Signals & systems Ch.3 Fourier Transform of Signals and LTI System

Signals & systemsCh.3 Fourier Transform of

Signals and LTI System

04/21/23

Signals and systems in the Frequency domain

04/21/23 KyungHee University 2

Time [sec]

Frequency [sec-1, Hz]

Fourier transform

3.1 Introduction

Orthogonal vector => orthonomal vector

What is meaning of magnitude of H?


)2/1,2/1()2/1,2/1( 1 vvo

1,0,][110 1101 jiforjivvvvvvvv jioo

),(),( 1010 HHHhhh Fourier

110 vHvHh o

110 vHvHh o

110 vhHvhH o

Any vector in the 2- dimensional space can be represented by weighted sum of 2 orthonomal vectors

Fourier Transform(FT)

Inverse FT

3.1 Introduction cont’

CDMA?


1 1 1 1( , , , )2 2 2 2ov

Orthogonal

? 1 0 1 10 1 1o ov v v v v v

[ ] , 0,1,2,3i jv v i j for i j

Any vector in the 4- dimensional space can be represented by weighted sum of 4 orthonomal vectors

Orthonormal

function?][)()(

2)cos()( 00 jidttvtv

TwheretkAtv

T jik

][)()(1

)()(2

)( **0

0 kmdttvtvT

tvtvT

whereetvT kmkm

tjkk

T

tkmj

T km dteT

dttvtvT

0)(* 1)()(

1

1

1 1 1 1( , , , )2 2 2 2

v

2

1 1 1 1( , , , )2 2 2 2

v

3

1 1 1 1( , , , )2 2 2 2

v

3.2 Complex Sinusoids and Frequency Response of LTI Systems


cf) impulse response

.][ ][][][ )(

k

knj

k

ekhknxkhny

,)( ][][ njj

k

kjnj eeHekheny

.][)(

k

kjj ekheH

How about for complex z? (3.1) nznx ][

,)( )( )()( )( tjjtjtj ejHdehedehty

.)()(

dehjH j

How about for complex s? (3.3) stetx )(

.)()( )})(arg{( jHtjejHty

Magnitude to kill or not? Phase delay

Fourier transform


2

1[ ] ( )

2j j nx n X e e d

( ) [ ]j j n

n

X e x n e

1

( ) ( )2

j tx t X j e d

( ) ( ) j tX j x t e dt

discrete time

Continuous time

Time domain frequency domain

1( ) ( )

2stx t X s e ds

j ( ) ( ) stX s x t e dt

Laplace transform

11[ ] ( )

2nx n X z z dz

j ( ) [ ] n

n

X z x n z

z-transform

(periodic) -

(discrete)

(discrete) -

(periodic)

3.6 DTFT: Discrete-Time Fourier Transform


(discrete) (periodic)

deeXnx njnj )(

2

1][

n

njj enxeX ][)(

)(][ jDTFT eXnx

n

nx ][

n

nx2

][

(3.31)

(3.32)

(a-periodic) (continuous)


Example 3.17 DTFT of an Exponential Sequence

Find the DTFT of the sequence ][][ nunx n

Solution :

n n

njnnjnj eenueX0

][)( 1,1

1)(

0

j

n

nj

ee

sincos1

1)(

jeX j

2/122/1222 )cos21(

1

)sin)cos1((

1)(

jeX

)cos1

sinarctan()}(arg{

jeX

x[n] = nu[n]. magnitude

phase

= 0.5 = 0.9

= 0.5 = 0.9

3.6 DTFT Example 3.17


Example 3.18 DTFT of a Rectangular Pulse

Mn

Mnnx

,0

,1][ ].[nxLet Find the DTFT of

Solution : (square) (sinc)

Figure 3.30 Example 3.18. (a) Rectangular pulse in the time domain. (b) DTFT in the frequency domain.



,4,2,0,12

,4,2,0,1

1

)(

)12(

2

9

2

0

)(

Me

ee

eeeeX

j

Mjmj

M

m

mjmjM

m

Mmjj

2/2/

2/)12(2/)12(

)2/2/2/

2/)12(2/)12(2/)12(

)(

)()(

jj

MjMj

jjj

MjMjMjMjj

ee

ee

eee

eeeeeX

)2/sin(

)2/)12(sin()(

MeX j

12)2/sin(

)2/)12(sin(lim

,,4,2,0

M

M

)2/sin(

)2/)12(sin()(

MeX j



Example 3.19 Inverse DTFT of a Rectangular Spectrum

Find the inverse DTFT of

W

WeX j

,0

,1)(

Solution : (sinc) (square)

0),sin(1

2

1

2

1][

nWnnW

We

njdenx njW

W

nj

W

Wnn

xn

)sin(1

lim]0[0

)sin(1

][ Wnn

nx )/(][

Wnncsi

Wnx

Figure 3.31 (a) Rectangular pulse in the frequency domain. (b) Inverse DTFT in the time domain.



Example 3.20 DTFT of the Unit ImpulseFind the DTFT of ][][ nnx

Solution : (impulse) - (DC) 1][)(

nj

n

j eneX

1][ DTFTn

Example 3.21 Find the inverse DTFT of a Unit Impulse Spectrum.

Solution : (impulse train) (impulse train)

denx nj)(

2

1][

k

j keX )2()(

),(2

1 DTFT

3.6 DTFT Example 3.20-21


Example 3.23 Multipath Channel : Frequency Response

]1[][][ naxnxny

Solution : jj aeeH 1)( })arg{(1 ajea })arg{sin(})arg{cos(1 aajaa

2/122/1222 }))arg{cos(21(}))arg{(sin}))arg{cos(1(()( aaaaaaaeH j

1,1

1)(

aae

eHj

jinv

( arg{ })

1 1( )

1 1 cos( arg{ }) sin( arg{ })inv j

j aH e

a e a a j a a

)(

1

}))arg{cos(21(

1)(

2/12

j

jinv

eHaaaeH

|)(| jeH (a) a = 0.5ej2/3. (b) a = 0.9ej2/3.

|)(| jinv eH (a) a = 0.5ej2/3. (b) a = 0.9ej2/3.


3.7 CTFT


(continuous aperiodic) (continuous aperiodic)

Inverse CTFT (3.35)

dwejwXtx jwt)(

2

1)(

CTFT (3.26)

dtetxjwX jwt)()(

)()( jwXtx CTFT

Condition for existence of Fourier transform:

dttx

2)(

3.7 CTFT Example 3.24


Example 3.24 FT of a Real Decaying ExponentialFind the FT of )()( tuetx at

Solution : 0,0

adte at Therefore, FT not exists.

jwae

jwa

dtedtetuejwXa

tjwa

tjwajwtat

11

)()(,0

0

)(

0

)(

22

1)(

wajwX

)/arctan()}(arg{ awjwX

LPF or HPF? Cut-off from 3dB point?

)()( tuetx at



Example 3.25 FT of a Rectangular Pulse

0

00

,0

,1)(

Tt

TtTtx Find the FT of x(t).

Solution : (square) (sinc)

0),sin(21

)()( 0

0

0

0

0

wwTw

ejw

dtedtetxjwXT

T

jwtT

T

jwtjwt

000

2)sin(2

lim,0 TwTw

wForw

Example 3.25. (a) Rectangular pulse. (b) FT.

)sin(2

)( 0wTw

jwX )/(2)( 00 wTincsTjwX

w

wTjwX

)sin(2)( 0

0/)sin(,

0/)sin(,0)}(arg{

0

0

wwT

wwTjwX



Example 3.26 Inverse FT of an Ideal Low Pass Filter!!Fine the inverse FT of the rectangular spectrum depicted in Fig.3.42(a) and given by

Ww

WwWjwX

,0

,1)(

Solution : (sinc) -- (square)

)()()sin(1

)(

,/)sin(1

lim,0

0),sin(1

2

1

2

1)(

0

Wtncsi

WtxorWt

ttx

WWtt

twhen

tWtt

etj

dwetx

t

W

W

jwtW

W

jwt

3.7 CTFT Example 3.27-28


Example 3.27 FT of the Unit Impulse

)()( ttx

Solution : (impulse) - (DC)

1)()( dtetjwX jwt 1)( FTt

Example 3.28 Inverse FT of an Impulse Spectrum

Find the inverse FT of )(2)( wjwX

Solution : (DC) (impulse)

1)(22

1)(

dwewtx jwt

)(21 wFT



Example 3.29 Digital Communication SignalsRectangular (wideband)

2/,0

2/)(

0

0,

Tt

TtAtx

r

r

Separation between KBS and SBS. Narrow band

2/,0

2/)),/2cos(1)(2/()(

0

00

Tt

TtTtAtx

c

c

Figure 3.44 Pulse shapes used in BPSK communications. (a) Rectangular pulse. (b) Raised cosine pulse.



Figure 3.45 BPSK (a) rectangular pulse shapes

(b) raised-cosine pulse shapes.

Solution : 22/

2/

2

0

0

0

1r

T

T rr AdtAT

P

crc

T

T

c

T

T cc

AAA

dtTtTtT

A

dtTtAT

P

3

8

8

3

)]/4cos(2/12/1/2cos(21[4

))/2cos(1)(4/(1

2

2/

2/ 000

2

2/

2/

20

2

0

0

0

0

0

the same power constraints

w

wTjwX r

)sin(2)( 2/0

f

fTjfX r

)sin()( 0'

2/

2/ 0

0

0

))/2cos(1(3

8

2

1)(

T

T

jwtc dteTtjwX

2/

2/

)/2(2/

2/

)/2(2/

2/

0

0

00

0

00

0 3

2

2

1

3

2

2

1

3

2)(

T

T

tTwjT

T

tTwjT

T

jwtc dtedtedtejwX

)2/sin(

2 02/

2/

0

0

Tdte

T

T

tj

0

00

0

000

/2

)2/)/2sin((

3

2

/2

)2/)2sin((

3

2)2/sin(

3

22)(

Tw

TTw

Tw

TTw

w

wTjwX c

)/1(

))/1(sin(

3

25.0

)/1(

))/1(sin(

3

25.0

)sin(

3

2)(

0

00

0

000'

Tf

TTf

Tf

TTf

t

fTjfX c



rectangular pulse.

One sinc

Raised cosine pulse 3 sinc’s

The narrower main lobe, the narrower bandwidth. But, the more error rate as shown in the time domain

Figure 3.47 sum of three frequency-shifted sinc functions.

Fourier transform


2

1[ ] ( )

2j j nx n X e e d

( ) [ ]j j n

n

X e x n e

1( ) ( )

2j tx t X j e d

( ) ( ) j tX j x t e dt

Discrete time

Continuous time

Time domain frequency domain

3.9.1 Linearity Property


)()()(][][][

)()()()()()(

jjjDTFT

FT

ebYeaXeZnbynaxnz

jwbYjwaXjwZtbytaxtz

)(2

1)(

3

2)( tytxtz

)(tx

)(ty

)2/sin(2

1)4/sin(

2

3][)(

)2/sin())/(1(][)(

)4/sin())/(1(][)(

2;

2;

2;

kk

kk

kZtz

kkkYty

kkkXtx

FS

FS

FS

3.9.1 Symmetry Properties

Real and Imaginary Signals


dtetx

dtetxjwX

jwt

jwt

)(

)()((3.37)

(real x(t)=x*(t)) (conjugate symmetric)

dtetxjwX twj )()()( )()( jwXjwX (3.38)

3.9.2 Symmetry Properties of FT

EVEN/ODD SIGNALS


(even) (real)(odd) (pure

imaginary)

)(txe )}(Re{ jX

)(txo )}(Im{ jXj

For even x(t), .* jwXdexdtetxjwX jwtjw

real

3.10 Convolution Property


(convolution) (multiplication)

dtxhtxthty *

.2

1

dwejwXtx tjw

.

2

1

dwejwXtx jwt

But given

ddweejwXhty jwjwt

2

1

change the order of integration

.

2

1dwejwXdeh jwtjw

,

2

1dwejwXjwHty jwt

40.3.* jwHjwXjwYtxthty FT

3.10 Convolution Property Example 3.31


Example 3.31 Convolution problem in the frequency domain

tttx sin1Input to a system with impulse response .2sin1 ttth

Find the output .* thtxty

Solution:

w

wjwXtx FT

,0

,1 .

2,0

2,1

w

wjwHth FT

,* jwHjwXjwYthtxty FT

,,0

,1

w

wjwY .sin1 ttty



Example 3.32 Find inverse FT’S by the convolution property

Use the convolution property to find x(t), where

.sin4 2

2w

wjwXtx FT

.sin2

ww

jwZ ,1,0

1,1jwZ

t

ttz FT

Ex 3.32 (p. 261). (a) Rectangular z(t). (b) txtztz )(*)(

3.10.2 Filtering


Continuous time Discrete time(periodic with 2π

LPF

HPF

BPF

Figure 3.53 (p. 263)

Frequency dependent gain (power spectrum) .log20log20 jeHorjwH

,222

jwXjwHjwY kill or not (magnitude)



Example 3.34 Identifying h(t) from x(t) and y(t)

The output of an LTI system in response to an input is .Find frequency response and the impulse response of this system.

tuetx t2 tuety t

Solution:

2

1

jwjwX .

1

1

jwjwY But

jwX

jwYjwH

.1

11

1

1

1

1

1

2

jwjwjw

jw

jw

jwjwH .tuetth t

2

1

s

sX .1

1

s

sY But note sX

sYsH

.1

11

1

1

1

1

1

2

sss

s

s

ssH .tuetth t

as

tue Lat

1



EXAMPLE 3.35 Equalization(inverse) of multipath channel

jwYjwHjwX inv , jjinvj eYeHeXor

Consider again the problem addressed in Example 2.13. In this problem, a distorted received signal y[n] is expressed in terms of a transmitted signal x[n] as .1,1 anaxnxny .

,0

1,

0,1

otherwise

na

n

nh

.* nnhnh inv ,1 jjinv eHeH

Then .1

j

jinv

eHeH

.1 jjDTFT aeeHnh .1

1

j

jinv

aeeH

.nuanh ninv

3.11 Differentiation and Integration Properties


.

2

1dwejwXtx jwt

dwjwejwXtx

dt

d jwt

21 .jwjwXFT

EXAMPLE 3.37 The differentiation property implies that

.jwa

jwtue

dt

d FTat

. tuaetetuaetuedt

d atatatat

jwa

jw

jwa

atue

dt

d FTat

1

dssesXtx

dt

d st

21 .ssXL

.sa

stue

dt

d Lat

sa

s

sa

atue

dt

d Lat

1

3.11 Differentiation and Integration Properties


예제 한 두개

3.11.2 DIFFERENTIATION IN FREQUENCY


,

dtetxjwX jwt Differentiate w.r.t. ω, ,

dtetjtxjwXdw

d jwt

Then, .jwXdw

dtjtx FT

Example 3.40 FT of a Gaussian pulse

Use the differentiation-in-time and differentiation-in-frequency properties for the FT of the Gaussian pulse, defined by and depicted in Fig. 3.60.

22

21 tetg

48.3.2 22

ttgettgdt

d t

Figure 3.60 (p. 275) Gaussian pulse g(t).

,)( jwjwGttgtgdt

d FT jwGdw

dtjtg FTand

jwGdw

d

jttg FT 1 .jwwGjwG

dw

d

.22wcejwG Then (But, c=?) .1210 22

dtejG t

.21 22 22 wFTt ee

Laplace transform and z transform


as

tue Lat

1

2)(

1

astute Lat

11

1][

znu zn

21

1

)1(][

z

znun zn

3.11.3 Integration


;t

dxty 52.3.1

jwXjw

jwY

53.3.01

wjXjwXjw

dx FTt

.t

dtu .1w

jwjwUtu FT Ex) Prove

54.3.sgn2

1

2

1ttu Note where a=0)()( tuetu at

.

0,1

0,0

0,1

sgn

t

t

t

t We know .21 wFT

.2sgn ttdt

d .2jwjwSgn

Fig. a step fn. as the sum of a constant and a signum fn.

0,0

0,2

)(

jjS

jtu FT 1

since linear

Differentiation and Integration Properties


Common Differentiation and Integration Properties.

jDTFT

FTt

FT

FT

eXd

dnjnx

jXjXj

dx

jXd

dtjtx

jXjtxdt

d

01

3.12.1 Time-Shift Property


dtettxetzjZ tjtj 0

jXedexedex tjjtjtj 000

)](arg[]arg[ 0

0 jXtjXe tj

Table 3.7 Time-Shift Properties of Fourier Representations

jnjDTFT

tjFT

eXennx

jXettx

0

0

0

0

Fourier transform of time-shifted z(t) = x(t-t0)

Note that x(t-t0) = x(t) * δ(-t0) and 0)( 0tjFT ett

3.12 Time-and Frequency-Shift Properties


Example)

Figure 3.62 1Ttxtz jXejZ Tj 1

0sin2

TjX

TejZ Tj

sin2

1

3.12.2 Frequency-Shift Property


dejX

dejZtz

tj

tj

2

12

1

txedejXe

dejXtz

tjtjtj

tj

2

12

1

jXettx tjFT 00

Recall

Table 3.8 Frequency-Shift Properties

00 0

00 0

;0

;0

FTj t

FSjk t

jDTFTj n

DTFSjk n

e x t X j

e x t X k k

e x n X e

e x n X k k

3.12.2 Frequency-Shift Property


Example 3.42 FT by Using the Frequency-Shift Property

t

tetz

tj

,0

,10

Solution: We may express as the product of a complex sinusoid and a rectangular pulse

tz tje 10

t

ttx

,0

,1 )(10 txetz tj

sin2 jXtx FT

10sin10

210)(10

jXtxetz FTtj

3.12 Shift Properties Ex. 3.43


Example 3.43 Using Multiple Properties to Find an FT

23 tuetuedt

dtx tt

Sol) Let and tuetw t3 2 tuetv t

Then we may write tvtwdt

dtx

By the convolution and differentiation properties

jVjWjjX

The transform pair

ja

tue FTat

1

jjW

3

1

222 tueetv t

j

eejV

j

1

22

jj

ejejX

j

31

22

3.12 Shift Properties Ex. 3.43


Example 3.43 Using Multiple Properties to Find an FT

23 tuetuedt

dtx tt

Sol) Let and tuetw t3 2 tuetv t

Then we may write tvtwdt

dtx

By the convolution and differentiation properties

X s s W s V s

The transform pair

1Laplaceate u ta s

1

3W s

s

222 tueetv t

2

2

1 3

sseX s e

s s

2

2

1

seV j e

s

s

3.13 Inverse FT: Partial-Fraction Expansions

3.13.1 Inverse FT by using


1FTate u ta s

1 01

1 1 0

MM

N NN

B sb s b s bX s

s a s a s a A s

0

M Nk

kk

B sX s f s

A s

Nkfordk ,,1N roots,

0

1

Mk

kk

N

kk

b sX j

s d

partial fraction

1

Nx

k k

CX s

s d

10sdte u t for d

s d

1 1

k

N Nd t s k

kk k k

Cx t C e u t X s

s d

3.13 Inverse FT: Partial-Fraction Expansions

3.13.1 Inverse FT by using


ja

tue FTat

1

jA

jB

ajajaj

bjbjbjX N

NN

MM

01

11

01

NM

k

kk jA

jBjfjX

0

~

Let then ,jv 0011

1 avavav N

NN

Nkfordk ,,1N roots,

N

kk

M

k

kk

dj

jbjX

1

0

partial fraction

N

k k

x

dj

CjX

1

01

dfordj

tue FTdt

N

k

N

k k

kFTtdk dj

CjXtueCtx k

1 1

Inverse FT: Partial-Fraction Expansions


1Laplaceate u ts a

'( ) ( )Laplacex t sX s 2 ( ) 3 '( ) ''( ) 2 ( ) 3 '( )y t y t y t x t x t 22 ( ) 3 ( ) ( ) 2 ( ) 3 ( )Laplace Y s sY s s Y s X s sX s

2

( ) 2 3( )

( ) 2 3 1 2

Y s s A BH s

X s s s s s

11

( 1)( )( 1)

2ss

B sH s s A A

s

2( ) ( ) ( )t th t Ae Be u t

( )x t( )h t

( ) ( )* ( )y t x t h t( ) ( ) ( )Y s X s H s( )X s

3.13.2 Inverse DTFT

3.13.2


11

)1(1

01

jNjN

NjN

jMjMj

eee

eeeX

N

k

jk

jNjN

NjN edeee

11

11 11

012

21

1

NNNNN vvvv

N

kj

k

kj

ed

CeX

1 1

jk

DTFTnk ed

nud1

1where

Then

N

k

nkk nudCnx

1

Inverse FT: Partial-Fraction Expansions


1

1[ ]

1Zna u n

az

1[ ] ( )Z

ox n n z X z 2 [ ] 3 [ 1] [ 2] 2 [ ] 3 [ 1]y n y n y n x n x n 1 2 12 ( ) 3 ( ) ( ) 2 ( ) 3 ( )Z transform Y z z Y z z Y z X z z X z

1

1 2 1 1

( ) 2 3( )

( ) 2 3 1 1 2

Y z z A BH z

X z z z z z

1

111

1

(1 )( )(1 )

1 2zz

B zH z z A A

z

[ ] ( 2 ) [ ]nh n A B u n

[ ]x n[ ]h n

[ ] [ ]* [ ]y n x n h n( ) ( ) ( )Y z X z H z( )X z

3.13.2 Inverse DTFT by z-transform

3.13.2


1

1 0( 1) 1

1 1 1

MM

N NN N

z zX z

z z z

1 1 11 1

1

1 1N

NNN N k

k

z z z d z

11 1

Nk

k k

CX z

d z

1

1

1n z

kk

d u nd z

where

Then

N

k

nkk nudCnx

1

3.13 Inverse FT Example 3.45


Example 3.45 Inversion by Partial-Fraction Expansion

2

6

1

6

11

56

5

jj

j

j

ee

eeX

Solution:

jjjj

j

e

C

e

C

ee

e

3

11

2

11

6

1

6

11

56

5

21

2

Using the method of residues described in Appendix B, We obtain

4

3

11

56

5

6

1

6

11

56

5

2

11

22

21

jj e

j

j

e

jj

j

j

e

e

ee

eeC

1

3

11

56

5

6

1

6

11

56

5

3

11

33

22

jj e

j

j

e

jj

j

j

e

e

ee

eeCAnd

Hence, nununx nn 3/12/14 2011.5.4

3.13 Inverse FT Example 3.45


Example 3.45 Inversion by Partial-Fraction Expansion

1

1 2

55

61 1

16 6

zX z

z z

Solution:1

1 2

1 2 1 1

55

61 1 1 1

1 1 16 6 2 3

z C C

z z z z

Using the method of residues described in Appendix B, We obtain

1 1

1 1

11

1 2 1

2 2

5 55 51 6 61 4

1 1 12 1 16 6 3z z

z zC z

z z z

1 1

1 1

12

1 2 1

3 3

5 55 51 6 61 1

1 1 13 1 16 6 3z z

z zC z

z z z

And

Hence, nununx nn 3/12/14 2011.5.4

3.14 Multiplication (modulation) Property


Given and

dvejvXtx jvt

21

dejZtz tj

2

1

dvdejZjvXtztxty tvj

22

1)()(

Change of variable to obtain

dedvvjZjvXty tj

2

1

2

1

jZjXjYtztxty FT 2

1

dvvjZjvXjZjX Where

jjjDTFT eZeXeYnznxny21

(3.56)

(3.57)

denotes periodic convolution.

Here, and are -periodic. jeX jeZ 2

deZeXeZeX jjjj

Modulation property

23年 4月 21日 MediaLab , Kyunghee University

53

jZjXjYtztxty

nconvolutiotionmultiplica

jZjXjYtztxty

ionmultilicatnconvolutio

FT

FT

FT

FT

2

1

2

1*

3.14 Modulation Property Ex 3.46


Example 3.46 Truncating the sinc function

Sol) truncated by

2sin

1 n

nnh

otherwise

Mnnw

,0

,1

otherwise

Mnnhnht

,0

, jt

DTFTt eHnh

1( )

2jj j

tH e H e W e d

2sin

1 n

nnh

2/,0

2/,1jeH

2/sin

2/12sin

MeW j

/2

/2

1

2F d

2/,0

2/,

j

jjeW

eWeHF

Figure 3.66 The effect of Truncating the impulse response of a discrete-time system. (a) Frequency response of ideal system. (b) for near zero. (c) for slightly greater than (d) Frequency response of system with truncated impulse response.

F F

2/

3.15 Scaling Properties


dteatxetzjZ tjtj

0,/1

0,/1

/

/

adexa

adexajZ

aj

aj

dexajZ aj //1

ajXaatxtz FT //1 (3.60)

3.15 Scaling Properties Example 3.48


Example 3.48 SCALING A RECTANGULAR PULSE

Let the rectangular pulse

1,0

1,1)(

t

ttx

2,0

2,1)

2()(

t

ttxty

Solution :

).sin(2

)( ww

jwX 2/1).2/()( atxty

).2sin(2

)2sin()2

2(2

)2(2)(

ww

ww

wiXjwY



Example 3.49 Multiple FT Properties for x(t) when

}.)3/(1

{)(2

wj

e

dw

djjwX

wj

Solution)jw

jwStuetsFT

t

1

1)()()(

)}.3/({)( 2 jwSedw

djjwX wj

we define )3/()( jwSjwY ).(3)3(3)3(3)( 33 tuetuetsty tt

Now we define )()( 2 jwYejwW wj

).2(3)2()( )2(3 tuetytw t

Finally, since ),.()( jwWdw

djjwX

).2(3)()( )2(3 tutettwtx t



Example 3.49 Multiple FT Properties for x(t) when

.13/

)(2

s

e

ds

dsX

s

Solution)s

sStuetsL

t

1

1)()()(

)}.3/({)( 2 sSedw

dssX s

we define )3/()( sSsY ).(3)3(3)3(3)( 33 tuetuetsty tt

Now we define )()( 2 sYesW s

).2(3)2()( )2(3 tuetytw t

Finally, since ).()( sWds

dsX

).2(3)()( )2(3 tutettwtx t

3.16 Parseval’s Relationships


? |)(|?|)(| 22 에너지쓴주자가바이올린꽝언제 jwXtx

.|)(|2

1|)(|

)()(2

1

.])(2

1)[(

.)(2

1)(

).()(|)(|

,|)(|

22

*

*

**

*2

2

dwjwXdttx

thatconcludeSo

dwjwXjwXW

dtdwejwXtxW

dwejwXtx

txtxtxthatNote

dttxW

x

jwtx

jwt

x

Representation

Parseval Relation

FT

DTFT

deXnx

dwjwXdwjwX

j

n

22

22

|)(|2

1|][|

|)(|2

1|)(|

Table 3.10 Parseval Relationships for the Four Fourier Representations

3.16 Parseval’s Relationships Example 3.50


Example 3.50 Calculate the energy in a signal

n

WnnxLet

)sin(

][ 2

22 2

sin ( )E = | [ ]| .

n n

Wnx n

n

Use the Parseval’s theorem

Solution)

.|)(|

2

1 2 deXE j

./12

1

WdEW

W

||,0

||,1{)(][

W

WeXnx jDTFT

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Signals & systems Ch.3 Fourier Transform of Signals and LTI System