1
SOLUTIONS MANUAL
to accompany
Digital Signal Processing:
A Computer-Based Approach
Second Edition
Sanjit K,Mitra
Prepared by
Rajeev Gandhi,Serkan Hatipoglu,Zhihai He,Luca Lucchese,
Michael Moore,and Mylene Queiroz de Farias
2
Chapter 2 (2e)
2.1 (a) un xn yn[] [] [] { }=+=?351 28140
(b) vn xn wn[] [] [] { }=? =15 8 0 6 20 0 2
(c) sn yn wn[] [] [] { }=? =53 2 999 3
(d) rn yn[], [] {,,,, }==45 0 315 45 135 18 405 9
2.2 (a) From the figure shown below we obtain
x[n]
y[n]
v[n] v[n–1]
z
–1
α
β
γ
vn xn vn[] [] [ ]=+?α 1 and y n v n v n v n[] [] []( )[]=?+?=+?βγ βγ1 1 1, Hence,
vn xn vn[][] [ ]?=?+?11 2α and y n v n[]( )[ ]?=+?12βγ, Therefore,
yn vn xn vn[]()[]()[]()[]=+?=+?+ +?βγ βγ αβγ11 2 =+?+ +
+
()[]()
[]
()
βγ αβγ
βγ
xn
yn
1
1
=+?+?( )[] []βγ αxn yn11.
(b) From the figure shown below we obtain
x[n]
y[n]
z
–1
αβ
γ
z
–1
z
–1
z
–1
x[n–1] x[n–2]
x[n–3]
x[n–4]
yn xn xn xn xn xn[] [ ] [ ] [ ] [] [ ]=?+?+?( )++?( )γβ α213 4.
(c) From the figure shown below we obtain
v[n]
x[n] y[n]
z
–1
z
–1
v[n–1]
–1
d
1
3
vn xn dvn[] [] [ ]=
1
1 and y n d v n v n[] [] [ ]=+?
1
1, Therefore we can rewrite the second
equation as y n d x n d v n v n d x n d v n[] [] [ ] [ ] [] [ ]=
( )
+?= +?
( )
11 1 1
2
1 1 1 1 (1)
=+?
( )

( )
dxn d xn dvn
11
2
1
2[] [ ] [ ] =+?
( )

( )
dxn d xn d d vn
11
2
11
2
1112[] [ ] [ ]
From Eq,(1),y n d x n d v n[] [] [ ]?=?+?
( )
11 2
11
2
,or equivalently,
dyn d xn d d vn
11
2
11
2
1112[] [] [ ]?=?+?
( )
, Therefore,
yn dyn dxn d xn d d vn d xn d d vn[] [] [] [] [ ] [] [ ]+?= +?
( )

( )
+?+?
( )
11 1
2
11
2
1
2
11
2
112112
=+?dxn xn
1
1[ ] [ ],or yn dxn xn dyn[] [] [ ] [ ]=+11.
(d) From the figure shown below we obtain
v[n]
x[n]
y[n]
v[n–1]
–1
d
1
z
–1
z
–1
v[n–2]
d
2
u[n]
w[n]
vn xn wn[] [] [],=? wn dvn d un[] [ ] [],=?+
12
1 and u n v n x n[] [ ] [].=?+2 From these equations
we get w n d x n d x n d x n d w n d w n[] [] [ ] [ ] [ ] [ ]= +?+
21 2 1 2
1 2 1 2, From the figure we also
obtain y n v n w n x n w n w n[] [ ] [] [ ] [] [ ],=?+ =?+2 2 2 which yields
dyn dxn dwn dwn
111 1
1313[] [] [] [],?=?+ and
dyn dxn dwn dwn
222 2
2424[] [] [] [],?=?+ Therefore,
yn dyn d yn xn dxn d xn[] [] [][] [] []+?+?=?+?+?
12 1 2
12 34
++?+?
( )
+?+?
( )
wn dwn d wn wn dwn d wn[] [] [] [] [] []
12 1 2
12234
=?+ +?xn d xn dxn[] [] []21
21
or equivalently,
yn d xn dxn xn dyn d yn[] [] [ ] [ ] [ ] [ ].= +?+
21 1 2
12 1 2
2.3 (a) xn[] { },=?3 2 0 1 4 5 2 Hence,x n n[]{ },.?=≤≤25410 23 3 3
Thus,xn xnxn n
ev
[] ([] [ ]) {/ / / /},,=+?=?≤≤
1
2
5 2 3 2 2 1 2 3 2 5 2 3 3 and
xn xnxn n
od
[] ([] [ ]) {/ / / /},.==≤≤
1
2
12 72 2 0 2 72 12 3 3
(b) y n[] { }=0 7 1 3 4 9 2, Hence,y n n[]{ },.?=≤≤294 3170 3 3
Thus,yn ynyn n
ev
[] ([] [ ]) { / / },,=+?=≤≤
1
2
1 8 5 2 3 5 2 8 1 3 3 and
yn ynyn n
od
[ ] ( [ ] [ ]) { / / },==≤≤
1
2
113201321 3 3.
4
(c) wn[] { },=5 4 3 6 5 0 1 Hence,w n n[]{ },.?=≤≤10 5634 5 3 3
Thus,wn wnwn n
ev
[] ( [] [ ]) { },,=+?=≤≤
1
2
2 2 1 6 1 2 2 3 3 and
wn wnwn n
od
[] ( [] [ ]) { },,==≤≤
1
2
3240 4 23 3 3
2.4 (a) xn gngn[] [][]=, Hence xn gngn[] [][]?=,Since g[n] is even,hence g[–n] = g[n],
Therefore x[–n] = g[–n]g[–n] = g[n]g[n] = x[n],Hence x[n] is even.
(b) u[n] = g[n]h[n],Hence,u[–n] = g[–n]h[–n] = g[n](–h[n]) = –g[n]h[n] = –u[n],Hence
u[n] is odd.
(c) v[n] = h[n]h[n],Hence,v[–n] = h[n]h[n] = (–h[n])(–h[n]) = h[n]h[n] = v[n],Hence
v[n] is even.
2.5 Yes,a linear combination of any set of a periodic sequences is also a periodic sequence and
the period of the new sequence is given by the least common multiple (lcm) of all periods,
For our example,the period = lcm (,,)NN N
123
,For example,if N
1
3=,N
2
6=,and N
3
12=,
then N = lcm(3,5,12) = 60.
2.6 (a) xn xnx n A A
pcs
nn
[ ] { [ ] *[ ]} { *( *) },=+?= +
1
2
1
2
αα and
xn xnx n A A
pca
nn
[ ] { [ ] *[ ]} { *( *) },==?
1
2
1
2
αα?≤≤NnN.
(b) hn j j j j j[] { }=?+? + +?+2543563 72?≤ ≤22n
,and hence,
hn j j j j j*[ ] { }?= +723 5643 25,?≤ ≤22n, Therefore,
hn hnh n j j j j
pcs
[] {[] *[ ]} {,,,,,,}= +?=?+? +
1
2
4 5 1 5 3 5 2 5 3 5 2 4 5 1 5 and
hn hnh n j jj j j
pca
[]{[]*[]}{...,,.}== ++
1
2
25 35 05 6 05 25 35?≤ ≤22n.
2.7 (a) xn A
n
[],{}=
{}
<αα α where A and are complex numbers,with 1
Since for n < 0,α
n
can become arbitrarily large hence {x[n]} is not a bounded sequence.
(b) yn A
n
[],{}=<αμ α α[n] where A and are complex numbers,with 1
In this case yn A[]≤? n hence {y[n]} is a bounded sequence.
(c) hn C
n
[],{}=>βμ β β[n] where C and are complex numbers,with 1
Since β
n
becomes arbitrarily large as n increases hence {h[n]} is not a bounded sequence.
5
(d) { [ ]} sin( ).gn n
a
= 4 ω Since?≤ ≤44gn[ ] for all values of n,{g[n]} is a bounded
sequence.
(e) { [ ]} cos ( ).vn n
b
= 3
22
ω Since?≤ ≤33vn[ ] for all values of n,{v[n]} is a bounded
sequence.
2.8 (a) Recall,xn xnxn
ev
[] [] [ ].=+?( )
1
2
Since x[n] is a causal sequence,thus x[–n] = 0?>n0,Hence,
xn x n x n
ev ev
[] [] [ ]=+? = 2x n
ev
[],?>n0,For n = 0,x[0] = x
ev
[0].
Thus x[n] can be completely recovered from its even part.
Likewise,xn xnxn
xn n
n
od
[] []– [ ]
[],,
,.
=?( ) =
>
=
1
2
1
2
0
00
Thus x[n] can be recovered from its odd part?n except n = 0.
(b) 2y n y n y n
ca
[] [] *[ ]=,Since y[n] is a causal sequence yn y n
ca
[] []=?2 n > 0.
For n = 0,Im{ [ ]} [ ]yy
ca
00=, Hence real part of y[0] cannot be fully recovered from yn
ca
[],
Therefore y[n] cannot be fully recovered from yn
ca
[].
2y n y n y n
cs
[] [] *[ ]=+?,Hence,yn y n
cs
[] []=?2 n > 0,
For n = 0,Re{ [ ]} [ ]yy
cs
00=, Hence imaginary part of y[0] cannot be recovered from yn
cs
[],
Therefore y[n] cannot be fully recovered from yn
cs
[].
2.9 xn xnxn
ev
[] [] [ ].=+?( )
1
2
This implies,xn xnxnxn
ev ev
[–] [–] [] [].=+( ) =
1
2
Hence even part of a real sequence is even.
xn xnxn
od
[] []– [–].= ( )
1
2
This implies,xn xnxn xn
od od
[– ] [– ] – [ ] – [ ].= ( ) =
1
2
Hence the odd part of a real sequence is odd.
2.10 RHS of Eq,(2.176a) is xnxnN xnx n xnNxNn
cs cs
[] [ ] [] *[ ] [ ] *[ ].+?= +?( )+?+?( )
1
2
1
2
Since x[n] = 0?<n0,Hence
xnxnN xnxNn x n
cs cs pcs
[] [ ] [] *[ ] [],+?= +?( ) ≤≤
1
2
= 0 n N –1.
RHS of Eq,(2.176b) is
xnxnN xnx n xnNxnN
ca ca
[] [ ] [] *[ ] [ ] *[ ]+?=( )+( )
1
2
1
2
=( ) =≤≤
1
2
xn x N n x n
pca
[ ] *[ ] [ ],0 n N –1.
6
2.11 xn xnx n
pcs N
[] [] *[ ]=+<?>
( )
1
2
for 0 n N –1≤≤,Since,xn xNn
N
[][]<? > =?,it follows
that x,1 n N –1.
pcs
[] [] *[ ]nxnxNn=+?( ) ≤≤
1
2
For n = 0,x = Re{x[0]}.
pcs
[] [] *[]000
1
2
=+( )xx
Similarly xn xnx n
pca N
[] [] *[ ]=?<?>
( )
1
2
=( ) ≤≤
1
2
xn x N n[ ] *[ ],1 n N –1,Hence,
for n = 0,x = jIm{x[0]}.
pca
[] [] *[]000
1
2
=?( )xx
2.12 (a) Given xn
n
[],
=?∞
∞
∑
<∞ Therefore,by Schwartz inequality,
xn xn xn
nnn
[] [] [],
2
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑
≤
<∞
(b) Consider x n
nn
otherwise
[]
/,,
,.
=
≥
{
11
0
The convergence of an infinite series can be shown
via the integral test,Let a f x
n
= ( ),where f(x) is a continuous,positive and decreasing
function for all x ≥1,Then the series a
n
n=
∞
∑
1
and the integral f x dx()
1
∞
∫
both converge or
both diverge,For a n
n
=1 /,f(x) = 1/n,But
1
0
1
1
x
dx x
∞
∞
∫
= ( ) =∞? =∞ln, Hence,
xn
n
nn
[]
=?∞
∞
=
∞
∑∑
=
1
1
does not converge,and as a result,x[n] is not absolutely
summable,To show {x[n]} is square-summable,we observe here a
n
n
=
1
2
,and thus,
fx
x
(),=
1
2
Now,
1111
1
1
2
1
1
x
dx
x
∞
∞
∫
=?
=?
∞
+=,Hence,
1
21
n
n=
∞
∑
converges,or in other
words,x[n] = 1/n is square-summable.
2.13 See Problem 2.12,Part (b) solution.
2.14 xn
n
n
n
c
2
1[]
cos
,.=
π
≤≤∞
ω
Now,
cos
.
ω
c
nn
n
n nπ
≤
π
=
∞
=
∞
∑∑
2
1
22
1
1
Since,
1
6
2
1
2
n
n=
∞
∑
=
π
,
cos
.
ω
c
n
n
nπ
≤
=
∞
∑
2
1
1
6
Therefore x n
2
[ ] is square-summable.
Using integral test we now show that x n
2
[ ] is not absolutely summable,
cos
cos
cos
cosint( )
ω
ω
ω
ω
c
c
c
c
x
x
dx
x
x
x
xx
π
=
π

∞
∞
∫
1
1
1
where cosint is the cosine integral function.
Since
cosω
c
x
x
dx
π
∞
∫
1
diverges,
cosω
c
n
n
nπ
=
∞
∑
1
also diverges.
7
2.15 xn x n x n
n
ev od
n
2
2
[] [] []
=?∞
∞
=?∞
∞
∑∑
=+
()
=++ =+
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
=?∞
∞
∑∑ ∑ ∑∑
xn xn xnxn xn xn
ev
n
od
n
ev od
n
ev
n
od
n
22 22
2[] [] [] [] [] []
as xnxn
ev
n
od
=∞
∞
∑
=
–
[] [] 0 since x n x n
ev od
[ ] [ ] is an odd sequence.
2.16 x n kn N[ ] cos( / ),=≤≤2π 0 n N –1,Hence,
EknN
x
n
N
=
=
∑
cos ( / )
2
0
1
2π =+( )
=
∑
1
2
0
1
14cos( / )πkn N
n
N
=+
=
∑
N
n
N
kn N
2
1
2
0
1
4cos( / ).π
Let C kn N
n
N
=
=
∑
cos( / ),4
0
1
π and SknN
n
N
=
=
∑
sin( / ).4
0
1
π
Therefore CjS e
jknN
n
N
+=
=
∑
4
0
1
π /
=
=
e
e
jk
jkN
4
4
1
1
0
π
π /
,Thus,C C jS=+{}=Re 0,
As C C jS=+{}=Re 0,it follows that E
x
N
=
2
.
2.17 (a) xn n
1
[] [].=μ Energy = μ
22
1[],n
nn=?∞
∞
=?∞
∞
∑∑
==∞
Average power = lim ( [ ]) lim lim,
K
nK
K
K
n
K
KK
n
K
K
K→∞
=?
→∞
=
→∞+
=
+
=
+
=
1
21
1
21
1
21
1
2
0
μ
(b) xn nn
2
[] [].=μ Energy = n n n
nn
μ[],( ) ==∞
=?∞
∞
=?∞
∞
∑∑
2
2
Aveerage power = lim ( [ ]) lim,
K
nK
K
K
n
K
K
nn
K
n
→∞
=?
→∞
=
+
=
+
=∞
1
21
1
21
22
0
μ
(c) xn Ae
o
jn
o
3
[],=
ω
Energy = Ae A
o
jn
n
o
n
o
ω
=?∞
∞
=?∞
∞
∫
∑
==∞
2
2
,Average power =
lim lim lim,
K
o
jn
nK
K
K
o
nK
K
K
o
o
K
Ae
K
A
KA
K
A
o
→∞
=?
→∞
=?
→∞+
=
+
=
+
=
∑∑
1
21
1
21
2
21
22
2
2ω
(d) xn A
n
M
Ae Ae
o
jn jn
oo
[ ] sin,=
π
+
=+
2
1
φ
ωω
where ω
o
M
=
π2
,A
A
e
o
j
=?
2
φ
and
A
A
e
j
1
2
=
φ
,From Part (c),energy = ∞ and average power = AA AA A
oo
2
1
22
1
22
4
3
4
++ =,
2.18 Now,μ[]
,,
,.
n
n
n
=
≥
<
10
00
Hence,μ[]
,,
,.
=
<
≥
n
n
n
1
10
00
Thus,x[n] = μμ[] [ ].nn+1
2.19 (a) Consider a sequence defined by x n k
k
n
[] [].=
=?∞
∑
δ
8
If n < 0 then k = 0 is not included in the sum and hence x[n] = 0,whereas for n ≥ 0,k = 0 is
included in the sum hence x[n] = 1?≥ n0,Thus x n k
n
n
n
k
n
[] []
,,
,,
[].==
≥
<
=
=?∞
∑
δμ
10
00
(b) Since μ[]
,,
,,
n
n
n
=
≥
<
10
00
it follows that μ[–]
,,
,.
n
n
n
1
11
00
=
≥
≤
Hence,μμ δ[n] – [ ]
,,
[].n
n
n
n?=
=
≠
{
=1
10
00
2.20 Now xn A n[ ] sin=+
( )
ωφ
0
.
(a) Given xn[]=
{}
0 2 2 2 0 2 2 2, The fundamental period is N = 4,
hence ω
o
=π =π28 4/ /, Next from x A[ ] sin( )00==φ we get φ = 0,and solving
xA A[ ] sin( ) sin( / )1
4
42=
π
+= π =?φ we get A = –2.
(b) Given xn[]=
{}
2222,The fundamental period is N = 4,hence
ω
0
= 2π/4 = π/2,Next from xA[ ] sin( )02==φ and
xA A[ ] sin( / ) cos( )12 2=+==πφ φ it
can be seen that A = 2 and = /4 is one solution satisfying the above equations,φπ
(c) xn[]=?{}33,Here the fundamental period is N = 2,hence ωπ
0
=, Next from x[0]
= Asin( )φ=3 and x A A[ ] sin( ) sin( )13= + =? =?φπ φ observe that A = 3 and = /2φπ that A = 3
and φ = π/2 is one solution satisfying these two equations.
(d) Given xn[],,=?{}0150 15,it follows that the fundamental period of x[n] is N =
4,Hence ωπ π
0
24 2==/ /, Solving x A[ ] sin( )00==φ we get φ = 0,and solving
xA[ ] sin( / ),1215==π,we get A = 1.5.
2.21 (a)? [],
.
xn e
jn
1
04
=
π
Here,ω
o
=π04., From Eq,(2.44a),we thus get
N
rr
r
o
=
π
=
π
π
==
22
04
55
ω,
for r =1.
(b)? [ ] sin(,, ).xn n
2
06 06=π+π Here,ω
o
=π06., From Eq,(2.44a),we thus get
N
rr
r
o
=
π
=
π
π
==
22
06
10
3
10
ω,
for r = 3.
(c)? [ ] cos(,, ) sin(, ).xn n n
3
211 05 207=π?π+π Here,ω
1
11=π,and ω
2
07=π., From Eq,
(2.44a),we thus get N
rr
r
1
1
1
1
1
22
11
20
11
=
π
=
π
π
=
ω,
and N
rr
r
2
2
2
2
2
22
07
20
7
=
π
=
π
π
=
ω,
,To be periodic
9
we must have N N
12
=, This implies,
20
11
20
7
12
rr=, This equality holds for r
1
11= and r
2
7=,
and hence N = N N
12
20==.
(d) N
r
r
1
1
1
2
13
20
13
=
π
π
=
.
and N
r
r
2
2
2
2
03
20
3
=
π
π
=
.
,It follows from the results of Part (c),N = 20
with r
1
13= and r
2
3=,
(e) N
r
r
1
1
1
2
12
5
3
=
π
π
=
.
,N
r
r
2
2
2
2
08
5
2
=
π
π
=
.
and N N
32
=, Therefore,N N N N====
122
5 for
r
1
3= and r
2
2=,
(f)? []xn
6
=n modulo 6,Since? []xn
6
6+= (n+6) modulo 6 = n modulo 6 =? []xn
6
.
Hence N = 6 is the fundamental period of the sequence? []xn
6
.
2.22 (a) ω
o
=π014.,Therefore,N
rr
r
o
=
π
=
π
π
==
22
014
100
7
100
ω,
for r = 7.
(b) ω
o
=π024., Therefore,N
rr
r
o
=
π
=
π
π
==
22
024
25
3
25
ω,
for r = 3.
(c) ω
o
=π034.,Therefore,N
rr
r
o
=
π
=
π
π
==
22
034
100
17
100
ω,
for r = 17.
(d) ω
o
=π075., Therefore,N
rr
r
o
=
π
=
π
π
==
22
075
8
3
8
ω,
for r = 3.
2.23 xn x nT nT
ao
[ ] ( ) cos( )==? is a periodic sequence for all values of T satisfying?
o
TN r?=π2
for r and N taking integer values,Since,?
o
TrN=π2 / and r/N is a rational number,?
o
T
must also be rational number,For?
o
=18 and T = π/6,we get N = 2r/3,Hence,the smallest
value of N = 3 for r = 3.
2.24 (a)
xn n n n n n n[] [ ] [ ] [] [ ] [ ] [ ]=+?+++?+?+?3322 415223δδδδδδ
(b) yn n n n n n n[] [ ] [ ] [] [ ] [ ] [ ]=+++?+?+72 13419223δδ δδ δ δ
(c) wnn nnnnn[] [][][][][][]=? ++ ++ +++?5242316 51 3δ δδδδδ
2.25 (a) For an input x n
i
[ ],i = 1,2,the output is
yn xn xn xn xn
ii i i
[] [] [ ] [ ] [ ],=+?+?+?αα α α
12 3 4
1 2 4 for i = 1,2,Then,for an input
x n Ax n Bx n[] [] [],=+
12
the output is y n A x n Bx n A x n Bx n[] [] [] [ ] [ ]=+( ) +?+?( )αα
11 2 21 2
11
+?+?( ) +?+?( )
31 2 41 2
22 33Ax n Bx n Ax n Bx n[] [] [] []
=+ +?+?
( )
Axn xn xn xnαα α α
11 21 31 41
12 4[] [ ] [ ] [ ]
++?+?+?
( )
Bxn xn xn xnαα α α
12 22 32 42
12 4[] [ ] [ ] [ ] =+Ay n By n
12
[] [].
Hence,the system is linear.
10
(b) For an input x n
i
[ ],i = 1,2,the output is
yn bxn bxn bxn ayn ayn
ii i i i
[] [] [] [ ] [] [ ],= +?+?+?+?
01 2 1 2
1 2 1 2 i = 1,2,Then,for an
input x n A x n Bx n[] [] [],=+
12
the output is
yn Abxn bxn bxn ayn ayn[] [] [] [ ] [] [ ]= +?+?+?+?
( )
01 11 21 11 21
1212
+ +?+?+?+?
( )
Bbxn bxn bxn ayn ayn
02 12 22 12 22
[] [] [ ] [] [ ] =+Ay n By n
12
[] [],
Hence,the system is linear.
(c) For an input x n
i
[ ],i = 1,2,the output is
yn
xnL n L L
otherwise
i
i
[],
[ / ],,,,
,,
=
=± ±
02
0
K
Consider the input x n A x n Bx n[] [] [],=+
12
Then the output y[n] for
nLL=± ±02,,,K is
given by y[n] = x n L A x n L Bx n L A y n By n[/] [/] [/] [] []=+
12
,For all other values
of n,y n A B[],=?+?=0 0 0 Hence,the system is linear.
(d) For an input x n
i
[ ],i = 1,2,the output is y n x n M
ii
[] [ / ]=, Consider the input
xn Ax n Bx n[] [] [],=+
12
Then the output y n A x n M Bx n M A y n By n[] [ / ] [ / ] [] [].=+=+
12 12
Hence,the system is linear.
(e) For an input x n
i
[ ],i = 1,2,the output is yn
M
xn k
ii
k
M
[] [ ]=?
=
∑
1
0
1
,Then,for an
input x n A x n Bx n[] [] [],=+
12
the output is yn
M
Axnk Bxnk
i
k
M
[] [] []+?( )
=
∑
1
12
0
1
=?
+?
=+
=
=
∑∑
A
M
xn k B
M
xn k Ayn Byn
k
M
k
M11
1
0
1
2
0
1
12
[ ] [ ] [ ] [ ],Hence,the system is
linear.
(f) The first term on the RHS of Eq,(2.58) is the output of an up-sampler,The second
term on the RHS of Eq,(2.58) is simply the output of an unit delay followed by an up-
sampler,whereas,the third term is the output of an unit adavance operator followed by an
up-sampler We have shown in Part (c) that the up-sampler is a linear system,Moreover,
the delay and the advance operators are linear systems,A cascade of two linear systems is
linear and a linear combination of linear systems is also linear,Hence,the factor-of-2
interpolator is a linear system.
(g) Following thearguments given in Part (f),it can be shown that the factor-of-3
interpolator is a linear system.
2.26 (a) y[n] = n
2
x[n].
For an input x
i
[n] the output is y
i
[n] = n
2
x
i
[n],i = 1,2,Thus,for an input x
3
[n] = Ax
1
[n]
+ Bx
2
[n],the output y
3
[n] is given by y n n A x n Bx n A y n By n
3
2
12 12
[] [] [] [] []=+( ) =+.
Hence the system is linear.
Since there is no output before the input hence the system is causal,However,yn[] being
proportional to n,it is possible that a bounded input can result in an unbounded output,Let
x[n] = 1? n,then y[n] = n
2
,Hence yn[]→∞ →∞ as n,hence not BIBO stable.
Let y[n] be the output for an input x[n],and let y
1
[n] be the output for an input x
1
[n],If
xn xn n
10
[] [ ]=? then y n n x n n x n n
1
2
1
2
0
[] [] [ ]==?,However,y n n n n x n n[]()[]?=
00
2
0
.
Since y n y n n
10
[] [ ]≠?,the system is not time-invariant.
11
(b) yn x n[] []=
4
.
For an input x
1
[n] the output is y
i
[n] = x
i
4
[n],i = 1,2,Thus,for an input x
3
[n] = Ax
1
[n] +
Bx
2
[n],the output y
3
[n] is given by y n A x n Bx n A x n A x n
312
44
1
44
2
4
[] ( [] []) [] []=+ ≠+
Hence the system is not linear.
Since there is no output before the input hence the system is causal.
Here,a bounded input produces bounded output hence the system is BIBO stable too.
Let y[n] be the output for an input x[n],and let y
1
[n] be the output for an input x
1
[n],If
xn xn n
10
[] [ ]=? then y n x n x n n y n n
11
44
00
[] [] [ ] [ ].==?=? Hence,the system is time-
invariant.
(c) yn xn[] [ ]=+?
=
∑
β l
l 0
3
.
For an input x
i
[n] the output is yn xn
ii
[] [ ]=+?
=
∑
β l
l 0
3
,i = 1,2,Thus,for an input x
3
[n] =
Ax
1
[n] + Bx
2
[n],the output y
3
[n] is given by
y n Ax n Bx n Ax n Bx n[] [] [] [] []=+? +?( ) =+? +?
===
∑∑∑
ββ
12
0
3
1
0
3
2
0
3
ll l l
ll
≠+Ay n By n
12
[ ] [ ],Since β≠0 hence the system is not linear.
Since there is no output before the input hence the system is causal.
Here,a bounded input produces bounded output hence the system is BIBO stable too.
Also following an analysis similar to that in part (a) it is easy to show that the system is time-
invariant.
(d) yn xn[] [ ]
–
=+?
=
∑
β l
l 3
3
For an input x
i
[n] the output is yn xn
ii
[] [ ]
–
=+?
=
∑
β l
l 3
3
,i = 1,2,Thus,for an input x
3
[n] =
Ax
1
[n] + Bx
2
[n],the output y
3
[n] is given by
y n Ax n Bx n Ax n Bx n[] [] [] [] []=+? +?( ) =+? +?
=? =? =?
∑∑∑
ββ
12
3
3
1
3
3
2
3
3
ll l l
ll
≠+Ay n By n
12
[ ] [ ],Since β≠0 hence the system is not linear.
Since there is output before the input hence the system is non-causal.
Here,a bounded input produces bounded output hence the system is BIBO stable.
12
Let y[n] be the output for an input x[n],and let y
1
[n] be the output for an input x
1
[n],If
xn xn n
10
[] [ ]=? then
yn xn xn n yn n
11
3
3
10
3
3
0
[] [ ] [ ] [ ].
––
=+?=+=?
==
∑∑
ββll
ll
Hence the
system is time-invariant.
(e) yn x n[] [ ]=?α
The system is linear,stable,non causal,Let y[n] be the output for an input x[n] and y n
1
[]
be the output for an input x n
1
[ ],Then y n x n[] [ ]=?α and y n x n
11
[] [ ]=?α,
Let x n x n n
10
[] [ ]=?,then y n x n x n n
11 0
[] [ ] [ ]=?=αα,whereas y n n x n n[][]?=?
00
α,
Hence the system is time-varying.
(f) y[n] = x[n – 5]
The given system is linear,causal,stable and time-invariant.
2.27 y[n] = x[n + 1] – 2x[n] + x[n – 1].
Let y n
1
[ ] be the output for an input x n
1
[ ] and y n
2
[ ] be the output for an input x n
2
[ ],Then
for an input x n x n x n
312
[] [] []=+αβ the output y n
3
[ ] is given by
yn xn xn xn
33 33
12 1[] [ ] [] [ ]=+? +?
=+? +?++? +?αααβββxn xn xn x n x n x n
111222
12 1 12 1[] [] [] [] [] []
=+αβyn y n
12
[] [].
Hence the system is linear,If x n x n n
10
[] [ ]=? then y n y n n
10
[] [ ]=?,Hence the system is
time-inavariant,Also the system's impulse response is given by
hn
n
[]
,
,
,
,
=
=
2
1
0
0
n = 1,-1,
elsewhere.
Since h[n] ≠?0 n < 0 the system is non-causal.
2.28 Median filtering is a nonlinear operation,Consider the following sequences as the input to
a median filter,x n
1
345[] {,,}= and x n
2
222[] {,,}=,The corresponding outputs of the
median filter are y n n
1
42[] []==? and y
2
,Now consider another input sequence x
3
[n] =
x
1
[n] + x
2
[n],Then the corresponding output is y n
3
[ ] = 3,On the other hand,
yn yn
12
2[] []+=,Hence median filtering is not a linear operation,To test the time-
invariance property,let x[n] and x
1
[n] be the two inputs to the system with correponding
outputs y[n] and y
1
[n],If x n x n n
10
[] [ ]=? then
y n median x n k x n x n k
1111
[ ] { [ ],.......,[ ],......,[ ]}=? +
+?=?median x n k n x n n x n k n y n n{ [ ],.......,[ ],......,[ ]} [ ]
00 00
.
Hence the system is time invariant.
13
2.29 yn yn
xn
yn
[] [ ]
[]
[]
=?+
1
2
1
1
Now for an input x[n] = α μ[n],the ouput y[n] converges to some constant K as n →∞,
The above difference equation as n →∞ reduces to KK
K
=+
1
2
α
which is equivalent to
K
2
=α or in other words,K =α.
It is easy to show that the system is non-linear,Now assume y
1
[n] be the output for an
input x
1
[n],Then yn yn
xn
yn
11
1
1
1
2
1
1
[] [ ]
[]
[]
=?+
If x n x n n
10
[] [ ]=?,Then,yn yn
xn n
yn
11
0
1
1
2
1
1
[] [ ]
[]
[]
.=?+
Thus y n y n n
10
[] [ ]=?,Hence the above system is time invariant.
2.30 For an input x n
i
[ ],i = 1,2,the input-output relation is given by
yn xn y n yn
iii i
[] [] [ ] [ ].=+?
2
11 Thus,for an input Ax
1
[n] + Bx
2
[n],if the output is
Ay
1
[n] + By
2
[n],then the input-output relation is given by A y n By n
12
[] []+=
Ax n Bx n Ay n By n Ay n By n
12 1 2
2
12
11 11[] [] [] [] [] []++?( ) +?+? = A x n Bx n
12
[] []+
++?+?Ayn Byn AByn yn Ayn Byn
2
1
22
2
2
12 1 2
1121111[] [] [][] [] []
≠+?++?Ax n A y n Ay n Bx n B y n By n
1
2
1
2
12
2
2
2
2
11 1[ ] [ ] [ ] [ ] [ ] [ ],Hence,the system
is nonlinear.
Let y[n] be the output for an input x[n] which is related to x[n] through
yn xn y n yn[][][][]=+?
2
1 1, Then the input-output realtion for an input x n n
o
[]? is given
by ynn xnn ynn ynn
oo o o
[][][ ][ ]?=+
2
1 1,or in other words,the system is time-
invariant.
Now for an input x[n] = α μ[n],the ouput y[n] converges to some constant K as n →∞,
The above difference equation as n →∞ reduces to K K K=? +α
2
,or K
2
=α,i.e,
K =α.
2.31 As δμμ[] [] [ ]nnn=1,ΤΤΤ{ [ ]} { [ ]} { [ ]}δμμnnn=1?=hn sn sn[] [] [ ]1
For a discrete LTI system
yn hkxn k
k
[] [][ ]=?
=?∞
∞
∑
=( )?
=?∞
∞
∑
sk sk xn k
k
[] [ ] [ ]1 =
=?∞
∞
=?∞
∞
∑∑
skxn k sk xn k
kk
[][] [][]1
2.32 yn hmxn m
m
[] [ ]?[]
=?∞
∞
∑
,Hence,y n kN h m x n kN m
m
[] []?[]+= +?
=?∞
∞
∑
=?=
=?∞
∞
∑
hmxn m yn
m
[]?[ ] [ ],Thus,y[n] is also a periodic sequence with a period N.
14
2.33 Now δ[]nr?
*
δ[]ns? = δδ δ[][ ][ ]mr nsm nrs
m
=
=?∞
∞
∑
(a) yn
1
[]= xn
1
[]
*
hn
1
[] = 21053δδ[].[]nn( )
*
2133δδ δ[] [ ] [ ]nn n+( )
= 4δ[]n?1
*
δ[]n – δ[]n? 3
*
δ[]n + 2δ[]n?1
*
δ[]n?1 – 0.5 δ[]n? 3
*
δ[]n?1
– 6δ[]n?1
*
δ[]n? 3 + 1.5 δ[]n? 3
*
δ[]n? 3 = 4 δ[]n?1 – δ[]n? 3 + 2δ[]n?1
– 0.5 δ[]n? 4 – 6δ[]n? 4 + 1.5δ[]n? 6
= 4 δ[]n?1 + 2δ[]n?1 – δ[]n? 3 – 6.5δ[]n? 4 + 1.5δ[]n? 6
(b) yn
2
[]= xn
2
[]
*
hn
2
[] =++( )31 2δδ[][ ]nn
*
+?( )δδδ[].[][]nnn205 13 3
=?+?+?+?+05 1 3 1 15 2 3 3 9 4.[ ] [] [ ],[ ] [ ] [ ]δδδ δ δ δnnn n n n
(c) yn
3
[]= xn
1
[]
*
hn
2
[] =
21053δδ[].[]nn( )
*
+?( )δδδ[].[][]nnn205 13 3 =
+δ δ[][].[].[]–.[]n n2 2 3 6 25 4 0 5 5 1 5 6
(d) yn
4
[]= xn
2
[]
*
hn
1
[] =++( )31 2δδ[][ ]nn
*
2133δδ δ[] [ ] [ ]nn n+( ) =
22 1 13294δδδ δ δ[][][][]–[]nnn n n++ +
2.34 yn gmhn m
mN
N
[] [ ][ ]=?
=
∑
1
2
,Now,h[n – m] is defined for M n m M
12
≤? ≤, Thus,for
mN=
1
,h[n–m] is defined for M n N M
112
≤? ≤,or equivalently,for
MNnM N
11 21
+≤≤ +,Likewise,for m N=
2
,h[n–m] is defined for M n N M
122
≤? ≤,or
equivalently,for M N n M N
12 22
+≤≤+.
(a) The length of y[n] is M N M N
2211
1+? +–,
(b) The range of n for y n[]≠ 0 is min,max,MNMN n M NM N
1112 2122
++( )≤≤ + +( ),i.e.,
MNnM N
11 2 2
+≤≤ +.
2.35 y[n] = x
1
[n]
*
x
2
[n] = x n k x k
k
12
[][]?
=?∞
∞
∑
.
Now,v[n] = x
1
[n – N
1
]
*
x
2
[n – N
2
] = x n N k x k N
k
112 2
[][]
=?∞
∞
∑
,Let
kN m?=
2
,Then v[n] = x n N N m x m y n N N
m
112 2 12
[][], =
=?∞
∞
∑
2.36 g[n] = x
1
[n]
*
x
2
[n]
*
x
3
[n] = y[n]
*
x
3
[n] where y[n] = x
1
[n]
*
x
2
[n],Define v[n] =
x
1
[n – N
1
]
*
x
2
[n – N
2
],Then,h[n] = v[n]
*
x
3
[n – N
3
], From the results of Problem
2.32,v[n] = y n N N[]
12
,Hence,h[n] = y n N N[]
12*
x
3
[n – N
3
], Therefore,
using the results of Problem 2.32 again we get h[n] = g[n– N
1
– N
2
– N
3
],
15
2.37 y[n] = x[n]
*
h[n] =?
=?∞
∞
∑
xn khk
k
[ ] [ ],Substituting k by n-m in the above expression,we
get y n x m h n m
m
[] [ ][ ]=?
=?∞
∞
∑
= h[n]
*
x[n],Hence convolution operation is commutative.
Let y[n] = x[n]
*
hn h n
12
[] []+
( )
= =?+
( )
=?∞
∞
∑
xn k h k h k
k
[ ] [] []
12
=?+?
=?∞
=∞
=?∞
∞
∑∑
xn kh k xn kh k
k
k
k
[][] [][]
12
= x[n]
*
h
1
[n] + x[n]
*
h
2
[n],Hence convolution is
distributive.
2.38 x
3
[n]
*
x
2
[n]
*
x
1
[n] = x
3
[n]
*
(x
2
[n]
*
x
1
[n])
As x
2
[n]
*
x
1
[n] is an unbounded sequence hence the result of this convolution cannot be
determined,But x
2
[n]
*
x
3
[n]
*
x
1
[n] = x
2
[n]
*
(x
3
[n]
*
x
1
[n]), Now x
3
[n]
*
x
1
[n] =
0 for all values of n hence the overall result is zero,Hence for the given sequences
x
3
[n]
*
x
2
[n]
*
x
1
[n] ≠ x
2
[n]
*
x
3
[n]
*
x
1
[n],
2.39 w[n] = x[n]
*
h[n]
*
g[n],Define y[n] = x[n]
*
h[n] =?
∑
xk hn k
k
[ ] [ ] and f[n] =
h[n]
*
g[n] =?
∑
gk hn k
k
[ ] [ ],Consider w
1
[n] = (x[n]
*
h[n])
*
g[n] = y[n]
*
g[n]
=
∑∑
gm xk hn m k
km
[ ] [ ] [ ],Next consider w
2
[n] = x[n]
*
(h[n]
*
g[n]) = x[n]
*
f[n]
∑∑
xk gmhnkm
mk
[] [ ][ ],Difference between the expressions for w
1
[n] and w
2
[n] is
that the order of the summations is changed.
A) Assumptions,h[n] and g[n] are causal filters,and x[n] = 0 for n < 0,This implies
ym
for m
xk hm k form
k
m
[]
,,
[][ ],.
=
<
≥
=
∑
00
0
0
Thus,w n g m y n m g m x k h n m k
m
n
k
nm
m
n
[] [][ ] [] [][ ]=
==
=
∑∑∑
000
.
All sums have only a finite number of terms,Hence,interchange of the order of summations
is justified and will give correct results.
B) Assumptions,h[n] and g[n] are stable filters,and x[n] is a bounded sequence with
xn X[],≤ Here,y m h k x m k
k
[] [][ ]=?
=?∞
∞
∑
=?+
=
∑
hkxm k m
kk
k
kk
[][ ] [ ]
,
1
2
12
ε with
εε
kk n
mX
12
,
[],≤
16
In this case all sums have effectively only a finite number of terms and the error can be
reduced by choosing k
1
and k
2
sufficiently large,Hence in this case the problem is again
effectively reduced to that of the one-sided sequences,Here,again an interchange of
summations is justified and will give correct results.
Hence for the convolution to be associative,it is sufficient that the sequences be stable and
single-sided.
2.40 yn xn khk
k
[] [ ][].=?
=?∞
∞
∑
Since h[k] is of length M and defined for 0 1≤≤k M –,the
convolution sum reduces to y n x n k h k
k
M
[] [ ][].
()
=?
=
∑
0
1
y[n] will be non-zero for all those
values of n and k for which n – k satisfies 01≤?≤?nk N,
Minimum value of n – k = 0 and occurs for lowest n at n = 0 and k = 0,Maximum value
of n – k = N–1 and occurs for maximum value of k at M – 1,Thus n – k = M – 1
=+?nNM2,Hence the total number of non-zero samples = N + M – 1.
2.41 y[n] = x[n]
*
x[n] =?
=?∞
∞
∑
xn kxk
k
[][].
Since x[n – k] = 0 if n – k < 0 and x[k] = 0 if k < 0 hence the above summation reduces to
yn xn kxk
kn
N
[] [ ][]=?
=
∑
1
=
+≤≤?
≤≤?
nnN
Nn Nn N
10 1
222
,,
,.
Hence the output is a triangular sequence with a maximum value of N,Locations of the
output samples with values
N
4
are n =
N
4
– 1 and
7
4
N
– 1,Locations of the output samples
with values
N
2
are n =
N
2
– 1 and
3
2
N
– 1,Note,It is tacitly assumed that N is divisible by 4
otherwise
N
4
is not an integer.
2.42 yn hkxn k
k
N
[] [][ ]=?
=
∑
0
1
,The maximum value of y[n] occurs at n = N–1 when all terms
in the convolution sum are non-zero and is given by
yN hk k
NN
k
N
k
N
[] []
()
.?= = =
+
==
∑∑
1
1
2
10
1
2.43 (a) y[n] = g
ev
[n]
*
h
ev
[n] = hnkgk
ev ev
k
[][]?
=?∞
∞
∑
,Hence,y[–n] = h n k g k
ev ev
k
[][].
=?∞
∞
∑
Replace k by – k,Then above summation becomes
yn h nkg k
ev ev
k
[] [ ] []?=?+?
=?∞
∞
∑
= h n k g k
ev ev
k
[( )] [ ]
=?∞
∞
∑
= h n k g k
ev ev
k
[( )] [ ]?
=?∞
∞
∑
= y[n].
Hence g
ev
[n]
*
h
ev
[n] is even.
17
(b) y[n] = g
ev
[n]
*
h
od
[n] = hnkgk
od ev
k
[( )] [ ]?
=?∞
∞
∑
,As a result,
y[–n] = h n k g k
od ev
k
[( )] [ ]
=?∞
∞
∑
= h n k g k
od ev
k
[( )] [ ]
=?∞
∞
∑
=
=?∞
∞
∑
hnkgk
od ev
k
[( )] [ ].
Hence g
ev
[n]
*
h
od
[n] is odd.
(c) y[n] = g
od
[n]
*
h
od
[n] = hnkgk
od od
k
[][]?
=?∞
∞
∑
,As a result,
y[–n] = h n k g k
od od
k
[][]
=?∞
∞
∑
= h n k g k
od od
k
[( )] [ ]
=?∞
∞
∑
= h n k g k
od od
k
[( )] [ ].?
=?∞
∞
∑
Hence g
od
[n]
*
h
od
[n] is even.
2.44 (a) The length of x[n] is 7 – 4 + 1 = 4,Using xn
h
yn hkxn k
k
[]
[]
[] [][ ],=
{}
=
∑
1
0
1
7
we arrive at x[n] = {1 3 –2 12},
03≤≤n
(b) The length of x[n] is 9 – 5 + 1 = 5,Using xn
h
yn hkxn k
k
[]
[]
[] [][ ],=
{}
=
∑
1
0
1
9
we
arrive at x[n] = {1 1 1 1 1},0 4≤≤n.
(c) The length of x[n] is 5 – 3 + 1 = 3,Using xn
h
yn hkxn k
k
[]
[]
[] [][ ],=
{}
=
∑
1
0
1
5
we get
x[n] =?+? + +{}4 0 6923 0 4615 3 4556 1 1065jj j,.,,.,,0 2≤≤n.
2.45 y[n] = ay[n – 1] + bx[n],Hence,y[0] = ay[–1] + bx[0],Next,
y[1] = ay[0] + bx[1] = a y a b x b x
2
101[] [] []?+ +, Continuing further in similar way we
obtain y[n] = a y a b x k
nn
k
n
+?
=
+
∑
1
0
1[] [].
(a) Let y n
1
[ ] be the output due to an input x n
1
[ ],Then y n a y a b x k
nnk
k
n
1
1
1
0
1[] [ ] [].=?+
+?
=
∑
If x n
1
[ ] = x[n – n
0
] then
yn a y a bxk n
nn
kn
n
1
1
0
0
1[] [ ] [ ]=?+?
+?
=
∑
=?+
+
=
∑
ay a bxr
nnn
r
nn
1
0
0
0
1[] [].
However,y n n a y a b x r
nn nn r
r
nn
[] [] []?=?+
+
=
∑0
0
1
0
0
0
1.
Hence y n y n n
10
[] [ ]≠? unless y[–1] = 0,For example,if y[–1] = 1 then the system is time
variant,However if y[–1] = 0 then the system is time -invariant.
18
(b) Let y
1
[n] and y
2
[n] be the outputs due to the inputs x
1
[n] and x
2
[n],Let y[n] be the output
for an input αβxn n
1
[] []+ x
2
,However,
αβ α β α βyn yn ay ay a bxk a bxk
nn nk
k
n
nk
k
n
12
11
1
0
2
0
[] [] [ ] [ ] [] []+ =?+?+ +
++?
=
=
∑∑
whereas
ynay abxk abxk
nnk
k
n
nk
k
n
[] [ ] [] []=?+ +
+?
=
=
∑∑
1
1
0
2
0
1 αβ.
Hence the system is not linear if y[–1] = 1 but is linear if y[–1] = 0.
(c) Generalizing the above result it can be shown that for an N-th order causal discrete time
system to be linear and time invariant we require y[–N] = y[–N+1] =
L
= y[–1] = 0.
2.46 y n hk n k hk
step
k
n
k
n
[] [][ ] [],=?=
==
∑∑
μ
00
n ≥ 0,and y n
step
[],= 0 n < 0,Since h[k] is
nonnegative,y n
step
[ ] is a monotonically increasing function of n for n ≥ 0,
and is not
oscillatory,Hence,no overshoot.
2.47 (a) f[n] = f[n – 1] + f[n – 2],Let f[n] = αr
n
,then the difference equation reduces to
αα αrr r
nn n
=
12
0 which reduces to r r
2
10= resulting in r=
15
2
±
.
Thus,f[n] = αα
12
15
2
15
2
+
+
nn
.
Since f[0] = 0 hence αα
12
0+=,Also f[1] = 1 hence
αα αα
12 12
2
5
2
1
+
+
=,
Solving for α
1
and α
2
,we get =–αα
12
1
5
=, Hence,f[n] =
1
5
15
2
1
5
15
2
+
nn
.
(b) y[n] = y[n – 1] + y[n – 2] + x[n – 1],As system is LTI,the initial conditions are equal
to zero.
Let x[n] = δ[]n, Then,y[n] = y[n – 1] + y[n – 2] + δ[]n?1, Hence,
y[0] = y[– 1] + y[– 2] = 0 and y[1] = 1,For n > 1 the corresponding difference equation is
y[n] = y[n – 1] + y[n – 2] with initial conditions y[0] = 0 and y[1] = 1,which are the same as
those for the solution of Fibonacci's sequence,Hence the solution for n > 1 is given by
y[n] =
1
5
15
2
1
5
15
2
+
nn
Thus f[n] denotes the impulse response of a causal LTI system described by the difference
equation y[n] = y[n – 1] + y[n – 2] + x[n – 1].
2.48 y[n] = αyn xn[][]?+1, Denoting,y[n] = y
re
[n] + j y
im
[n],and α = a + jb,we get,
y n jy n a jb y n jy n x n
re im re im
[] [] ( )( [ ] [ ]) []+=+?+?+11.
19
Equating the real and the imaginary parts,and noting that x[n] is real,we get
yn ayn by n xn
re re im
[] [] [][],=+1 1 (1)
ynbyn ayn
im re im
[] [] []=?+?11
Therefore
yn
a
yn
b
a
yn
im im re
[] [] []?=1
1
1
Hence a single input,two output difference equation is
y n ay n
b
a
yn
b
a
yn xn
re re im re
[] [ ] [] [ ] []= +?+11
2
thus b y n a y n a b y n a x n
im re re
[] []( )[ ][]?=+ +?+?11 21
22
.
Substituting the above in Eq,(1) we get
yn ayn a byn xnaxn
re re re
[] [ ] ( ) [ ] [] [ ]=+?+21 2 1
22
which is a second-order difference equation representing y n
re
[ ] in terms of x[n].
2.49 From Eq,(2.59),the output y[n] of the factor-of-3 interpolator is given by
yn xn xn xn xn xn
uu u u u
[] [] [ ] [ ] [ ] [ ]=+?++( ) +?++( )
1
3
12
2
3
21 where x n
u
[ ] is the output of
the factor-of-3 up-sampler for an input x[n],To compute the impulse response we set x[n] =
δ[]n,in which case,x n n
u
[] [ ].=δ3 As a result,the impulse response is given by
hn n n n n n[][][][] [][]=+?++( ) +?++( )δδ δ δ δ3
1
3
33 36
2
3
36 33 or
=+++++?+?
1
3
2
2
3
1
1
3
1
2
3
2δδδδδ[] [][][] []nnnnn.
2.50 The output y[n] of the factor-of-L interpolator is given by
yn x n
L
xn xn L
L
xn xn L
uuu u u
[] [] [ ] [ ] [ ] [ ]=+?++?( ) +?++?( )
1
11
2
22
++
++ +( )K
L
L
xn L xn
uu
1
11[][] where x n
u
[ ] is the output of the factor-of-L up-
sampler for an input x[n],To compute the impulse response we set x[n] = δ[]n,in which
case,x n Ln
u
[] [ ].=δ As a result,the impulse response is given by
hn Ln
L
Ln L Ln L L
L
Ln L Ln L L[][][][()] [ ][()]=+?++?( ) +?++?( )δδ δ δ δ
1
1
2
22
++
++( )K
L
L
Ln L L Ln L
1
1[(–)][] =
1
1
2
2
L
nL
L
nLδδ[( )] [( )]+?+ +?+K
+
+
L
L
n
1
1δ[]++?+?+
δδ δ δ[] [ ] [ ] [ ( )]n
L
n
L
n
L
L
nL
1
1
2
2
1
1
2.51 The first-order causal LTI system is characterized by the difference equation
yn p xn pxn dyn[] [] [ ] [ ]=+
01 1
11,Letting x[n] = δ[]n we obtain the expression for its
impulse response h n p n p n d h n[] [] [ ] [ ]=+
01 1
11δδ, Solving it for n = 0,1,and 2,we get
hp[]0
0
=,h p d h p d p[] [ ],10
11 110
=? =? and h d h d p d p[] [],21
11110
=? ==
( )
Solving these
equations we get p h
0
0= [],d
h
h
1
2
1
=?
[]
[]
,and ph
hh
h
1
1
20
1
=?[]
[][]
[]
.
20
2.52 pxn k dyn k
kk
k
N
k
M
[] [].?=?
==
∑∑
00
Let the input to the system be xn n[] []=δ, Then,p n k d h n k
kk
k
N
k
M
δ[] []?=?
==
∑∑
00
,Thus,
pdhrk
rk
k
N
=?
=
∑
[]
0
,Since the system is assumed to be causal,h[r – k] = 0 k > r.?
pdhrk
rk
k
r
=?
=
∑
[]
0
=
=
∑
hkd
rk
k
r
[],
0
2.53 The impulse response of the cascade is given by h[n] = h
1
[n]
*
h
2
[n] where
hn n
n
1
[] []=α μ and h n n
n
2
[] []=β μ, Hence,h n n
knk
k
n
[] []=
=
∑
αβ μ
0
.
2.54 Now,h n n
n
[] []=α μ, Therefore y n h k x n k
k
[] [][ ]=?
=?∞
∞
∑
=?
=
∞
∑
α
k
k
xn k
0
[]
=+?
=
∞
∑
xn xn k
k
k
[] [ ]α
1
=+
=
∞
∑
xn xn k
k
k
[] [ ]αα
0
1 =+?xn yn[] [ ]α 1.
Hence,x[n] = y[n] – αyn[]?1, Thus the inverse system is given by y[n] = x[n] – αxn[].?1
The impulse response of the inverse system is given by g n[]=?
↑
1 α,
2.55 yn yn yn xn[][][ ][]=?+?+?121,Hence,x[n – 1] = y[n] – y[n – 1] – y[n – 2],i.e.
x[n] = y[n + 1] – y[n] – y[n – 1],Hence the inverse system is characterised by
y[n] = x[n + 1] – x[n] – x[n – 1] with an impulse response given by g[n] = 1 –1 –1
↑
.
2.56 yn p xn pxn dyn[] [] [ ] [ ]=+
01 1
1 1 which leads to xn
p
yn
d
p
yn
p
p
xn[] [] [ ] [ ]=+
1
11
0
1
0
1
0
Hence the inverse system is characterised by the difference equation
yn
p
xn
d
p
xn
p
p
yn
1
0
1
1
0
1
1
0
1
1
11[] [] [ ] [ ].=+
2.57 (a) From the figure shown below we observe
x[n]
y[n]
h
1
[n] h
2
[n]
h
3
[n] h
4
[n]
h
5
[n]
v[n]
↓
21
v[n] = (h
1
[n] + h
3
[n]
*
h
5
[n])
*
x[n] and y[n] = h
2
[n]
*
v[n] + h
3
[n]
*
h
4
[n]
*
x[n].
Thus,y[n] = (h
2
[n]
*
h
1
[n] + h
2
[n]
*
h
3
[n]
*
h
5
[n] + h
3
[n]
*
h
4
[n])
*
x[n].
Hence the impulse response is given by
h[n] = h
2
[n]
*
h
1
[n] + h
2
[n]
*
h
3
[n]
*
h
5
[n] + h
3
[n]
*
h
4
[n]
(b) From the figure shown below we observe
x[n] y[n]
h
1
[n ] h
2
[n] h
3
[n]
h
4
[n]
h
5
[n]
v[n]
↓
v[n] = h
4
[n]
*
x[n] + h
1
[n]
*
h
2
[n]
*
x[n].
Thus,y[n] = h
3
[n]
*
v[n] + h
1
[n]
*
h
5
[n]
*
x[n]
= h
3
[n]
*
h
4
[n]
*
x[n] + h
3
[n]
*
h
1
[n]
*
h
2
[n]
*
x[n] + h
1
[n]
*
h
5
[n]
*
x[n]
Hence the impulse response is given by
h[n] = h
3
[n]
*
h
4
[n] + h
3
[n]
*
h
1
[n]
*
h
2
[n] + h
1
[n]
*
h
5
[n]
2.58 hn h n[] []=
1
*
hn hn
23
[] []+
Now hn
1
[]
*
hn
2
[]
= 2231δδ[][]nn +( )
*
δδ[][ ]nn?+ +( )12 2
= 22δ[]n?
*
δ[]n?1 – 31δ[]n +
*
δ[]n?1 + 2 2δ[]n?
*
22δ[]n +
– 31δ[]n +
*
22δ[]n + = 23δ[]n? – 3δ[]n + 4δ[]n – 63δ[]n +
= 23δ[]n? + δ[]n – 63δ[]n +, Therefore,
y[n] = 2 3δ[]n? + δ[]n – 63δ[]n + + 557321 31δδδδδ[][][][][]nnnnn?+?+ + +
= 5593213163δδδδδ[][][][][]nnnnn?+?+?+ +? +
2.59 For a filter with complex impulse response,the first part of the proof is same as that for a
filter with real impulse response,Since,y n h k x n k
k
[] [][ ]=?
=?∞
∞
∑
,
yn hkxn k
k
[] [][ ]=?
=?∞
∞
∑
≤?
=?∞
∞
∑
hk xn k
k
[] [ ].
Since the input is bounded hence 0 ≤≤xn B
x
[], Therefore,yn B hk
x
k
[] [].≤
=?∞
∞
∑
So if hk S
k
[]
=?∞
∞
∑
=<∞ then yn BS
x
[]≤ indicating that y[n] is also bounded.
22
To proove the converse we need to show that if a bounded output is produced by a bounded
input then S <∞,Consider the following bounded input defined by xn
hn
hn
[]
*[ ]
[]
=
.
Then y
hkhk
hk
hk S
kk
[]
*[ ] [ ]
[]
[],0 ===
=?∞
∞
=?∞
∞
∑∑
Now since the output is bounded thus S <∞.
Thus for a filter with complex response too is BIBO stable if and only if hk S
k
[]
=?∞
∞
∑
=<∞.
2.60 The impulse response of the cascade is g[n] = h
1
[n]
*
h
2
[n] or equivalently,
gk h k r h r
r
[] [ ] [].
–
=?
=∞
∞
∑ 12
Thus,
gk h k r h r
krk
[] [ –] []
–––=∞
∞
=∞
∞
=∞
∞
∑∑∑
=
12
≤
=∞
∞
=∞
∞
∑∑
hk h r
kr
12
[] [].
––
Since h
1
[n] and h
2
[n] are stable,hk
k
1
[]
∑
<∞ and hk
k
2
[]
∑
<∞,Hence,gk
k
[]
∑
<∞,
Hence the cascade of two stable LTI systems is also stable.
2.61 The impulse response of the parallel structure g[n] = h
1
[n]
+ h
2
[n], Now,
gk hk h k hk h k
kk k k
[] [] [] [] [].
∑∑ ∑ ∑
=+≤+
12 1 2
Since h
1
[n] and h
2
[n] are stable,
hk
k
1
[]
∑
<∞ and hk
k
2
[]
∑
<∞,Hence,gk
k
[]
∑
<∞,Hence the parallel connection of
two stable LTI systems is also stable.
2.62 Consider a cascade of two passive systems,Let y
1
[n] be the output of the first system which is
the input to the second system in the cascade,Let y[n] be the overall output of the cascade,
The first system being passive we have yn xn
nn
1
2
2
[] []
=?∞
∞
=?∞
∞
∑∑
≤,
Likewise the second system being also passive we have yn y n xn
nn n
[] [] []
2
1
2
2
=?∞
∞
=?∞
∞
=?∞
∞
∑∑ ∑
≤≤,
indicating that cascade of two passive systems is also a passive system,Similarly one can
prove that cascade of two lossless systems is also a lossless system.
2.63 Consider a parallel connection of two passive systems with an input x[n] and output y[n],
The outputs of the two systems are y n
1
[] and y n
2
[ ],respectively,Now,
yn xn
nn
1
2
2
[] [],
=?∞
∞
=?∞
∞
∑∑
≤ and yn xn
nn
2
2
2
[] [].
=?∞
∞
=?∞
∞
∑∑
≤
Let y n y n x n
12
[] [] []== satisfying the above inequalities,Then y n y n y n x n[] [] [] []=+=
12
2
and as a result,yn xn xn
nnn
[] [] [].
2 22
4
=?∞
∞
=?∞
∞
=?∞
∞
∑∑∑
=> Hence,the parallel
connection of two passive systems may not be passive.
23
2.64 Let p x n k y n d y n k
k
k
M
k
k
N
[][] []?= +?
==
∑∑
01
be the difference equation representing the causal
IIR digital filter,For an input xn n[] []=δ,the corresponding output is then y[n] = h[n],the
impulse response of the filter,As there are M+1 {p
k
} coefficients,and N {d
k
} coefficients,
there are a total of N+M+1 unknowns,To determine these coefficients from the impulse
response samples,we compute only the first N+M+1 samples,To illstrate the method,without
any loss of generality,we assume N = M = 3,Then,from the difference equation
reprsentation we arrive at the following N+M+1 = 7 equations:
hp[],0
0
=
hhdp[] [ ],10
11
+=
hhdhdp[] [] [],21 0
122
++ =
hhdhdhdp[] [] [] [],32 1 0
12 33
+++=
hhdhdhd[] [] [] [],43 2 1 0
123
+++=
hhdhdhd[] [] [] [],54 3 2 0
12 3
+++=
hhdhdhd[] [] [] [],65 4 3 0
123
+++=
Writing the last three equations in matrix form we arrive at
h
h
h
hhh
hhh
d
d
d
[]
[]
[]
[] [] []
[] [] []
[] [] []
,
4
5
6
321
432
543
0
0
0
1
2
3
+
=
and hence,
d
d
d
hhh
hhh
h
h
h
1
2
3
1
321
432
543
4
5
6
=
–
[] [] []
[] [] []
[] [] []
[]
[]
[]
.
–
Substituting these values of {d
i
} in the first four equations written in matrix form we get
p
p
p
p
h
hh
h
hh h
d
d
d
0
1
2
3
1
2
3
0000
1000
2100
321
1?
=
[]
[] [ ]
[] [] []
[] [] [] []
.
2.65
yn y x y y y
nnnn n
[] [] [] [] [] [] []
()
.=?+ =?+ =?+ =?+
+
==
∑∑ ∑
11 11
1
2
00 0
llll
l
μ
(a) For y[–1] = 0,y[n] =
nn()+1
2
(b) For y[–1] = –2,y[n] = –2 +
nn()+1
2
=
nn
2
4
2
+?
.
2.66 ynT y n T x d y n T T x n T
nT
nT
( ) (–) () (–) (–),
()
= ( ) +=( ) +?( )
∫
111
1
ττ Therefore,the difference
equation representation is given by y n y n T x n[][] [],=?+1 1 where y n y nT[] ( )= and
xn xnT[] ( ).=
24
2.67
yn
n
x
n
x
n
xn
nn
[] [] [] [],==+
==
∑∑
111
11
1
ll
ll
n ≥1,
yn
n
x
n
[] [],?=
=
∑
1
1
1
1
1
l
l
n ≥1,Hence,
xnyn
n
[] ( )[ ].l
l=
∑
=
1
1
11 Thus,the difference equation representation is given by
yn
n
n
yn
n
xn[] [ ] [].=
+
1
1
1
n≥1.
2.68 yn yn n n[], [ ] [],+?= ≥05 1 2 0μ with y[ ],?=12 The total solution is given by
yn y n y n
cp
[] [] []=+ where y n
c
[ ] is the complementary solution and y n
p
[ ] is the particular
solution.
yn
c
[ ] is obtained ny solving y n y n
cc
[], [ ]+?=0 5 1 0,To this end we set y n
c
n
[],=λ which
yields λλ
nn
+=
05 0
1
,whose solution gives λ=?05., Thus,the complementary solution is
of the form y n
c
n
[] (,).=?α 05
For the particular solution we choose y n
p
[],=β Substituting this solution in the difference
equation representation of the system we get ββμ+=05 2.[].n For n = 0 we get
β(.)105 2+= or β=43/.
The total solution is therefore given by y n y n y n
cp
[] [] []=+ = α(.),.?+≥05
4
3
0
n
n
Therefore y[–1] = α(.)?+=
05
4
3
2
1
or α=–.
1
3
Hence,the total solution is given by
y[n] = –(,),.
1
3
05
4
3
0?+≥
n
n
2.69 yn yn yn n
n
[], [ ], [ ] []+=01 1 006 2 2 μ with y[–1] = 1 and y[–2] = 0,The complementary
solution y n
c
[ ] is obtained by solving y n y n y n
cc c
[], [ ], [ ]+=0 1 1 0 06 2 0, To this end we
set y n
c
n
[],=λ which yields λλ λλλλ
nn nn
+=+

01 006 01 006 0
122
,–, (, –, ) whose
solution gives λ
1
03= –, and λ
2
02=,, Thus,the complementary solution is of the form
yn
c
nn
[] (,) (.)=? +αα
12
03 02,
For the particular solution we choose y n
p
n
[] ().=β2 ubstituting this solution in the difference
equation representation of the system we get ββ β μ2012 0622
12nn nn
n+=

(.) – (,) [],For
n = 0 we get ββ β+=

(.) – (,)012 0062 1
12
or β= =200 207 0 9662/..
The total solution is therefore given by y n y n y n
cp
[] [] []=+ =? + +αα
12
03 02
200
207
2(.) (.),
nnn
From the above y[ ] (, ) (, )?=? + + =

103 02
200
207
21
1
1
2
11
αα and
y[ ] (, ) (, )?=? + + =

203 02
200
207
20
1
2
2
22
αα or equivalently,–
10
3
5
107
207
12
αα+= and
100
9
25
50
207
12
αα+=– whose solution yields α
1
0 1017= –, and α
2
0 0356=,, Hence,the total
solution is given by y[n] = + +0 1017 0 3 0 0356 0 2 0 9662 2.(.).(.).(),
nnn
for n ≥ 0.
2.70 yn yn yn xn xn[], [ ], [ ] [] [ ]+=0 1 1 0 06 2 2 1 with xn n
n
[] []= 2 μ,and y[–1] = 1 and y[–2]
= 0,For the given input,the difference equation reduces to
yn yn yn n n n
nn
[], [ ], [ ] [] ( )[ ] [].+==
0 1 1 0 06 2 2 2 2 1
1
μμδ The solution of this
25
equation is thus the complementary solution with the constants determined from the given
initial conditions y[–1] = 1 and y[–2] = 0.
From the solution of the previous problem we observe that the complementary solution is of the
form y n
c
nn
[] (,) (.)=? +αα
12
03 02,
For the given initial conditions we thus have
y[ ] (, ) (, )?=? + =

103 021
1
1
2
1
αα and y[ ] (, ) (, )?=? + =

203 020
1
2
2
2
αα, Combining these
two equations we get
=
103 102
1009 1004
1
0
1
2
/,/.
/,/.
α
α
which yields α
1
018=?, and α
2
008=,,
Therefore,y n
nn
[],(.),(.)= +018 03 008 02,
2.71 The impulse response is given by the solution of the difference equation
yn yn n[], [ ] [].+?=05 1 δ From the solution of Problem 2.68,the complementary solution is
given by y n
c
n
[] (,).=?α 0 5 To determine the constant we note y[0] = 1 as y[–1] = 0,From
the complementary solution y[ ] (–, ),005
0
==αα hence α = 1,Therefore,the impulse
response is given by h n
n
[] (,).=?05
2.72 The impulse response is given by the solution of the difference equation
yn yn yn n[], [ ], [ ] []+=0 1 1 0 06 2 δ, From the solution of Problem 2.69,the complementary
solution is given by y n
c
nn
[] (,) (.)=? +αα
12
0 3 0 2, To determine the constants α
1
and α
2
,we
observe y[0] = 1 and y[1] + 0.1y[0] = 0 as y[–1] = y[–2] = 0,From the complementary
solution y[ ] (, ) (, ),00302 1
1
0
2
0
12
=? + =+=αααα and
y[ ] (, ) (, ) –,,,,10302030201
1
1
2
1
12
=? + = + =?αα α Solution of these equations yield α
1
06=,
and α
2
04=,, Therefore,the impulse response is given by h n
nn
[],(,),(.)=? +06 03 0402,
2.73 Let An
n
K
i
n
= ()λ, Then
A
A
n
n
n
n
K
i
+
=
+
1
1
λ, Now lim,
n
K
n
n→∞
+
=
1
1 Since λ
i
<1,there exists
a positive integer N
o
such that for all n > N
o
,0
1
2
1
1
<<
+
<
+
A
A
n
n
i
λ
,Hence A
n
n=
∞
∑
0
converges.
2.74 (a) xn[]=?{}3201452,?≤ ≤33n.rxnx
xx
n
[] [][ ]ll=?
=?
∑
3
3
,Note,
rxx
xx
[] [][],?=?=×=633236 rxxxx
xx
[] [][] [][] (),?=?+?=×?+×=5322322531
xxxx
xx
[] [][] [][] [][] (),?=?+?+?=×+×?+×=43122132052432
rxx xx
xx
[ ] [][] [][ ] [][ ] [][ ] ( ),?= +?+?+?=×+×+×?+×=?330211203215042133
xxxx xx
xx
[ ] [] [] [ ] [ ] [] [ ] [ ] [ ] [ ] [ ],?= + +?+?+=231201102 131
rxx xx xx
xx
[] [][] [][] [][] [][] [][] [][],?= + + +?++=132211001 12 2328
rxxxxxxxxxxxxxx
xx
[] [][] [][] [][] [][] [ ][ ] [ ][ ] [ ][ ],03322110011223359= + + + +++=
The samples of r
xx
[]l for 16≤≤l are determined using the property rr
xx xx
[] [ ]ll=?,Thus,
r
xx
[],l ={}6112 31128592811 32116
≤≤66l,
yn[]=071 349 2,?≤ ≤33n, Following the procedure outlined above we get
26
r
yy
[],l ={}0 14 61 43 52 10 160 10 52 43 61 14 0
≤≤66l,
wn[],=5436 501?≤ ≤33n,Thus,
r
ww
[],l =5 4 28 44 11 20 112 20 11 44 28 4 5
≤≤66l,
(b) rxny
xy
n
[] [][ ]ll=?
=?
∑
3
3
,Note,r x y
xy
[] [][],?=?=6330
rxyxy
xy
[] [][] [][],?=?+?=5322314 rxyxyxy
xy
[] [][] [][] [][],?=?+?+?=431221337
rxyxyxyxy
xy
[ ] [][] [][ ] [][ ] [][ ],?= +?+?+?=33021120327
rxyxyxyxyxy
xy
[ ] [] [] [ ] [ ] [] [ ] [ ] [ ] [ ] [ ],?= + +?+?+=231201102 134
rxyxyxyxyxyxy
xy
[] [][] [][] [][] [][] [][] [][],?= + + +?++=132211001 12 2327
rxyxyxyxyxyxyxy
xy
[] [][] [][] [][] [][] [ ][ ] [ ][ ] [ ][ ],03322110011223340= + + + +++=
rxyxyxyxyxyxy
xy
[] [ ] [] [] [ ] [ ] [] [ ] [ ] [ ] [ ] [ ] [ ],123120110213249=+++?++=
rxyxyxyxyxy
xy
[] [][] [][] [ ][] [ ][] [ ][ ],2130211203110= + +? +? +=
rxyxyxyxy
xy
[] [][] [ ][] [ ][] [ ][],303 12 21 3019=+?+?+?=?
rxyxyxy
xy
[] [ ][] [ ][] [ ][],41322316=? +? +? =? rxyxy
xy
[] [ ][] [ ][],5233231=? +? =
rxy
xy
[] [ ][],6336=? =? Hence,
r
xy
[],l {}0 14 37 27 4 27 40 49 10 19 6 31 6,
≤≤66l,
rxnw
xw
n
[] [] [ ]ll=?
=?
∑
3
3
={}10 17 6 38 36 12 35 6 1 29 15 2 3,
≤≤66l,
2.75 (a) xn n
n
1
[] [].=α μ
rxnx nn
xx
n
n
n
n
[] [] [ ] [] [ ]ll l
l
=?=
=?∞
∞
=?∞
∞
∑∑11
αμ α μ
=?
=
∞
∑
αμ
2
0
n
n
n
l
l[]
=
=
<
=
≥
=
∞
=
∞
∑
∑
α
α
α
α
α
α
2
0 2
2
2
1
0
1
0
n
n
n
n
for
for
l
l
l
l
l
l
l
,,
.
Note for
l≥ 0,r
xx
[],l
l
=
α
α1
2
and for
l< 0,r
xx
[],l
l
=
α
α1
2
Replacing l with?l in the
second expression we get
rr
xx xx
[] [].
()
=
=
=

ll
ll
α
α
α
α11
22
Hence,
r
xx
[]l is an even function
of l.
Maximum value of r
xx
[]l occurs at
l= 0 since
α
l
is a decaying function for increasing
l
when α<1.
(b) xn
nN
otherwise
2
10 1
0
[]
,,
,.
=
≤≤?
rxn
xx
n
N
[] [ ],ll=?
=
∑ 2
0
1
where xn
nN
otherwise
2
11
0
[]
,,
,.
=
≤≤?+
l
ll
27
Therefore,
r
for N
N for N
N for
N for N
for N
xx
[]
,(),
,–(–),
,,
,.
l
l
ll
l
ll
l
=
<
+≤<
=
<≤?
>?
01
10
0
01
01
It follows from the above
r
xx
[]l is a trinagular function of
l,and hence is even with a
maximum value of N at l = 0
2.76 (a) xn
n
M
1
[ ] cos=
π
where M is a positive integer,Period of x n
1
[ ] is 2M,hence
r
M
xnxn
M
n
M
n
M
xx
n
M
n
M
[ ] [ ] [ ] cos cos
()
ll
l
=+=
π
π+
=
=
∑∑
1
2
1
2
1
0
21
1
0
21
=
π
π
π
π
π
=
∑
1
2
0
21
M
n
M
n
MM
n
MM
n
M
cos cos cos sin sin
ll
=
π
π
=
∑
1
2
2
0
21
MM
n
M
n
M
cos cos,
l
From the solution of Problem 2.16 cos,
2
0
21 2
2
π
==
=
∑
n
M
M
M
n
M
Therefore,
r
n
M
xx
[ ] cos,l =
π
1
2
(b) x n n ulo n
2
6 0123450 5[] mod,.=={}≤≤ It is a peridic sequence with a period
6,Thus,,
rxnxn
xx
n
[] [] [ ],.lll=+≤
=
∑
1
6
05
22
0
5
It is also a peridic sequence with a period 6.
rxxxxxxxxxxxx
xx
[] ([][] [][] [][] [][] [][] [][]),0
1
6
00 11 22 33 44 55
55
6
22 22 22 22 22 22
=+++++=
rxxxxxxxxxxxx
xx
[] ([][] [][] [][] [][] [][] [][]),1
1
6
01 12 23 34 45 50
40
6
22 22 22 22 22 22
=+++++=
rxxxxxxxxxxxx
xx
[] ([][] [][] [][] [][] [][] [][]),2
1
6
02 13 24 35 40 51
32
6
22 22 22 22 22 22
=+++++=
rxxxxxxxxxxxx
xx
[] ([][] [][] [][] [][] [][] [][]),3
1
6
03 14 25 30 41 52
28
6
22 22 22 22 22 22
=+++++=
rxxxxxxxxxxxx
xx
[] ([][] [][] [][] [][] [][] [][]),4
1
6
04 15 20 31 42 53
31
6
22 22 22 22 22 22
=+++++=
rxxxxxxxxxxxx
xx
[] ([][] [][] [][] [][] [][] [][]),5
1
6
05 10 21 32 43 54
40
6
22 22 22 22 22 22
=+++++=
(c) xn
n
3
1[] ( ).=? It is a periodic squence with a period 2,Hence,
rxnx
xx
n
[] [] [ ],.lll=+≤
=
∑
1
2
01
33
0
1
rxxxx
xx
[] ([][] [][]),0
1
2
00 111
22 22
=+= and
rxxxx
xx
[] ([][] [][]),1
1
2
01 10 1
22 22
=+=? It is also a periodic squence with a period 2.
2.77 E X Y x y p x y dxdy
XY
{}()(,)+= +
∫∫
=+
∫∫ ∫∫
x p x y dxdy y p x y dxdy
XY XYy
(,) (,)
=
( )
+
( )
∫ ∫
x p x y dy dx y p x y dx dy
XY XY
(,) (,) =+
∫∫
xp x dx yp y dy
XY
() ()
28
= E{X} + E{Y}.
From the above result,E{2X} = E(X + X} = 2 E{X},Similarly,E{cX} = E{(c–1)X + X}
= E{(c–1)X} + E{X},Hence,by induction,E{cX} = cE{x}.
2.78 CEX=?
{}
()κ
2
,To find the value of κ that minimize the mean square error C,we
differentiate C with respect to κ
and set it to zero.
Thus
dC
d
EX EX EK EX K
κ
κ=?=?=?={ ( )} { } { } { },222220
which results in
κ = E{X},
and
the minimum value of C is σ
x
2
.
2.79 mean m E x xp x dx
xX
== =
∞
∞
∫
{} ( )
variance ==? =?
∞
∞
∫
σ
xx xX
Ex m x m p xdx
22 2
{( ) } ( ) ( )
(a) px
x
X
(),=
+
α
π
α
1
22
Now,m
xdx
x
x
=
+
∞
∞
∫
α
π
α
22
,Since
x
x
22
+α
is odd hence the
integral is zero,Thus m
x
= 0,
σ
α
π
α
x
xdx
x
2
2
22
=
+
∞
∞
∫
,Since this integral does not converge hence variance is not defined for
this density function.
(b) px e
x
x
()=
α
α
2
,Now,mxedx
x
x
==
∞
∞
∫
α
α
2
0,Next,
σ
α
α
αα
x
xx
xe dx xe dx
22 2
02
==
∞
∞
∞
∫∫
=
+
∞
∞
∫
α
αα
α
α
xe x
edx
x
x
2
0
0
2
=+
+
=
∞
∞
∫
α
αα
αα
α
α
0
222
0
2
0
2
xe
edx
x
x
.
(c) px
n
pp x
x
n
n
() ( )=
( )?
=
∑
l
l
l
l l
0
1 δ, Now,
mx
n
pp xdx
n
pp np
x
n
n
n
n
=
( )
=
( )
=
∞
∞
=
=
∫
∑∑
l
l
l
l
l
l
l
l
l
00
11δ()
σδ
xx
n
n
Ex m x
n
pp xdxnp
222 2
0
2
1=?=
( )
∞
∞
=
∫
∑
{} ( ) ( )
l
l
l
l l
=
( )
=?
=
∑
l
l
l
l
l2
0
22
11
n
pp npnpp
n
n
().
(d)
px
e
x
x
()
!
()=?
=
∞
∑
α
α
δ
l
l
l
l
0
,Now,
29
mx
e
xdx
e
x
=?==
∞
∞
=
∞
=
∞
∫
∑∑
α α
α
δ
α
α
l
ll
l
l
ll
l!
()
!
.
00
σ
αα
δα
α
xx
Ex m x
e
xdx
222 2
0
2
2
=?=
=
∞
∞
∞
∑
∫
{}
!
()
l
l
l
l
=?=
=
∞
∑
l
l
l
l
2
0
2
e
α
α
αα
!
.
(e) px
x
ex
x
x
() ()
/
=
α
μ
α
2
2
2
2
,Now,
m x e x dx x e dx
x
xx
===
∞
∞
∞
∫∫
11
2
2
2
2
2
2
2
2
0
2
2
2
α
μ
α
απ
αα//
() /,
σ
α
μ
απ
α
xx
x
Ex m x e xdx
222
2
3
2
2
2
2
1
2
=?=?
∞
∞
∫
{} ()
/
=?
π
2
2
2
α,
2.80 Recall that random variables x and y are linearly independent if and only if
Eax ay a
12
2
2
0+
>? a
1
,except when a
1
= a
2
= 0,Now,
Ea x Ea y Ea axy Ea a x y
1
2
2
2
2
2
12 12
+
+
{}
+
{}
()* * ( )** =
{}
+
{}
aEx a Ey
1
2
2
2
2
2
> 0
a and a
12
except when a
1
= a
2
= 0.
Hence if x,y are statistically independent they are also linearly independent.
2.81 σ
xx
Ex m
22
=?
{}
() =+?
{}
=+?Ex m xm Ex Em Exm
xx x x
22 2 2
22{} { } { }.
Since m m m E x m is a constant,hence E{m and E{xm
x
2
x
}}===,Thus,
σ
xxx
Ex m m Ex m
222222
2=+?=?{} {},
2.82 V = aX + bY,Therefore,E{V} = a E{X} + b E{Y} =+am bm
xy
,and
σ
vv
EV m
22
=?{( ) } =?+?EaX m bY m
xy
{( ( ) ( )) }.
2
Since X,Y are statistically independent hence σσσ
vxy
ab
22222
=+.
2.83 vn axn byn[] [] [].=+ Thus φ
vv
nEvmnvm[] {[ }[ ]}=+
=++++++ +
{}
Ea xm n xm b ym n ym abxm nym abxmym n
22
[ }[] [ }[] [ ][] [][ ],Since x[n] and
y[n] are independent,
φφφ
vv xx yy
nEaxmnxm Ebymnym a nb n[] [ }[ ] [ }[ ] [] []=+
{}
++
{}
=+
2222
.
φ
vx
n Evmnxm Eaxmnxm bymnxm[]{[][]} [][][][]=+ = +++ =+{}=aE x m n x m a n
xx
[][] [].φ
Likewise,φφ
vy yy
nb n[] [].=
2.84 φ
xy
Exn y n[] [ ] *[],ll=+
{}
φ
xy
Exn y n[– ] [ – ] *[ ],ll=
{}
φ
yx
Eyn x n[] [ ] *[].ll=+
{}
Therefore,φ
yx
Ey n xn*[ ] * [ ] [ ]ll=+
{}
= =Exn y n
xy
[ – ] *[ ] [– ].llφ
30
Hence φφ
xy yx
l[] *[]?=ll.
Since γφ
xy xy x y
mm[] [] ( )*ll=?, Thus,γφ
xy xy x y
mm[] [] ( )*ll=?, Hence,
γφ
xy xy x y
mm[] [] ( )*ll=?, As a result,γφ
xy xy x y
mm*[ ] *[ ] ( )*,ll=?
Hence,
γγ
xy yx
[] *[].?=ll
The remaining two properties can be proved in a similar way by letting x = y.
2.85
Exn xn[] [ ]
{}
≥l
2
0.
E xn E xn E xnx n E x nxn[] [ ] []*[ ] *[][ ]
22
0
{}
+?
{}

{}

{}
≥lll
202 0φφ
xx xx
[] []?≥l
φφ
xx xx
[] []0 ≥ l
Using Cauchy's inequality Ex Ey E xy
2 2 2
{}{}
≤ {},Hence,φφ φ
xx
[] [] []00
2
yy xy
≤ l,
One can show similarly
γγ γ
xx yy xy
[] [] []00
2
≤ l,
2.86 Since there are no periodicities in {x[n]} hence x[n],x[n+
l] become uncorrelated as
l→∞.,
Thus
lim
ll
ll
→∞ →∞
=?→γφ
xx xx x
m[] lim [],
2
0 Hence
lim [ ],
l
l
→∞
= φ
xx x
m
2
2.87
φ
XX
[],l
ll
ll
=
++
++
911 14
13 2
24
24
Now,
m
Xn XX[]
lim [ ] lim,
2
24
24
911 14
13 2
7==
++
++
=
→∞ →∞ll
l
ll
ll
φ
Hence,
m
Xn[]
.= m 7 EXn
XX
[] [],
2
09
( )
==φ Therefore,σφ
XXX X
m
2
2
0972=?=?=[],
M2.1 L = input('Desired length = '); n = 1:L;
FT = input('Sampling frequency = ');T = 1/FT;
imp = [1 zeros(1,L-1)];step = ones(1,L);
ramp = (n-1).*step;
subplot(3,1,1);
stem(n-1,imp);
xlabel(['Time in ',num2str(T),' sec']);ylabel('Amplitude');
subplot(3,1,2);
stem(n-1,step);
xlabel(['Time in ',num2str(T),' sec']);ylabel('Amplitude');
subplot(3,1,3);
stem(n-1,ramp);
xlabel(['Time in ',num2str(T),' sec']);ylabel('Amplitude');
M2.2 % Get user inputs
A = input('The peak value =');
L = input('Length of sequence =');
N = input('The period of sequence =');
FT = input('The desired sampling frequency =');
DC = input('The square wave duty cycle = ');
% Create signals
T = 1/FT;
31
t = 0:L-1;
x = A*sawtooth(2*pi*t/N);
y = A*square(2*pi*(t/N),DC);
% Plot
subplot(211)
stem(t,x);
ylabel('Amplitude');
xlabel(['Time in ',num2str(T),'sec']);
subplot(212)
stem(t,y);
ylabel('Amplitude');
xlabel(['Time in ',num2str(T),'sec']);
0 20 40 60 80 100
-10
-5
0
5
10
Time in 5e-05 sec
Amplitude
0 20 40 60 80 100
-10
-5
0
5
10
Time in 5e-05 sec
Amplitude
M2.3 (a) The input data entered during the execution of Program 2_1 are
Type in real exponent = -1/12
Type in imaginary exponent = pi/6
Type in the gain constant = 1
Type in length of sequence = 41
32
0 10 20 30 40
-1
-0.5
0
0.5
1
Time index n
Amplitude
Real part
0 10 20 30 40
-1
-0.5
0
0.5
1
Time index n
Amplitude
Imaginary part
(b) The input data entered during the execution of Program 2_1 are
Type in real exponent = -0.4
Type in imaginary exponent = pi/5
Type in the gain constant = 2.5
Type in length of sequence = 101
0 10 20 30 40
-0.5
0
0.5
1
1.5
Time index n
Amplitude
Real part
0 10 20 30 40
-0.5
0
0.5
1
1.5
Time index n
Amplitude
Imaginary part
M2.4 (a) L = input('Desired length = ');
A = input('Amplitude = ');
omega = input('Angular frequency = ');
phi = input('Phase = ');
n = 0:L-1;
x = A*cos(omega*n + phi);
stem(n,x);
xlabel('Time index');ylabel('Amplitude');
title(['\omega_{o} = ',num2str(omega)]);
(b)
33
0 10 20 30 40 50 60 70 80 90 100
-2
-1
0
1
2
Time index n
Amplitude
ω
o
= 0.14π
0 5 10 15 20 25 30 35 40
-2
-1
0
1
2
Time index n
Amplitude
ω
o
= 0.24π
0 10 20 30 40 50 60 70 80 90 100
-2
-1
0
1
2
Time index n
Amplitude
ω
o
= 0.68π
34
0 5 10 15 20 25 30 35 40
-2
-1
0
1
2
Time index n
Amplitude
ω
o
= 0.75π
M2.5 (a) Using Program 2_1 we generate the sequence? []
.
xn e
jn
1
04
=
π
shown below
0 5 10 15 20 25 30 35 40
-1
-0.5
0
0.5
1
Time index n
Amplitude
Real part
0 5 10 15 20 25 30 35 40
-1
-0.5
0
0.5
1
Time index n
Amplitude
Imaginary part
(b) Code fragment used to generate? [ ] sin(,, )xn n
2
06 06=π+π is:
x = sin(0.6*pi*n + 0.6*pi);
35
0 5 10 15 20 25 30 35 40
-2
-1
0
1
2
Time index n
Amplitude
(c) Code fragment used to generate? [ ] cos(,, ) sin(, )xn n n
3
211 05 207=π?π+π is
x = 2*cos(1.1*pi*n - 0.5*pi) + 2*sin(0.7*pi*n);
0 5 10 15 20 25 30 35 40
-4
-2
0
2
4
Time index n
Amplitude
(d) Code fragment used to generate? [ ] sin(, ) cos(,, )xn n n
4
3 13 4 03 045=π? π+π is:
x = 3*sin(1.3*pi*n) - 4*cos(0.3*pi*n+0.45*pi);
0 5 10 15 20 25 30 35 40
-6
-4
-2
0
2
4
6
Time index n
Amplitude
(e) Code fragment used to generate
[ ] sin(,, ) sin(, ) cos(, )xn n n n
5
5 1 2 0 65 4 0 8 0 8=π+π+π?π is:
x = 5*sin(1.2*pi*n+0.65*pi)+4*sin(0.8*pi*n)-cos(0.8*pi*n);
36
0 5 10 15 20 25 30 35 40
-10
-5
0
5
10
Time index n
Amplitude
(f) Code fragment used to generate? [] modx n n ulo
6
6= is,x = rem(n,6);
0 5 10 15 20 25 30 35 40
-1
0
1
2
3
4
5
6
Time index n
Amplitude
M2.6 t = 0:0.001:1;
fo = input('Frequency of sinusoid in Hz = ');
FT = input('Samplig frequency in Hz = ');
g1 = cos(2*pi*fo*t);
plot(t,g1,'-')
xlabel('time');ylabel('Amplitude')
hold
n = 0:1:FT;
gs = cos(2*pi*fo*n/FT);
plot(n/FT,gs,'o');hold off
M2.7 t = 0:0.001:0.85;
g1 = cos(6*pi*t);g2 = cos(14*pi*t);g3 = cos(26*pi*t);
plot(t/0.85,g1,'-',t/0.85,g2,'--',t/0.85,g3,':')
xlabel('time');ylabel('Amplitude')
hold
n = 0:1:8;
gs = cos(0.6*pi*n);
plot(n/8.5,gs,'o');hold off
M2.8 As the length of the moving average filter is increased,the output of the filter gets more
smoother,However,the delay between the input and the output sequences also increases
(This can be seen from the plots generated by Program 2_4 for various values of the filter
length.)
37
M2.9 alpha = input('Alpha = ');
yo = 1;y1 = 0.5*(yo + (alpha/yo));
while abs(y1 - yo) > 0.00001
y2 = 0.5*(y1 + (alpha/y1));
yo = y1; y1 = y2;
end
disp('Square root of alpha is'); disp(y1)
M2.10 alpha = input('Alpha = ');
yo = 0.3; y = zeros(1,61);
L = length(y)-1;
y(1) = alpha - yo*yo + yo;
n = 2;
while abs(y(n-1) - yo) > 0.00001
y2 = alpha - y(n-1)*y(n-1) + y(n-1);
yo = y(n-1); y(n) = y2;
n = n+1;
end
disp('Square root of alpha is'); disp(y(n-1))
m=0:n-2;
err = y(1:n-1) - sqrt(alpha);
stem(m,err);
axis([0 n-2 min(err) max(err)]);
xlabel('Time index n');
ylabel('Error');
title(['\alpha = ',num2str(alpha)])
The displayed output is
Square root of alpha is
0.84000349056114
0 5 10 15 20 25
-0.04
-0.02
0
0.02
0.04
0.06
Time index n
Error
α = 0.7056
M2.11 N = input('Desired impulse response length = ');
p = input('Type in the vector p = ');
d = input('Type in the vector d = ');
[h,t] = impz(p,d,N);
n = 0:N-1;
stem(n,h);
xlabel('Time index n');ylabel('Amplitude');
38
0 10 20 30 40
-1
-0.5
0
0.5
1
1.5
Time index n
Amplitude
M2.12 x =?
[]
3201452 y=
[]
071 349 2,w=
[]
5436 501.
(a) rn
xx
[]= [6 11 2 -3 11 28 59 28 11 -3 2 11 6],
r n
yy
[]= [0 -14 61 43 -52 10 160 10 -52 43 61 -14 0],
r n
ww
[]= [–5 4 28 –44 –11 –20 112 –20 –11 –44 28 4 –5].
-6 -4 -2 0 2 4 6
0
10
20
30
40
50
60
Lag index
Amplitude
r
xx
[n]
-6 -4 -2 0 2 4 6
-50
0
50
100
150
Lag index
Amplitude
r
yy
[n]
-6 -4 -2 0 2 4 6
-50
0
50
100
Lag index
Amplitude
r
ww
[n]
(b) rn
xy
[]= [ –6 31 –6 –19 10 49 40 27 4 27 37 14 0].
39
rn
xw
[]= [3 –2 –15 29 1 6 –35 12 36 38 6 –17 –10].
-6 -4 -2 0 2 4 6
-50
0
50
100
Lag index
Amplitude
r
xy
[n]
-6 -4 -2 0 2 4 6
-50
0
50
Lag index
Amplitude
r
xw
[n]
M2.13 N = input('Length of sequence = ');
n = 0:N-1;
x = exp(-0.8*n);
y = randn(1,N)+x;
n1 = length(x)-1;
r = conv(y,fliplr(y));
k = (-n1):n1;
stem(k,r);
xlabel('Lag index');ylabel('Amplitude');
gtext('r_{yy}[n]');
-30 -20 -10 0 10 20 30
-10
0
10
20
30
Lag index
Amplitude
r
yy
[n]
M2.14 (a) n=0:10000;
phi = 2*pi*rand(1,length(n));
A = 4*rand(1,length(n));
x = A.*cos(0.06*pi*n + phi);
stem(n(1:100),x(1:100));%axis([0 50 -4 4]);
xlabel('Time index n');ylabel('Amplitude');
mean = sum(x)/length(x)
var = sum((x - mean).^2)/length(x)
40
0 20 40 60 80 100
-4
-2
0
2
4
Time index n
Amplitude
0 20 40 60 80 100
-4
-2
0
2
4
Time index n
Amplitude
0 20 40 60 80 100
-4
-2
0
2
4
Time index n
Amplitude
0 20 40 60 80 100
-4
-2
0
2
4
Time index n
Amplitude
mean =
0.00491782302432
var =
2.68081196077671
From Example 2.44,we note that the mean = 0 and variance = 8/3 = 2.66667.
M2.15 n=0:1000;
z = 2*rand(1,length(n));
y = ones(1,length(n));x=z-y;
mean = mean(x)
var = sum((x - mean).^2)/length(x)
mean =
0.00102235365812
var =
0.34210530830959
Using Eqs,(2,129) and (2.130) we get m
X
=
=
11
2
0,and σ
X
2
2
11
22
1
3
=
+
=
()
,It should
be noted that the values of the mean and the variance computed using the above MATLAB
program get closer to the theoretical values if the length is increased,For example,for a length
100001,the values are
mean =
9.122420593670135e-04
41
var =
0.33486888516819