信号与系统-连续时域分析运算：Ch3_Soln

42
Chapter 3 (2e)
3.1 X(e
jω
) = x n e
jn
n
[]
=?∞
∞
∑
ω
where x[n] is a real sequence,Therefore,
X e xne xn e xn n
re
jjn
n
jn
nn
( ) Re [ ] [ ]Re [ ]cos( ),
ωω ω
ω=
=
( )
=
=?∞
∞
=?∞
∞
=?∞
∞
∑∑ ∑
and
X e xne xn e xn n
im
jjn
n
jn
nn
( ) Im [ ] [ ]Im [ ]sin( ).
ωω ω
ω=
=
( )
=?
=?∞
∞
=?∞
∞
=?∞
∞
∑∑ ∑
Since cos( )ωn and
sin( )ωn are,respectively,even and odd functions of ω,X e
re
j
()
ω
is an even function of ω,
and X e
im
j
()
ω
is an odd function of ω.
Xe X e X e
j
re
j
im
j
() () ()
ωωω
=+
22
,Now,X e
re
j2
()
ω
is the square of an even function and
Xe
im
j2
()
ω
is the square of an odd function,they are both even functions of ω,Hence,
Xe
j
()
ω
is an even function of ω.
arg ( ) tan
()
()
Xe
Xe
Xe
j im
j
re
j
ω
ω
ω
{}
=
1
,The argument is the quotient of an odd function and an even
function,and is therefore an odd function,Hence,arg ( )Xe
jω
{}
is an odd function of ω.
3.2 Xe
ee
e
e
ej
j
jj
()
cos
cos sin
cos
––
–
–
–
ω
ωω
ω
ω
ω
αα
α
α
α
αωα
αωαω
αωα
=
=
=
+
=

+
1
1
1
1
1
1
1
12
1
12
22
Therefore,Xe
re
j
()
cos
cos
ω
αω
αωα
=
+
1
12
2
and Xe
im
j
()–
sin
cos
ω
αω
αωα
=
+12
2
.
Xe Xe X e
ee
jjj
jj
() ()*()
cos
.
–
ωωω
ωω
αα αωα
2
2
1
1
1
1
1
12
=? =
=
+
Therefore,Xe
j
()
cos
.
ω
αωα
=
+
1
12
2
tan ( )
()
()
–
sin
cos
.θω
αω
αω
ω
ω
==
Xe
Xe
im
j
re
j
1
Therefore,θω
αω
αω
( ) tan –
sin
cos
.==
1
1
3.3 (a) yn n y n y n
ev od
[] [] [] [],== +μ where yn ynyn n n
ev
[] [] [ ] [] [ ]=+?( )=+?( )
1
2
1
2
μμ = +
1
2
1
2
δ[],n
and yn ynyn n n n n
od
[] [] [ ] [] [ ] [] []–.=( )=( )=?
1
2
1
2
1
2
1
2
μμ μ δ
Now,Ye k k
ev
j
kk
() () ().
––
ω
δω δω=π +π
+=π +π+
=∞
∞
=∞
∞
∑∑
1
2
1
2
1
2
22 2
Since yn n n
od
[] [] [],=?+μδ
1
2
1
2
yn n n
od
[][] [].=+?μδ11
1
2
1
2
As a result,
43
ynyn n n n n n n
od od
[] [ ] [] [ ] [ ] [] [] [ ].=+ = +?111 1
1
2
1
2
1
2
1
2
μμ δ δ δ δ Taking the DTFT of
both sides we then get Ye e Ye e
od
jj
od
jj
() (),
–ωω ω ω
=+
( )
1
2
1 or
Ye
e
ee
od
j
j
jj
(),
ω
ω
ωω
=
+
=

1
2
1
2
1
1
1
1
Hence,
Ye Y e Y e
e
k
j
ev
j
od
j
j
k
() () () ( ).
ωωω
ω
δω=+=
+π + π
=?∞
∞
∑
1
1
2
(b) Let x[n] be the sequence with the DTFT X e k
j
o
k
() ( )
ω
πδ ω ω π=?+
=?∞
∞
∑
2 2, Its inverse
DTFT is then given by xn e d e
o
jn j n
o[] ( ),=?=
∫
1
2
2
π
πδ ω ω ω
ω
π
π
ω
3.4 Let X e k
j
k
() ( ).
ω
δω=π+π
=?∞
∞
∑
2 2 Its inverse DTFT is then given by
xn e d
jn
[] ( ),
–
=
π
π=
π
π
=
π
π
∫
1
2
2
2
2
1δω ω
ω
3.5 (a) Let y[n] = g[n – n
o
],Then Y e y n e
jjn
n
() []
ωω
=
=?∞
∞
∑
=?
=?∞
∞
∑
gn n e
o
jn
n
[]
ω
=

=?∞
∞
∑
egne
jn jn
n
o
ωω
[] = e Ge
jn j
o
ω ω
().
(b) Let h n e g n
jn
o[] []=
ω
,then H e h n e e g n e
jjn
n
jn jn
n
o( ) [] []
ωωωω
==
=?∞
∞
=?∞
∞
∑∑
=

=?∞
∞
∑
gn e
jn
n
o[]
()ωω
= G e
j
o()
()ωω?
.
(c) Ge gne
jj
n
() []
ωω
=
=?∞
∞
∑
,Hence
dGe
d
jng n e
j
jn
n
()
[]
ω
ω
ω
( )
=?
=?∞
∞
∑
.
Therefore,j
dGe
d
ng n e
j
jn
n
()
[]
ω
ω
ω
( )
=
=?∞
∞
∑
,Thus the DTFT of ng[n] is j
dGe
d
j
()
ω
ω
( )
.
(d) y[n] = g[n]
*
h[n] = g k h n k
k
[][ ]?
=?∞
∞
∑
,Hence Y e g k h n k e
jjn
kn
() [][ ]
ωω
=?
=?∞
∞
=?∞
∞
∑∑
==
=?∞
∞
=?∞
∞
∑∑
gk He e He gk e
jjk
k
jj
k
[]() () []
ωω ω ω
= He Ge
jj
()()
ωω
.
(e) y[n] = g[n]h[n],Hence Y e g n h n e
jj
n
( ) [][]
ωω
=
=?∞
∞
∑
44
Since gn Ge e d
jjn
[] ( )=
∫
1
2π
θ
θθ
π
π
we can rewrite the above DTFT as
Ye hne Ge e d
jjjn
n
() [] ()
ωωθ
π
π
π
θ=
=?∞
∞
∫
∑
1
2
=

=?∞
∞
∑
∫
1
2π
θ
θω
π
π
Ge hne d
jjn
n
() []
()
=
∫
1
2π
θ
θωθ
π
π
Ge He d
jj
()( )
()
.
(f) yn gn h n gn H e e d
nn
jjn
[] [] *[] [] *( )==
=?∞
∞
=?∞
∞
∑∑
∫
1
2π
ω
ωω
π
π
=
1
2π
ω
ωω
π
π
He gne d
j
n
jn
*( ) [ ]
=?∞
∞
∑
∫
=
1
2π
ω
ωω
π
π
He Ge d
jj
*( ) ( )
∫
,
3.6 DTFT{x[n]} = X e x n e
jjn
n
() []
ωω
=
=?∞
∞
∑
.
(a) DTFT{x[–n]} =?= =
=?∞
∞
=?∞
∞
∑∑
xne xme Xe
jn
n
jm
m
j
[] [] ( ).
ωωω
(b) DTFT{x*[-n]} =?=?
=?∞
∞
=?∞
∞
∑∑
xne xne
jn
n
jn
n
*[ ] [ ]
ωω
using the result of Part (a).
Therefore DTFT{x*[-n]} = X e
j
*( ).
ω
(c) DTFT{Re(x[n])} = DTFT
xn x n
Xe X e
jj
[] *[]
() *( )
+
=+
{}
2
1
2
ωω
using the result of Part
(b).
(d) DTFT{j Im(x[n])} = DTFT j
xn x n
j
Xe X e
jj
[] *[]
() *( )

=?
{}
2
1
2
ωω
.
(e) DTFT x n DTFT
xn x n
Xe X e Xe X e
cs
jj j
re
j
[]
[] *[ ]
() *( )Re() ().
{}
=
+?
=+
{}
=
{}
=
2
1
2
ωω ωω
(f) DTFT x n DTFT
xn x n
Xe X e jX e
ca
jj
im
j
[]
[] *[ ]
() *( ) ().
{}
=

=?
{}
=
2
1
2
ωω ω
3.7 Xe xne
jjn
n
() []
ωω
=
=?∞
∞
∑
where x[n] is a real sequence,For a real sequence,
Xe X e
jj
() *( ),
–ωω
= and IDFT X e x n
j
* ( ),*[ ].
– ω
{}
=?
45
(a) Xe Xe Xe
re
jj j
() () *( ).
–ωω ω
=+
{}
1
2
Therefore,IDFT X e
re
j
()
ω
{}
=+
{}
=+?{}=+?{}=
1
2
1
2
1
2
IDFT X e X e x n x n x n x n x n
jj
ev
( ) *( ) [] *[] [] [] [].
–ωω
(b) jX e X e X e
im
jj j
() () *( ).
–ωω ω
=?
{}
1
2
Therefore,IDFT jX e
im
j
()
ω
{}
=?
{}
={}={}=
1
2
1
2
1
2
IDFT X e X e x n x n x n x n x n
jj
od
( ) *( ) [] *[] [] [] [].
–ωω
3.8 (a) Xe xne
jjn
n
() []
ωω
=
=?∞
∞
∑
,Therefore,X e x n e X e
jj
n
j
*( ) [ ] ( ),
–ωωω
==
=?∞
∞
∑
and hence,
X e xne Xe
jjn
n
j
*( ) [ ] ( ).
––ωωω
==
=?∞
∞
∑
(b) From Part (a),X e X e
jj
() *( ).
–ωω
= Therefore,X e X e
re
j
re
j
() ( ).
–ωω
=
(c) From Part (a),X e X e
jj
() *( ).
–ωω
= Therefore,X e X e
im
j
im
j
() ( ).
–ωω
=?
(d) Xe X e X e
j
re
j
im
j
() () ()
ωωω
=+
22
=+Xe X e
re
j
im
j22
() ()
––ωω
=
Xe
j
()
ω
.
(e) arg ( ) tan
()
()
tan
()
()
arg ( )Xe
Xe
Xe
Xe
Xe
Xe
j im
j
re
j
im
j
re
j
jω
ω
ω
ω
ω
ω
==? =?

11
3.9 x[n] =
1
2π
ω
ωω
π
π
Xe e d
jjn
()
∫
,Hence,xn Xe e d
jjn
*[ ] *( )
–
=
∫
1
2π
ω
ωω
π
π
.
(a) Since x[n] is real and even,hence X e X e
jj
() *()
ωω
=, Thus,
x[– n] =
1
2π
ω
ωω
π
π
Xe e d
jjn
()
∫
,
Therefore,xn xn x n Xe nd
j
[ ] [ ] [ ] ( )cos( ),=+?( )=
∫
1
2
1
2π
ωω
ω
π
π
Now x[n] being even,X e X e
jj
() ( )
–ωω
=, As a result,the term inside the above integral is even,
and hence xn Xe nd
j
[ ] ( )cos( )=
∫
1
0
π
ωω
ω
π
(b) Since x[n] is odd hence x[n] = – x[– n].
Thus x[n] =
1
2
xn x n[] [ ]() =
j
Xe nd
j
2π
ωω
ω
π
π
( )sin( )
∫
,Again,since x[n] = – x[– n],
46
Xe Xe
jj
() ( )
–ωω
=?, The term inside the integral is even,hence xn
j
Xe nd
j
[ ] ( )sin( )=
∫
π
ωω
ω
π
0
3.10 xn n n A
eee e
n
n
o
n
jnj jnj
[ ] cos( ) [ ] [ ]=+=
+

αωφμ α μ
ωφ ω φ
00
2
=
A
ee n
A
ee n
jj
n
jj
oo
22
φω φ ω
αμ αμ
( )
+
( )

[ ] [ ],Therefore,
Xe
A
e
ee
A
e
ee
jj
jj
j
jj
oo
()
ωφ
ωω
φ
ωω
αα
=
+

2
1
1 2
1
1
.
3.11 Let x[n] = αμ α
n
n[],.<1 From Table 3.1,DTFT{x[n]} = Xe
e
j
j
(),
–
ω
ω
α
=
1
1
(a) Xe n e e e e
jnjn
n
njn
n
jnjn
n
1
1
1
0
1() [ ]
ωωωωω
αμ α α α=+= =+
=?∞
∞
=?
∞

=
∞
∑∑∑
=+
=
α
α
α
α
α
ω
ω
ω
ω
1
1
1
1
1
e
e
e
e
j
j
j
j––
.
(b) xn n n
n
2
[] [].=αμ Note that x n n x n
2
[] [].= Hence,using the differentiation-in-frequency
property in Table 3.2,we get Xe j
dX e
d
e
e
j
jj
j
2
2
1
()
()
()
ω
ωω
ω
ω
α
α
==
.
(c) xn
nM
otherwise
n
3
0
[]
,,
,.
=
≤
α
Then,Xe e e
jnjn
n
M
njn
nM
3
0
1
()
ωω ω
αα=+
=

=?
∑∑
=
1
1
1
1
11
1
+
+? +

α
α
α
α
α
ω
ω
ω
ω
ω
MjM
j
MjM
MjM
j
e
e
e
e
e
()
.
(d) Xe e e e e
jn
n
jn n
n
jn j j
4
30
22
1()
––ωωωωω
αααα==
=
∞
=
∞

∑∑
=

1
1
1
22
α
αα
ω
ωω
e
ee
j
jj
.
(e) Xe ne ne e e
jn
n
jn n
n
jn j j
5
20
22 1
2()
––ωωωωω
αααα==
=?
∞
=
∞

∑∑
=

α
α
αα
ω
ω
ωω
e
e
ee
j
j
jj
()
.
1
2
2
22 1
(f) Xe e e e
e
jn
n
jn m
m
jm m
m
jm
j
6
1
10
1
1
1
1
1()
–ωωωω
ω
αα α
α
===?=
=?∞
=
∞
=
∞
∑∑∑
=
e
e
j
j
ω
ω
α
.
47
3.12 (a) yn
NnN
otherwise
1
1
0
[]
,–,
,.
=
≤≤
Then Ye e e
e
e
N
jjn
nN
N
jN
jN
j1
21
1
1 2
1
2
()
()
()
sin( )
sin( / )
()
ωωω
ω
ω
ω
ω
==
=
+
[]
=?
+
∑
(b) yn
n
N
NnN
otherwise
2
1
0
[]
,,
,.
=
<<
Assume N to be odd,Now yn
N
yn
20
1
[] []=
*
where
yn
N
n
N
otherwise
0
1
1
2
1
2
0
[]
,,
,.
=
≤≤
Thus Ye
N
Ye
N
N
jj
20
2
2
2
112
2
() ()
sin /
sin ( / )
.
ωω
ω
ω
=? =?
( )
Note,The above result also holds for N even.
(c) y n
nN NnN
otherwise
3
2
0
[]
cos( / ),,
,.
=
≤≤
π
Then,
Ye e e e e
jjnNjn
nN
N
jn N jn
nN
N
3
22
1
2
1
2
()
(/ ) (/ )ωπωπω
=+

=?
=?
∑∑
=+

=?
+
=?
ππ
∑∑
1
2
1
2
22
ee
jn
nN
N
jn
nN
N
NN
ωω–
=
+
( )
( )
+
++
( )
+
( )
π
π
π
π
1
2
2
1
2
2
2
1
2
2
2
1
2
2
sin ( )( )
sin ( ) /
sin ( )( )
sin ( ) /
ω
ω
ω
ω
N
N
N
N
.
3.13 Denote xn
nm
nm
n
m
n
[]
()!
!( )!
[],.=
+?
<
1
1
1αμ α We shall prove by induction that
DTFT x n
m
[]
{}
= Xe
e
m
j
jm
()
()
.
–
ω
ω
α
=
1
1
From Table 3.1,it follows that it holds for m = 1.
Let m = 2,Then xn
n
n
nnxnnxnxn
n
211
1
1[]
()!
!(
[] ( ) [] [] [].=
+
=+ = +αμ Therefore,
Xe
e
eee
j
j
jjj
2
22
1
1
1
1
1
()
() ()
–
–––
ω
ω
ωωω
α
ααα
=
+
=
using the differentiation-in-frequency
property of Table 3.2.
Now asuume,the it holds for m,Consider next xn
nm
nm
n
m
n
+
=
+
1
[]
()!
!( )!
[]αμ
=
+
+?
=
+
= +
nm
m
nm
nm
n
nm
m
xn
m
nx n x n
n
mmm
()!
!( )!
[],[] [] []
1
1
1
αμ, Hence,
Xe
m
j
d
d
ee
e
ee
m
j
jm jm
j
jm jm
+
+
=
+
=
+
1
1
11
1
1
11
1
1
()
()()()()
––
–
––
ω
ωω
ω
ωω
ω
αα
α
αα
=
+
1
1
1
()
.
–
α
ω
e
jm
48
3.14 (a) Xe k
a
j
k
() ( )
ω
δω π=+
=?∞
∞
∑
2, Hence,x[n] =
1
2
() e d
jn
π
=
π
π
∫
δω ω
ω
1.
(b) Xe
e
e
e
b
j
jN
j
jn
n
N
()
()
ω
ω
ω
ω
=
=
+
=
∑
1
1
1
0
,Hence,x[n] =
10
0
,,
,.
≤≤
nN
otherwise
(c)
Xe e
c
j
N
j
N
N
( ) cos( )
ωω
ω=+ =+
=
=?
∑∑
12 2
0
l
l
l
l
,Hence x[n] =
30
10
0
,,
,.
n
nN
otherwise
=
<≤
(d) Xe
je
e
d
j
j
j
()
()
,
ω
ω
ω
α
α
α=
<
1
1
2
, Now we can rewrite X e
d
j
()
ω
as
Xe
d
d e
d
d
Xe
d
j
j o
j
()
()
()
ω
ω
ω
ω α ω
=
=
( )
1
1
where Xe
e
o
j
j
()
ω
ω
α
=
1
1
.
Now x n n
o
n
[] []=α μ, Hence,from Table 3.2,x n jn n
d
n
[] []=? α μ,
3.15 (a) He
ee e e
ee e e
j
jj j j
jj j j
1
22
22
12
2
3
2
1
3
2
3
2
()
––
––ω
ωω ω ω
ωω ω ω
=+
+
+
+
=+ + + +,
Hence,the inverse of H e
j
1
()
ω
is a length-5 sequence given by
hn n
1
1511115 2 2[],,,.=?≤≤
(b) He
ee e e e e
e
j
jj j j j j
j
2
22 2 2
2
32
2
4()
––/–/
–/ω
ωω ω ω ω ω
ω
=+
+
+
+
+
=++ ++
( )
1
2
2344 3 2
223
ee ee e
jj jj jωω ω ω ω–– –
,Hence,the inverse of H e
j
2
()
ω
is a length-6
sequence given by h n n
2
115 2 2151 2 3[],,,.=?≤≤
(c) He j
ee e e e e
j
j
jj j j j j
3
22 2 2
34
2
2()
––/–/
ω
ωω ω ω ω ω
=+
+
+
+
=+++
( )
1
2
2202 2
32 23
eee eee
jjj j jjωωω ω ωω–––
,Hence,the inverse of H e
j
3
()
ω
is a
length-7 sequence given by h n n
3
05 1 1 0 1 1 05 3 3[], – –,,.=≤≤
(d) He j
ee e e e e
j
e
j
jj j j j j
j
4
22 2 2
2
42
2
3()
––/–/
/ω
ωω ω ω ω ω
ω
=+
+
+
+
=?+?+?
1
2
3
2
1
2
33
1
2
3
2
32 2
eee e e
jjj j jωωω ω ω––
,Hence,the inverse of H e
j
4
()
ω
is a length-6
sequence given by h n n
4
075 025 15 15 025 075 3 2[]......,.=
[]
≤≤
3.16 (a) He e e e e
jjjj
2
22
12
3
2
12
3
4
5
2
3
4
( ) cos ( cos ),
––ωωωω
ωω=+ + + = + + + +
49
Hence,the inverse of H e
j
1
()
ω
is a length-5 sequence given by
hn n
1
075 1 25 1 075 2 2[],,,,.=?≤≤
(b) He e
jj
2
2
13
4
2
12 2( ) cos ( cos ) cos( / )
/ωω
ωωω=+ + +
=+++++
1
2
5
4
11
4
11
4
5
4
1
2
32 2
ee e e e
jj j j jωωω ω ω––
,Hence,the inverse of H e
j
2
()
ω
is a length-6
sequence given by h n n
2
05 125 275 275 125 05 3 2[]......,.=
[]
≤≤
(c) He j
j
3
34 1 2( ) cos ( cos ) sin( )
ω
ωωω=+ ++
[]
=+++
1
4
7
4
0
7
4
1
4
32 2 3
ee e e e e
jj j j j jωω ω ω ω ω–– –
,Hence,the inverse of H e
j
3
()
ω
is a
length-7 sequence given by h n n
3
025 1 175 0 175 1 025 3 3[],,,,,.=≤≤
(d) He j e
4
2
42
3
2
12 2( ) cos ( cos ) sin( / )
–/
ωωω=+ ++
=++?+?
1
2
3
4
1
4
9
2
9
2
1
4
3
4
22
ee e e e
jj j j jωω ω ω ω–– –
,Hence,the inverse of H e
j
4
()
ω
is a
length-6 sequence given by hn n
4
3
8
1
8
9
4
9
4
1
8
3
8
23[],.=
≤≤
3.17 Ye Xe X e
jj j
() ( ) ().
ωωω
==
( )
33
Now,X e x n e
jj
n
() [],
ωω
=
=?∞
∞
∑
Hence,
Ye yne X e xn e xm e
n
n
jj jm
mn
() [] () []( ) [/],
ω ω ω
( )
==
=?∞
∞

=?∞
∞
=?∞
∞
∑∑∑
33
3
Therefore,
yn
xn n
otherwise
[]
[],,,,
,.
=
=±±
{
036
0
K
3.18 Xe xne
jjn
n
() [],
ωω
=
=?∞
∞
∑
Xe xne
jjn
n
() [],
/(/)ωω22
=
=?∞
∞
∑
and X e x n e
jnjn
n
() [](),
/(/)
=?
=?∞
∞
∑
ωω22
1 Thus,
Ye yne Xe X e xn xn e
jn
n
jj njn
n
( ) [] ( ) ( ) [] []( ),
// (/)ωωωω ω
==+?
{}
=+?
( )
=?∞
∞
=?∞
∞
∑∑
1
2
1
2
1
22 2
Thus,
yn xn xn
x n for n even
for n odd
n
[] [] []( )
[],
.
=+?
( )
=
{
1
2
1
0
.
3.19 From Table 3.3,we observe that even sequences have real-valued DTFTs and odd sequences
have imaginary-valued DTFTs.
(a) Since?=nn,xn
1
[ ] is an even sequence with a real-valued DTFT.
(b) Since ( ),?=?nn
33
xn
2
[ ] is an odd sequence with an imaginary-valued DTFT.
50
(c) Since sin( ) sin( )?=?ωω
cc
n n and ωω
cc
nn(),?=?,x n
3
[ ] is an even sequence with a real-
valued DTFT.
(d) Since x n
4
[ ] is an odd sequence it has an imaginary-valued DTFT.
(e) Since x n
5
[ ] is an odd sequence it has an imaginary-valued DTFT.
3.20 (a) Since Y e
j
1
()
ω
is a real-valued function of ω,its inverse is an even sequence.
(b) Since Y e
j
2
()
ω
is an imaginary-valued function of ω,its inverse is an odd sequence.
(c) Since Y e
j
3
()
ω
is an imaginary-valued function of ω,its inverse is an odd sequence.
3.21 (a) He
LLP
j
()
ω
is a real-valued function of ω,Hence,its inverse is an even sequence.
(b) He
BLDIF
j
()
ω
is a real-valued function of ω,Hence,its inverse is an even sequence.
3.22 Let u[n] = x[–n],and let X e
j
()
ω
and U e
j
()
ω
denote the DTFTs of x[n] and u[n],respectively.
From the convolution property of the DTFT given in Table 3.2,the DTFT of y[n] = x[n]
*
u[n]
is given by Y e
j
()
ω
= X e
j
()
ω
Ue
j
()
ω
,From Table 3.3,U e X e
jj
() ( ).
ωω
=
But from Table 3.4,
Xe X e
jj
()*().
=
ωω
Hence,Y e
j
()
ω
= X e
j
()
ω
Xe
j
*( )
ω
= Xe
j
()
ω
2
which is real-valued function
of ω.
3.23 From the frequency-shifting property of the DTFT given in Table 3.2,the DTFT of
xne
jn
[]
/?π 3
is given by X e
j
()
(/)ω+π 3
,A sketch of this DTFT is shown below.
X(e
j(ω+π/3)
)
–π /3 π /3–2π /3 2π /3–ππ0
ω
1
3.24 The DTFT of x n n
n
[] – [– –]=αμ 1 is given by
Xe e e e
e
jn
n
jn n
n
jn j
j
n
n
()
–
–––ωωωω
ω
ααα
α
=? =? =?
=∞
=
∞
=
∞
∑∑ ∑
1
1
1
0
.
For α>1,Xe e
ee
jj
jj
()
(/)
.
–ωω
ωω
α
αα
=?
=
1
1
1
1
1
Xe
j
()
cos
.
ω
ααω
2
2
1
12
=
+?
From Parseval's relation,
1
2
2
2
π
=
π
π
=?∞
∞
∫
∑
Xe d xn
j
n
() [].
ω
ω
51
(a) Xe
j
()
cos
.
ω
ω
2
1
54
=
+
Hence,α=?2,Therefoe,x n n
n
[] ( ) [ ].=21μ
Now,42 4
2
0
2
2
Xe d Xe d xn
jj
n
() () []
ωω
π
π
π
=?∞
∞
∫∫
∑
==π=π?
=?∞
∑
42
2
1
()
n
n
=π
=π
=
π
=
∞
=
∞
∑∑
4
1
4
1
4
4
3
10
n
n
n
n
.
(b) Xe
j
()
,cos
.
ω
ω
2
1
325 3
=
Hence,α=15,and therefore,x n n
n
[] (.) [ ].=15 1μ Now,
Xe d Xe d xn
jj
n
() () []
ωω
2
0
2
2
1
2
π
π
π
=?∞
∞
∫∫
∑
==π=π
=?∞
∑
(.)15
2
1
n
n
=π
=
∞
∑
4
9
1
n
n
=
π
=
π
=
π
=
∞
∑
4
9
4
9
4
9
9
5
4
5
0
n
n
.
(c) Using the differentiation-in-frequency property of the DTFT,the inverse DTFT of
Xe j
d
d e
e
e
j
j
j
j
()
()
ω
ω
ω
ω
ω α
α
α
=
=
1
11
2
is x n n n
n
[] [ ].=αμ 1 Hence,the inverse DTFT of
1
1
2
()?
α
ω
e
j
is?+()[ ].nn
n
11αμ
Ye
j
()
( cos )
.
ω
ω
2
2
1
54
=
Hence,α=2 and y n n n
n
[] ( ) [ ].=? +12 1μ Now,
42 4
2
0
2
2
Xe d Xe d xn
jj
n
() () []
ωω
π
π
π
=?∞
∞
∫∫
∑
==π=π+?
=?∞
∑
412
22
1
()n
n
n
=π
=π =π
=
∞
∑
1
4
94
916
4
0
2
n
n
n
/
/
..
3.25 (a) Xe xn
j
n
() [],
0
31234113= =+++?=
=?∞
∞
∑
(b) Xe xne
j
n
jn
() [],
π
=?∞
∞
π
==+?+=
∑
312 3 4111
(c) Xe d x
j
() [],
ω
ω
π
π
∫
=π =?π20 4
(d) Xe d xn
j
n
() [],
ω
ω
2
2
28
π
π
=?∞
∞
∫
∑
=π = π (Using Parseval's relation)
52
(e)
dX e
d
dnxn
j
n
()
[],
ω
ω
ω
2
2
2 378
π
π
=?∞
∞
∫
∑
=π? = π (Using Parseval's relation with differentiation-in-
frequency property)
3.26 (a) Xe xn
j
n
() [],
0
241532438==?+?+?+=
=?∞
∞
∑
(b) X e x n
jn
n
() [](),
π
=?∞
∞
=++?+=
∑
1 2415324314
(c) Xe d x
j
() [],
ω
ω
π
π
∫
=π =?π20 4
(d) Xe d xn
j
n
() [],
ω
ω
2
2
2 168
π
π
=?∞
∞
∫
∑
=π = π (Using Parseval's relation)
(e)
dX e
d
dnxn
j
n
()
[],
ω
ω
ω
2
2
2 1238
π
π
=?∞
∞
∫
∑
=π? = π
3.27 Let G e
j
1
()
ω
denote the DTFT of g n
1
[].
(b) g n gn gn
211
4[] [] [ ].=+? Hence,the DTFT of g n
2
[ ] is given by
G e Ge e Ge e Ge
jjjj jj
21
4
1
4
1
1() () ()( )().
ωωωω ωω
=+ =+

(c) gn g n gn
31 1
34[ ] [ ( )] [ ].=+? Now,the DTFT of g n
1
[]? is given by G e
j
1
()
ω
.
Hence,the DTFT of g n
3
[ ] is given by G e e G e e G e
jj jjj
3
3
1
4
1
() ( ) ().
ωωωωω
=+

(d) gn gn g n
411
7[] [] [( )].=+ Hence,the DTFT of g n
4
[ ] is given by
Ge Ge e Ge
jjjj
41
7
1
() () ( ).
ωωωω
=+

3.28 Y e X e X e X e
jjjj
() () () (),
ωωωω
=
123
i.e.,
yne x ne x ne x ne
n
jn
n
jn
n
jn
n
jn
[] [] [] []
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑∑
=
ωωωω
123
(a) Therefore,setting ω=0 we get y n x n x n x n
nnnn
[] [] [] []
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑∑
=
123
.
(b) Setting ω=π we get ()[] () [] () [] () []?=?
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑∑
1111
123
n
n
n
n
n
n
n
n
yn x n x n x n,
53
3.29 (a)
xn x n x n
ev od
[] [] [].=+
Now,for a causal x[n],from the results of Problem 2.4(),we
observe
xn xnnx nhnx n
ev
[] [][] [][] [] [][],=?=?20 0μδ δ (1)
xn x n n x n
od
[] [][] [][].=+μδ (2)
Taking the DTFT of both sides of Eq,(2) we get
Xe He x
jj
() ()[],
ωω
=?0 (3)
where H e DTFT x n n X e
j
ev re
j
( ) [][] ( )
–
ωθ
μ=
{}
=
π
π
π
∫
2
1
m(),
()
ed
j ωθ
θ
(4)
Note,The DTFT of x
ev
[n] is X e
re
j
()
ω
,and the DTFT of μ[]n is m ()e
jω
.
Now,from Table 3.1,m ()e
jω
=
+π + π =?
+π + π
=?∞
∞
=?∞
∞
∑∑
1
1
2
1
22 2
2
e
k
j
k
j
kk
ω
δω
ω
δω() cot ().
Substituting the above in Eq,(4) we get
He X e
j
kd
j
re
j
k
( ) ( ) cot ( )
–
ωθ
θ
δθ θ=
π
+π + π
π
π
=?∞
∞
∫
∑
11
22 2
2
=+
π
π
π
π
π
π
∫∫
Xe Xe d
j
Xe d
re
j
re
j
re
j
( ) () ()cot
––
ωθ θ
θ
ωθ
θ
1
22 2
.
Substituting the above in Eq,(3) we get
Xe X e jX e He x
j
re
j
im
jj
() () () ()[]
ωω ωω
=+ =?0
=+
π
π
π
π
π
π
∫∫
Xe Xe d
j
Xe d x
re
j
re
j
re
j
( ) () ()cot []
––
ωθ θ
θ
ωθ
θ
1
22 2
0
=?
π
π
π
∫
Xe
j
Xe d
re
j
re
j
( ) ( )cot
–
ωθ
ωθ
θ
22
,(5)
since
1
2
0
π
=
π
π
∫
Xe d x
re
j
() [],
–
θ
θ as x[n] is real,Comparing the imaginary part of both sides of
Eq,(5) we therefore get Xe Xe d
im
j
re
j
( ) – ( )cot,
–
ωθ
ωθ
θ=
π
π
π
∫
1
22
(b) Taking the DTFT of both sides of Eq,(2) we get
Xe Ge x
jj
() ()[],
ωω
=+0 (6)
where,G e DTFT x n n
j
Xe
j
od im
j
( ) [][] ( )
–
ωθ
μ=
{}
=
π
π
π
∫
2 m(),
()
ed
j ωθ
θ
(7)
as j X e
im
j
()
ω
is the DTFT of x
od
[n],Substituting the expression for m ()e
jω
given above in
Eq,(7) we get Ge
j
Xe
j
kd
j
im
j
k
( ) ( ) cot ( )
–
ωθ
θ
δθ θ=
π
+π + π
π
π
=?∞
∞
∫
∑
1
22 2
2
54
=+
π
+
π
π
π
π
π
∫∫
jX e
j
Xed Xe d
im
j
im
j
im
j
( ) () ()cot
––
ωθ θ
θ
ωθ
θ
2
1
22
Substituting the above in Eq,(6) we get
Xe X e jX e Ge x
j
re
j
im
jj
() () () ()[]
ωω ωω
=+ =+0
=+
π
+
π
+
π
π
π
π
∫∫
jX e
j
Xed Xe dx
im
j
im
j
im
j
( ) () ()cot []
––
ωθ θ
θ
ωθ
θ
2
1
22
0
=+
π
+
π
π
∫
jX e X e d x
im
j
im
j
( ) ( )cot [ ]
–
ωθ
ωθ
θ
1
22
0 (8)
as
1
2
0
π
=
π
π
∫
Xed
im
j
()
–
θ
θ since X e
im
j
()
ω
is an odd function of ω,Comparing the real parts of
both sides of Eq,(8) we finally arrive at Xe X e d x
re
j
im
j
( ) ( )cot [ ].
–
ωθ
ωθ
θ=
π
+
π
π
∫
1
22
0
3.30 SW e
N
kln
n
N
jnklN
n
N
==

=
=
∑∑
() ()/
0
1
2
0
1
π
If k – l ≠ rN then S =
1
1
11
1
0
2
22
=
=

e
ee
jnkl
jnklN jnklN
π
ππ
()
()/ ()/
.
If k – l = rN then S = W e N
N
rnN
n
N
jnr
n
N
n
N
=
=
=
∑∑∑
==
0
1
2
0
1
0
1
1
π
.
Hence,W
N for
N
kln
n
N

=
∑
=
()
,
0
1
k – l = rN,r an integer,
0,otherwise.
3.31?[]?[]
[]yn xr hn r
r
N
=?
=
∑
0
1
,Then?[]?[]
[]y n kN x r h n kN r
r
N
+= +?
=
∑
0
1
,Since
[]h n is periodic in n
with a period N,
[]h n kN r+? =
[]hn r?, Therefore?[]?[]
[]y n kN x r h n r
r
N
+=?
=
∑
0
1
=?[]y n,hence
[]y n is also periodic in n with a period N.
3.32?[]xn =?{}0 1 0 2 3 and
[],hn =?{}2010 2
Now,?[]?[]
[]?[]
[]?[]
[]?[]
[]?[]
[]?[]
[]yxrhrxhxhxhxhxh
r
00014233241
0
4
=?=++++
=
∑
= –4.
Similarly?[]?[]
[]?[]
[]?[]
[]?[]
[]?[]
[]yxrhrxhxhxhxh
r
1101102332
0
3
=?=+++
=
∑
= 5.
Continuing the process we can show that?[],y2 4=?[],y3 9=? and?[],y4 6=
55
3.33?[]xn ={}2 1 2 3 2 and
[],hn ={}12 30 3
Now,?[]?[]
[]?[]
[]?[]
[]?[]
[]?[]
[]?[]
[]yxrhrxhxhxhxhxh
r
00014233241
0
4
=?=++++
=
∑
= –8.
Similarly?[]?[]
[]?[]
[]?[]
[]?[]
[]?[]
[]yxrhrxhxhxhxh
r
1101102332
0
3
=?=+++
=
∑
= 3.
Continuing the process we can show that?[],y2 15=[],y3 16= and?[],y4 8=? 3.34 Since
[]
[]ψψ
kk
nrN n+=,hence all the terms which are not in the range 0,1,....N-1 can be
accumulated to
[],ψ
k
n where 0 k N –1≤≤, Hence in this case the Fourier series representation
involves only N complex exponential sequences,Let
[]
[]
/
xn
N
Xke
jknN
k
N
=
=
∑
1
2
0
1
π
then
[]
[]
/()/
xne
N
Xke
n
N
jrnN j krnN
k
N
n
N
=

=
=
∑∑∑
=
0
1
22
0
1
0
1
1
ππ
=
1
2
0
1
0
1
N
Xk e
jkrnN
n
N
k
N
[]
()/π?
=
=
∑∑
.
Now from Eq,(3.28),the inner summation is equal to N if k = r,otherwise it is equal to 0,
Thus?[]
/
xne
n
N
jrnN
=
∑
0
1
2π
=
[]Xr.
[]?[]?[]?[]
()/ / /
Xk N xne xne e xne
n
N
jkNnN
n
N
jknN j n
n
N
jknN
+= = =
=
+
=

=
∑∑ ∑
l
ll
0
1
2
0
1
22
0
1
2ππππ
=
[].Xk
3.35 (a)? [ ] cos,
//
xn
n
ee
jn jn
1
44
4
1
2
=
π
=+
{}
π?π
The period of? []xn
1
is N = 8,
[]
// //
Xk e e e e
jn
n
jkn jn
n
jkn
1
28
0
7
28 28
0
7
28
1
2
=+
π
=
π?π
=
π
∑∑
=+
π?
=
π +
=
∑∑
1
2
218
0
7
218
0
7
ee
jnk
n
jnk
n
()/ ()/
,Now,from Eq,(3.28) we observe
e
for k
otherwise
jnk
n
π?
=
∑
=
=
{
218
0
7
81
0
()/
,,
and e
for k
otherwise
jnk
n
π +
=
∑
=
=
{
218
0
7
87
0
()/
,,
,.
Hence,
[]
,,,
,.
Xk
k
otherwise
1
417
0
=
=
{
(b)? [ ] sin cos,
// //
xn
nn
j
ee ee
jn jn jn jn
2
33 44
3
3
4
1
2
3
2
=
π
+
π
=?
{}
++
{}
π?π π?π
The period of
sin
π
n
3
is 6 and the period of cos
π
n
4
is 8,Hence,the period of? []xn
2
is the gcm of
(6,8) and is 24,
[]
// //
Xk
j
ee e e
jn
n
jkn jn
n
jkn
2
824
0
23
224 824
0
23
224
1
2
=?
π
=
π?π
=
π
∑∑
56
++
π
=
π?π
=
π
∑∑
3
2
624
0
23
224 624
0
23
224
ee e e
jn
n
jkn jn
n
jkn// //
=?
++
π?
=
π +
=
π?
=
π +
=
∑∑ ∑∑
1
2
3
2
2324
0
23
2324
0
23
2424
0
23
2424
0
23
j
ee ee
jnk
n
jnk
n
jnk
n
jnk
n
( )/ ( )/ ( )/ ( )/
.
Hence
[]
,,
,,,
,.
Xk
jk
k
otherwise
2
12 3
12 21
36 4 20
0
=
=
=
=
3.36 Since?[]pn is periodic with period N,then from Eq,(3.168a) 0f Problem 3.34,
[]
[]
/
pn
N
Pke
jknN
k
N
=
=
∑
1
2
0
1
π
where using Eq,(3.168b) we get
[]?[]
/
Pk pne
jknN
n
N
=
=
∑
2
0
1
π
= 1,
Hence?[]
/
pn
N
e
jknN
k
N
=
=
∑
1
2
0
1
π
.
3.37
[] ( ) ( ) [],–,
/
//
Xk Xe Xe xne k
j
kN
j k N j kn N
n
=== ∞<∞
=π
π?π
=?∞
∞
∑
ω
ω 2
22
(a)
[]( )( )( )
[].
()/ / /
Xk N Xe Xe e Xe Xk
j kNN jkNj jkN
+= = = =
π+ π π π
l
ll2222
(b)?[]
[]
/
xn
N
Xke
jknN
k
N
=
=
∑
1
2
0
1
π
=
1
22
0
1
N
xe e
j k N j kn N
k
N
[]
//
l
l
l
π
=?∞
∞
=
∑∑
π
=
π?
=?∞
∞
=
∑∑
1
2
0
1
N
xe
jkn N
k
N
[]
()/
l
l
l
,Let l=+?n r N, Then?[] [ ]xn
N
xn r N e
jkr
k
N
r
=+?
π
=
=?∞
∞
∑∑
1
2
0
1
.
But e N
jkr
k
N
π
=
∑
=
2
0
1
,Hence,?[] [ ].xn xn r N
r
=+?
=?∞
∞
∑
3.38 (a)
[]?[]?[]?[]
//
Gk gne xnyne
jknN
n
N
jknN
n
N
==
=
=
∑∑
2
0
1
2
0
1
ππ
,Now,?[]
[]
/
xn
N
Xre
jrnN
r
N
=
=
∑
1
2
0
1
π
Therefore,
[]
[]?[]
()/
Gk
N
Xryne
jkrnN
r
N
n
N
=

=
=
∑∑
1
2
0
1
0
1
π
=
1
2
0
1
0
1
N
Xr yne
jkrnN
n
N
r
N
[]?[]
()/
=
=
∑∑
π
=
1
0
1
N
XrYk r
r
N
[]
[]?
=
∑
.
(b)
[]
[]
[]
/
hn
N
XkYke
jknN
k
N
=
=
∑
1
2
0
1
π
=
1
0
1
2
0
1
N
xrYke
r
N
jknrN
k
N
[]
[]
()/
=
=
∑∑
π
57
=?[]
[]
()/
xr
N
Yke
jknrN
k
N
r
N
1
2
0
1
0
1
π?
=
=
∑∑
=?[]?[]xr yn r
r
N
=
∑
0
1
.
3.39 (a) yn gn hn[] [] []=+αβ,Therefore
Yk ynW gnW hnW Gk Hk
N
nk
n
N
N
nk
n
N
N
nk
n
N
[] [] [] [] [] []== + =+
=
=
=
∑∑∑
0
1
0
1
0
1
αβαβ
(b) xn g n n
N
[] [ ]=<? >
0
,Therefore X[k] = gnn W
NN
nk
n
N
[]<? >
=
∑ 0
0
1
= gN n n W
N
nk
n
n
[]+?
=
∑ 0
0
1
0
+ gn n W
N
nk
nn
N
[]?
=
∑ 0
1
0
= gnW gnW
N
nn Nk
nNn
N
N
nn k
n
Nn
[] []
() ()+?
=?
+
=

∑∑
+
0
0
0
0
1
0
1
= WgnW
N
nk
N
nk
n
N
0
0
1
[]
=
∑
= WGk
N
nk
0
[].
(c) un W gn
N
kn
[] []=
0
,Hence U k u n W g n W
N
nk
n
N
N
kk n
n
N
[] [] []
()
==
=
=
∑∑
0
1
0
1
0
=
≥
<
=
+?
=
∑
∑
Wgnifkk
Wgnifk
N
kk n
n
N
N
Nkk n
n
N
()
()
[],,
[],.
0
0
0
1
0
0
1
0
Thus,U k
Gk k ifk k
GN k k ifk k
[]
[],,
=
≥
+? <
00
= Gkk
N
[]<? >
0
.
(d) h[n] = G[n],Therefore,Hk hnW GnW
N
nk
n
N
N
nk
n
N
[] [] []==
=
=
∑∑
0
1
0
1
=
=
=
∑∑
grW W
nr kr
r
N
n
N
[]
0
1
0
1
= gr W
krn
n
N
r
N
[]
()+
=
=
∑∑
0
1
0
1
.
The second sum is non-zero only if k = r = 0 or else if r = N – k and k ≠ 0,Hence,
H[k] =
Ng if k
Ng N k if k
Ng k
N
[],,
[],,
[]
00
0
=
>
=<?>
.
(e) un gmh n m
N
m
N
[] [ ][ ]=<?>
=
∑
0
1
,Therefore,U k g m h n m W
N
m
N
n
N
N
nk
[] [ ][ ]=<?>
=
=
∑∑
0
1
0
1
= gm h n m W
N
n
N
m
N
N
nk
[] [ ]<? >
=
=
∑∑
0
1
0
1
=
=
∑
gmHkW
m
N
N
mk
[][]
0
1
= H[k]G[k].
58
(f) v[n] = g[n]h[n],Therefore,V k g n h n W
n
N
N
nk
[] [][]=
=
∑
0
1
==
=
=
∑∑
1
0
1
0
1
N
hnGr W W
nk nr
r
N
n
N
[] []
11
0
1
0
1
0
1
N
Gr hnW
N
GrH k r
krn
n
N
r
N
N
r
N
[] [ ] [] [ ]
()?
=
=
=
∑∑∑
=<?>.
(g) xn
N
XkW
nk
k
N
[] [],=
1
=
∑
0
1
Thus xn
N
XkW
nk
k
N
*[ ] *[ ],=
1
=
∑
0
1
Therefore,
xn
N
XrW X W
n
N
nr
r
N
n
N
n
N
[] [] *[]
2
0
1
2
0
1
0
1
0
1
1
=
=
=
=
∑∑∑∑
=
l
l
l
=
=
=
=
∑∑∑
1
2
0
1
0
1
0
1
N
Xr X W
nr
n
NN
r
N
[] []
*()
l
l
l
.
Since the inner sum is non-zero only if
l= r,we get xn
N
Xk
n
N
k
N
[] [],
2
0
1
2
0
1
1
=
=
∑∑
=
3.40 X[k] = xnW
nk
n
N
[]
=
∑
0
1
.
(a) X
*
[k] = x n W
nk
n
N
*[ ],
=
∑
0
1
Replacing k by N – k on both sides we obtain
X
*
[N – k] = x n W x n W
nN k
n
N
n
N
nk
*[ ] *[ ],
()
=
=
∑∑
=
0
1
0
1
Thus x
*
[n]? X
*
[N – k] = X
*
[< –k >
N
].
(b) X
*
[k] = x n W
nk
n
N
*[ ],
=
∑
0
1
Replacing n by N – n in the summation we get
X
*
[k] = x N n W x N n W
Nnk
n
N
nk
n
N
*[ ] *[ ],
()
=?

=
=
∑∑
0
1
0
1
Thus x
*
[N – n] = x
*
[< –n >
N
]? X
*
[k].
(c) Re{x[n]} =
1
2
{x[n] + x
*
[n]},Now taking DFT of both sides and using results of part (a)
we get Re{x[n]}?
1
2
{X[k] + X
*
[< –k >
N
]}.
(d) j Im{x[n]} =
1
2
{x[n] – x*[n]} this imples j Im{x[n]}?
1
2
{X[k] – X
*
[< –k >
N
]}.
(e) x
pcs
[n] =
1
2
{x[n] + x
*
[<–n >
N
]} Using linearity and results of part (b) we get
x
pcs
[n]?
1
2
{X[k] + X
*
[k]} = Re{X[k]}.
(f) x
pca
[n] =
1
2
{x[n] – x
*
[< –n >
N
]},Again using results of part (b) and linearity we get
x
pca
[n]?
1
2
{X[k] –X
*
[k]} = j Im {X[k]}.
59
3.41 X[k] = Re{X[k]} + j Im{X[k]} = x n e
jknN
n
N
[],
–/2
0
1
π
=
∑
(a) xn xnx n
pe N
[] [] [ ].=+<?>
{}
1
2
From Table 3.6,x n X k
N
DFT
*[ ] *[ ].<? >? Since x[n] is
real,x n x n X k
NN
DFT
[]*[]*[].<? > = <? >? Thus,X k Xk X k Xk
pe
[ ] [ ] *[ ] Re{ [ ]}.=+{}=
1
2
(b) xn xnx n
po N
[] [] [ ].=?<?>
{}
1
2
As a result,Xk XkXk j Xk
po
[] [] *[] Im{[]}.=?{}=
1
2
3.42 Since for a real sequence,x[n] = x*[n],taking DFT of both sides we get X[k] =
X*[<– k>
N
],This implies,Re{X[k]} + j Im{X[k]} = Re{X[<– k>
N
} – j Im{X[<– k>
N
}.
Comparing real and imaginary parts we get
Re{X[k]} = Re{X[<– k>
N
} and Im{X[k]} = – Im{X[<– k>
N
}.
Also Xk Xk Xk[ ] Re{ [ ]} Im{ [ ]}= ( ) +( )
22
=<>
( )
+<>
( )
Re{[ – ]} –Im{[ – ]}Xk Xk
NN
22
=<>Xk
N
[– ]
and arg{ [ ]} tan
Im{ [ ]}
Re{ [ ]}
tan
–Im{ [ ]}
Re{ [ ]}
Xk
Xk
Xk
Xk
Xk
N
N
=
=
<? >
<? >
11
=? <? >arg{ [ ] }Xk
N
.
3.43 (a) xn xn
18 1
11100011[] [].<? > =
[]
= Thus,x n
1
[ ] is a periodic even sequence,
and hence it has a real-valued 8-point DFT.
(b) xn
28
1 1 100001[],<? > = Thus,x n
2
[ ] is neither a periodic even or a
periodic odd sequence,Hence,its 8-point DFT is a complex sequence.
(c) xn xn
38 3
0 1 100011[] [].<? > =
[]
=? Thus,x n
3
[ ] is a periodic odd
sequence,and hence it has an imaginary-valued 8-point DFT.
(d) xn xn
48 4
01100011[] [].<? > =
[]
= Thus,x n
4
[ ] is a periodic even sequence,
and hence it has a real-valued 8-point DFT.
3.44 (a) Now,X[N/2] = xnW xn
N
nN
n
N
n
n
N
[] ( ) []
/2
0
1
0
1
1
=
=
∑∑
=?, Hence if x[n] = x[N – 1 – n] and N is
even,then ()[]?=
=
∑
10
0
1
n
n
N
xn or X[N/2] = 0.
(b) X[0] = xn
n
N
[]
=
∑
0
1
so if x[n] = – x[N – 1 – n],then X[0] = 0.
60
(c )
XxnWxnW xnW
n
N
n
n
n
nN
N
n
N
[ ] [] [] []
/
2
0
1
2
0
1
2
2
1
2
2
l
ll l
==+
=
=
=
∑∑∑
=++=++
=
=
=
∑∑ ∑
xnW xn W xn xn W
n
n
N
n
n
n
N
n
NN N
[] [ ] ([] [ ])
0
1
2
2
0
1
2
0
1
2
2
22
ll l
.
Hence if x[n] = – x[n +
N
2
] = – x[n+M],then X[2
l] = 0,for
l = 0,1,.,,,,M – 1.
3.45 X m xnW xnW xnW
N
mn
n
N
N
mn
n
N
N
mn
n
N
N
[ ] [] [] []2
2
0
1
2
0
2
1
2
2
1
==+
=
=
=
∑∑∑
=++=++
=
+
=
=
=
∑∑ ∑∑
xnW xn
N
WxnWxn
N
WW
N
mn
n
N
N
mn
N
n
N
N
mn
n
N
N
mn
n
N
N
mN
[] [ ] [] [ ]
()
2
0
2
1
2
2
0
2
1
2
0
2
1
2
0
2
1
22
=++
=≤≤?
=
∑
xn xn
N
Wm
N
N
mn
n
N
[] [ ],.
2
00
2
1
2
0
2
1
This implies xn xn
N
[] [ ],++=
2
0
3.46 (a) Using the circular time-shifting property of the DFT given in Table 3.5 we observe
DFT x n m W X k
N N
km
[[]<? >
{}
=
1
1
and DFT x n m W X k
N N
km
[[].<? >
{}
=
2
2
Hence,
Wk DFTxn W Xk W Xk W W Xk
N
km
N
km
N
km
N
km
[] [] [] [] [].= {}=+=+
( )
αβ αβ
12 1
A proof of the
circular time-shifting property is given in Problem 3,39.
(b) gn xn xn xn W xn
n
N
N
n
[] [] ( ) [] [] [],=+?
( )
=+
1
2
1
1
2
2
Using the circular frequency-shifting
property of the DFT given in Table 3.5,we get Gk DFTgn Xk X k
N
N
[] [] [] [ ].= {}=+<?>
1
22
(c) Using the circular convolution property of the DFT given in Table 3.5 we get
Yk DFTyn Xk Xk X k[] [] [] [] [].= {}=?=
2
A proof of the circular convolution property is given in
Problem 3,39.
3.47 (a) DFT x n
N
WXk Xk
N
k
N
[ ] [] [].?
==?
2
2
Hence,
uk DFT un DFT xn xn
N
Xk Xk[] [] [] [ ] [] [],= {}=+?
=?=
2
0
(b) Vk DFT vn DFT xn xn
N
Xk Xk Xk[] [] [] [ ] [] [] [].= {}=
=+=
2
2
61
(c) yn xn W xn
n
N
N
n
[] ( ) [] [].=? =1
2
Hence,Y k DFT y n DFT W x n X k
N
N
N
n
N
[] [] [] [ ]= {}=
=<?>
2
2
using the circular frequency-shifting property of the DFT given in Table 3.5.
3.48 (a) From the circular frequency-shifting property of the DFT given in Table 3.5,
IDFT X k m W x n
N N
mn
[][]<? >
{}
=
1
1
and IDFT X k m W x n
N N
mn
[][].<? >
{}
=
2
2
Hence,
w n IDFT W k IDFT X k m X k m
NN
[] [] [ [= {}=<?>+<?
{}
αβ
12
=+=
( )

αβ αβW xn W xn W W xn
N
mn
N
mn
N
mn
N
mn
12 12
[] [] [].
(b) Gk Xk Xk Xk W Xk
k
N
N
k
[] [] ( ) [] [] [].=+?
( )
=+
1
2
1
1
2
2
Using the circular time-shifting
property of the DFT given in Table 3.5,we get g n IDFT G k x n x n
N
N
[] [] [] [ ].= {}=+<?>
1
22
(c) Using the modulation property of the DFT given in Table 3.5 we get
y n IDFT Y k N x n x n N x n[] [] [] [] [].= {}= =?
2
3.49 (a) X m xnW xnW xnW
N
mn
n
N
N
mn
n
N
N
mn
n
N
N
[ ] [] [] []2
2
0
1
2
0
2
1
2
2
1
==+
=
=
=
∑∑∑
=++=++
=
+
=
=
=
∑∑ ∑∑
xnW xn
N
WxnWxn
N
WW
N
mn
n
N
N
mn
N
n
N
N
mn
n
N
N
mn
n
N
N
mN
[] [] [] []
()
2
0
2
1
2
2
0
2
1
2
0
2
1
2
0
2
1
22
=++
=?( ) =≤≤?
=
=
∑∑
xn xn
N
WxnxnW m
N
N
mn
n
N
N
mn
n
N
[] [ ] [] [],.
2
00
2
1
2
0
2
1
2
0
2
1
(b)
X xnW xnW xnW xnW xnW
N
n
n
N
N
n
n
N
N
n
n
N
N
N
n
n
N
N
N
n
n
N
N
[ ] [] [] [] [] []4
4
0
1
4
0
4
1
4
4
2
1
4
2
3
4
1
4
3
4
1
l
lll l l
==+ + +
=
=
=
=
=
∑∑∑∑ ∑
= ++++++
++ +
=
∑
xnW xn
N
Wxn
N
Wxn
N
W
N
n
N
n
N
N
n
N
N
n
N
n
N
[] [] [] [ ]
() () ( )
4
4
4
4
2
4
3
4
0
4
1
42
3
4
l
ll l
=++++++
=
∑
xn xn
N
Wxn
N
Wxn
N
WW
N
N
N
N
N
N
n
N
N
n
[][] [] [ ]
42
3
4
23
0
4
1
4ll ll
=?+?( ) =
=
∑
xn xn xn xn W
n
N
N
n
[] [] [] []
0
4
1
4
0
l
as
WW W
N
N
N
N
N
Nlll
===
23
1.
62
3.50 (a) XN k xnW xnW X k
N
Nkn
n
N
N
kn
n
N
[] [] [] *[].
()
= = =
=
=
∑∑
0
1
0
1
(b) XxnWxn
N
n
N
n
N
[] [] []0
0
0
1
0
1
==
=
=
∑∑
which is real.
(c) X
N
xnW xn
N
Nn
n
N
n
n
N
[ ] [] ( ) []
(/)
2
1
2
0
1
0
1
==?
=
=
∑∑
which is real.
3.51 (a) Hk DFThn DFTg n W Gk e Gk
k
j
k
[ ] [ ] [{ ] [ ] [ ]= {}=?>
{}
==
π
3
77
3
6
7
=+?+ +?+ +
π
π
π
π
π
12 23 12 0 84 32 25
6
7
12
7
24
7
30
7
36
7
je j e j e j e j e j
jj jj j
,( ),( ),,( ),( ),( )
(b) h n IDFT H k IDFT G k W g n e g n
n
j
n
[] [] [ ] [] []= {}=<?>
{}
==
π
4
77
4
8
7
=
[ ]
ππππππ
31 24 45 6 3 7
8 7 16 7 24 7 32 7 40 7 42 7
.,.,.,,,,.
//////
ee ee ee
jj jj jj
3.52 Yk ynW
MN
nk
n
MN
[] []=
=
∑
0
1
= xnW
MN
nk
n
N
[]
=
∑
0
1
,Thus,Y[kM] = xnW
N
nk
n
N
[]
=
∑
0
1
= X[k].
Hence,X[k] = Y[kM].
3.53 Note X[k] is the MN-point DFT of the sequence x n
e
[ ] obtained from x[n] by appending it
with M(N-1) zeros,Thus,the length-MN sequence y[n] is given by
yn x n N
e
M
MN
[] [ ],=<?>
=
∑
l
l
0
1
01≤≤?nMN, Taking the MN-point DFT of both sides we
get
Yk W Xk W Xk
MN
Nk
M
M
k
M
[] [] [].=
==
=
=
∑∑
l
l
l
l0
1
0
1
3.54 (a) Xxn
n
[] [],013
0
11
==
=
∑
(b) Xxn
n
n
[] ( ) [],61 13
0
11
=? =?
=
∑
(c) Xk x
k
[] [],
=
∑
=? =
0
11
12 0 36
(d) The inverse DFT of e X k
jk?π(/)
[]
46
is x n[]<?>4
12
,Thus,
eXkx x
jk
k
π
=
∑
=?<?>=? =?
(/)
[] [ ] [],
46
0
11
12
12 0 4 12 8 48
63
(e) From Parseval's relation,Xk xn
kn
[] [],
2
0
11
2
0
11
12 1500
==
∑∑
=? =
3.55 XX X j[] *[ ] *[],88 623
14
=<?>= =?+ XX X j[] *[ ] *[],99 563
14
=<?>= =?
XX X j[] *[ ] *[],10 10 4 2 2
14
=<?>= = XX X j[] *[ ] *[],11 11 3 1 5
14
=<?>= =+
XX X j[] *[ ] *[],12 12 2 3 4
14
=<?>= =? XX X j[] *[ ] *[],13 13 1 1 3
14
=<?>= =
(a) xXk
k
[] [],,0
1
14
32
14
2 2857
0
13
===
=
∑
(b) xXk
k
k
[] ( ) [],,7
1
14
1
12
14
0 8571
0
13
=? =? =?
=
∑
(c) xn X
n
[] [],
=
∑
==
0
13
012
(d) Let g n e x n W x n
jn n
[] [] [].
(/)
==
π?47
14
4
Then DFT g n DFT W x n X k
n
{ [ ]} { [ ]} [ ]==<?>
14
4
14
4
=
[ ]
XXXXXXXXXXXXXX[] [] [] [] [] [] [] [] [] [] [] [] [] [10 11 12 13 0 1 2 3 4 5 6 7 8 9
Thus,g n e x n X j
n
jn
n
[] [] [ ] –,
(/)
=
π
=
∑∑
===?
0
13
47
0
13
10 2 2
(e) Using Parseval's relation,xn Xk
nk
[] [],,
2
0
13
2
0
13
1
14
498
14
35 5714
==
∑∑
===
3.56 Now y n g k h n k
c
k
[] [][ ]=<?>
=
∑ 6
0
6
,Hence,
yghghghghghghgh
c
[] [][] [][] [][] [][] [][] [][] [][]000162534435261=++++++,
yghghghghghghgh
c
[] [ ][] [][ ] [ ][ ] [][] [ ][ ] [][] [ ][ ],101102635445362++++
yghghghghghghgh
c
[] [][] [][] [][] [][] [][] [][] [][],202112036455463=++++++
yghghghghghghgh
c
[] [][] [][] [][] [][] [][] [][] [][],303122130465564=+++
yghghghghghghgh
c
[] [][] [][] [][] [][] [][] [][] [][],404132231405665=++++++
yghghghghghghgh
c
[] [][] [][] [][] [][] [][] [][] [][],505142332415066+++
yghghghghghghgh
c
[] [][] [][] [][] [][] [][] [][] [][].606152433425160=++++++
Likewise,y n g k h n k
L
k
[] [][ ].=?
=
∑
0
6
Hence,
ygh
L
[] [][],000=
yghgh
L
[] [ ] [] [] [ ],10110=+
yghghgh
L
[] [][] [][] [][],2 021120=++
yghghghgh
L
[] [][] [][] [][] [][],303122130=+++
yghghghghgh
L
[] [][] [][] [][] [][] [][],40413223140=++++
yghghghghghgh
L
[] [][] [][] [][] [][] [][] [][],5051423324150=+++++
yghghghghghghgh
L
[] [][] [][] [][] [][] [][] [][] [][],606152433425160+++++
64
yghghghghghgh
L
[] [][] [][] [][] [][] [][] [][],7162534435261=+++++
yghghghghgh
L
[] [][] [][] [][] [][] [][],82635445362=++++
yghghghgh
L
[] [][] [][] [][] [][],936455463=+++
yghghgh
L
[ ] [][] [][] [][],10 4 6 5 5 6 4=++
yghgh
L
[ ] [][] [][],11 5 6 6 5
ygh
L
[ ] [][].12 6 6=
Comparing y
10
[n] with y
L
[n] we observe that
yy y
cLL
[] [] [],007=+
yy y
cLL
[] [] [],118=+
yy y
cLL
[] [] [],229=+
yy y
cLL
[] [] [ ],3310=+
yy y
cLL
[] [] [ ],441=+
yy y
cLL
[] [] [ ],5512=+
yy
cL
[] [].66=
3.57 Since x[n] is a real sequence,its DFT satisfies X k X k
N
[] *[ ]=<?> where N = 11 in this case.
Therefore,XX X j[ ] *[ *[ ],111032
11
=<?> =+
XX X j[ ] *[ *[ ],33858
11
=<?> =?+
XX X j[ ] *[ *[ ],55696
11
=<?> =+
XX X j[ ] *[ *[ ],77425
11
=<?> =?
XX X j[ ] *[ *[ ],99213
11
=<?> =
3.58 The N-point DFT X[k] of a length-N real sequence x[n] satisfy X k X k
N
[] *[ ].=<?> Here N
= 11,Hence,the remaining 5 samples are X X X j[] *[ ] *[ ],,,1 1 10 31 52
11
=<?>= =
XX X j[] *[ ] *[],,,44 74102
11
=<?>= = X X X j[] *[ ] *[],,66 5659
11
=<?>= =?
XX X j[] *[ ] *[],,,88 35341
11
=<?>= =+ X X X j[] *[ ] *[],,99 2322
11
=<?>= =?+
3.59 A length-N periodic even sequence x[n] satisfying x n x n
N
[] *[ ]=<?> has a real-valued N-
point DFT X[k],Here N = 10,Hence,the remaining 4 samples of x[n] are given by
xx x j[] *[ ] *[],,6 6 4 2 87 2
10
=<?>= =? xx x j[] *[ ] *[],,,77 32146
10
=<?>= =
xx x j[] *[ ] *[],,,8 8 2 3 25 1 12
10
=<?>= = and x x x j[] *[ ] *[],,,99 0708
10
=<?>= =+
3.60 As x[n] is a real-valued sequence of length 498,its 498-point DFT X[k] satisfy
Xk X k X k[] *[ ] *[ ]=<?>=?
498
498 (See Table 3.7).
(a) From the specified DFT samples we observe that X k X[] *[ ]
1
412= implying
k
1
498 412 86=?=,X k X[] *[ ]
2
309= implying k
2
498 309 189=?=,Xk X[] *[ ]
3
112= implying
k
3
498 112 386=?=,Xk X[] *[]
4
11= implying k
4
498 11 487=?=.
(b) dc value of {x[n]} = X[0] = 2.
(c) xn XkW X X W X W
kn
k
nn
[] [] ([] Re[] Re[]==+?
{}
+?
{}
=

∑
1
498
1
498
0 2 11 2 86
498
0
497
498
11
498
86
65
+?
{}
+?
{}
+?

2 112 2 189 249
498
112
498
189
498
249
Re [ ] Re [ ] [ ] )XW XW XW
nnn
=?+
π
π
π
+
π
1
249
045 7
11
249
31
11
249
22
86
249
15
86
249
{, cos, sin, cos, sin
nn nn
+
π
+
π
π
π
3
112
249
07
112
249
47
189
249
19
189
249
cos, sin, cos, sin }.
nn nn
(d) xn Xk
nk
[] [],,
2
0
497
2
0
497
1
498
0 2275
==
∑∑
==
3.61 As x[n] is a real-valued sequence of length 316,its 316-point DFT X[k] satisfy
Xk X k X k[] *[ ] *[ ]=<?>=?
316
316 (See Table 3.7).
(a) From the specified DFT samples we observe that X X k[] *[ ]17
4
= implying
k
4
316 17 299=?=,X k X[] *[ ]
1
210= if ε=0,implying k
1
316 210 106=?=,Xk X[] *[ ]
2
179=
if δ=0 implying k
2
316 179 137=?=,and X k X[] *[ ]
3
110= if γ=0 implying
k
3
316 110 206=?=.
(b) Now,X x n
n
[] []0
0
315
=
=
∑
which is a real number as x[n] are real numbers implying α=0,Since
the length N = 316 is an even number,X X N x n
n
n
[] [/] ()[]158 2 1
0
315
==?
=
∑
is also a real number
implying β=0,We have already shown in Part (a),δεγ===0.
(c) The dc value of {x[n]} is X[0] = 3.
(d) xn X X W X W X W
nnn
[] {[] Re[] Re[ ] Re[ ]=+
( )
+
( )
+
( )

1
316
0 2 17 2 106 2 137
316
17
316
106
316
137
+
( )
=+
π
+?
π
2 110
1
316
3215
17
158
223
106
158
316
110
Re [ ] } { (, )cos (, )sinXW
nn
n
+
π
+
π
242
137
158
2172
110
158
13(, )cos (, )sin }
nn
=?+
π
π
+
π
+
π
1
316
10 3
17
158
46
106
158
84
137
158
344
110
158
{ cos, sin, )cos, sin }.
nn n n
(e) xn Xk
nk
[] [],,
2
0
315
2
0
315
1
316
0 6521
==
∑∑
==
3.62 {[ ]},,,,,,,,xn ={}452302340 7≤≤n, Let X[k] denote the 8-point DFT of
x[n],Consider the 8-point DFT Y k W X k W X k
kk
[] [] [].==
4
3
8
6
Using the circular time-shifting
property of the DFT given in Table 3.5 we observe that the IDFT y[n] of Y[k] is given by
yn x n[] [ ].=<?>6
8
Therefore,y x x[] [ ] [],0622
8
=<?>= = yx x[] [ ] [],116 33
8
=<?>= =?
yx x[] [ ] [],226 40
8
=<?>= = yx x[] [ ] [],336 52
8
=<?>= =? yx x[] [ ] [],446 63
8
=<?>= =
66
yx x[] [ ] [],556 74
8
=<?>= = yx x[] [ ] [],666 04
8
=<?>= =? yx x[] [ ] [],776 15
8
=<?>= = Thus,
{[ ]},,,,,,,,yn ={}23023445 0 7≤≤n.
3.63 { [ ]},,,,,xn =?112300,05≤≤n,Let X[k] denote the 6-point DFT of x[n],
Consider the 6-point DFT Y k W X k W X k
kk
[] [] [].==
3
2
6
4
Using the circular time-shifting
property of the DFT given in Table 3.5 we observe that the IDFT y[n] of Y[k] is given by
yn x n[] [ ].=<?>4
6
Therefore,y x x[] [ ] [],0422
6
=<?>= = yx x[] [ ] [],1333
6
=<?>= =
yx x[] [ ] [],2240
6
=<?>= = yx x[] [ ] [],3150
6
=<?>= = yx x[] [ ] [],40 01
6
=<>= =
yx x[] [ ] [],51 11
6
=<>= =? Thus,{ [ ]},yn =?{}23001 1 0 5≤≤n.
3.64 (a) ygh
L
[] [][],0006==?
yghgh
L
[] [ ][] [][ ],1011016=+=
yghghgh
L
[] [][] [][] [][],20211200=++=
yghghgh
L
[] [][] [][] [][],303122119=++=?
yghgh
L
[] [][] [][],413222=+=
ygh
L
[] [][],5234==
(b) y ghghghgh ghghgh
ce e e e
[] [][] [][] [][] [][] [][] [][] [][],0001322310013224=+++=++=?
y ghgh ghgh ghgh gh
ceee
[] [ ][] [][ ] [ ][] [][ ] [ ][] [][ ] [ ][],10110233201102320=+++=
y ghghghgh ghghgh
ce e e e
[] [][] [][] [][] [][] [][] [][] [][],2021120330211200=+++=++=
y ghghghgh ghghgh
ce eee
[] [][] [][] [][] [][] [][] [][] [][],30312213003122119=+++?
(c)
G
G
G
G
jj
jj
j
j
e
e
e
e
[]
[]
[]
[]
,
0
1
2
3
1111
11
111 1
11
3
2
4
0
3
72
1
72
=

=

+
H
H
H
H
jj
jj
j
j
[]
[]
[]
[]
.
0
1
2
3
1111
11
111 1
11
2
4
0
1
1
25
5
25
=

=
+
Y
Y
Y
Y
GH
GH
GH
GH
j
j
c
c
c
c
e
e
e
e
[]
[]
[]
[]
.
[] []
[] []
[] []
[] []
.
0
1
2
3
00
11
22
33
3
439
5
439
=
=

+
Therefore
y
y
y
y
jj
jj
j
j
c
c
c
c
[]
[]
[]
[]
.
0
1
2
3
1
2
1111
11
111 1
11
3
439
5
439
4
20
0
19
=

+
=
3.65 We need to show Ng[n] h[n] = N
h[n] g[n].
Let x[n] = Ng[n] h[n] = g m h n m
N
m
N
[][ ]<? >
=
∑
0
1
and y[n] =
N
h[n] g[n] = h m g n m
N
m
N
[][ ]<? >
=
∑
0
1
67
= h m g n m h m g N n m
m
n
mn
N
[][ ] [][ ]?+ +?
==+
∑∑
01
1
= h n m g m h N n m g m
m
n
mn
N
[ ][] [ ][]?+ +?
==
∑∑
01
1
= h n m g m
N
m
N
[][]<? >
=
∑
0
1
= x[n].
Hence circular convolution is commutative.
3.66 (a) Let g[n] = N
x [n]
1
x [n]
2
= xmx n m
N
m
N
12
0
1
[] [ ].<? >
=
∑
Thus,
gn x m x n m x n x m
n
N
N
n
N
m
N
n
N
m
N
[] [] [ ] [] [].
=
=
=
=
=
∑∑ ∑
=<?>
0
1
12
0
1
0
1
1
0
1
1
0
1
Similarly we can show that if
y[n] =
N
x [n]
3
g[n],then y n g m x n m g n x m
n
N
N
n
N
m
N
n
N
m
N
[] [ ] [ ] [] [ ]
=
=
=
=
=
∑∑∑∑∑
=<?>
0
1
3
0
1
0
1
0
1
3
0
1
=
=
=
=
∑∑ ∑
xn x m xm
n
N
m
N
m
N
1
0
1
2
0
1
3
0
1
[] [] [].
(b) ()[] [] [ ]()?= <?>?
=
=
=
∑∑∑
11
0
1
12
0
1
0
1
n
n
N
N
n
N
m
N
n
gn x m x n m
= x m x N n m x n m
m
N
nn
nm
N
n
m
1
0
1
22
1
0
1
11[] [ ]() [ ]().
=
=
=
∑
+ +
Replacing n by N+n–m in the first sum and by n–m in the second we obtain
( ) [] [ ] []( ) []( )?=
+?
=
=
+ +
=
=
∑∑ ∑∑
111
0
1
1
0
1
22
0
1
0
1
n
n
N
m
N
nNm nm
n
N
n
m
gn xm xn xn
=?
=
=
∑∑
() [] () []11
1
0
1
2
0
1
n
n
N
n
n
N
x n x n, Similarly we can show that if y[n] =
N
x [n]
3
g[n],then
()[] ()[] () [] () [] () [] () [?=?
=?
=
=
=
=
=
∑∑∑ ∑∑
111 111
0
1
0
1
3
0
1
1
0
1
2
0
1
3
n
n
N
n
n
N
n
n
N
n
n
N
n
n
N
n
yn gn xn xn xn xnn
n
N
]
=
∑
0
1
.
3.67 yn
n
N
xn
xn
ee xnWxnW
jnN jnN
N
n
N
n
[] cos []
[]
[] [],
//
=
=+
( )
=+

2
2
1
2
1
2
22
π
ππ
l
ll l l
Hence Y[k] =
1
2
1
2
Xk Xk
NN
[][].<+> + <?>ll
3.68 yn x n n
N
[] [ ],.=≤≤?40 1
4
Therefore,Yk ynW x nW
N
nk
n
N
nk
n
NN
[] [] [ ],
//
==
=
=
∑∑4
0
1
4
0
1
44
4
68
Now,xn
N
XmW
N
XmW
N
mn
m
N
N
mn
m
N
[ ] [] []
/
4
11
4
0
1
4
0
1
==
=
=
∑∑
,Hence,
Yk
N
XmW W
N
Xm W
N
mn
N
nk
m
N
n
N
N
kmn
n
N
m
N
[] [] []
//
/
()?
=
=
=
=
∑ ∑44
0
1
0
4
1
4
0
4
1
0
1
,Since,
W
mkk k k k
elsewhere
N
kmn
n
N
NNNNN
/
()
,,,,,,
4
0
4
1
44
2
4
3
4
4
4
0
=
∑
=
=+ + + +
Thus,
Yk Xk Xk Xk Xk
NNN
[] [] [ ] [ ] [ ].=
1
++++++
( )
4
4
2
4
3
4
Vk j j j j j j j j[],,,,,,=?+ +?+ + +2 3 1 5 4 7 2 6 1 3 4 3 8 6, Hence,
Vk jj j j j j j j*[ ],,,,,,,<? > = +? +
8
2 3 6 3 8 4 1 3 2 6 4 7 1 5, Therefore,,
Xk j j j j j j[ ],.,,.,,.,,.,.,,.,= + + +
[ ]
2 05 05 05 05 3 35 1 3 35 05 05 05 05 and
Yk j j j j j j[],.,,.,,.,,.,.,,.,=? + + +3 55 05 75 35 25 3 25 75 35 55 05
3.69 vn xn jyn[] [] [].=+ Hence,Xk Vk V k[] [] * ]=+<?>
{}
1
2
8
is the 8-point DFT of x[n],and
Yk
j
Vk V k[] [] * ]=?<?>
{}
1
2
8
is the 8-point DFT of y[n],Now,
Vk j j j j j j j j[],,,,,,,=?+ +?+ + +2315 4726 134 386
Vk jj j j j j j j*[ ],,,,,,,<? > = +? +
8
2 3 6 3 8 4 1 3 2 6 4 7 1 5, Therefiore,
Xk j j j j j j[],,.,,.,,.,,.,.,,.,= + + +
[ ]
02 05 05 05 05 3 35 1 3 35 05 05 05 05
Yk j j j j j j[],.,,.,,.,.,.,,.,=? + + +3 55 05 75 35 25 3 25 75 35 55 05
3.70 vn g n jhn j j j
e
[] [] [],,,=+=?+?3 2 2 4 4, Therefore,
V
V
V
V
jj
jj
j
j
j
j
j
j
[]
[]
[]
[]
,
0
1
2
3
1111
11
111 1
32
24
4
3
12
15
24
=

+
=
+
+
i.e.,{ [ ]},,,Vk j j j=+?+
[]
3121524.
Thus,{ *[ ]},,,Vk j j j<? > = +
[]
4
3241512
Therefore,Gk Vk V k j j
e
[] [] *[ ],,,=+<?>
{}
=+
[]
1
2
372172
4
and
Hk
j
Vk V k j j[] [] *[ ],,,=?<?>
{}
=? +?
[]
1
2
12 552 5
4
,
3.71 vn gn jhn j j j j[] [] [],,,=+ =?++ +2 1 2 3 3 4 2, Therefore,
V
V
V
V
jj
jj
j
j
j
j
j
j
j
j
[]
[]
[]
[]
,
0
1
2
3
1111
11
111 1
2
12
33
42
2
17
10 6
1
=

+
+

+
=
+

+
i.e.,{ [ ]},,,.Vk j j j j=++
[]
21 7 10 61
69
Thus,{ *[ ]},,,Vk j j j j<? > = +?
[]
4
21 10 61 7.
Therefore,Gk Vk V k j j
e
[] [] *[ ],,,=+<?>
{}
=+
[]
1
2
01 3 101 3
4
and
Hk
j
Vk V k[] [] *[ ],,,=?<?>
{}
=?
[]
1
2
24 64
4
.
3.72 (a) Let p n IDFT P k[] [].= {} Thus,
p
p
p
p
jj
jj
j
j
[]
[]
[]
[]
.
0
1
2
3
1
4
1111
11
111 1
4
17
2
17
1
4
8
12
4
16
2
3
1
4
=

+
=
=
Similarly,let d n IDFT D k[] [].= {}
d
d
d
d
jj
jj
j
j
[]
[]
[]
[]
.
.
.
.
.0
1
2
3
1
4
1111
11
111 1
45
15
55
15
1
4
2
8
4
12
05
2
1
2
=

+
=
=
,Therefore,
Xe
ee e
ee e
j
jj j
jj j
()
.
.
ω
ωω ω
ωω ω
=
++
+?+

23 4
05 2 3
23
23
(b) Let p n IDFT P k[] [].= {} Thus,
p
p
p
p
jj
jj
j
j
[]
[]
[]
[]
.
0
1
2
3
1
4
1111
11
111 1
7
72
9
72
3
3
4
5
=

+
=
Similarly,let
d n IDFT D k[] [].= {} Thus,
d
d
d
d
jj
jj
j
j
[]
[]
[]
[]
.
0
1
2
3
1
4
1111
11
111 1
0
46
4
46
1
2
3
4
=

+
=
Therefore,
Xe
ee e
ee e
j
jj j
jj j
(),
ω
ωωω
ωωω
=
+? +
+

33 4 5
12 3 4
23
23
3.73 Xe xne
jjn
n
N
() []
ωω
=
=
∑
0
1
and
[] []
/
Xk xne
jknM
n
N
=
=
∑
2
0
1
π
.
Now?[]
[] [ ]
/
xn
M
XkW
M
xme W
M
nk
k
M
jkmM
m
N
M
nk
k
M
==
=
=
=
∑∑∑
11
0
1
2
0
1
0
1
π
=
1
2
0
1
0
1
M
xm e
jkmnM
k
M
m
N
[]
()/
=
=
∑∑
π
= x n rM
r
[].+
=?∞
∞
∑
.
Thus?x[n] is obtained by shifting x[n] by multiples of M and adding the shifted copies,Since
the new sequence is obtained by shifting in multiples of M,hence to recover the original
sequence take any M consecutive samples,This would be true only if the shifted copies of x[n]
did not overlap with each other,that is,if only if M ≥ N.
3.74 (a) Xe xne
jjn
n
() [],
ωω
=
=
∑
0
7
Therefore,X k x n e
jkn
n
1
210
0
7
[] [],
/
=
π
=
∑
Hence,
70
xn Xke xme e
jkn
k
jkm
mk
jkn
11
210
0
9
210
0
7
0
9
210
1
10
1
10
[] [] [ ]
///
==
π
=
π
==
π
∑∑∑
+
π?
===?∞
∞
∑
1
10
10
210
0
9
0
7
xm e xn r
jknm
kmr
[] [ ]
()/
using the result of Problem 3.74.
Since M = 10 and N = 8,M > N,and hence x[n] is recoverable from x n
1
[ ],In fact
{[]}xn
1
111111110 0={} and x[n] is given by the first 8 samples of x n
1
[].
(b) Here,X k x n e
jkn
n
2
26
0
7
[] [],
/
=
π
=
∑
Hence,
xn Xke xme e
jkn
k
jkm
mk
jkn
21
26
0
5
26
0
7
0
5
26
1
6
1
10
[] [] [ ]
///
==
π
=
π
==
π
∑∑∑
+
π?
===?∞
∞
∑
1
10
6
26
0
6
0
7
xm e xn r
jknm
kmr
[] [ ].
()/
Since M = 6 and N = 8,M < N,the sequence x[n]
is not recoverable from x n
2
[ ],In fact,x n
2
2 21111[],={} As there is time-domain
aliasing,x[n] is not recoverable from x n
2
[].
3.75 Since
FF
=
1
1
N
thus
F =
NF
1
,Thus
y[n] =
F {
F {
F {
F {
F {
F {x[n]}}}}}}}
= N
F
-1
{
F {N
F
-1
{
F {N
F
-1
{
F {x[n]}}}}}} = N
3
x[n].
3.76 y[n] = x[n]
*
h[n] = x k h n k h k x n k h k x n k
kkk
[][ ] [][ ] [][ ]?=?=?
===
∑∑∑
0
39
0
39
12
27
.
u[n] =
Nx[n] h[n]
= h k x n k h k x n k
kk
[][ ] [][ ].<?> = <?>
==
∑∑40
0
39
40
12
27
Now for n ≥ 27,x n k x n k[][]<?> =?
40
,Thus u[n] = y[n] for 27 39≤≤n.
3.77 (a) Overlap and add method,Since the impulse response is of length 55 and the DFT size to
be used is 64,hence the number of data samples required for each convolution will be 64 –
54 = 10,Thus the total number of DFT's required for length-1100 data sequence is
1100
10
110
=, Also the DFT of the impulse response needs to be computed once,Hence,the
total number of DFT's used are = 110 + 1 = 111,The total number of IDFT's used are = 110.
(b) Overlap and save method,In this since the first 55 – 1 = 54 points are lost,we need to
pad the data sequence with 54 zeros for a total length of 1154,Again each convolution will
result in 64 – 54 = 10 correct values,Thus the total number of DFT's required for the data are
thus
1154
10
116
=, Again 1 DFT is required for the impulse response,Thus
The total number of DFT's used are = 116 + 1 = 117.
The total number of IDFT's used are = 116.
71
3.78 (a) yn
xn L n L L N L
elsewhere
[]
[ / ],,,,.....,( ),
,.
=
=?
02 1
0
Yk ynW xnW xnW
NL
nk
n
NL
NL
nLk
n
N
N
nk
n
N
[] [] [] [],===
=
=
=
∑∑∑
0
1
0
1
0
1
For k ≥ N,let k = krN
0
+ where kk
N0
=< >, Then,
Y[k] = Y[ krN
0
+ ] = xnW
N
nk rN
n
N
[]
()
0
0
1
+
=
∑
= xnW
N
nk
n
N
[]
0
0
1
=
∑
= X[k
0
] = X[<>k
N
].
(b) Since Y[k] = X[<>k
7
] for k = 0,1,2,.....,20,a sketch of Y[k] is thus as shown below.
0123456
k
1
2
3
4
7 820
Y[k]
3.79 xn xn xn
0
21 2[] [ ] [ ],=++ xn xn xn
1
21 2[] [ ] [ ],=+? yn yn yn
1
21 2[] [ ] [ ],=++ and
yn yn yn
0
21 2[] [ ] [ ],=+? 01
2
≤≤?n
N
,Since x[n] and y[n] are real,symmetric sequences,it
follows that x
0
[n] and y
0
[n] are real,symmetric sequences,and x
1
[n] and y
1
[n] are real,anti-
symmetric sequences,Now consider,the (N/2)-length sequence
un x n y n jx n y n[] [] [] [] [].=++ +
( )
01 10
Its conjugate sequence is given by
un xn yn jxn yn*[ ] [ ] [ ] [ ] [ ],=+? +
( )
01 10
Next we observe that
un x n y n jx n y n
NNN NN
[– ] [– ] [– ] [– ] [– ]
/// //
<> =<> +<> + <> +<>
( )
20 21 2 1 20 2
=?+?+
( )
xn yn j xn yn
01 10
[ ] [ ] [ ] [ ], Its conjugate sequence is given by
un xnynjxnyn
N
*[ ] [ ] [ ] [ ] [ ],
/
<? > = +
( )
20 1 1 0
By adding the last 4 sequences we get
4
022
xn un u n u n u n
NN
[] [] *[] [ ] *[ ].
//
= + +<?> + <?>
From Table 3.6,if U[k] = DFT{u[n]},then U k DFT u n
N
*[ ] { *[ ]},
/
<? > =
2
Uk DFTu n
N
*[ ] { *[ ]},
/
=<?>
2
and U k DFT u n
NN
[ ] { [ ]}.
//
<? > = <? >
22
Thus,
XkDFTxnUkUkUkUk
00
1
4
[] { []} [] *[ ] [ ] *[].= = + <?> + <?> +
( )
Similarly,
jxnununu n u n
NN
4
122
[] [] *[] [ ] *[ ].
//
=<?> + <?> Hence,
Xk DFTxn Uk U k U k U k
j
NN11
1
4
22
[] { []} [] *[ ] [ ] *[].
//
==?<?>?<?>+
( )
Likewise,
4
12 2
yn un u n u n u n
NN
[] [] [ ] *[] *[ ].
//
=?<?>+?<?> Thus,
Yk DFTyn Uk U k U k U k
NN11
1
4
22
[] { []} [] [ ] *[ ] *[].
//
= =? <?> + <?>?
( )
Finally,
jyn un u n u n u n
NN
4
02 2
[] [] [ ] *[] *[ ].
//
=+<?><?> Hence,
YkDFTynUkUkUkUk
j
NN00
1
4
22
[] { []} [] [ ] *[ ] *[].
//
= = + <?>? <?>?
( )
72
3.80 gn xn xn hn xn xn n
N
[] [][],[] []–[],–.=++( ) =+( ) ≤≤
1
2
1
22
2 2 1 2 2 1 0 1 Solving for x[2n] and
x[2n+1],we get x[2n] = g[n] + h[n] and x[2n + 1] = g[n] – h[n],01
2
≤≤n
N
–, Therefore,
Xz xnz x nz z x n z
n
n
N
n
n
n
n
NN
() [] [ ] [ ]
–– –
== ++
=
=
=
∑∑ ∑
0
1
0
1
1
0
1
221
=+( ) ++( ) =+ +?
=
=
=
=
∑∑ ∑∑
gn hn z z gn hn z z gnz z hnz
n
n
n
n
n
n
n
n
NN NN
[] [] [] [] ( ) [] ( ) [],
––––
0
1
1
0
1
1
0
1
1
0
1
22 22
11
Hence,Xk Xz W G k W H k
zW N
k
NN
k
N
N
k[]()()[]()[],
–
/
–
/
==+<>+?<>
=
11
22
01≤≤kN–.
3.81 gn ax n a x n[] [ ] [ ]=++
12
2 2 1 and h n a x n a x n[] [ ] [ ]=++
34
2 2 1,with a a a a
14 23
≠, Solving for x[2n]
and x[2n+1],we get x[2n] =
agn ahn
aa a a
42
14 23
[] []
,
and x[2n + 1] =
+
agn ahn
aa a a
31
14 23
[] []
,Therefore
Xz xnz x nz z x n z
n
n
N
n
n
n
n
NN
() [] [ ] [ ]
–– –
== ++
=
=
=
∑∑ ∑
0
1
0
1
1
0
1
221
=
agn ahn
aa a a
zz
agn ahn
aa a a
z
n
n
n
n
NN
42
14 2
0
1
1 31
14 2
0
1
33
22
[] [] [] []
––
+
+
=
=
∑∑
=? +?+
=
=
∑∑
1
14 23
43
1
0
1
21
1
0
1
22
aa a a
a az gnz a az hnz
n
n
n
n
NN
( ) [] ( ) [],
––
Hence,
Xk
aa a a
aaW Gk aaW Gk
N
nk
NN
nk
N
[] ( )[]( )[],
//
=?<>+?+<>

1
14 23
43 2 21 2
01≤≤kN–.
3.82 Xkab xn j
nakb
N
GDFT
n
N
[,,] [ ]exp
()()
.=?
π+ +
=
∑
0
1
2
xn
N
Xkab j
nakb
N
k
N
[ ] [,,]exp
()()
=
π+ +
=
∑
12
0
1
=?
π+ +
π+ +
=
=
∑∑
12 2
0
1
0
1
N
xr j
rakb
N
j
nakb
N
r
N
k
N
[ ]exp
()( )
exp
()()
=
π+ +
=
=
∑∑
12
0
1
0
1
N
xr j
narakb
N
r
N
k
N
[ ]exp
()()
=
π? +
=
=
∑∑
12
0
1
0
1
N
xr j
nrkb
N
r
N
k
N
[ ]exp
()( )
=
π? +
=
=
∑∑
12
0
1
0
1
N
xr j
nrkb
N
k
N
r
N
[ ] exp
()( )
==
1
N
xn N xn[] [],
73
as from Eq,(3.28),exp
()( )
,,
,.
j
nrkb
N
Nifnr
otherwise
k
N
2
0
0
1
π? +
=
=
{
=
∑
3.83 Xz xnz xnz
n
n
n
n
() [] [],==
=?∞
∞
=
∞
∑∑
0
Therefore,
lim ( ) lim [ ] lim [ ] lim [ ] [ ].
zz
n
n
zz
n
n
Xz xnz x xnz x
→∞ →∞
=
∞
→∞ →∞
=
∞
==+=
01
00
3.84 Gz
zz zz
zzzz
()
(.)(.)(,,)
(..)( )
.=
+? ++
+ ++
04 091 03 04
06 06 3 5
2
22
G(z) has poles at zj=±03
204
2
.
.
and zj=±
3
2
11
2
.
Hence,there are possible ROCs.
(i) R
1
,z ≤ 06., The inverse z-transform g[n] in this case is a left-sided sequence.
(ii) R
2
,06 5,≤≤z, The inverse z-transform g[n] in this case is a two-sided sequence.
(iii) R
3
,z ≥ 5,The inverse z-transform g[n] in this case is a right-sided sequence.
3.85 (a) (i) x n n
n
1
04[] (.) []=μ is a right-sided sequence,Hence,the ROC of its z-transform is
exterior to a circle,Thus,Xz xnz z
z
z
n
n
nn
n
11
0
1
04
1
104
04() [] (.)
.
,.===
>
=?∞
∞
=
∞
∑∑
The ROC of X
1
(z) is given by R
1
,z > 04..
(ii) x n n
n
2
06[] (,) []=? μ is a right-sided sequence,Hence,the ROC of its z-transform is exterior to
a circle,Thus,Xz xnz z
z
z
n
n
nn
n
22
0
1
06
1
106
06() [] (,)
.
,.==?=
+
>
=?∞
∞
=
∞
∑∑
The ROC of X
2
(z) is given by R
2
,z > 06..
(iii) x n n
n
3
03 4[] (.) [ ]=?μ is a right-sided sequence,Hence,the ROC of its z-transform is
exterior to a circle,Thus,X z x n z z
n
n
nn
n
33
4
03() [] (.)==
=?∞
∞
=
∞
∑∑
Xz xnz z
z
z
z
n
n
nn
n
33
4
44
1
03
03
103
03() [] (.)
(.)
.
,..===
>
=?∞
∞
=
∞
∑∑
The ROC of X
3
(z) is given by
R
3
,z > 03..
74
(iv) x n n
n
4
03 2[] (,) [ ]=μ is a left-sided sequence,Hence,the ROC of its z-transform is
interior to a circle,Thus,
Xz xnz z z
az
z
z
n
n
nn
n
mm
m
44
2
2
1
1
03 03
2
103
03() [] (,) (,)
.
,..
–
==?=?=?
+
+
<
=?∞
∞
=∞
=
∞
∑∑ ∑
The ROC of X
4
(z) is given by R
4
,z < 03..
(b) (i) Now,the ROC of X
1
(z) is given by R
1
,z > 04,and the ROC of X
2
(z) is given by
R
2
,z > 06., Hence,the ROC of Y
1
(z) is given by R
1
∩R
2
= R
2
,z > 06.
(ii) The ROC of Y
2
(z) is given by R
1
∩R
3
= R
1
,z > 04..
(iii) The ROC of Y
3
(z) is given by R
1
∩R
4
=?,Hence,the z-transform of the sequence
y
3
[n] does not converge anywhere in the z-plane.
(iv) The ROC of Y
4
(z) is given by R
2
∩R
3
= R
2
,z > 06..
(v) The ROC of Y
5
(z) is given by R
2
∩R
4
=?,Hence,the z-transform of the sequence
y
5
[n] does not converge anywhere in the z-plane.
(vi) The ROC of Y
6
(z) is given by R
3
∩R
4
=?,Hence,the z-transform of the sequence
y
6
[n] does not converge anywhere in the z-plane.
3.86 (a) Z{ [ ]} [ ] [ ],δδδnnz
n
n
===
=?∞
∞
∑
0 1 which converges everywhere in the z-plane,
(b) Z {[]} [] ( ),αμ αμ α
α
nnn
n
n
n
zz
z
===
=?∞
∞
=
∞
∑∑
1
0
1
1
1
>z α
(c) See Example 3.29.
(d) x[n] = r n n
n
sin( ) [ ]ωμ
0
=
r
j
ee n
n
jn jn
2
00
ωω
μ?
( )
[ ],Using the results of (iii) and the
linearity property of the z-transform we obtain
Z{ r n n
n
sin( ) [ ]ωμ
0
} =

1
2
1
1
1
2
1
100
11
j re z j re z
jjωω
=
( )
+
( )
+
=
+

r
j
eez
rz e e r z
rz
rzrz
jj
jj
2
1
12
00
00
1
122
0
1
0
122
ωω
ωω
ω
ω
sin( )
cos( )
,?>zr
3.87 (a) xn n n n
nn n n
1
605 03 605 603[] (.) (.) [] (.) [] (.) []=?
[]
=?μμμ,Therefore,
Xz
zz
zz
1 11
11
6
105
6
103
61 03 1 05
105 103
()
..
(., )
(,)(,)
=
=
+

=

12
105 103
1
11
.
(,)(,)
,
z
zz
z > 05.
75
(b) xn n n
nn
2
603 605 1[] (.) [] (.) [ ].=μμ Now,
Z{(.) []}
.
,?=
603
6
103
1
n
n
z
μ z > 03,and
Z{(.) [ ]}
.
,=
605 1
6
105
1
n
n
z
μ z < 05., Therefore,
Xz
zz
z
zz
2 11
1
11
6
105
6
103
12
105 103
()
..
.
(,)(,)
=
=

,03 05...<<z
(c) xn n n
nn
3
603 1 605 1[] (.) [ ] (.) [ ].=μμ Now,
Z{(,) [ ]}
.
,603 1
6
103
1
n
n
z
μ =
z < 03,
and
Z{(.) [ ]}
.
,=
605 1
6
105
1
n
n
z
μ z < 05., Therefore,
Xz
zz
z
zz
3 11
1
11
6
105
6
103
12
105 103
()
..
.
(,)(,)
=
=

,z < 03..
Note,X z X z X z
123
() () (),== but their ROCs are all different.
3.88 (a) xn n n
nn
1
[] [] []=+αμ βμ with βα>, Note that x n
1
[ ] is a right-sided sequence,Hence,the
ROC of its z-transform is exterior to a circle,Now,
Z{ [ ]},αμ
α
n
n
z
=
1
1
1
with ROC given
by z >α and
Z{ [ ]},βμ
β
n
n
z
=
1
1
1
with ROC given by z >β,Hence,
Xz
zz
zz
zz
1 11
12
11
1
1
1
1
1
11
()
()
()()
,=
+
=
+ +

αβ
αβ
αβ
z >β
(b) xn n n
nn
2
1[] [ ] []=+αμ βμ, Note that x n
2
[ ] is a two-sided sequence,Now,
Z{ [ ]},αμ
α
n
n
z
=
1
1
1
1
with ROC given by z <α and
Z{ [ ]},βμ
β
n
n
z
=
1
1
1
with ROC
given by z >β,Since the regions z <α and z >β do not intersect,the z-transform of x n
2
[]
does not converge.
(c) xn n n
nn
3
1[] [] [ ]=+αμ βμ, Note,is a two-sided sequence,Now,
Z{ [ ]},αμ
α
n
n
z
=
1
1
1
with ROC given by z >α and
Z{ [ ]},βμ
β
n
n
z
=
1
1
1
1
with ROC given by z <β,Since the
regions z >α and z <β intersect,the z-transform X z
3
( ) of x n
3
[ ] converges and is given by
Xz
zz
zz
zz
3 11
12
11
1
1
1
1
1
11
()
()
()()
,=
+
=
+ +

αβ
αβ
αβ
Its ROC is an annular region given by
αβ<<z
76
3.89 (a) Let y n g n h n[] [] [].=+αβ Then,
Yz gn hn z gnz gnz
n
n
n
n
n
n
() [] [] [] []=+( ) =+
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑
αβ α β = +αβGz Hz( ) ( ),In this case,Y(z)
will converge wherever both G(z) and H(z) converge,Thus,the ROC of Y(z) is given by
RR
gh
∩ where
R
g
is the ROC of G(z) and
R
h
is the ROC of H(z).
(b) yn g n[] [ ].=? Then,Y z g n z g n z g n z G z
n
n
n
n
n
n
() [ ] [] [](/) (/).=?= = =
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑
1 1 Y(z) will
converge wherever G(1/z) converges,Hence,if
R
g
is the ROC of G(z),then the ROC of Y(z) is
given by 1/
R
g
is the ROC of G(z).
(c) yn gn n[] [ ]=?
0
,Hence Y(z) = y n z g n n z g m z
n
n
n
n
mn
m
[] [ ] [ ]
()?
=?∞
∞
=?∞
∞
+
=?∞
∞
∑∑ ∑
=
0
0
==

=?∞
∞
∑
zgmzzGz
nm
m
n
00[] ().
In this case the ROC of Y(z) is the same as that of G(z) except for the possible addition or
elimination of the point z = 0 or z = ∞ (due to the factor z
n?
0
).
(d) yn gn
n
[] []=α, Hence,Y z y n z g n z G z
n
n
n
n
() [] []( ) (/ ).== =
=?∞
∞

=?∞
∞
∑∑
αα
1
The ROC of Y(z) is
αR
g
.
(e) y[n] = ng[n],Hence Y(z) = ng n z
n
n
[]
=?∞
∞
∑
.
Now G z g n z
n
n
() []=
=?∞
∞
∑
,Thus,
dG z
dz
ng n z
n
n
()
[]=?

=?∞
∞
∑
1
z
dG z
dz
ng n z
n
n
()
[]=?
=?∞
∞
∑
.
Thus Y(z) = – z
dG z
dz
()
.
(f) y[n] = g[n]
*
h[n] = g k h n k
k
[][ ]?
=?∞
∞
∑
,Hence,
Yz ynz gkhnkz gk hnkz
n
nk
n
n
n
kn
() [] [][ ] [] [ ]==?
=?
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
∑∑ ∑∑∑
= g k H z z H z G z
k
k
[] () () ()
=?∞
∞
∑
=,
In this case also Y(z) will converge wherever both H(z) and G(z) converge,Thus ROC of Y(z)
is
RR
gh
∩,
(g) y[n] = g[n]h[n],Hence,Y(z) = g n h n z
n
n
[][]
=?∞
∞
∑
,From Eq,(3.),
77
gn
j
Gvv dv
n
C
[] ()=
∫
1
2
1
π
,Thus,Yz hn
j
Gvv dv z
n
n
C
n
() [] ()=
=?∞
∞

∑
∫
1
2
1
π
=
=
=?∞
∞

∑
∫∫
1
2
1
2
11
ππj
Gv hnz v dv
j
GvHz vv dv
n
nn
CC
() [] () (/ ),
(h) gn h n
j
Gv h nv v dv
j
GvH v v dv
n
n
n
[] *[] () *[] () *(/*)
=?∞
∞
=?∞
∞
∑∑
==
1
2
1
2
1,
3.90
Xz xn() {[]}= Z with an ROC given by
R
x
,Using the conjugation property of the z-transform
given in Table 3.9 we observe that
Z{*[]} *(*)xn Xz= whose ROC is given by
R
x
,Now,
Re( [ ]) ( [ ] *[ ]).xn xn x n=+
1
2
Hence,
Z Re( [ ]) ( ) *( )xn Xz X z{}=+( )
1
2
whose ROC is also
R
x
,
Likewise,Im( [ ]) ( [ ] *[ ]).xn
j
xn x n=?
1
2
Thus
Z Im( [ ]) ( ) *( )xn
j
Xz X z{}=?( )
1
2
with an ROC given
by
R
x
.
3.91 {[ ]}xn =
↑
3 0 1 2 3 4 1 0 1, Then,
[] () () ( ),//
/
Xk Xz Xz Xe
ze ze
j
k
jk j k== =
==
=π
ππ326
26
ω
ω
Note that
[]X k is a periodic sequence of
period 6,Hence,from the results of Prpblem 3.37,the inverse of the discrete Fourier series
[]Xk
is given by?[] [ ] [ ] [] [ ],xn xn r xn xn xn
n
=+=?+++
=?∞
∞
∑
6 6 6 for 0 5≤≤n, Let
yn xn xn xn[] [ ] [] [ ],=?+++66?≤ ≤35n, It follows { [ ]},xn?=?
↑
6 00000030 1
and { [ ]},xn+=
↑
6 301000000 Therefore,
{ [ ]},yn =
↑
4 0 4 2 3 4 4 0 0 Hence,{?[]},xn =
↑
234400
3.92 Xz xnz
n
n
() [],=
=
∑
0
11
Xk Xz xne
o ze
jkn
n
jk[] () [],/
/
==
=
π
=
π
∑
29
29
0
11
Therefore,
xn Xke xre e
oo
k
jkn jkr
rk
jkn
[] [] []
///
==
=
π?π
==
π
∑∑∑
1
9
1
9
0
8
29 29
0
11
0
8
29
=
π?
==
∑∑
1
9
29
0
11
0
8
xre
jknr
rk
[]
()/
π?
==

==
∑ ∑
1
9
1
9
29
0
8
0
11
9
0
8
0
11
xr e xr W
jknr
kr
rnk
kr
[] []
()/ ()
78
But,from Eq,(3.28),W
for r n i
otherwise
rnk
k
9
0
8
99
0

=
∑
=
=
{
()
,
,.
Hence,
xn
x x for n
x x for n
x x for n
x n for n
o
[]
[] [],,
[] [ ],,
[] [ ],,
[],,
=
+=
≤≤
09 0
110 1
211 2
38
i.e.,x n
o
[],={}9174 3 201 4,0 8≤≤n.
3.93 (a) Xz xnz
n
n
() [],=
=?∞
∞
∑
Hence,X z x n z x m z
n
n
m
m
m even
() [] [/],
22
2==
=?∞
∞
=?∞
∞
∑∑
If we define a new
sequence
gm
xm m
otherwise
[]
[ / ],,,,
,,
=
=±±
{
2024
0
K
,we can then express X z g n z
n
n
() [],
2
=
=?∞
∞
∑
Thus,the
inverse z-transfoprm of X z()
2
is given by g[n],For x n n
n
[] (.) [],= 04 μ
gn
n
otherwise
n
[]
(.),,,,
,,
/
=
=
04 024
0
2
K
(b) Yz z Xz Xz z Xz()( )() () ().=+ = +

1
12 2 12
Therefore,
yn Yz Xz z Xz[ ] { ( )} { ( )} { ( )}== +

ZZ Z
11212
= g[n] + g[n–1],where g[n] is the inverse z-
transform of X z()
2
,Now,gn
xn n
otherwise
[]
[ / ],,,,
,,
=
=±±
{
2024
0
K
and
gn
xn n
otherwise
[]
[( ) / ],,,
,,
=
=±
{
1
12 1 3
0
K
,Hence,
yn gn gn
xn n
xn n
[] [] [ ]
[ / ],,,,
[( ) / ],,,,
=+?=
=±±
=±
{
1
2024
12 1 3 5
K
K
,For x n n
n
[] (.) [],= 04 μ therefore,
yn
n
n
n
n
n
[]
(.),,,,,
(.),,,,,
,.
/
()/
=
=
=
<
04 024
04 135
00
2
12
K
K
3.94 (a) xn n
n
1
1[] [ ],=+αμ α<1,Therefore,X z n z z
n
n
nn
n
n
1
1
1() [ ]=+=
=?∞
∞
=?
∞
∑∑
αμ α
==
>
=
∞

∑
z
z
zz
z
n
n
n
α
α
αα
α
0
11
1
1()
,,The ROC of X z
1
( ) includes the unit circle since α<1.
On the unit circle Xe Xz
ee
j
ze
jj
j
11
1
1
() ()
()
,
ω
ωω
ω
αα
==
=

which is the same as the DTFT of
xn
1
[].
(b) xn n n
n
2
[] [],=?αμ α<1,Therefore,X z n n z n z
n
n
nn
n
n
2
0
() []=? =?
=?∞
∞
=
∞
∑∑
αμ α
=
>
α
α
α
z
z
z
1
12
1()
,,The ROC of X z
2
( ) includes the unit circle since α<1,On the unit circle,
Xe Xz
e
e
j
ze
j
j
j
22 2
1
() ()
()
,
ω
ω
ω
ω
α
α
==
=
which is the same as the DTFT of x n
2
[].
79
(c) xn
nM
otherwise
n
3
0
[]
,,
,.
=
<
α
Therefore,X z z z
n
nM
nn
n
M
n
3
1
0
()=+
=?
=
∑∑
αα
=
+

+? +

α
α
α
α
α
MM
MM M M
z
z
z
z
z
1
1
1
1
11
11
11
()
,Since x n
3
[ ] is a finite-length sequence,the ROC is the
whole z-plane except possibly the origin,On the unit circle
Xe Xz e
e
e
e
e
j
ze
MjM
MjM
j
MjM
j
j() (),
()
ωω
ω
ω
ω
ω
ω α
α
α
α
α
==
+
=

+? +

1
1
1
1
1
11
1
(d) xn n
n
4
31[] [ ],.=?<αμ α Note that we can express x n x n
4
4
1
4[] [ ],=?α where
xn n
n
1
1[] [ ],=+αμ is the sequence considered in Part (a),Therefore,
Xz zXz
z
z
z
4
44
1
33
1
1
() (),.==
>
α
α
α
α The ROC of X z
4
( ) includes the unit circle since α<1,
On the unit circle,Xe
e
e
j
j
j3
33
1
(),
ω
ω
ω
α
α
=
which is the same as the DTFT of x n
4
[].
(e) xn n n
n
5
21[] [ ],.=? + <αμ α Therefore,
Xz nz z z nz
n
n
nn
n
n
5
2
22 1
0
2()=? =+?
=?
∞

=
∞
∑∑
ααα α=? + +
>

zz
z
z
z()
()
,.2
1
21
1
12
αα
α
α
α
The ROC of X z
5
( ) includes the unit circle since α<1,On the unit circle,
Xe e e
e
e
jj j
j
j5
21
2
2
1
() ( )
()
,
ωω ω
ω
ω
αα
α
α
=? + +

which is the same as the DTFT of x n
5
[].
(f) xn n
n
6
11[] [ ],.=>αμ α Therefore,
Xz z z
z
z
n
n
nn
n
n
6
10
1
1
1
(),.==?=
<
=?∞
=?∞
∑∑
ααα
α
αα The ROC of X z
6
( ) includes the unit
circle since α>1,On the unit circle,Xe
e
j
j6 1
1
1
(),
ω
ω
α
α=
which is the same as the DTFT of
xn
6
[].
3.95 (a) yn
NnN
otherwise
1
1
0
[]
,–,
,.
=
≤≤
Therefore,Yz z z
z
z
n
nN
N
N
N
1
21
1
1
1
()
()
()
.
()
==
=?
+
∑
Since y n
1
[ ] is a
finite-length sequence,the ROC of its z-transform is the whole z-plane except possibly the origin,
and therefore includes the unit circle,On the unit circle,
Ye e e
e
e
N
jjn
nN
N
jN
jN
j1
21
1
1 2
1
2
()
()
()
sin( )
sin( / )
()
ωωω
ω
ω
ω
ω
==
=
+
[]
=?
+
∑
,which is the same DTFT of y n
1
[].
80
(b) yn
n
N
NnN
otherwise
2
1
0
[]
,,
,.
=
≤≤
Now y
2
[n] = y
0
[n]
*
y
0
[n] where
yn
NnN
otherwise
0
12 2
0
[]
,/ /,
,.
=
≤≤
Therefore,Yz Yz z
z
z
N
N
20
2
12
12
1
1
() ()
()
()
.
()
==
+
Since y n
2
[ ] is a
finite-length sequence,the ROC of its z-transform is the whole z-plane except possibly the origin,
and therefore includes the unit circle,On the unit circle,
Ye Ye
N
jj
20
2
2
2
1
2
2
() ()
sin
sin ( / )
ωω
ω
ω
==
+
which is the same DTFT of y n
2
[].
(c) yn
nN NnN
otherwise
3
2
0
[]
cos( / ),,
,.
=
≤≤
π
Therefore,
Yz e z e z
jn N n
nN
N
jn N n
nN
N
3
22
1
2
1
2
()
(/ ) (/ )
=+

=?
=?
∑∑
ππ
=
+
+?+

+?+
ez e z
ez
eze z
ez
jN jN NN
jN
jN jN NN
jN
(/) ( )(/ ) ( )
(/ )
(/) ( )(/ ) ( )
(/ )
ππ
π
ππ
π
22121
21
221221
21
2
1
1 2
1
1
,Since
yn
3
[ ] is a finite-length sequence,the ROC of its z-transform is the whole z-plane except possibly
the origin,and therefore includes the unit circle,On the unit circle,
Ye
NN
j
N
N
N
N
()
sin ( )( )
sin ( ) /
sin ( )( )
sin ( ) /
ω
ω
ω
ω
ω
=
+
( )
( )
+
++
( )
+
( )
π
π
π
π
1
2
2
1
2
2
2
1
2
2
2
1
2
2
which is the same DTFT of y n
3
[].
3.96 (a) xn n
n
1
1[] [ ].=αμ Note,is a left-sided sequence,Hence,the ROC of its z-transform is
interior to a circle,Therefore,X z n z z z
n
n
nnn
n
mm
m
1
1
1
1)[]= =? =?
=?∞
∞

=?∞
=
∞
∑∑∑
αμ α α
=?
=
∞
∑
(/ )z
m
m
α
1
=?
=
<
z
z
z
z
z
/
(/ )
,/,
α
αα
α
1
1 The ROC of X z
1
( ) is thus given by z <α.
(b) xn n
n
2
1[] [ ].=+αμ Note,is a right-sided sequence,Hence,the ROC of its z-transform is
exterior to a circle,Therefore,X z n z z
n
n
nn
n
n
2
1
1() [ ]
–
=+=
=∞
∞
=?
∞
∑∑
αμ α
=+ =+
<
=
∞

∑
ααα
α
α
1
0
1
1
1
1
1zzz
z
z
n
n
n
,/, Simplifying we get Xz
z
z
2
1
()
/
(/ )
=
α
α
whose
ROC is given by z >α.
81
(c) xn n
n
3
[] [ ].=?αμ Note,is a left-sided sequence,Hence,the ROC of its z-transform is
interior to a circle,Therefore,X z n z z z
n
n
nn
n
nm
m
m
3
0
0
() [ ]=?= =
=?∞
∞
=?∞

=
∞
∑∑∑
αμ α α
=
<
1
1
1
1
1
α
α
z
z,,Therefore,the ROC of X z
3
( ) is given by z <α.
3.97 vn n n
nn n
[] [] [ ].== +
ααμαμ 1 Now,
Z{ [ ]},αμ
α
α
n
n
z
z=
>
1
1
1
(see Table 3.8) and
Z{[ ]}αμ α α α

=?∞
=
∞
=
∞
= = =?
∑∑∑
nn
n
mm
m
mm
m
nzzz11
1
10
=
=
<
1
1
11
1
α
α
α
α
z
z
z
z,,
Therefore,Vz
z
z
z
z
zz
()
()
()()
,=
+
=

1
1
1
1
11
1
11
α
α
ααα
with the ROC of V(z) given by
αα<<z1/
3.98 (a) Yz
zz
zz
z
zz
K
A
z
B
z
1
1
11
11
1
113
1
1113
1113
()
()
()( /)
()
()(/)
(/)
,=
++
=
++
( )
=+
+
+
+

where
KYz
z
==
=
1
0
0(),A
z
z
z
=
+
( )
=
=?
()
(/)
,
1
113
3
1
1
1
1
and B
z
z
z
=
+
=?
=?
()
.
1
1
2
1
1
3
1
Thus,
Yz
zz
z
1
11
3
1
2
113
1()
(/)
,.=
+
+
+
>

Since,the ROC is exterior to a circle,the inverse
z-transform y n
1
[ ] of Y z
1
( ) is a right-sided sequence and is given by
yn n n
nn
1
31 213[] ( ) [] ( / ) [].=μμ
(b) Yz
zz
z
2
11
3
1
2
113
1
3
()
(/)
,.=
+
+
+
<

Since,the ROC is interior to a circle,the inverse
z-transform y n
2
[ ] of Y z
2
( ) is a left-sided sequence and is given by
yn n n
nn
2
31 1 213 1[] ( ) [ ] ( / ) [ ].= +μμ
(c) Yz
zz
z
3
11
3
1
2
113
1
3
1()
(/)
,.=
+
+
+
<<

Since,the ROC is an annular region in the
z-plane,the inverse z-transform y n
3
[ ] of Y z
3
( ) is a two-sided sequence and is given by
yn n n
nn
3
31 1 213[] ( ) [ ] ( / ) [].=μμ
3.99 (a) Xz
zz
zz
z
zz
zz
zzz
zz

a
()
()() ()()()()
.=
+
+?
=
+
+?
=
+
+?

43 3
23
43 3
12 13
433
12 13
12
2
3
12
112
345
112
z > 3,
82
Let Gz
zz
zz
()
()()
.=
+
+?

43 3
12 13
12
112
A partial-fraction expansion of G(z) yields
Gz
A
z
B
z
C
z
()
()
,=
+
+
+

12 13 13
11 12
where A
zz
z
z
=
+
==

=?
43 3
13
25 4
25 4
1
12
12
12
1
()
/
/
,
/
C
zz
z
z
=
+
+
==

=
43 3
12
10 3
53
2
12
1
13
1
/
/
/
,and
B
d
dz
zz
z
z
=

+
+
=
=

1
21 3
43 3
12
25 3
25 3
1
21 1
12
1
13
()!()
/
/
.
/
Therefore Xz z
z
z
z
z
z
a
()
()
.=
+
+
+
3
1
3
1
3
12
1
12
1
13
2
13
Since the ROC is z > 3,
the inverse z-transform x n
a
[ ]of X z
a
( ) is a right-sided sequence.
Thus,
Z
+
=?
1
1
1
12
2
z
n
n
()[],μ
Z
=
1
1
1
13
3
z
n
n
() [],μ and
Z
=
1
1
12
3
13
3
z
z
nn
n
()
() [].μ Thus,
xn
a
[] =+?+

()[]()[]()()[].2333
2
3
23 2
33 2nn n
nnnnμμ μ
(b) Xz z
z
z
z
z
z
b
()
()
,=
+
+
+
3
1
3
1
3
12
1
12
1
13
2
13
z < 2,Here the ROC is z < 2,
Hence,the inverse z-transform x n
b
[ ] of X z
b
( ) is a left-sided sequence,Thus,
Z
+
=
1
1
1
12
21
z
n
n
()[ ],μ
Z
=
1
1
1
13
31
z
n
n
() [ ],μ and
Z
=
1
1
12
3
13
31
z
z
nn
n
()
() [ ].μ Therefore,
xn n n n n
b
nn n
[] ( ) [ ] () [ ] ( )() [ ].=+++

2232
2
3
23 1
33 2
μμ μ
(c) Xz z
z
z
z
z
z
c
()
()
,=
+
+
+
3
1
3
1
3
12
1
12
1
13
2
13
23<<z, Here the ROC is an
annular region in the z-plane,Hence,the inverse z-transform x n
c
[ ] of X z
c
( ) is a two-sided
sequence,Now
Z
+
1
1
1
12z
is right-sided sequence as z > 2,whereas,
Z
1
1
1
13z
and
Z
1
1
12
2
13
z
z()
are left-sided sequences as z < 3,Thus,
Z
+
=?
1
1
1
12
2
z
n
n
()[],μ
83
Z
=
1
1
1
13
31
z
n
n
() [ ],μ and
Z
=
1
1
12
3
13
31
z
z
nn
n
()
() [ ].μ Therefore,
xn n n n n
c
nn n
[] ( ) [ ] () [ ] ( )() [ ].=++

233 2
2
3
23 1
33 2
μμ μ
3.100 Gz
Pz
Dz
Pz
zRz
()
()
()
()
()()
.==
1
1
λ
l
By definition the residue
ρ
l
at the pole
z =λ
l
is given by
ρ
λ
l
l
=
=
Pz
Rz
z
()
()
,Now,
Dz
dD z
dz
dzRz
dz
Rz z
dR z
dz
'( )
()
()()
() ( )
()
==
[]
=? +?
1
1
1
1
1
1
λ
λλ
l
ll
,Hence,
Dz Rz
zz
'( ) ( ),
==
=?
λλ
λ
ll
l
Therefore,
ρλ
λ
ll
l
=?
=
Pz
Dz
z
()
'( )
.
3.101 Gz
Pz
Dz
ppz pz
ddz dz
M
M
N
N
()
()
()
.==
+++
+++

01
1
01
1
L
L
Thus,G
p
d
(),∞=
0
0
Now a partial-fraction expansion
of G(z) in z
1
is given by Gz
z
N
(),=
=
∑
ρ
λ
l
l
l
1
1
1
from which we obtain G
N
(),∞=
=
∑
ρ
l
l 1
Therefore,
G
p
d
N
(),∞= =
=
∑
ρ
l
l 1
0
0
3.102 Hz
rzrz
zr()
cos( )
,=
+
>>

1
12
0
122
θ
,By using partial-fraction expansion we write
Hz
ee
e
re z
e
re z j
e
re z
e
re z
jj
j
j
j
j
j
j
j
j
()
sin( )
.=
( )
=

1
11
1
2 11
11 11
θθ
θ
θ
θ
θ
θ
θ
θ
θ
θ
Thus,
h[n] =?
{}

1
2j
ere n re e n
j n jn n j jn
sin( )
[] []
θ
μμ
θθ θθ
=
+?+
re e
j
n
njn jn
sin( )
[]
() ()
θ
μ
θθ11
2
=
rn
n
n
sin ( )
sin( )
[].
+( )1 θ
θ
μ
3.103 G z g n z
n
n
() []=
=?∞
∞
∑
with a ROC given by R
g
.
(a) Therefore G z g n z
n
n
*( ) *[ ]( *)=
=?∞
∞
∑
and G z g n z
n
n
*( *) *[ ]=
=?∞
∞
∑
.
Thus the z-transform of g n*[ ] is G z*( *).
(b) Replace n by –m in the summation,This leads to G z g m z
m
m
() [ ]=?
=?∞
∞
∑
,Therefore
84
Gz gmz
m
m
(/ ) [ ]1 =?
=?∞
∞
∑
,Thus the z-transform of g[–n] is G(1/z),Note that since z has been
replaced by 1/z,the ROC of G(1/z) will be
1/R
g
.
(c) Let y n g n h n[] [] []=+αβ,Then,
Yz gn hn z gnz hnz Gz Hz
n
n
n
n
n
n
() [] [] [] [] () ()=+( ) =+=+
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑
αβ α β αβ
In this case Y(z) will converge wherever both G(z) and H(z) converge,Thus the ROC of Y(z)
is
RR
gh
∩,where is
R
g
the ROC of G(z) and
R
h
is the ROC of H(z).
(d) yn gn n[] [ ]=?
0
,Hence Y(z) = y n z g n n z g m z
n
n
n
n
mn
m
[] [ ] [ ]
()?
=?∞
∞
=?∞
∞
+
=?∞
∞
∑∑ ∑
=
0
0
==

=?∞
∞
∑
zgmzzGz
nm
m
n
00[] ().
In this case the ROC of Y(z) is the same as that of G(z) except for the possible addition or
elimination of the point z = 0 or z = ∞ (due to the factor z
n?
0
).
(e) yn gn
n
[] []=α, Hence,Y z y n z g n z G z
n
n
n
n
() [] []( ) (/ ).== =
=?∞
∞

=?∞
∞
∑∑
αα
1
The ROC of Y(z) is
αR
g
.
(f) y[n] = ng[n],Hence Y(z) = ng n z
n
n
[]
=?∞
∞
∑
.
Now G z g n z
n
n
() []=
=?∞
∞
∑
,Thus,
dG z
dz
ng n z
n
n
()
[]=?

=?∞
∞
∑
1
z
dG z
dz
ng n z
n
n
()
[]=?
=?∞
∞
∑
.
Thus Y(z) = – z
dG z
dz
()
.
(g) y[n] = g[n]
*
h[n] = g k h n k
k
[][ ]?
=?∞
∞
∑
,Hence,
Yz ynz gkhnkz gk hnkz
n
nk
n
n
n
kn
() [] [][ ] [] [ ]==?
=?
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
∑∑ ∑∑∑
= g k H z z H z G z
k
k
[] () () ()
=?∞
∞
∑
=,
In this case also Y(z) will converge wherever both H(z) and G(z) converge,Thus ROC of Y(z)
is
RR
gh
∩,
(h) y[n] = g[n]h[n],Hence,Y(z) = g n h n z
n
n
[][]
=?∞
∞
∑
,From Eq,(3.107),
85
gn
j
Gvv dv
n
C
[] ()=
∫
1
2
1
π
,Thus,Yz hn
j
Gvv dv z
n
n
C
n
() [] ()=
=?∞
∞

∑
∫
1
2
1
π
=
=
=?∞
∞

∑
∫∫
1
2
1
2
11
ππj
Gv hnz v dv
j
GvHz vv dv
n
nn
CC
() [] () (/ ),
(i) gn h n
j
Gv h nv v dv
j
GvH v v dv
n
n
n
[] *[] () *[] () *(/*)
=?∞
∞
=?∞
∞
∑∑
==
1
2
1
2
1,
3.104 xn x n jx n
re im
[] [] [],=+ where xn xn xn
re
[] [] *[],=+( )
1
2
and xn
j
xn x n
im
[] [] *[].=?( )
1
2
From
Table 3.9,
Z xn Xz*[ ] *( *),{}= with an ROC
R
x
,Therefore,
Z xn Xz Xz
re
[ ] ( ) *( *),
{}
=+{}
1
2
and
Z xn
j
Xz X z
im
[ ] ( ) *( *),
{}
=?{}
1
2
3.105 (a) Expnading in a power series we get Xz
z
z
n
n
1 3
3
0
1
1
()=
=
=
∞
∑
,z >1.
Thus,x
1
[n] =
13 0
0
,,
,.
if n k and n
elsewhere
=≥
{
Using partial fraction,we get
Xz
zz
jz jz
1 3
1
3
1
1
3
1
2
3
2
1
1
3
1
2
3
2
1
1
11
11
()
() ()
=
=
+
++
+
+?

,Therefore,
x
1
[n] =
1
3
1
3
1
2
3
2
1
3
1
2
3
2
μμμ[] [] []njnjn
nn
+
+
+
=
π π
++=+π
1
3
1
3
23
1
3
23
1
3
2
3
23μμμμ μ[ ] [ ] [ ] [ ] cos( / ) [ ]
//
ne ne n n n n
jn jn
.
Thus x
1
[n] =
13 0
0
,,
,.
if n k and n
elsewhere
=≥
{
(b) Expnading in a power series we get Xz
z
z
n
n
2 2
2
0
1
1
(),=
=
=
∞
∑
z >1.
Thus,x
2
[n] =
12 0
0
,,
,.
if n k and n
elsewhere
=≥
Using partial fraction,we get Xz
zzz
2 2
1
2
1
1
2
1
1
111
()=
=
+
+

,Therefore,
x
2
[n] =
1
2
1
2
1μμ[] ( ) []nn
n
+?
Thus,x
2
[n] =
12 0
0
,,
,.
if n k and n
elsewhere
=≥
3.106 (a) Xz z z
1
1
1( ) log,=?
( )
>
αα,Expanding log 1
1
( )
αz in a power series we get
86
Xz z
zz
n
z
n
n
n
1
1
22 33
1
23
( ),....= =?

=
∞
∑
α
αα α
.
Therefore,,xn
n
n
n
1
1[] [ ]=
α
μ,
(b) Xz
z
zz
2
1
1
1( ) log log ( ),=
=?
( )
<
α
α
αα,Expanding log ( )1
1
( )
αz in a power
series we get
Xz z
zz z
n
n
n
2
1
23
1
23
() ( )
() () ()
.= =?

=
∞
∑
α
αα α
L
Therefore,xn
n
n
n
2
1[] [ ]=
α
μ,
(c) Xz
z
zz
3 1
1
1
1
1( ) log log( ),.=
= >
α
αα Expanding X z
3
( ) in a power series we get
Xz z
zz
n
z
n
n
n
1
1
22 33
1
23
( ),....=+ + +=

=
∞
∑
α
αα α
.
Therefore,xn
n
n
n
3
1[] [ ]=?
α
μ,
(d) Xz
z
zz
4 1
1
1( ) log log ( ),.=
=
( )
<
α
αα
αα Expanding X z
4
( ) in a power series we
get Xz z
zz z
n
n
n
4
1
23
1
23
() ( )
() () ()
.=+++=

=
∞
∑
α
αα α
L
Therefore,xn
n
n
n
4
1[] [ ].=?
α
μ
3.107 Xz
z
z
()
()
=
α
α
1
12
1
where
xn Xz[] {()}=
Z
1
is a causal sequence,Now,from Table 3.8,
Z{ [ ]},αμ
α
n
n
z
=
1
1
1
But,
d
dz z
z
z
1
11
1
2
12
=
α
α
α()
,Thus,Xz z
d
dz z
(),=
1
1
1
α
Therefore,x n n n
n
[] [].=αμ
3.108 Hz
zz
k
A
z
B
z
()
(,)(,),,
,=
+?
=+
+
+

12
11 1 1
2
104 102 104 102
where kH==
=()
.(,)
,0
2
04 02
25
A
zz
z
z
=
=
( )

=?

=?
11
1
25
12
102
251 2 25
10225
10
1
()
.
.(.
.(,)
,
.
B
zz
z
z
=
+
=
( )
+
=?

=
11
1
5
12
104
51 10
1045
15
1
()
.,()
,
Thus,Hz
zz
()
..
.=?
+

25
10
104
15
102
11
Using the M-file residuz we also arrive at the same
partial-fraction expansion.
87
Therefore,h n n n n
nn
[] [] (,) [] (.) [].=25 10 0 4 15 0 2δμμ
3.109 From Eq,(3.177),for N = 3,we get
D
3
0
1
0
2
1
1
1
2
2
1
2
2
1
1
1
=

zz
zz
zz
,The determinant of D
3
is given by
det( )D
3
0
1
0
2
1
1
1
2
2
1
2
2
0
1
0
2
1
1
0
1
1
2
0
2
2
1
0
1
2
2
0
2
1
1
0
1
1
2
0
2
2
1
0
1
2
2
0
2
1
1
1
1
0
0
==

=

zz
zz
zz
zz
zzzz
zzzz
zzzz
zzzz
=?
( )
( )
+
+
=?
( )
( )
( )
=?
( )

≥>≥
∏
zzzz
zz
zz
zzzzzz zz
k
k
1
1
0
1
2
1
0
1 1
1
0
1
2
1
0
1 1
1
0
1
2
1
0
1
2
1
1
111
20
1
1
l
l
.
From Eq,(3.177),for N = 4,we get
D
4
0
1
0
2
0
3
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
3
3
1
1
1
1
=

zzz
zzz
zzz
zzz
,The determinant of D
4
is given by
det( )D
4
0
1
0
2
0
3
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
3
3
0
1
0
2
0
3
1
1
0
1
1
2
0
2
1
3
0
3
2
1
0
1
2
2
0
1
1
1
1
1
0
0
==

zzz
zzz
zzz
zzz
zzz
zzzzzz
zzzz
2
2
3
0
3
3
1
0
1
3
2
0
2
3
3
0
3
0
zz
zzzzzz

=

=?
( )
( )
( )

zzzzzz
zzzzzz
zzzzzz
zzzzzz
1
1
0
1
1
2
0
2
1
3
0
3
2
1
0
1
1
2
0
2
2
3
0
3
3
1
0
1
1
2
0
2
3
3
0
3
1
1
0
1
2
1
0
1
3
1
0
1
1 zzzzzzz
zzzzzz
zzzzzz
1
1
0
1
1
2
1
1
0
1
0
2
2
1
0
1
2
2
2
1
0
1
0
2
3
1
0
1
3
2
3
1
0
1
0
2
1
1

+++
+++
=?
( )
( )
( )
+++

zzzzzz
zz zzzz
zz zzzzz
zz zz
1
1
0
1
2
1
0
1
3
1
0
1
1
1
0
1
1
2
1
1
0
1
0
2
2
1
1
1
2
1
1
1
2
1
1
2
0
1
3
1
1
1
3
1
1
1
0
0
()( )
(
1
3
1
1
2
0
1
)( )zzz

++
=?
( )
( )
( )
++

zzzzzz
zz zzzzz
zz zzzzz
1
1
0
1
2
1
0
1
3
1
0
1 2
1
1
1
2
1
1
1
2
1
1
1
0
1
3
1
1
1
3
1
1
1
3
1
1
1
0
1
()( )
()( )
=?
( )
( )
( )
( )
( )
++
++

zzzzzzzzzz
zzz
zzz
1
1
0
1
2
1
0
1
3
1
0
1
2
1
1
1
3
1
1
1 2
1
1
1
0
1
3
1
1
1
0
1
1
1
=?
( )
( )
( )
( )
( )
( )
=?
( )

≥>≥
∏
zzzzzzzzzzzz zz
k
k
1
1
0
1
2
1
0
1
3
1
0
1
2
1
1
1
3
1
1
1
3
1
2
111
30
l
l
.
Hence,in the general case,det(D
N
) =
zz
k
Nk

≥>≥
( )
∏
11
10
l
l
,It follows from this expression
that the determinant is non-zero,i.e,D
N
is non-singular,if the sampling points z
k
are distinct.
88
3.110 XXz
NDFT
[] ( ),0 4412812
0
==++?= XXz
NDFT
[] ( ),142316
1
==?++=
XXz
NDFT
[] ( ),242820
2
==?++=
NDFT
[] ( ),3441852
3
==?++=
Iz z z z z z z
0
1
1
2
1
1
3
1123
11 1
11
6
1
6
() ( )( )( ),= =? +?

Thus,I
0
1
2
10?
( )
=,
Iz z z z z z z
1
1
2
1
1
2
1
1
3
1123
111 1
1
3
1
4
1
12
() ( )( )( ),=+ = +

Thus,I
1
1
1
2
()=,
Iz z z z z z z
2
1
2
11
1
3
1123
111 1
5
6
1
3
1
6
() ( )( )( ),=+ = +

Thus,I
2
2
2
3
( )=? and
Iz z z z z z z
3
1
2
11
1
2
1123
111 1
1
4
1
4
() ( )( )( ),=+ = +

Thus,I
3
1
3
5
2
=, Therefore,
Xz I z I z I z I z z z z() ()
/
()
/
()
/
(),=+? + =?++

12
10
6
12
20
23
52
52
42 3
012 3
123
3.111 xn
xn n N
Nn N
e
[]
[],,
,.
=
≤≤?
≤≤?
01
021
y[n] = x
e
[n] + x
e
[2N –1– n],Therefore,
Yk ynW xnW x N nW
N
nk
n
N
N
nk
n
N
N
nk
nN
N
[] [] [] [ ]==+?
=
=
=
∑∑∑2
0
21
2
0
1
2
21
21
=+ = +
=

=
=

∑∑ ∑
xnW xnW xn W W W
N
nk
n
N
N
Nnk
n
N
N
nk
n
N
N
k
N
nk
[] [] []( )
()
2
0
1
2
21
0
1
2
0
1
22
.
Thus,C
x
[k] = W
N
k
2
2/
Y[k] = xn W W
N
kn
n
N
N
kn
[]( )
(/) (/
2
12
0
1
2
12+
=
+
∑
+
= 2
21
2
01
0
1
xn
nk
N
kN
n
N
[ ]cos
()
,.
=
∑
+
≤≤?
π
3.112 Y[k] =
WCk kN
kN
WCNkN kN
N
k
x
N
k
x
2
2
2
2
01
0
2121
≤≤?
=
+≤?
/
/
[],,
,,
[],.
Thus,
yn
N
YkW
N
nk
k
N
[] []=
=
∑
1
2
2
0
21
=
1
2
1
2
2
2
12
0
1
2
12
1
21
N
CkW
N
CNkW
x N
nk
k
N
x N
nk
kN
N
[] [ ]
(/) (/)?+
=
+
=+
∑∑

=
1
2
1
2
2
12
0
1
2
12 2
1
1
N
CkW
N
CkW
x N
nk
k
N
x N
nNk
k
N
[] []
(/) (/)( )?+
=
+?
=
∑∑
=
1
2
1
2
2
12
0
1
2
12
1
1
N
CkW
N
CkW
x N
nk
k
N
x N
nk
k
N
[] []
(/) (/)?+
=
+
=
+
=
C
NN
Ck
kn
N
x
x
k
N
[]
[ ]cos
()
0
2
12
2
1
1
+
+
=
∑
π
,
Hence,
C
NN
Ck
kn
N
x
x
k
N
[]
[ ]cos
()
0
2
121
2
1
1
+
+
=
∑
π
,where w k
k
kN
[]
/,,
,.
=
=
≤≤?
12 0
11 1
89
Moreover,xn
yn n N
elsewhere
N
wkC k
kn
N
nN
elsewhere
x
k
N
[]
[],,
,,
[] []cos
()
,,
,.
=
≤≤?
=
+
≤≤?
=
∑
01
0
121
2
01
0
0
1
π
(b) From Eq,(3.163),xn
wkC k n N
elsewhere
N
x
nk
N
k
N
[]
[] []cos,,
,.
()
=
( )
≤≤?
+π
=
∑
1 21
2
0
1
01
0
Hence,
2
21
2
0
1
xn
nm
N
m
N
[ ]cos
()+π
=
∑
=
1
21
2
0
1
0
1
21
2
N
wkC k
x
nk
N
m
N
k
N
nm
N
[ ] [ ]cos cos
() ()+π
=
=
+π
( ) ( )∑∑
=
1
21
2
0
1
0
1
21
2
N
wkC k
x
nk
N
m
N
k
N
nm
N
[ ] [ ] cos cos
() ()
.
+π
=
=
+π
( ) ( )∑∑
(10)
Now,cos
()
cos
()
,,
/,,
,.
21
2
0
1
21
2
0
2
0
nk
N
m
N
nm
N
Nifkm
elsewhere
+π
=
∑
+π
=
==
=
Thus,Eq,(10) reduces to
2
21
2
0
1
xn
nm
N
m
N
[ ]cos
()+π
=
∑
=
1
1
00 0
11
N
x
N
x
wC N m
wmC m N m N
[] [],,
[] [],,
=
≤≤?
=
Cm
Cm m N
x
x
[],,
[],,
00
11
=
≤≤?
= C
x
[m],0 1≤≤?mN,
3.113 y[n] = αβgn hn[] []+, Thus,
C
y
[k] = yn
kn
N
gn hn
kn
N
n
N
n
N
[ ]cos
()
[] []cos
()π
αβ
π21
2
21
2
0
1
0
1
+
=+( )
+
=
=
∑∑
=
= α
π
β
π
gn
kn
N
hn
kn
N
n
N
n
N
[ ]cos
()
[ ]cos
()21
2
21
2
0
1
0
1
+
+
+
=
=
∑∑
= αβCk Ck
gh
[] []+,
3.114 Ck xn
kn
N
x
n
N
[ ] [ ]cos
()
=
+
=
∑
π 21
2
0
1
Ck x n
kn
N
x
n
N
*
[ ] *[ ]cos
()
=
+
=
∑
π 21
2
0
1
.
Thus the DCT coefficients of x*[n] are given by C k
x
*
[].
3.115 Note that cos
()
cos
()
,,
/,
,.
,
ππkn
N
mn
N
Nifkm
N if k m and k
otherwise
n
N
21
2
21
2
0
20
0
0
1
+
+
=
==
=≠
=
∑
Now,xnx n
N
kmCmCk
nk
N
nm
N
xx
m
N
k
N
[]*[] [][ ]
*
[ ] [ ]cos
()
cos
()
=
+
+
=
=
∑∑
121
2
21
2
2
0
1
0
1
αα
ππ
Thus,xn
N
kmCmCk
nk
N
nm
N
n
N
xx
n
N
m
N
k
N
[] [][ ]
*
[ ] [ ] cos
()
cos
()
2
0
1
2
0
1
0
1
0
1
2
21
2
=
=
=
=
∑∑∑∑
=
+
+
αα
ππ
90
Now,using the orthogonality property mentioned above xn
N
kC k
n
N
x
k
N
[] [] []
2
0
1
2
0
1
1
2
=
=
∑∑
=α,
3.116 Xk xn
nk
N
nk
N
DHT
n
N
[ ] [ ] cos sin=
+
=
∑
22
0
1
ππ
,Now,
Xk
nk
N
mk
N
DHT
[ ] cos sin
22ππ
+
=
+
+
=
∑
xn
nk
N
nk
N
mk
N
mk
N
n
N
[ ] cos sin cos sin
22 2 2
0
1
πππ π
,Therefore,
Xk
mk
N
mk
N
DHT
k
N
[ ] cos sin
22
0
1
ππ
+
=
∑
=
+
+
=
=
∑∑
xn
nk
N
nk
N
mk
N
mk
N
k
N
n
N
[ ] cos sin cos sin
22 2 2
0
1
0
1
ππ π π
It can be shown that cos cos
,,
/,,
/,,
,,
22
0
20
2
0
0
1
ππnk
N
mk
N
Nifmn
Nifmn
NifmNn
elsewhere
k
N
=
==
=≠
=?
=
∑
sin sin
/,,
/,,
,,
22
20
2
0
0
1
ππnk
N
mk
N
Nifmn
NifmNn
elsewhere
k
N
=
=≠
=?
=
∑
and
sin cos cos sin,
22 22
0
0
1
0
1
ππ ππnk
N
mk
N
nk
N
mk
N
k
N
k
N
=
=
=
=
∑∑
Hence,xm
N
Xk
mk
N
mk
N
DHT
k
N
[ ] [ ] cos sin,=
+
=
∑
122
0
1
ππ
3.117 (a) yn x n n
xn n N n n
xn n n n N
N
[] ( )
[],,
[],.
=<? >=
+ ≤≤?
≤
0
00
00
01
1
Yk yn
nk
N
nk
N
DHT
n
N
[ ] [ ] cos sin=
+
=
∑
22
0
1
ππ
= xn n N
nk
N
nk
N
n
n
[ ] cos sin?+
+
=
∑ 0
0
1
22
0
ππ
+?
+
=
∑
xn n
nk
N
nk
N
nn
N
[ ] cos sin
0
1
22
0
ππ
.
Replacing n – n
0
+N by n in the first sum and n – n
0
by n in the second sum we get
Yk xn
nnk
N
nnk
N
DHT
nNn
N
[ ] [ ] cos
()
sin
()
=
+
+
+
=?
∑
22
00
1
0
ππ
+
+
+
+
=
∑
xn
nnk
N
nnk
N
n
n
[ ] cos
()
sin
()22
00
0
1
0
ππ
91
= xn
nnk
N
nnk
N
n
N
[ ] cos
()
sin
()22
00
0
1
ππ+
+
+
=
∑
= cos [ ] cos sin
2
22
0
0
1
π
ππ
nk
N
xn
nk
N
nk
N
n
N
+
=
∑
+
=
∑
sin [ ] cos sin
2
22
0
0
1
0
π
ππ
nk
N
xn
nk
N
nk
N
n
n
= cos [ ] sin [ ]
22
00
ππnk
N
Xk
nk
N
Xk
DHT DHT
+
,
(b) The N-point DHT of x[< –n >
N
] is X
DHT
[–k].
(c) xn
N
XkX
n
N
DHT DHT
N
k
N
2
0
1
2
0
1
0
1
1
[] [] []
=
=
=
∑∑∑
= l
l
×
cos sin cos sin
22 22
0
1
ππ ππnk
N
nk
N
n
N
n
N
n
N
+
+
=
∑
ll
.
Using the orthogonality property,the product is non-zero if k =
l and is equal to N.
Thus xn
N
Xk
n
N
DHT
k
N
2
0
1
2
0
1
1
[] []
=
=
∑∑
=,
3.118 cos
21
2
πnk
N
WW
N
nk
N
nk?
=+
( )
,and sin
21
2
πnk
Nj
WW
N
nk
N
nk?
=?
( )
.
Xk xn
ee ee
j
DHT
jnkN jnkN jnkN jnkN
n
N
[] []
////
=
+
+

=
∑
2222
0
1
22
ππππ
Therefore Xk XNkXkjXNkjXk
DHT
[] [ ] [] [ ] []=?++()
1
2
.
3.119 y[n] =
N
x[n] g[n],Thus,Yk yn
nk
N
nk
N
DHT
n
N
[ ] [ ] cos sin=
+
=
∑
22
0
1
ππ
= xr g n r
nk
N
nk
N
N
n
N
r
N
[ ] [ ] cos sin<?>
+
=
=
∑∑
0
1
0
1
22ππ
.
Fro m results of Problem 3.117
Yk xGk
k
N
Gk
k
N
DHT DHT DHT N
N
[ ] [ ] [ ]cos [ ]sin=
+<?>
=
∑
l
ll
l
22
0
1
ππ
= Gkx
k
N
Gkx
k
N
DHT
N
DHT N
N
[ ] [ ]cos [ ] [ ]sinl
l
l
l
ll0
1
0
1
+<?>
=
=
∑∑
92
=
1
2
GkXkX k
DHT DHT DHT N
[] [] [ ]+<?>
()
+
1
2
GkXkXk
DHT N DHT DHT N
[][][]<? >? <? >
()
.
or Yk X kGkG k
DHT DHT DHT DHT N
[] [] [] [ ]=+<?
( )
1
2
+ <?>? <?>
()
1
2
XkGkGk
DHT N DHT DHT N
[][][].
3.120 (a) H
2
11
11
=
,H
4
1111
111 1
11 1 1
1111
=

,and H
8
11111111
11111111
11 1111 11
11111 111
1111 1111
1 11 1 11 11
11 111111
1111 111 1
=

.
(b) From the structure of H
2
,H
4
and H
8
it can be seen that
H
HH
HH
4
22
22
=
,and H
HH
HH
8
44
44
=
.
(c) XHx
HT N
=, Therefore xHXHXHX== =
NHT N
T
HT N HT
NN
1*
,Hence,
xn X k
HT
bnbk
k
N
ii
i
[] []( ),
() ()
=?
=
∑
=
∑
1
0
1
0
1
l
where b r
i
( ) is the i
th
bit in the binary representation of r.
3.121
Xz xnz xnA V xnA V V V
n
n
N
nn
n
N
nn n
n
N
( ) [] [] []
//()/
ll
lll
== =
=
=

=
∑∑ ∑
0
1
0
1
22 2
0
1
=?
=
∑
Vgnhn
n
N
l
l
2
2
0
1
/
[ ] [ ],where g n x n A V
nn
[] []
/
=
2
2
and h[n] = V
n?
2
2/
.
A block-diagram representation for the computation of
Xz()
l
using the above scheme is thus
precisely Figure P3.6.
3.122
z
l
l
=α, Hence,
AV e e
jj
00
00

=
lllθφ
α, Since α is real,we have
A
0
1=,V
0
1= /α,θ
0
0= and φ
0
0=,
3.123 (i) N = 3.
Xz x x z x z() [] [] []=+ +

01 2
12
,and H z h h z h z() [] [] []=+ +

01 2
12
Yz h x h x x h z h x h x h x z
L
() [][] [][] [][] [][] [][] [][]=++( ) +++( )

00 10 10 20 11 02
12
93
++( ) +

hx h x z h x z[][] [][] [][],12 21 22
34
On the other hand,
Yz hx hx hx hx hx hx z
c
() [][] [][] [][] [][] [][] [][]=++( )+++( )
00 12 20 01 10 22
1
+++( )
hx hx hx z[ ] [ ] [] [] [ ] [ ],02 11 20
2
It is easy to see that in this case Y z Y z z
cL
() ()mod( )=?
1
3
.
(ii) N = 4.
Xz x x z x z x z() [] [] [] []=+ + +

01 2 3
123
and H z h h z h z h z() [] [] [] []=+ + +

01 2 3
123
.
Yzhx hx hx z hx hx hx z
L
() [][] [][] [][] [][] [][] [][]=++( ) +++( )

00 10 01 02 11 20
12
+ +++( ) +++( )hx hx hx hx z hx hx hx z[][] [][] [][] [][] [][] [][] [][]03 12 21 30 13 22 31
34
++( ) +

hx hx z hxz[][] [][] [][]23 32 33
56
,
whereas,Y z h x h x h x h x
c
() [][] [][] [][] [][]=+++( )00 13 22 31
++++( )
hx hx hx hx z[ ] [] [] [ ] [ ] [] [] [ ]01 10 23 32
1
+h x h x h x h x z[ ] [ ] [] [] [ ] [ ] [] []02 11 20 33
2
++ +( )
+ h x h x h x h x z[][] [][] [][] [][]03 12 21 30
3
+++( )
.
Again it can be seen that Y z Y z z
cL
() ()mod( )=?
1
4
.
(ii) N = 5.
Xz x x z x z x z x z() [] [] [] [] []=+ + + +

01 2 3 4
1234
and
Hz h h z h z h z h z() [] [] [] [] [],=+ + + +

01 2 3 4
1234
Yzhx hx hx z hx hx hx z
L
() [][] [][] [][] [][] [][] [][]=++( ) +++( )

00 10 01 02 11 20
12
+ +++( )
hx hx hx hx z[][] [][] [][] [][]03 12 21 30
3
+++++( )
hx hx hx hx hx z[][] [][] [][] [][] [][]04 13 22 31 40
4
+ +++( ) +++( )

hx h x hx h x z hx hx h x z[][] [][] [][] [][] [][] [][] [][]14 23 32 41 24 33 42
56
++( ) +

hx hx z hxz[][] [][] [][]43 34 44
78
,
whereas,Y z h x h x h x h x h x
c
() [][] [][] [][] [][] [][]= ++++( )00 41 32 23 14
+++++( )
hx hx hx hx hx z[ ] [] [] [ ] [ ] [ ] [] [] [ ] [ ]01 10 24 33 42
1
+++++( )
hx hx hx hx hx z[ ] [ ] [] [] [ ] [ ] [] [ ] [ ] []02 11 20 34 43
2
+ ++++( )
hx hx hx hx hx z[][] [][] [][] [][] [][]03 12 21 30 44
3
+++++( )
hx hx hx hx hx z[][] [][] [][] [][] [][]04 13 22 31 40
4
.
Again it can be seen that Y z Y z z
cL
() ()mod( )=?
1
5
.
3.124 (a) Xz xnz
n
n
() []=
=?∞
∞
∑
,Let
( ) log( ( ))Xz Xz=?=Xz e
Xz
()
()
,Thus,X e e
jXe
j
()
()ω
ω
=
94
(b)?[ ] log ( )xn Xe e d
jjn
=
( )
∫
1
2π
ω
ωω
π
π
,If x[n] is real,then X e X e
jj
() *( )
ωω
=
,Therefore,
log ( ) log *( ),Xe X e
jjωω
( )
=
( )
.
*[ ] log *( )xn Xe e d
jjn
=
( )
∫
1
2π
ω
ωω
π
π
=
1
2π
ω
ωω
π
π
log ( )Xe e d
jjn
( )
∫
=
1
2π
ω
ωω
π
π
log ( )Xe e d
jjn
( )
∫
=?[]xn.
(c)? []
[]?[]
log ( )xn
xn x n
Xe
ee
d
ev
j
jn jn
=
+?
=
( )
+
∫
2
1
22π
ω
ω
ωω
π
π
=
( )
∫
1
2π
ωω
ω
π
π
log ( ) cos( )Xe nd
j
,
and similarly,? []
[]?[]
log ( )xn
xn x n j
Xe
ee
j
d
od
j
jn jn
=

=
( )
∫
22 2π
ω
ω
ωω
π
π
=
( )
∫
j
Xe nd
j
2π
ωω
ω
π
π
log ( ) sin( ),
3.125 xn a n b n[] [] [ ]=+?δδ1 and X(z) = a bz+
1
,Also,
( ) log( ) log( ) log( / ) log( ) ( )
/
Xz a bz a b az a
ba
n
z
n
n
n
n
=+ = ++ = +?
( )

=
∞
∑
111
1
1 1, Therefore,
[]
log( ),,
()
(/)
,,
,.
xn
aifn
ba
n
for n
elsewhere
n
n
=
=
>
0
10
0
1
3.126 (a)
( ) log( ) log logXz K z z
k
k
N
k
k
N
=+?
( )
+?
( )
==
∑∑
11
1
αγ
α
γ

( )

( )
==
∑∑
log log11
1
βδ
β
δ
k
k
N
k
k
N
zz.
( ) log( ),Xz K
n
z
n
z
n
z
n
z
k
n
n
nk
N
k
n
n
nk
N
k
n
n k
n
n
nk
N
nk
N
= + +
=
∞
==
∞
=
=
∞
==
∞
=
∑∑∑ ∑∑∑
αγβδ
α
γ
δ
β
11111111
Thus,?[]xn =
log( ),,
,,
,.
Kn
nn
n
nn
n
k
n
n
N
k
n
n
N
k
n
n
N
k
n
n
N
=
>
<
==
=
=
∑∑
∑∑
0
0
0
11
11
βα
γδ
β
α
γ
δ
95
(b)?[]xn N
r
n
n
< as n→∞,where r is the maximum value of α
k
,β
k
,γ
k
and δ
k
for all values
of k,and N is a constant,Thus?[]xn is a decaying bounded sequence as n→∞.
(c) From Part (a) if α
k
= β
k
=0 then?[]xn = 0 for all n > 0,and is thus anti-causal.
(d) If γ
k
= δ
k
= 0 then?[]xn = 0 for all n < 0 and is thus a causal sequence.
3.127 If X(z) has no poles and zeros on the unit circle then from Problem 3.95,γ
k
= δ
k
= 0 and
[]xn = 0 for all n < 0.
() log ()Xz Xz= ( ) therefore
dX z
dz X z
dX z
dz
()
()
()
=
1
,Thus,z
dX z
dz
zX z
dX z
dz
()
()
()
=,
Taking the inverse z-transform we get n x n k x k x n k
k
[]?[][ ]=?
=?∞
∞
∑
,n ≠ 0.
Since x[n] = 0 and?[]xn = 0 for n < 0,thus xn
k
n
xk xn k
k
n
[]?[][ ]
=
∑
0
,n≠0.
Or,xn
k
n
xk xn k xnx
k
n
[]?[][ ]?[][]=?+
=
∑
0
1
0, Hence,?[]
[]
[]
[][ ]
[]
xn
xn
x
k
n
xk xn k
x
k
n
=?
=
∑
00
0
1
,n≠0.
For n = 0,?[]
( ) ( ) log( [ ])xXz Xz x
zz
00===
=∞ =∞
,Thus,
[]
,,
log( [ ]),,
[]
[]
[][ ]
[]
,.
xn
n
xn
xn
x
k
n
xk xn k
x
n
k
n
=
<
=
>
=
∑
00
00
0
0
1
3.128 This problem is easy to solve using the method discussed in Section 4.13.2.
x[n] h[n] g[n]
v[n]
y[n]
hn n
n
[], []= 06 μ, Thus,Hz
z
()
.
.=
1
106
1
gn n
n
[], []= 08 μ, Thus,Gz
z
()
.
.=
1
108
1
From Eq,(4.212) we get ΦΦ
vv xx
zHzHz z() () ( ) (),=
1
(A)
and ΦΦ Φ
yy vv xx
z GzGz z GzGz HzHz z() () ( ) () () ( ) () ( ) ().==
111
(B)
Now,HzHz
zz
z
zz z z
() ( )
(, )(, ),,,
.
.
.
.
.

=

=
+?
=
+
1
1
1
12 1 1
1
1 06 1 06 06 136 06
1 5625
106
1 5625
1 1 6667
Thus,using Eq,(A) we get
96
Φ
vv x x
zHzHz
zz
() () ( ),
..
,==

12 2
11
1 5625
1
106
1
1 1 667
σσ 0 6 1 6667..<<z
Taking the inverse transform of the above we get
φσμμ σ
vv x
nn
x
n
nnn[],, [], [ ],,,=?
( )
=1 5625 0 6 1 6667 1 1 5625 0 6
22
As m
x
= 0 and m
v
= 0,we have σφ σ
vvv x
22
0 1 5625==[],,
Next we observe GzGz HzHz
zzzz
() ( ) () ( )
(, )(, )(, )(, )

=

11
11
1
106 106 108 108
=
+
+
+

15 0240
106
5 4087
1 1 6667
26 7094
108
17 094
1125
1111
.
.
.
.
.
.
.
.
.
zzzz
Using Eq,(B) and taking the inverse we get
φσ μ μ
yy x
nn
nnn[], (.) [], (,) [ ](=
2
15 024 0 6 5 4087 1 6667 1
++26 7094 0 8 17 094 1 25 1.(.)[].(.)[ ])μμ.
As m
v
= 0 and m
y
= 0,we have σφ σ σ
vyy x x
222
0 15 024 26 7094 11 6854==?+ =[] (,, ),,
M3.1 (a) r = 0.9,θ = 0.75,The various plots generated by the program are shown below:
0 0.2 0.4 0.6 0.8 1
-2
0
2
4
6
8
Real part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-8
-6
-4
-2
0
2
Imaginary part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
0
2
4
6
8
Magnitude Spectrum
Magnitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-2
-1.5
-1
-0.5
0
0.5
Phase Spectrum
Phase in radians
Normalized frequency
(b) r = 0.7,θ = 0.5,The various plots generated by the program are shown below:
97
0 0.2 0.4 0.6 0.8 1
0
1
2
3
4
Real part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-3
-2.5
-2
-1.5
-1
-0.5
0
Imaginary part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
0
1
2
3
4
5
Magnitude Spectrum
Magnitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-1.5
-1
-0.5
0
Phase Spectrum
Phase in radians
Normalized frequency
M3.2 (a) Ye
e
ee
j
jN
jN j N
1
21
1
1
()
–
.
– ()
––()
ω
ω
ωω
=
+
+
For example,for N = 6,Ye
e
ee
j
j
jj
1
13
67
1
()
–
.
–
––
ω
ω
ωω
=
0 0.2 0.4 0.6 0.8 1
-5
0
5
10
15
Real part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-5
0
5
10
x 10
-15
Imaginary part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
0
5
10
15
Magnitude spectrum
Magnitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-4
-2
0
2
4
Phase spectrum
Phase,radians
Normalized frequency
(b) Ye
ee
ee e
j
jN j N
jN jN jN
2
121
12
2
()
–
.
– () – ()
––() – ()
ω
ωω
ωωω
=
+
+
++
For example,for N = 6,
98
Ye
ee
eee
j
jj
jjj
2
714
678
12
2
()
–
.
––
–––
ω
ωω
ωωω
=
+
+
0 0.2 0.4 0.6 0.8 1
0
10
20
30
40
50
Real part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-5
0
5
10
15
x 10
-14
Imaginary part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
0
10
20
30
40
50
Magnitude spectrum
Magnitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-4
-2
0
2
4
6
8
x 10
-14
Phase spectrum
Phase,radians
Normalized frequency
(c) Ye e e e
jjnNjnN
nN
N
jn
3
1
2
22
()
– //
–
–ωω
=+
( )
ππ
=
∑
=
( )
π
=
∑
cos
–
–n
N
nN
N
jn
e
2
ω
=
( )
π
=
+
∑
ee
jN n
N
nN
N
jN nωω
cos
–
– ()
2
=
( )
π
=
+
∑
cos
.
–
– ()
–
n
N
nN
N
jN n
jN
e
e
2
ω
ω
For example,for N = 6,Ye
e
e
j
n
N
n
jn
j
3
2
6
6
6
6
()
cos
.
–
– ()
–
ω
ω
ω
=
( )
π
=
+
∑
99
0 0.2 0.4 0.6 0.8 1
-2
0
2
4
6
8
Real part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
x 10
-15
Imaginary part
Amplitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
0
2
4
6
8
Magnitude spectrum
Magnitude
Normalized frequency
0 0.2 0.4 0.6 0.8 1
-4
-2
0
2
4
Phase spectrum
Phase,radians
Normalized frequency
M3.3 (a)
0 0.2 0.4 0.6 0.8 1
-0.5
0
0.5
1
ω/π
Amplitude
Real part
0 0.2 0.4 0.6 0.8 1
-1
-0.5
0
0.5
1
ω/π
Amplitude
Imaginary part
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ω/π
Magnitude
Magnitude spectrum
0 0.2 0.4 0.6 0.8 1
-3
-2
-1
0
1
2
3
ω/π
Phase,radians
Phase spectrum
(b)
100
0 0.2 0.4 0.6 0.8 1
-1
-0.5
0
0.5
1
ω/π
Amplitude
Real part
0 0.2 0.4 0.6 0.8 1
-0.5
0
0.5
1
1.5
ω/π
Amplitude
Imaginary part
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ω/π
Magnitude
Magnitude spectrum
0 0.2 0.4 0.6 0.8 1
-4
-2
0
2
4
ω/π
Phase,radians
Phase spectrum
M3.4 N = input('The length of the sequence = ');
k = 0:N-1;
gamma = -0.5;
g = exp(gamma*k);
% g is an exponential sequence
h = sin(2*pi*k/(N/2));
% h is a sinusoidal sequence with period = N/2
[G,w] = freqz(g,1,512);
[H,w] = freqz(h,1,512);
% Property 1
alpha = 0.5;
beta = 0.25;
y = alpha*g+beta*h;
[Y,w] = freqz(y,1,512);
% Plot Y and alpha*G+beta*H to verify that they are equal
% Property 2
n0 = 12; % S equence shifted by 12 samples
y2 = [zeros([1,n0]) g];
[Y2,w] = freqz(y2,1,512);
G0 = exp(-j*w*n0).*G;
% Plot G0 and Y2 to verify they are equal
% Property 3
w0 = pi/2; % the value of omega0 = pi/2
r=256; %the value of omega0 in terms of number of samples
k = 0:N-1;
101
y3 = g.*exp(j*w0*k);
[Y3,w] = freqz(y3,1,512);
k = 0:511;
w = -w0+pi*k/512; % creating G(exp(w-w0))
G1 = freqz(g,1,w');
% Compare G1 and Y3
% Property 4
k = 0:N-1;
y4 = k.*g;
[Y4,w] = freqz(y4,1,512);
% To compute derivative we need sample at pi
y0 = ((-1).^k).*g;
G2 = [G(2:512)' sum(y0)]';
delG = (G2-G)*512/pi;
% Compare Y4,delG
% Property 5
y5 = conv(g,h);
[Y5,w] = freqz(y5,1,512);
% Compare Y5 and G.*H
% Property 6
y6 = g.*h;
[Y6,w] = freqz(y6,1,512,'whole');
[G0,w] = freqz(g,1,512,'whole');
[H0,w] = freqz(h,1,512,'whole');
% Evaluate the sample value at w = pi/2
% and verify with Y6 at pi/2
H1 = [fliplr(H0(1:129)') fliplr(H0(130:512)')]';
val = 1/(512)*sum(G0.*H1);
% Compare val with Y6(129) i.e sample at pi/2
% Can extend this to other points similarly
% Parsevals theorem
val1 = sum(g.*conj(h));
val2 = sum(G0.*conj(H0))/512;
% Comapre val1 with val2
M3.5 N = 8; % Number of samples in sequence
gamma = 0.5;
k = 0:N-1;
x = exp(-j*gamma*k);
y = exp(-j*gamma*fliplr(k));
% r = x[-n] then y = r[n-(N-1)]
% so if X1(exp(jw)) is DTFT of x[-n],then
% X1(exp(jw)) = R(exp(jw)) = exp(jw(N-1))Y(exp(jw))
[Y,w] = freqz(y,1,512);
X1 = exp(j*w*(N-1)).*Y;
m = 0:511;
w = -pi*m/512;
X = freqz(x,1,w');
% Verify X = X1
% Property 2
k = 0:N-1;
y = exp(j*gamma*fliplr(k));
[Y,w] = freqz(y,1,512);
102
X1 = exp(j*w*(N-1)).*Y;
[X,w] = freqz(x,1,512);
% Verify X1 = conj(X)
% Property 3
y = real(x);
[Y3,w] = freqz(y,1,512);
m = 0:511;
w0 = -pi*m/512;
X1 = freqz(x,1,w0');
[X,w] = freqz(x,1,512);
% Verify Y3 = 0.5*(X+conj(X1))
% Property 4
y = j*imag(x);
[Y4,w] = freqz(y,1,512);
% Verify Y4 = 0.5*(X-conj(X1))
% Property 5
k = 0:N-1;
y = exp(-j*gamma*fliplr(k));
xcs = 0.5*[zeros([1,N-1]) x]+0.5*[conj(y) zeros([1,N-1])];
xacs = 0.5*[zeros([1,N-1]) x]-0.5*[conj(y) zeros([1,N-1])];
[Y5,w] = freqz(xcs,1,512);
[Y6,w] = freqz(xacs,1,512);
Y5 = Y5.*exp(j*w*(N-1));
Y6 = Y6.*exp(j*w*(N-1));
% Verify Y5 = real(X) and Y6 = j*imag(X)
M3.6 N = 8;
k = 0:N-1;
gamma = 0.5;
x = exp(gamma*k);
y = e xp(gamma*fliplr(k));
xev =0.5*([zeros([1,N-1]) x]+[y zeros([1,N-1])]);
xod = 0.5*([zeros([1,N-1]) x]-[y zeros([1,N-1])]);
[X,w] = freqz(x,1,512);
[Xev,w] = freqz(xev,1,512);
[Xod,w] = freqz(xod,1,512);
Xev = exp(j*w*(N-1)).*Xev;
Xod = exp(j*w*(N-1)).*Xod;
% Verify real(X)= Xev,and imag(X)= Xod
r = 0:511;
w0 = -pi*r/512;
X1 = freqz(x,1,w0');
% Verify X = conj(X1)
% real(X)= real(X1) and imag(X)= -imag(X1)
% abs(X)= abs(X1) and angle(X)= -angle(X1)
M3.7 N = input('The size of DFT to be computed =');
for k = 1:N
for m = 1:N
D(k,m) = exp(-j*2*pi*(k-1)*(m-1)/N);
end
end
diff = inv(D)-1/N*conj(D);
% Verify diff is N*N matrix with all elements zero
103
M3.8 (a)
clf;
N = input('The value of N = ');
k = -N:N;
y1 = ones([1,2*N+1]);
w = 0:2*pi/255:2*pi;
Y1 = freqz(y1,1,w);
Y1dft = fft(y1);
k = 0:1:2*N;
plot(w/pi,abs(Y1),k*2/(2*N+1),abs(Y1dft),'o');
xlabel('Normalized frequency');ylabel('Amplitude');
(b) Add the statement y2 = y1 - abs(k)/N; below the statement y1 =
ones([1,2*N+1]); and replace y1,Y1,and Y1dft in the program with y2,Y2
and Y2dft,respectively.
(c) Replace the statement y1 = ones([1,2*N+1]); with y3 =
cos(pi*k/(2*N));,and replace y1,Y1,and Y1dft in the program with y3,Y3
and Y3dft,respectively.
The plots generated for N = 3 is shown below where the circles denote the DFT samples.
(a) (b)
0 0.5 1 1.5 2
0
2
4
6
8
Normalized frequency
Amplitude
0 0.5 1 1.5 2
0
1
2
3
4
Normalized frequency
Amplitude
(c)
0 0.5 1 1.5 2
0
1
2
3
4
Normalized frequency
Amplitude
M3.9 g = [3 4 -2 0 1 -4];
104
h = [1 -3 0 4 -2 3];
x = [2+j*3 3-j -1+j*2 j*3 2+j*4];
v = [-3-j*2 1+j*4 1+j*2 5+j*3 1+j*2];
k = 0:4;
z = sin(pi*k/2);
y = 2.^k;
G = fft(g); H = fft(h); X = fft(x);
V = fft(v); Z = fft(z); Y = fft(y);
y1 = ifft(G.*H); y2 = ifft(X.*V); y3 = ifft(Z.*Y);
M3.10 N = 8; % N is length of the sequence(s)
gamma = 0.5;
k = 0:N-1;
g = exp(-gamma*k); h = cos(pi*k/N);
G = fft(g); H=fft(h);
% Property 1
alpha=0.5; beta=0.25;
x1 = alpha*g+beta*h;
X1 = fft(x1);
% Verify X1=alpha*G+beta*H
% Property 2
n0 = N/2; % n0 is the amount of shift
x2 = [g(n0+1:N) g(1:n0)];
X2 = fft(x2);
% Verify X2(k)= exp(-j*k*n0)G(k)
% Property 3
k0 = N/2;
x3 = exp(-j*2*pi*k0*k/N).*g;
X3 = fft(x3);
G3 = [G(k0+1:N) G(1:k0)];
% Verify X3=G3
% Property 4
x4 = G;
X4 = fft(G);
G4 = N*[g(1) g(8:-1:2)]; % This forms N*(g mod(-k))
% Verify X4 = G4;
% Property 5
% To calculate circular convolution between
% g and h use eqn (3.67)
h1 = [h(1) h(N:-1:2)];
T = toeplitz(h',h1);
x5 = T*g';
X5 = fft(x5');
% Verify X5 = G.*H
% Property 6
x6 = g.*h;
X6 = fft(x6);
H1 = [H(1) H(N:-1:2)];
T = toeplitz(H.',H1); %,' is the nonconjugate transpose
G6 = (1/N)*T*G.';
% Verify G6 = X6.'
105
M3.11 N = 8; % sequence length
gamma = 0.5;
k = 0:N-1;
x = exp(-gamma*k);
X = fft(x);
% Property 1
X1 = fft(conj(x));
G1 = conj([X(1) X(N:-1:2)]);
% Verify X1 = G1
% Property 2
x2 = conj([x(1) x(N:-1:2)]);
X2 = fft(x2);
% Verify X2 = conj(X)
% Property 3
x3 = real(x);
X3 = fft(x3);
G3 = 0.5*(X+conj([X(1) X(N:-1:2)]));
% Verify X3 = G3
% Property 4
x4 = j*imag(x);
X4 = fft(x4);
G4 = 0.5*(X-conj([X(1) X(N:-1:2)]));
% Verify X4 = G4
% Property 5
x5 = 0.5*(x+conj([x(1) x(N:-1:2)]));
X5 = fft(x5);
% Verify X5 = real(X)
% Property 6
x6 = 0.5*(x-conj([x(1) x(N:-1:2)]));
X6 = fft(x6);
% Verify X6 = j*imag(X)
M3.12 N = 8;
k = 0:N-1;
gamma = 0.5;
x = exp(-gamma*k);
X = fft(x);
% Property 1
xpe = 0.5*(x+[x(1) x(N:-1:2)]);
xpo = 0.5*(x-[x(1) x(N:-1:2)]);
Xpe = fft(xpe);
Xpo = fft(xpo);
% Verify Xpe = real(X) and Xpo = j*imag(X)
% Property 2
X2 = [X(1) X(N:-1:2)];
% Verify X = conj(X2);
% real(X) = real(X2) and imag(X) = -imag(X2)
% abs(X)= abs(X2) and angle(X) = -angle(X2)
106
M3.13 Using the M-file fft we obtain X[ ],013= Xj[],,,1 12 8301 5 634=+ Xj[],2 4 3 4641=? +,
Xj[]387=? +,X j[],4 13 1 7321=+,X j[],,5 4 1699 7 366=+,X[6]= –13,X j[],,7 4 1699 7 366=?,
[],8 13 1 7321=?,X j[]987=,X j[],10 4 3 4641=,X j[],,11 12 8301 5 634=?.
(a) X[0] = 13,(b) X[6] = – 13,(c) Xk
k
[],=
=
∑
36
0
11
(d) eXk
jk
k
π
=
∑
=?
(/)
[],
46
0
11
48
(e) Xk
k
[],
12
0
11
1500=
=
∑
M3.14 Using the M-file ifft we obtain x x x x[].,[].,[].,[].,0 2 2857 1 0 2643 2 0 7713 3 0 4754==?==?
xx[].,[].,4 1 1362 6 1 6962=? = xx[],,[],,[],,[],,6 3 5057 7 0 8571 8 2 0763 9 0 6256==?=?
[],,[],,10 1 9748 11 1 0625== xx[],,[],,12 1 5224 13 0 4637
(a) x[0] = 2.2857,(b) x[7] = – 0.8571,(c) xn
n
[],=
=
∑
12
0
13
(d) exn j
jn
n
(/)
[],
47
0
13
22
π
=
=
∑
(e) xn
n
[],,
2
0
13
35 5714=
=
∑
M3.15
function y = overlapsave(x,h)
X = length(x); %Length of longer sequence
M = length(h); %length of shorter sequence
flops(0);
if (M > X) %Error condition
disp('error');
end
%clear all
temp = ceil(log2(M)); %Find length of circular convolution
N = 2^temp; %zero padding the shorter sequence
if(N > M)
for i = M+1:N
h(i) = 0;
end
end
m = ceil((-N/(N-M+1)));
while (m*(N-M+1) <= X)
if(((N+m*(N-M+1)) <= X)&((m*(N-M+1)) > 0))
for n = 1:N
x1(n) = x(n+m*(N-M+1));
end
end
if(((m*(N-M+1))<=0)&((N+m*(N-M+1))>=0)) %underflow adjustment
for n = 1:N
x1(n) = 0;
end
for n = m*(N-M+1):N+m*(N-M+1)
if(n > 0)
x1(n-m*(N-M+1)) = x(n);
end
end
end
107
if((N+m*(N-M+1)) > X) %overflow adjustment
for n = 1:N
x1(n) = 0;
end
for n = 1:(X-m*(N-M+1))
x1(n) =x (m*(N-M+1)+n);
end
end
w1 = circonv(h,x1); %circular convolution using DFT
for i = 1:M-1
y1(i) = 0;
end
for i = M:N
y1(i) = w1(i);
end
for j = M:N
if((j+m*(N-M+1)) < (X+M))
if((j+m*(N-M+1)) > 0)
yO(j+m*(N-M+1)) = y1(j);
end
end
end
m = m+1;
end
disp('Number of Floating Point Operations')
flops
%disp('Convolution using Overlap Save:');
y = real(yO);
function y = circonv(x1,x2)
L1 = length(x1); L2 = length(x2);
if L1 ~= L2,
error('Sequences of unequal lengths'),
end
X1 = fft(x1);
X2 = fft(x2);
X_RES = X1.*X2;
y = ifft(X_RES);
The MATLAB program for performing convolution using the overlap-save method is
h = [1 1 1]/3;
R = 50;
d = rand(1,R) - 0.5;
m = 0:1:R-1;
s = 2*m.*(0.9.^m);
x = s + d;
%x = [x x x x x x x];
y = overlapsave(x,h);
k = 0:R-1;
plot(k,x,'r-',k,y(1:R),'b--');
xlabel('Time index n');ylabel('Amplitude');
legend('r-','s[n]','b--','y[n]');
108
0 10 20 30 40 50
0
2
4
6
8
Time index n
Amplitude
s[n]
y[n]
M3.16 (a) Using the M-file roots we determine the zeros and the poles of G z
1
( ) which are
given by z z j z j z
12 3 4
3 5616 0 4500 0 7730 0 4500 0 7730 0 5616=? =? + = =.,.,,.,,.,and
pjp p pj
12
1 0000 1 7321 1 0000 1 7321 0 6000 0 3742 0 6000 0 3742=?+ = =+ =?..,..,..,...
Next using the M-file conv we determine the quadratic factors corresponding to the complex
conjugate zeros and poles resulting in the following factored form of G z
1
(),
Gz
zzzz
zz z z
1
1112
12 1 2
4
3
1 3 5616 1 0 5616 1 0 9 0 8
1 2 4 1 12 05
()
(,)(,)(., )
()(.)
=?
+?++
++? +

.
(b) Using the M-file roots we determine the zeros and the poles of G z
2
( ) which are
given by z z z z
123 4
2 1 05 03=? =? =? =,,.,.,and
pjp jpp
1234
1 0482 1 7021 1 0482 1 7021 0 6094 0 3942=? + = =? =?..,..,.,..
Next using the M-file conv we determine the quadratic factors corresponding to the complex
conjugate zeros and poles resulting in the following factored form of G z
2
():
Gz
zz z z
z
2
11 1 1
2
2
5
12 1 105 103
1 0 6094 1 0 3942 1 2 0963 3 9957
()
()()(.)(.)
(,)(,)(., )
=?
+++?
++ +

M3.17 Using Program 3_9 we arrive at we get
Residues
3.0000 2.0000
Poles
-1.0000 -0.3333
Constant
[]
Hence,Yz
z
z
1
1
1
3
1
3
1
2
1
(),=
+
+
which is same as that determined in Problem 3.98.
M3.18 (a) A partial-fraction expansion of Xz
zz
zz
zzz
zz z
a
()
()()
.=
+
+?
=
+
+

43 3
23
433
14 3 18
12
2
345
12 3
using Program 3_9 we arrive at
Xz z z
zz z
a
(),,,
..
()
.
.=?+?
+
+

0 2361 0 1389 0 1667
0 1852
13
0 0741
13
0 125
12
21
1121
Since the ROC is
109
given by z > 3,the inverse z-transform is a right-sided sequence and is thus given by
xn n n n
a
[], [ ], [ ], []=+0 2361 2 0 1389 1 0 1667δδδ
+?0 1852 3 0 0741 3 0 125 2,()[], ()[], ()[].
nn
n nμμμ
A more compact solution is obtained by writing X z
a
( ) as Xz z
zz
zz z
a
(),=
+
+

3
12
12 3
43 3
14 3 18
and making a partial-fraction expansion of the function inside the brackets,
Xz z
zz z
z
zz
z
z
z
a
()
() ()
=
+
+
+
=
+
+
+

3
1121
3
11
2
1
12
1
13
2
13
1
12
1
13
1
12
2
13
The inverse z-trasform of the above function is given by
xn n n n n
a
nn
[]()[]()[]()()[]=?++

3323
2
3
23 2
2
μμ μ
=?+?
[]
() ( ) ( ) [ ].3232 3nnμ
(b) Here the ROC is z < 2,hence,the inverse z-transform of
Xz z
zz
z
z
z
b
()
()
=
+
+
+

3
11
2
1
12
1
13
1
12
2
13
is thus given by
xn n n n n
b
nn n
[] () [ ] ( ) [ ] ( )() [ ]=+++
3222
2
3
23 1
33 2
μμ μ
=+

() ( ) ( ) [ ].3232 2μ
(c) In this case,the ROC is given by 23<<z, Hence,the inverse z-transform of
Xz z
zz
z
z
z
c
()
()
=
+
+
+

3
11
2
1
12
1
13
1
12
2
13
is thus given by
xn n n n n
c
nn n
[] () [ ] ( ) [ ] ( )() [ ]= + + +3223
2
3
23 1
33 2
μμ μ
= + +

() ( )[ ] ( ) [ ].323 22 3nn n
M3.19 (a) Using the statement
[num,den] = residuez([10/4 -8/2],[-1/4 -1/2],0); we get
num = –3.5 –1.25 –0.25 and den = 1 0.75 0.125.
Hence the transfer function is given by Xz
zz
1
12
35 125 025
1 0 75 0 125
()
..,
..
.=?
++
++

(b) Using the statement [r,p,k]=residuez([–3 –1],[1 0 – 0.25]); we
first obtain a partial-fraction expansion of?
+
3
1025
1
2
z
z.
which yields
r = – 2.5 0.5,p = 0.5 –0.5 and k = 0.
Therefore we can write
Xz
zzz zz
2
111 11
35
2
105
25
105
05
105
35
45
105
05
105
(),
.
.
.
.
.
.
.
.
.
.
=?
+
+
=?
+
+

,Next we use the
statement [num,den] = residuez([-4.5 0.5],[0.5 -0.5],3.5); which
110
yields num = –0.5 –2.5 –0.075 and den = 1 0 0.25,Therefore,
Xz
zz
z
2
12
2
05 25 0875
1025
()
..,
.
.=?
++

(c) We first make a partial-fraction expansion of
3
1081
2
+
.z
using the statement [r,p,k] =
residuez(3,[1 0 0.81],0); which yields
3
1081
15
109
15
109
211
+
=
+
+

.
.
.
.
.
.
zjzjz
Hence,we can write Xz
jz jz
z
z
3
11
3
1
3
1
2
15
109
15
109
43
1
59
1
2
2
()
.
.
.
.
//
.=
+
+
+
+
+

Using Program
3_10 we then obtain Xz
zzz
zzz
3
12
1234
2 2222 3 1111 0 70333 0 72
1 1 3333 1 2544 1 08 0 36
()
..,,
.
=
++?
++++

.
(d) We first make a partial-fraction expansion of
1
6
1
5
6
1
1
6
2
1
z
zz

++
using the statement
[r,p,k] = residuez([0 1 0]/6,[1 5/6 1/6],0); which yields
–
1
1
1
1
1
1
2
1
3
1
+
+
+

zz
,Hence,we can write Xz
zzz
4
5
1
2
1
3
1
4
10 5
1
1
1
1
1
21
1
()
/
–,=+
++
+
+

Using
Program 3_10 we then arrive at Xz
zz z
zz z
4
12 3
12 3
6 6 7667 2 4 0 2667
1 1 2333 0 5 0 0667
()
...
=
+++

.
M3.20 (a) From Part (a) of M3.19 we observe Xz
zz
1
4
1
2
1
2
25
1
4
1
1
1
()
.
–,=? +
++

Hence its inverse
z-transform is given by x n n n n
nn
1
2 2 5 0 25 4 0 5[] [],(, ) []– (.)[].=? +δμμ Evaluating for values of
n = 01,,K we get x
1
035[],=?,x
1
1 1 375[],,= x
1
2 0 8438[],,=? x
1
3 0 4609[],,= x
1
4 0 2402[],,=?
x
1
5 0 1226[],,= x
1
6 0 0619[],,=? x
1
7 0 0311[],,= x
1
8 0 0156[],,=? x
1
9 0 0078[],=
Using Program 3_11 we get
Coefficients of the power series expansion
Columns 1 through 6
-3.5000 1.3750 -0.8438 0.4609 -0.2402 0.1226
Columns 7 through 10
-0.0619 0.0311 -0.0156 0.0078
(b) From Part (b) of M3.19 we observe Xz
zz
2 11
35
45
105
05
105
(),
.
.
.
.
.=?
+
+

Hence,its inverse
z-transform is given by x n n n n
nn
2
35 4505 05 05[],[],(.)[],(.)[].=? +?δμ μ Evaluating for values
of n = 01,,K we get x
2
005[],,=? x
2
125[],,=? x
2
201[],,=? x
2
3 0 625[],,=? x
2
4025[],,=?
x
2
5 0 1562[],,=? x
2
6 0 0625[],,=? x
2
7 0 0391[],,=? x
2
8 0 0156[],,=? x
2
9 0 0098[],,=?
Using Program 3_11 we get
111
Coefficients of the power series expansion
Columns 1 through 6
-0.5000 -2.5000 -0.1000 -0.6250 -0.2500 -0.1562
Columns 7 through 10
-0.0625 -0.0391 -0.0156 -0.0098
(c) From Part (c) of M3.19 we observe Xz
z
z
z
3
2
3
1
3
1
2
3
1081
43
1
59
1
2
2
()
.
//
.=
+
+
+
+
Hence,
its inverse z-transform is given by
xn n n n n n
n
nn
3
1
309 2 1
4
3
2
3
5
9
3
2
1
2
3
1[ ] (, ) cos( / ) [ ] [ ] ( ) [ ]=π+
+?
+?
+
+
μμ μ
=π
++?
309 2
4
3
2
3
5
9
1
2
3
(, ) cos( / ) [ ] [ ] ( ) [ ].
n
nn
nn nn nμμ μ Evaluating for values of n = 0,
1,.,,,we get x
3
0 2 2222[],,= x
3
1 0 14815[],,=
x
3
2 2 2819[],=–,x
3
3 0 26337[],,=?
x
3
4 2 2536[],=, x
3
5 0 26337[],= –, x
3
6137[],=?, x
3
7 0 18209[],,=? x
3
8 1 4345[],=,
x
3
9 0 10983[],=?, Using Program 3_11 we get
Coefficients of the power series expansion
Columns 1 through 6
2.2222 0.14815 -2.2819 -0.26337 2.2536 -0.26337
Columns 7 through 10
-1.37 -0.18209 1.4345 -0.10983
(d) From Part (d) of M3.19 we observe Xz
zzz
4
5
1
2
1
3
1
4
10 5
1
1
1
1
1
21
1
()
/
–,=+
++
+
+

Hence,its
inverse z-transform is given by x n n n n n
nn
4
4225 12 13[] [] ( / ) [] ( / ) [] ( / ) [].=+ +?δμμμ
Evaluating for values of
n = 01,,K we get x
4
06[],= x
4
1 0 6333[],,=? x
4
2 0 1811[],,=
x
4
3004[],,=? x
4
4 0 0010[],,= x
4
5 0 0067[],,= x
4
6 0 0061[],,=? x
4
7 0 0041[],,= x
4
8 0 0024[],,=?
x
4
9 0 0014[],=, Using Program 3_11 we get
Coefficients of the power series expansion
Columns 1 through 6
6.0000 -0.6331 0.1808 -0.0399 0.0011 0.0066
Columns 8 through 10
-0.0060 0.0040 -0.0024 0.0014
M3.21 % As an example try a sequence x = 0:24;
% calculate the actual uniform dft
% and then use these uniform samples
% with this ndft program to get the
% the original sequence back
% [X,w] = freqz(x,1,25,'whole');
% use freq = X and points = exp(j*w)
freq = input('The sample values = ');
points = input('Frequencies at which samples are taken = ');
L = 1;
112
len = length(points);
val = zeros(size(1,len));
L = poly(points);
for k = 1:len
if(freq(k) ~= 0)
xx = [1 -points(k)];
[yy,rr] = deconv(L,xx);
F(k,:) = yy;
down = polyval(yy,points(k))*(points(k)^(-len+1));
F(k,:) = freq(k)/down*yy;
val = val+F(k,:);
end
end
coeff = val;