ee3tp4 13b fourierseriesproperties v3

8/10/2019 EE3TP4 13b FourierSeriesProperties v3

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Fourier Series Expansion


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x t=k=

cke

jk 0

t Fourier Series

(Complex ExponentialForm)

x t=A0

k=1

Ak

cos k0

tk

Fourier Series

(Trigonometric Form)

These are 3

different

forms for the

same

expression.

x t=a0k=1

[ akcos k0 tbksin k0 t ]

Fourier Series(Trigonometric Form)

T = 2/0

0= fundamental frequency (rad/sec)


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=k=

ck

ejk

0t

x t=a0k=1


How to find cand c! from aand "#

akcos(k0 t)+bksin(k0 t)=ak[1

$e

jk 0 t+1

$ejk 0 t]+bk[

1

$%e

jk 0 t1

$%ejk 0 t]

=[a

k$+

bk

$% ]ejk 0 t+ [

ak

$b

k$% ]e

jk 0 t

ck ck

ck=1

$(akj bk)

ck=1

$(ak+ j bk)

ak=ck+ck

bk=j (ckck)


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&imilarly "y expanding out t'e t'term a"oe using Eulers

Formula you can see t'at

ck=A

k

$ e j k

ck=Ak

$ e

jk

Ak=$ck

k=ck

x t=A0

k=1

Ak

cos k0

tk =

k=

ck

ejk

0t


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c0=A0

ck=1

$Ake

jk k=1* $* +*

ck=1$

Akejk k=1* $* +*

c0=a0

ck=1

$akjbk k=1* $* +*

ck=1

$ akjbk k=1* $* +*

Exponential Form

x t=k=

cke

jk 0t

Trig Form: Amplitude & Phase

x t=A0

k=1

Akcos k

0t

k

Trig Form: Sine-Cosinex t=a

0

k=1


a0=c

0

ak=ckck, k=1*$*+*

bk=jckck, k=1*$*+*

A0=c0

Ak=$ck k=1* $* +*

k, ck k=1* $* +*

A0

=a0

Ak=ak$bk

$

k=tan1bka

k

a0=A0

ak=Akcosk

bk=A

ksin

k

Three (Eui!alent" Forms of FS and Their #elationships


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x t=k=

cke

jk 0

t

ck= 1

Tt0t0Tx tejk 0 tdt

Complex Exponential Form


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Fourier Series Trigonometric Form

f(t)=a0+k=1 [akcos(k0 t)+bksin (k0 t)]

ak=$

T

Tf(t)cos(k0 t)dt (k0)

bk= $TT f(t)sin(k0 t)dt (k0)

a0

=1

TT

ftdt

-e can derie t'ese results in t'e same way as t'e

complex exponential case using ort'ogonal functions.


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f(t) = 1 for t = [-T/2, T/2]


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Fourier Series Properties


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x t=k=

cke

jk 0

t Fourier Series


x t=A0

k=1

Ak

cos k0

tk

Fourier Series


These are 3

different

forms for the

sameexpression.

x t=a0k=1



T = 2/0



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=

k=

ck

ejk

0t

x t=a0k=1


How to find cand c

!from a

and "

#

akcos(k0 t)+bksin(k0 t)=ak[1

$

ejk 0 t+

1

$

ejk 0 t]+bk[

1

$%

ejk 0 t

1

$%

ejk 0 t]

=[ak

$+

bk

$%]e jk 0 t+ [

ak

$

bk

$%]ejk 0 t

ck ck

ck=1

$(akj bk)

ck=1

$(ak+ j bk)

ak=ck+ck

bk=j (ckck)

ecall from last time


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Fourier &eries roperties

if x (t)=a0+k=1

[ akcos (k0 t)+bksin (k0 t) ] is real*t'en "ot' akand bkare real 2nd since

ck=akj bk

$

ck=ak+ j bk

$

ck=ck3

t'en

ck=ck ck=ckie*

4Hermitian &ymmetry5

6agnitude is an een function 'ase is an odd function


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ecall t'e full wae rectifier example t'at we did -e deried

and plotted t'e complex exponential F& coefficients for t'e

rectified sinusoid

The t$o-sided spe%trum after the re%tifier:

A$

+

$A

0

0$ 0+ 07 0800$0+0708

0

0$ 0+ 07 08

0

0$

0+

07

08

ck

kc

x (t)=Asin ($9

T1t)

:ote t'e Hermitian symmetry.

t

x(t)

A


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Fourier Series

x t=k=

cke

jk 0

t

ck=

1

T

t0

t0Tx te

jk 0

tdt

T = 2/0


T'is is true for a

ery wide class of

periodic signals.

(see ;iric'let

Conditions)


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Time s'ifting a periodic signal creates a new periodic signal. How

are t'e Fourier series coefficients of t'e two signals related#


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hat has the time shift

done to the FS %oeffi%ents'


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Fourier Series

x t=k=

cke

jk 0

t

ck=

1

T

t0

t0Tx te

jk 0

tdt

T = 2/0



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ftcn

ftt0cn ejn0 t0


Time &'ifting


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x t=k=

cke

jk 0

t Fourier Series


x t=A0

k=1

Ak

cos k0

tk

Fourier Series


These are 3

different

forms for the

sameexpression.

x t=a0k=1



T = 2/0



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Can we use t'e

differentiation property

to 'elp find t'e F&coefficients off(t) ?


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6atla" example


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ftcn

ftt0cn ejn0 t0


Time &'ifting

Time ;ifferentiation

ftcndft

dt jn0cn

E l


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Find t'e Fourier series coefficients* dn* of !(t)

Example

dn=1

TT/$

T/$!tejn0 tdt

dn=1TT/7

T/7tT

7ejn0 tdt

-e 'ae to integrate t'is "y parts. =nstead* we can use

t'e differentiation t'eorem.

E l


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Example


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ee3tp4 13b fourierseriesproperties v3

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