtetfT

dttnjtntfT

tdtntfjtdtntfT

tdtntfT

jtdtntf

T

jbac

0

0

00

00

)(1

]sin)[cos(

1

sin)(cos)(1

sin)(2

2cos)(

2

2

1

2


89/110

In fact, in many cases, the complex

Fourier series is easier to obtain rather

than the trigonometrical Fourier series

In summary, the relationship between the

complex and trigonometrical Fourier series

are:

nnnn

n Ajbac

21

2 2

nnn

jbac

T

dttfT

ac0

00 )(1

T

tjn

n dtetfT

c0

)(1

nn ccor

22

22

nnn

n

Abac


90/110

The average power and the rms value in

the term of Fourier complex coefficient are

1

22

0

2

1 2n

n

n

n cccP

1

22

0

2

rms 2n

n

n

n cccF


91/110

Example 8

For the given function,

(i) Obtain the complex Fourier series

(ii) Plot the amplitude and the phase spectra

of the complex Fourier series for5

n

5(iii) Calculate the average power and the rms value

of the signal

2

4

4

2

0

2e

1

)(tv

t


92/110

(i) Since , . Hence

Solution

2

1

2

1

2

1

)(1

22

0

2

0

0

0

ee

dte

dttv

T

c

t

t

T

10 2T

T


93/110

)1(21

)1(21

)1(21

12

1

2

1

2

1

)(1

222)1(2

2

0

)1(

2

0

)1(

2

0

0

0

jne

jnee

jne

jn

e

dtedtee

dtetvT

c

njjn

tjn

tjnjntt

T

tjn

n

since 1012sin2cos2 njne nj


94/110

jnt

nn

tjn

n ejn

eectv

)1(2

1)(

2

0

Therefore, the complex Fourier series ofv(t) is

0

2

0

2

0 2

1

)1(2

1c

e

jn

ec

nnn

*Notes: Even though c0 can be found by substituting

cn with n = 0, sometimes it doesnt works (as shown

in the next example). Therefore, it is always better to

calculate c0 alone.

ne2 1


95/110

nn

n

n

ec nn

1

2

1

2

tan1

85

1tan0

12

1

(ii)

The complex frequency spectra are


96/110

kW19.20

)7.166.209.26381.60(285 222222

2

1

n

ncP

(ii) The average power is (assumeR= 1 )

and since 2rm s1 VP

V09.142k19.20rms V


97/110

Example 9

Obtain the complex Fourier series of the function in

Example 1. Then plot the complex frequency spectra

for4 n4 and calculate the rms value of the signal.


98/110

Solution

2

11

2

1)(

11

00

0 dtdttfTc

T 0

)1(22

1

012

1)(

1

1

0

2

1

1

00

jntjn

tjn

T

tjn

n

enj

jne

dtedtetfT

c


99/110

)1(2

jn

n en

jc

Butnjn nnjne )1(cossincos

Thus,

even,0

odd,/]1)1[(

2 n

nnj

nj n

Therefore,

odd

0

2

1)( 0

n

nn

tjn

n

tjn

n en

jectf

*Here notice that .00 cc nn

00c For even


100/110

2

10 c

0,0 nnc Forneven,

Fornodd,

0,90

0,90,1

n

n

nc nn

0.5


101/110

692.0)11.032.0(25.0

222

2

rms

nncF

The average rms value is


102/110

Summary

Sine-cosine form

Amplitude-phase form

Exponential (or complex) form

1

000 )sincos()(n

nn tnbtnaatf

,cos)(2

00

T

n

tdtntfT

a

T

n tdtntf

T

b0

0sin)(2

,)(1

00

T

dttfT

a,2

0

T

1

00 )cos()(n

nn tnAatf

n

tjn

nectf0)(

nnnn

jbaA

,00

ac ,2

nnn

jbac

*nn cc


103/110

Prob.17.37, pg.803

If the sawtooth waveform in Fig.17.9 is the

voltage source vs(t) in the circuit of

Fig.17.22, find the response vo(t).


104/110

Prob.17.37, pg.803

If the periodic current waveform in

Fig.17.73(a) is applied to the circuit in

Fig.17.73(b), find vo.

3

)(tis

0 1 2 3

1

3

t

(a) (b)


105/110

Problem

1 2 30 4

1

5

2

3

6

1oR

1sR F1C

)(tvo

)(tvs

)(tvs

t

(i) Compute the output voltage vo(t).(ii) Plot the amplitude and phase spectra ofvo(t)

for the first three nonzero harmonics.

(ii) Calculate the average power dissipated in

the resistor Ro.


106/110

QUIZ 1

)(tvs

)(tvo

H1

H1

1

1

Given the Fourier series expansion of thevoltage signal vs(t) is

Find vo

(t) in the amplitude-phase form.

1

sin10

5)(

n

s nt

n

tv


107/110

Quiz1

Determine the Fourier series of the following

waveform.

)(tv

4


108/110

QUIZ 2

Find i(t) ifv(t) is the signal as given in QUIZ 1.

odd1odd1

)90cos(2

2

1sin

2

2

1)(

nnnn

tn

n

tn

n

tv


109/110

QUIZ 3

Find the output voltage vo(t) if the currentsource is(t) is given by

1

)90cos(2

1)(ns

ntn

ti

)(tis

)(tvo5 F5.0

3


110/110

QUIZ 4

Using time differentiation technique, find theFourier transform off(t).

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Ee602 Fourier Series