mitra solutions 3rd ed 3
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Chapter 3
3.1 Now,.)()( =
dtetxjX
tjaa .)(
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(d)
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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
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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)(
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.12
2)(2
2
1][ =
=
=
denx
nj
3.13nj
n
nj eeY
=
=)( with .1
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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
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,)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
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