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

04/20/23

Signals & systemsCh.3 Fourier Transform of

Signals and LTI System

Signals and systems in the Frequency domain

04/20/23 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

3.1 Introduction cont’

• Fourier Series (FS)


i) If

ii) If ,km

1111

)()(1

0)(*

TT

tkmj

T km dtT

dteT

dttvtvT

02)(

11

0

)(2)( 0

T

tT

kmj

T

tkmj ekmj

dteT

,km

Any periodic function(signal) with period T can be represented by

weighted sum of orthonormal functions.

T kkk

kk dttvtfFwheretvFTtftf )()()()()(

What is meaning of magnitude of Fk?

Think about equalizer in audio

system.

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

3.3 Fourier Representations for Four Classes of signals

• 3.3.1 Periodic Signals: Fourier Series Representations


“Any periodic function can be represented by weighted sum of basic periodic function.” Fourier

said

Time Property

Periodic Non-periodic

Continuous (t)

Fourier Series (FS)

Fourier Transform (CTFT)

Discrete [n]Discrete-Time Fourier Series

(DTFS)

Discrete-Time Fourier

Transform (DTFT)

(periodic) -

(discrete)

(discrete) -

(periodic)

NekAnx

k

njk 2,][][ˆ 0

0

TekAtx

k

tjk 2,][)(ˆ 0

0 CTFS (3.5)

DTFS (3.4)

. 00000 2)( njknjknjnjknjNnkNj eeeeee Periodic

?

yes! With the period of

N

3.3 Fourier Representations for Four Classes of signals cont’

• 3.3.2 Non-periodic Signals: Fourier-Transform Representations


(aperiodic) - (continuous)(continuous) - (aperiodic)

Inverse continuous time Fourier Transform

(CTFT)

.)(2

1)(ˆ

dejXtx tj

(aperiodic) - (continuous)(discrete) - (periodic)

Inverse discrete time Fourier Transform

(DTFT)

.)(2

1][ˆ

deeXnx njj

3.4 Discrete-Time Periodic Signals: The Discrete-Time

Fourier Series


][][ 0; kXnx DTFS

(periodic) - (discrete)(discrete) - (periodic)

Inverse DFT (3.10)

DFT

.,][][1

0

0

N

k

njkekXnx

1

0

0][1

][N

n

njkenxN

kX

Example 3.2 Determining DTFS Coefficients

Find the frequency-domain representation of the signal depicted in Fig. 3.5

Figure 3.5 Time-domain signal for Example 3.2.


Fourier Series cont’


5/45/205/25/42

2

5/2 ]2[]1[]0[]1[]2[5

1][

5

1][ jkjkjjkjk

n

njk exexexexexenxkX

Just inner product to orthonormal vectors in the 5 dimensional space!!

),,,,(])2[],1[],0[],1[],2[(5

1 cx

5

1 X[k] 5/45/205/25/4*

k jkjkjjkjk eeeeexxxxx

),,,,(c 5/85/405/45/82

jjjjj eeeee

),,,,( c 5/125/605/65/123

jjjjj eeeee

),,,,(),,,,(5

1 cc

5

1 5/125/605/65/125/85/405/45/8*

32 jjjjjjjjjj eeeeeeeeee

][cc 5

1 *

ji ji




.)5/2sin(15

1

2

1

2

11

5

1][ 5/25/2 kjeekX jkjk

531.0232.05

)5/4sin(

5

1]2[ jejX

760.0276.05

)5/2sin(

5

1]1[ jejX

02.05

1]0[ jeX DC!!

760.0276.05

)5/2sin(

5

1]1[ jejX

.232.0

5

)5/4sin(

5

1]2[ 531.0jejX

Figure 3.6 Magnitude and phase of the DTFS coefficients for the signal in Fig. 3.5.

Recall “weighted sum of orthonormal vectors in the 5 dimensional space”




Example 3.3 Computation of DTFS Coefficients by Inspection

11 00 cacav

1 0 cc

andorthnormal

)a,(a)v,(v 10DFT

10

Determine the DTFS coefficients of , using the method of

inspection. njj

njj

njnj

eeeeee

nx 3333

2

1

2

1

2 ][

Figure 3.8 Magnitude and phase of DTFS coefficients for Example 3.3.




Example 3.4 DTFS Representation of an Impulse Train

Find the DTFS Coefficients of the N-periodic impulse train

,][][

l

lNnnx

Figure 3.9 A discrete-time impulse train with period N.

.1

][1

][1

0

/2

Nen

NkX

N

n

Njkn

How about in other period?

Draw X[k] in the k-axis.

.`1

][1

][2/3

2/

/2

NeNn

NkX

N

Nn

Njkn

(impulse train) - (impulse train)




Example 3.5 The Inverse DTFS of periodic X[k]

Use Eq. (3.10) to determine the time-domain signal x[n] from the DTFS coefficients depicted in Fig. 3.10

Figure 3.10 Magnitude and phase of DTFS coefficients for Example 3.5.

.1)3/9/4cos(4)3/29/6cos(2

212

][ ][

9/63/29/43/9/43/9/63/2

4

4

9/2

nn

eeeeeeee

ekXnx

njjnjjnjjnjj

k

njk




Example 3.6 DTFS Representation of a Square Wave

Find the DTFS coefficients for the N-periodic square wave

. ,0

- 1,][

MNnM

MnMnx

Figure 3.11 Square wave for Example 3.6.

.1

1

][1

][ 000

11

M

Mn

njkMN

Mn

njkMN

Mn

njk eN

eN

enxN

kX

.1

1

][2

0

2

0

)( 000

M

m

mjkMjkM

m

Mmjk eeN

eN

kX (3.15) 등비수열의 합

,2,0, ,12

11

][2

0

NNkN

M

NkX

M

m

DC=mean= 평균

,,2,0, ,1

1][

0

00 )12(

NNke

e

N

ekX

jk

MjkMjk

1

1

1

1

1][

2/2/

2/)12(2/)12(

2/

2/

)12(

00

00

0

0

0

00

jkjk

MjkMjk

jk

jk

jk

MjkMjk

ee

ee

e

e

N

e

ee

NkX




.,2,0, ,)2/sin(

)2/)12(sin(1][

0

0 NNkk

Mk

NkX

.

,2,0, ,/)12(

,2,0, ,)/sin(

)/)12(sin(1

][

NNkNM

NNkNk

NMk

NkX

.12

)/sin(

)/)12(sin(1lim

,2,0, N

M

Nk

NMk

NNNk

DC, also

.)/sin(

)/)12(sin(1][

Nk

NMk

NkX

Then for all k.

M=0 (impulse train)

2M+1 = N? ?)/sin(

)sin(1][

Nk

k

NkX

(square) - (sinc)




Example 3.7 Building a Square Wave form DTFS Coefficients

.)/sin(

)/)12(sin(1][

Nk

NMk

NkX

1

0 )cos(][2]0[][][ 0

kk

njk nkkXXekXnx

symmetric

Evaluate one period of the Jth term in Eq. (3.18) and the 2J+1 term approximation

][ˆ nxJ for J=1, 3, 5, 23, and 25.

,)cos(][][ˆ0

0

J

kJ nkkBnx (3.18)




Example 3.8 Numerical Analysis of the ECG

Evaluate the DTFS representations of the two electrocardiogram (ECG) waveforms depicted in Figs. 3.15(a) and (b).

Figure 3.15 Electrocardiograms for two different heartbeats and the first 60 coefficients of their magnitude spectra.

(a) Normal heartbeat. (b) Ventricular tachycardia. (c) Magnitude spectrum for the

normal heartbeat. (d) Magnitude spectrum for

ventricular tachycardia.

3.5 Continuous-Time Periodic Signals: The Fourier Series


Any periodic function can be represented by weighted sum of basic periodic functions.

(periodic) (discrete)(continuous) (aperiodic)

Inverse FT where (3.19) Recall “orthonormal”!!

.,][)( 0

k

tjkekXtx

T tjk dtetx

TkX

0

0)(1

][ FT (3.20)

3.5 Continuous-Time Periodic Signals: The Fourier Series cont’


Example 3.9 Direct Calculation of FS CoefficientsDetermine the FS coefficients for the signal depicted in Fig. 3. 16.

)(tx

Solution :

24

1

)2(2

1

2

1

2

1][

420

)2(2

0

)2(2

0

2

jk

ee

jkdtedteekX tjktjktjkt

X[0]=?

FT

Figure 3.16 Time-domain signal for Example 3.9.

Figure 3.17 Magnitude and phase for Ex. 3.9.



Example 3.10 FS Coefficients for an Impulse Train. Determine the FS

coefficients for the signal defined by

t

lttx )4()(

Solution :

6

2

)2/(2

2

)2/()2/( )4(4

1)(

4

1)(

4

1][ dtetdtetdtetxkX tjktjk

T

tjk

(impulse train) (impulse train)

Example 3.11 Calculation of FS Coefficients by Inspection

Determine the FS representation of the signal using the method of inspection.

)4/2/cos(3)( ttx

Solution :

k

tjkekXtx 2/][)( (3.21)

2/4/2/4/)4/2/()4/2/(

2

3

2

3

23)( tjjtjj

tjtj

eeeeee

tx

otherwise

ke

ke

kX j

j

,0

1,2

3

1,2

3

][ 4/

4/

(3.22)

Figure 3.18 Magnitude and phase for Ex. 3.11.



20/)2/1(][ jkk ekX .2T

Example 3.12 Inverse FS : Find the time-domain signal x(t) corresponding to the FS coefficients . Assume that the fundamental period is Solution : 0

1

20/

0

20/

1

20/

0

20/

)2/1()2/1(

)2/1()2/1()(

l

tjljll

k

tjkjkk

k

tjkjkk

k

tjkjkk

eeee

eeeetx

등비수열의 합

1)2/1(1

1

)2/1(1

1)(

)20/()20/(

tjtj ee

tx

)20/cos(45

3)(

ttx

Figure 3.20 FS coefficients for Problem 3.9.

Figure 3.21 Square wave for Example 3.13.



Example 3.13 FS for a Square Wave Determine the FS representation of the square wave depicted in Fig. 3.21.

0,)sin(2

)2

(2

0,11

)(1

][2/

2

kTkw

Tkw

j

ee

Tkw

kT

Te

Tjkwdte

Tdtetx

TkX

o

ooTjkwTjkw

o

o

otjkw

o

T

T

tjkwT

T

tjkw

oooo

oo

o

oo

For k=0, o

o

T

T

o

T

Tdt

TX

21]0[

T

T

Tkw

Tkw ooo

k

2)sin(2

00lim

)2

(sin2)sin(2

][ 00

0 T

Tkc

T

T

Tkw

TkwkX oo

u

uuc

)sin(

)(sin where

Figure 3.22 The FS coefficients, X[k], –50 k 50, for three square waves. In Fig. 3.21 Ts/T = (a) 1/4 . (b) 1/16. (c) 1/64.



To/T=1/2? (DC) (impulse)To 0? (impulse train) (impulse train)T ? (aperiodic) (continuous)

Figure 3.23 Sinc function sinc(u) = sin(u)/(u)

(square) (sinc)(sinc) (square)



Example 3.14 Square-Wave Partial-Sum Approximation

)cos(][)(0

^

tkwkBtx o

J

k

j

JkJ

1T 4/1/0 TT J

)(^

tx j ,9,7,3,1J

Let the partial-sum approximation to the FS in Eq.(3.29), be given by

This approximation involves the exponential FS Coefficients with indices

.

Consider a square wave and . Depict one period of the

th term in this sum, and find for and 99.

Solution :

evenk

oddkk

k

kB k

,0

,)1))(/(2(

0,2/1

][ 2/)1(

3.6 Discrete-Time Non-periodic Signals : The Discrete-Time Fourier

Transform


(discrete) (periodic)

deeXnx njnj )(

2

1][

n

njj enxeX ][)(

)(][ jDTFT eXnx

n

nx ][

n

nx2

][

(3.31)

(3.32)

3.6 Discrete-Time Non-periodic Signals : The Discrete-Time Fourier Transform cont’


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 Discrete-Time Non-periodic Signals : The Discrete-Time Fourier Transform

cont’


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.


cont’


,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


cont’


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

)/(][

WnncsiW

nx

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


cont’


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


cont’


Example 3.22 Two different moving-average systems

])1[][(2

1][1 nxnxny ])1[][(

2

1][2 nxnxny

]1[2

1][

2

1][1 nnnh ]1[

2

1][

2

1][2 nnnh

solution :

Figure 3.36 jj eeH

2

1

2

1)(1

jj eeH2

1

2

1)(2

)2

cos(2

222

2

jjj

je

eee )

2sin(

22

222

jjj

jje

eee

)2

cos()(1

jeH )

2sin()(2

jeH

2)}(arg{ 1

jeH

0,2/2

0,2

2/)}(arg{ 2

jeH


cont’


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{sin(})arg{cos(1

1

1

1)(

})arg{(

ajaeaeH

ajjinv

)(

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 Continuous-Time Non-periodic Signals : The Fourier

Transform


(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)(


Transform cont’


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


Transform cont’


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


Transform cont’


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


Transform cont’


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


Transform cont’


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.


Transform cont’


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


Transform cont’


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.

9.1 Linearity and Symmetry Properties of FT


][][][][][][

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

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

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

0

0

;

;

kbYkaXkZnbynaxnz

ebYeaXeZnbynaxnz

kbYkaXkZtbytaxtz

jwbYjwaXjwZtbytaxtz

DTFS

jjjDTFT

wFS

FT

)(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

9.1 Linearity and Symmetry Properties of FT cont’

• 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)

9.1 Linearity and Symmetry Properties of FT cont’

• 3.9.2 SYSMMEYRY PROPERTIES : 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

• 3.10.1 CONVOLUTION OF NON-PERIODIC SIGNALS


(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 cont’


Example 3.31 Convolution problem in the frequency domain tttx sin1 be the input 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 , where

tx

.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

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

jw

jwX .1

1

jw

jwY But jwX

jwYjwH

.1

11

1

1

1

1

1

2

jwjwjw

jw

jw

jwjwH .tuetth t



EXAMPLE 3.35 Equalization 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

jjinv

eHeH

.1 jjDTFT aeeHnh .1

1

j

jinv

aeeH

.nuanh ninv


• 3.10.3 Convolution of periodic signals : Cyclic convolution


Convolution in just one period Or better to derive in the frequency domain

dtzxtztxtyT

0 44.3.

2;

kZkXTkYtztxty TFS

Figure 3.56 (p. 268) Square wave for Example 3.36.

.

2

2sin2

k

kkX

EXAMPLE 3.36 Convolution of 2 periodic signals

Evaluate the periodic convolution of the sinusoidal signal

tttz 4sin2cos2 .

,0

2,21

2,21

1,1

otherwize

kj

kj

k

kZ



Figure 3.57 (p. 270) z(t) in Eq. (3.45) when J = 10.

1

0

N

k

knzkxnznxny kZkXNkYNDTFS

2

;

otherwise

JkJkZ

,0

,1

45.3.sin

12sin

t

Jttz

,ˆ kZkXkX j

,,0

1,1

otherwise

kkZkXkY

.2cos2 tty

3.11 Differentiation and Integration Properties

• 3.11.1 DIFFERENTIATION IN TIME


.

2

1dwejwXtx jwt

dwjwejwXtx

dt

d jwt

2

1 .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

3.11 Differentiation and Integration Properties cont’


Example 3.38 Resonance in MEMS accelerometerThe MEMS accelerometer introduced in Section 1.10 is described by the differential equation

.22

2

txtywtydt

d

Q

wty

dt

dn

n

.1

22n

n wjwQ

wjw

jwH

Fig 3.58 (p. 273) ||log20 10 jwH

n = 10,000 rad/sQ = 2/5, Q = 1, and Q = 200.Resonance in 10,000 rad/s



Example 3.39 Use the differentiation property to find the FS of the triangular wave depicted in Fig. 3.59(a)

.

0,2

sin4

0,0

0;

k

k

k

k

kZtz wFS

???][ kY

Signals for (a) Triangular wave y(t). (b) )()(

tzdt

tdy

.

0,2sin2

0,2

22

; 0

k

jk

kT

kT

kYty wFS

,

2j

T

kj

where

If we differentiate z(t) once more???


• 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

.jwwGjwGdw

d

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

dtejG t

.21 22 22 wFTt ee


• 3.11.3 Integration


;

tdxty 52.3.

1jwX

jwjwY

53.3.01

wjXjwXjw

dx FTt

.

tdtu .1

wjw

jwUtu 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

j

tu FT 1

since linear



Common Differentiation and Integration Properties.

01

0; 0

jXjXj

dx

eXd

dnjnx

jXd

dtjtx

kXjktxdt

d

jXjtxdt

d

FTt

jDTFT

FT

FS

FT

3.12 Time-and Frequency-Shift Properties

• 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

kXennx

eXennx

kXettx

jXettx

njkDTFS

jnjDTFT

tjkFS

tjFT

000

0

000

0

;0

0

;0

0

3.12 Time-and Frequency-Shift Properties cont’


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

0

;

0;

000

000

kkXnxe

exnxe

kkXtxe

jXtxe

DTFSnjk

jDTFTnj

FStjk

FTtj



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



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

3.13 Inverse FT by Using 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

3.13 Inverse FT by Using Partial-Fraction Expansions cont’

• 3.13.2 Inverse Discrete-Time Fourier Transform


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

3.13 Inverse FT by Using Partial-Fraction Expansions cont’


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

3.14 Multiplication (modulation) Property


Given and

dvejvXtx jvt

2

1

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 eZeXeYnznxny2

1

(3.56)

(3.57)

denotes periodic convolution.

Here, and are -periodic. jeX jeZ 2

deZeXeZeX jjjj

3.14 Multiplication (modulation) Property cont’


Example) Windowing in the time domain

jWjXjYtwtxty FT

2

1)()(

0sin2

TjW

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/



Example 3.46 Truncating the sinc function

Sol) truncated by

2sin

1 n

nnh

otherwise

Mnnw

,0

,1

otherwise

Mnnhnht

,0

, jt

DTFTt eHnh

deWeHeH jjjt 2

1

2sin

1 n

nnh

2/,0

2/,1jeH

2/sin

2/12sin

MeW j

2/

2/2

1

dFeH j

t

2/,0

2/,

j

jjeW

eWeHF



Example 3.47 Radar Range Measurement: RF Pulse Train

Solution) )5002sin( t

otherwise

kj

kj

kS

,0

500,21

500,21

k

TkekP jkT 00sin

00 )5002sin()()( ttptx ][*][][ kSkPkX

500

500sin

2

1

500

500sin

2

1 0050000500 0000

k

Tkejk

Tkej

kX TkjTkj

kZkXkYnznxny NDTFT /2; (3.59)

1

0

N

m

mkZmXkZkX



Figure 3.69 FS magnitude spectrum of FR pulse train for The result is depicted as a continuous curve, due to the difficulty of displaying 1000 steps

10000 k

Table 3.9 Multiplication Properties

kZkXnznx

eZeXnznx

kZkXtztx

jZjXtztx

DTFS

jjDTFT

FS

FT

0

0

;

;

2

1

2

1

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 cont’


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

3.15 Scaling Properties cont’


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

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

FS

DTFT

DTFS

1

0

21

0

2

22

22

22

|][||][|1

|)(|2

1|][|

|][||)(|1

|)(|2

1|)(|

N

k

N

K

j

n

K

T

O

kXnxN

deXnx

kXdttxT

dwjwXdwjwX

Table 3.10 Parseval Relationships for the Four Fourier Representations

3.16 Parseval’s Relationships cont’


Example 3.50 Calculate the energy in a signal

n

WnnxLet

)sin(

][

nn n

Wnnx .

)(sin|][|=∞E

22

22

Use the Parseval’s theorem

Solution)

.|)(|

2

1 2 deXE j

./12

1

WdEW

W

||,0

||,1{)(][

W

WeXnx jDTFT

3.17 Time –Bandwidth Product


./)sin(2)(||,0

||,1{)( wwTjwX

Tt

Tttx o

FT

o

o

Figure 3.72 (p. 305) Rectangular pulse illustrating the inverse relationship between the time and frequency extent of a signal.

.2/1

.]|)(|

|)(|[]

|)(|

|)(|[ 2/1

2

22

2/1

2

22

wd

wd

BT

dwjwX

dwjwXwB

dttx

dttxtT

3.17 Time –Bandwidth Product cont’


Example 3.51 Bounding the Bandwidth of a Rectangular Pulse

oTt

TttxLet

||,0

||,1)(

Use the uncertainty principle to place a lower bound on the effective bandwidth of x(t).

Solution:

).2/(3

.3/|)3/1))(2/(1(2/13

2/12

ow

oToTo

oT

To

T

Td

TB

TtTdt

dttT

o

o

o

3.18.1 The Duality Property of the FT


.)()(.)(2

1)( dtetxjwXdwejwXtx jwtjwt

Fig. 3.73 Duality of rectangular pulses and sinc functions

),(2)(

,)(2

1)(

).()(

.)(2

1)(

)66.3(.,

.)(2

1)(

wytz

dtetzwy

wzty

dwewzty

thatimpliesEqthenwandtvchooseweif

dvezvy

FT

jwt

FT

jwt

jv

).(2)(

),()(

wfjtF

jwFtfFT

FT

3.18.1 The Duality Property of the FT cont’


Example 3.52 By using Duality, find the FT of

jttx

1

1)(

Solution)

)()( jwFtf FT ).(2)( wfjtF FT

jwjwFtuetf FTt

1

1)()()(

jtjtF

1

1)(

3.18.2 & 3.18.3

• 3.18.2 The Duality Property of the DTFS

• 3.18.3 The Duality Property of the DTFT and FS


.][][1

0

N

K

njk oekXnx

1

0

][1

][N

n

onjkenxN

kX

][][ /2; kXnx NDTFS ][1

][ /2; kxN

kX NDTFS

k

tjkwekZtz 0][)(

dtetzkZ jkt)(

2

1][

n

njj enxeX .][)(

)(2)()()( wfjtFjwFtf FTFT FT

][)/1(][][][ /2;/2; kxNnXkXnx NDTFSNDTFS DTFS

][)()(][ 1; kxeXeXnx FSjtjDTFS DTFS

Table 3.11 Duality Properties of Fourier Representations

.)(

2

1][ jnj eeXnx )(][ jDTFT eXnx ].[)( 1; kxeX FSjt

3.18.3 The Duality Property of the DTFT and FS cont’


Example 3.53 FS-DTFT Duality

Use the duality property and the results of Example 3.39 to determine the inverse DTFT of the triangular spectrum depicted in Fig 3.75(a).

)( jeX

Solution:

0,)2/sin(40,

][][2

12/ n

k

kjn

kYekZ kjw

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