8/12/2019 Mitra Solutions 3rd Ed 3

1/43

Chapter 3

3.1 Now,.)()( =

dtetxjX

tjaa .)(

8/12/2019 Mitra Solutions 3rd Ed 3

2/43

(d)

8/12/2019 Mitra Solutions 3rd Ed 3

3/43

3.6 (a) = +

dexdtettx o

tja

tjoa

)()()( obtained using a change of variable

Therefore.= ott ).()()( = =

jXedexedtettx a

tjja

tjtjoa

oo

(b) ( ).)()()( )( otj

atjtj

a jXdtetxdteetx oo ==

(c) .)(2

1)(

=

dejXtx tjaa Therefore .)()(2 =

dejXtx tjaa

Interchanging t and we get .)()(2 dtejtXx tjaa =

(d) For a positive real constant the CTFT of is given bya )(atxa

).()()( 1)/(1

aaa

ajaa

tja jXdexdteatx

= = In a similar manner we

can show that for a negative constant the CTFT of is given bya )(atxa ).(1

aaajX

Therefore .)( 1

aaa

CTFT

a jXatx

(e) Differentiating both sides of

=

dejXtx tjaa )(

2

1)( get

.)(2

1)(

=

dejXj

dt

tdx tja

a Therefore ).()( CTFT jXj

dt

tdxa

a

3.7 ,)()()( )(

== a

ja

tjaa ejXdtetxjX where { .)(arg)( }= jXaa Thus,

.)()( dtetxjX tj

aa

= If is a real function of then it follows from the

definition of and the expression for

)(txa

)( jXa )( jXa that and)( jXa )( jXa are

complex conjugates. Therefore )()( = jXjX aa and ).()( = aa Or in

other words, for a real , the magnitude spectrum )( jXa is an even function of and

the phase spectrum is an odd function of)(a . 3.8

Not for sale 47

8/12/2019 Mitra Solutions 3rd Ed 3

4/43

3.9 where.)()()( =

dxthtx HT )(thHT is the impulse response of the Hilbert

transformer. Taking the CTFT of both sides we get )()()( = jXjHjX HT where

and)( jX )( jHHT denote the CTFTs of and)( tx ),(thHT respectively. Rewriting

( ) ).()()()()()( +=+= jXjjXjjXjXjHjX npnpHT As the magnitude andphase of are an even and odd function, is seen to be real signal. Consider

the complex signal

)( jX )( tx)()()( txjtxty += . Its CTFT is then given by

).(2)()()( =+= jXjXjjXjY p

3.10 The total energy [ ] .12/12

10

2

2

122

==

==== ttt

x edtedte

The total energy can also be computed using using the Parsevals theorem

.22

1

2

1 =

+

dx

Therefore, the 80% bandwidth c can be found by evaluating

+ d

c

c

22

1

2

1

2/1

1tantantantan 111

2

111

2

1

=

=

=

= ccc

c

c

8.0)2(tan 12 =

= c . Therefore, .5388.1tan2

8.0

2

1 =

= c

3.11 where],[][][][ nynynnyodev

+== ][])[][(][2

1

2

1

2

1nnynyny

ev

+=+= and

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

1

2

1

2

1

2

1nnnnnynynyod === Now,

.2

1)2(

2

1)2(2

2

1)( + +=+

+=

=

=

kk

jev kkeY Since

],[][][2

1

2

1nnnyod = ].1[]1[]1[ 2

1

2

1= nnnyod As a result,

( ).2

1

2

1

2

1]1[][][]1[]1[][]1[][ +== + nnnnnnnyny odod

Taking the DTFT of both sides of the above equation, we get

( ) += jjodjjod eeYeeY 1)()( 21 or .211 11121)( ==

+ jj

j

eeejod eY

Hence, .)2()()()(1

1 ++=+=

=

kj

e

jod

jev

jkeYeYeY

3.12 The inverse DTFT of is given by +=

=

k

jkeX )2(2)(

Not for sale 48

8/12/2019 Mitra Solutions 3rd Ed 3

5/43

.12

2)(2

2

1][ =

=

=

denx

nj

3.13nj

n

nj eeY

=

=)( with .1

8/12/2019 Mitra Solutions 3rd Ed 3

6/43

3.16 ][2

][)sin(][00

0 nj

eeeeAnnAnx

jnjjnjnn

=+=

( ) ( ) ].[2

][2

00 neej

Anee

j

A njjnjj = Therefore, the DTFT of is given

by

][nx

.1

1

21

1

2)(

00

=

jj

j

jj

jj

eee

j

A

eee

j

AeX

3.17 Let with]1[][ nnx n= .1

8/12/2019 Mitra Solutions 3rd Ed 3

7/43

8/12/2019 Mitra Solutions 3rd Ed 3

8/43

8/12/2019 Mitra Solutions 3rd Ed 3

9/43

,)1(

11

)1()(

222

=+

=

j

j

j

jj

ee

e

eeX and it also holds for .2=m

Now, assume that it holds for Consider next.m ][)!(!

)!(][1 n

mn

mnnx

nm

+=+

].[][1][][)!1(!)!1( nxnxn

mnx

mmnn

mnmn

mmn

mmmn +=

+=

+

+= Hence,

mjmj

j

mjmj

jm

ee

e

eed

dj

meX

)1(

1

)1()1(

1

)1(

11)(

11 +

+

+

=

+

=

.)1(

1

1+=

mje

3.21 (a) Hence,.)2()( +=

=

k

ja keX .1)(

2

1][ =

=

denx nj

a

(b) .1

1)(

1

0=

=

=

N

n

njj

j

Njjj

b eee

eeeX Let .nm = .)(

1

0=

+

=

N

m

mjjjb eeeX

Consider the DTFT Its inverse is given by

Therefore, by the time-shifting property of the DTFT, the

inverse DTFT of is given by

.)(1

0=

+

=

N

m

mjj eeX

=

otherwise.,0,0)1(,1

][ nN

nx

)()( = jjjb eXeeX

=+=

otherwise.,0,1,1

]1[][ nN

nxnxb

(c) Hence,.2)cos(21)(0

+= +==

=

N

N

jN

jc eeX

l

l

l

l