CHAPTER 4APPLICATIONS OF FOURIER REPRESENTATIONS TO

MIXED SIGNAL CLASSES

jeX

kX

jX

kX

DTFT:periodic-non

][DTFS:periodic :DT

)(FT:periodic-non

][FS:periodic :CT

• What about the Fourier representation of a mixture of a) periodic and non-periodic signals b) CT and DT signals?

(periodic)

assuch )()( tyHtx Examples: )()()cos( 0 tytht

][sampler)( nytx

• We will go through: a) FT of periodic signals, which we have used FS:

TekXtx

k

tjk 2 ,][)( 0

0

dtetxT

kXtxT tjkFS 0

; 00 )(1

][)(

We can take FT of x(t).

b) Convolution and multiplication with mixture of periodic and non-periodic signals.

c) Fourier transform of discrete-time signals.

FT of periodic signals

Chapter 3: for CT periodic signals, FS representations.What happens if we take FT of periodic signals?

(*) ][)( 0

k

tjkekXtx

FS representation of periodic signal x(t):

)(21 :shift Freq.

)(21 :1 of FT

00

ke FTtjk

FT

Take FT of equation (*)

kk

FTtjk kkXekXtx )(][2][)( 00

Note:a) FT of a periodic signal is a series of impulses spaced by the fundamental frequency 0.b) The k-th impulse has strength 2X[k].c) FT of x(t)=cos(0t) can be obtained by replacing

versaor vise ),(2 with 00 ke tjk

FS and FT representation of a periodic continuous-time signal.

E Example 4.1, p343:

)cos()( of ransformFourier t Find 0ttx

1,0

1,2/1)cos( 0 k

kt FS

)()()cos( Thus, 000 FTt

E Example 4.2, p344:

n

nTttp )()( train impulseunit theof FT theDetermine

p(t) is periodic with fundamental period T, fundamental frequency 0. FS coefficients:

k

T

T

tjk

kT

jP

kT

dtetT

kP

)(2

)(

,1

)(1

][

0

2/

2/

0

Relating DTFT to DTFS

N-periodic signal x[n] has DTFS expression

00

1

0

,),(2

)( ][][

0

0

kke

ekXnx

DTFTnjk

N

k

njk

Extending to any interval:

m

DTFTnjk mke )2(2 00

This, DTFT of x[n] given in (*) is expressed as:

m

N

k

jDTFTN

k

njk mkkXeXekXnx )2(][2][][ 0

1

0

1

0

0

Since X[k] is N periodic and N0=2, we have

k

jDTFTN

k

njk kkXeXekXnx )(][2][][ 0

1

0

0

Note:a) DTFS DTFT:

b) DTFT DTFS:

Also, replace sum intervals from 0~N-1 for DTFS to - ~ for DTFT

2by scale then )( 00 ke njk

)21/(by scale then )( 00 njkek

E Problem 4.3(c), p347:

. x[n]of DTFT theFind ].10[][

k

knnx

Fundamental period? :DTFS e Us/5.,10 0 N

kenxkXn

njk

,

10

1][

10

1][

9

0

5

Use note a) last slide:

n

DTFT knx 510

12][

Question: if we take inverse DTFS of X[k], we get

9

0

9

0

5

5

10

1

][][

n

njkn

njk

e

ekXnx

n

knnx ].10[][

expression original theequal toseemnot does which

Exercise: use Matlab to verify.

Convolution and multiplication with mixture of periodicand non-periodic signals

)(th)(tx )(ty ][nh][nx ][ny

For periodic inputs:

FT Use

)()()(periodic-nonperiodic

thtxty

DTFT Use

][][][periodic-nonperiodic

nhnxny

1) Convolution of periodic and non-periodic signals

tscoefficien FS :][

,][2)()(

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

0

kX

kkXjXtx

jHjXjYthtxty

k

FT

FT

k

k

FT

kkXjkH

jHkkXjYthtxty

00

0

][)(2

)(][2)()()()(

E Problem 4.4(a), p350: LTI system

has an impulse response

)(th)(tx )(ty

).(output system thedemermine toFT theuse

),4sin()cos(1)( signalinput For

ty

tttx

)./()sin()4cos(2)( tttth

2

4

2

4 )(

)sin()( 44

rectrectjH

eet

tth tjtj

2

)sin()( rect

t

tth FT

1

)4()4()()()(2 )( j jX

Because h(t) is an ideal bandpass filter with a bandwidth 2 centered at 4the Fourier transform of the output signal is thus

)4()4( )( j jY

which has a time-domain expression given as: )4sin()( tty

k

jkDTFT kkXeHnhnxny 0

periodic

][2][][][ 0

For discrete-time signals:

2) Multiplication of periodic and non-periodic signals

k

FT

kkXjG

jXjGjY

txtgty

0

21

periodic

][)(

)()()(

)()()(

k

FT

k

tjk

kkXjX

ekXtx

0][2)(

][)(

:Note0

Carrying out the convolution yields:

k

FT kGkXjYtxtgty 0][)()()()(

DT case:

1

0

)(

periodic

0][)(][][][N

k

kjjDTFT eZkXeYnznxny

E Problem 4.7, p357(b):

Consider the LTI system and input signal spectrum X(ej) depicted by the figure below. Determine an expression for Y(ej), the DTFT of the output y[n] assuming that z[n]=2cos(n/2).

..,0

1,1][

:][ of tscoefficien FS . with periodic :][

][][][][

20

wo

kkZ

nznz

nznxnxny

Thus,

)()(22 jjjj eXeXeXeY

E Example 4.6, p353: AM Radio

(a) Simplified AM radio transmitter & receiver.(b) Spectrum of message signal. Analyze the system in the frequency domain.

Signals in the AM transmitter and receiver. (a) Transmitted signal r(t) and spectrum R(j). (b) Spectrum of q(t) in the receiver. (c) Spectrum of receiver output y(t).

))(())(()(

)cos()()(

21

21

cc

FTc

jMjMjR

ttmtr

In the receiver, r(t) is multiplied by the identical cosine used in the transmitter to obtain:

))2(()())2((

))(())(()(

)cos()()(

41

21

41

21

21

cc

cc

FTc

jMjMjM

jRjRjG

ttrtg

After low-pass filtering:

)()( 21 jMjY