Digital Signal Processing Tutorial:Chapt 2 z transform

10/7/2009
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Chapt 02
Z-Transform
Digital Signal Processing
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Z-Transform
Syllabus
•Definitions and Properties of z-transform
•Rational z-transforms
•Inverse z-transform
•One sided z-transform
•Analysis of LTI systems in z-domain

10/7/2009
Z-Transform definition
Z-transform is mainly used for analysis of discrete signal and discrete
LTI system. Z.T of discrete time single x (n) is defined by the following
expression.
∑
∞
−∞=
−
=
n
n
znxzX )()(
where, X(z) z-transform of x(n)
z complex variable = rejw
From the above definition of Z.T. it is clear that ZT is power series & it
exist for only for those values of z for which X(z) attains finite value
( convergence) ,which is defined by Region of convergence. (ROC)
Region of Convergence: (ROC)
Region of Convergence is set of those values of z for which power
series x (z) converges. OR for which power series, x (z) attains finite value.

10/7/2009
Z-Transform of Finite duration signal
Find the Z - Transform and mention the Region of Convergence
(ROC) for the following discrete time sequences.
1. x (n) = { 2 1 2 3}
2. x (n) = { 2, 1, 2 3 }
3. x (n) = { 1 2 1 -2 3 1}
1st example Is of causal signal
2nd example Is of anti-causal signal
3rd example Is of non-causal signal

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Solution(1)
∑
∞
−∞=
−
=
n
n
znxzX )()(
321
)3()2()1()0()( −−−
+++= zxzxzxxzX
x (n) = { 2 1 2 3}
Z.T. is defined as
ROC is a set of those values of z for which x (z) is not infinite
In this case x(z) is finite for all values of z, except |z| = 0.
Because at z = 0, x(z) = ∞.
Thus ROC is entire z-plane except |z| = 0.
ROC
321
3212)( −−−
+++= zzzzX
321
322)( −−−
+++= zzzzX
Z=0

10/7/2009
Solution(2)
∑
∞
−∞=
−
=
n
n
znxzX )()(
321
)3()2()1()0()( zxzxzxxzX −+−+−+=
x (n) = { 2 1 2 3}
Z.T. is defined as
In this case X(z) is finite for all values of z, except |z| = ∞.
Because at z = ∞, X(z) = ∞.
Thus ROC is entire z-plane except |z| = ∞.
ROC
321
2123)( zzzzX +++=
32
223)( zzzzX +++=
Z= ∞

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Solution(3)
∑
∞
−∞=
−
=
n
n
znxzX )()(
x (n) = { 1 2 1 -2 3 1}
Z.T. is defined as
In this case X(z) is finite for all values of z, except |z| = ∞.
Because at z = ∞, X(z) = ∞.
Thus ROC is entire z-plane except |z| = 0 &|z| = ∞.
ROC
Z= ∞
32112
)3()2()1()0()1()2()( −−−
++++−+−= zxzxzxxzxzxzX
32112
132121)( −−−
++−++= zzzzzzX
32112
3212)( −−−
++−++= zzzzzzX
Z=0

10/7/2009
Quiz
Find the Z - Transform and mention the Region of Convergence
(ROC) for the following discrete time sequences.
1. x (n) =
2. x (n) =
)2( −nδ
)(nδ
Ans (1) z-2 ROC- entire z-plane except |z|= 0
Ans (2) 1 ROC- entire z-plane

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z-Transform of infinite duration signal
Find the z-transform for following discrete time sequences. Also mention
ROC for all the cases.
( ) ( )nUanx n
=
( ) ( )1−−−= nUanx n
)1()()( −−+= nUbnUanx nn
1
2
3

10/7/2009
Solution(1)
Sequence is causal( ) ( )nUanx n
=
∑
∞
−∞=
−
=
n
n
znxzX )()(
Z.T. for the given sequence x (n) is defined as
∑
∞
−∞=
−
=
n
nn
znUa )(
∑
∞
=
−
=
0
)(
n
nn
zazX
( )∑
∞
=
−
=
0
1
)(
n
n
azzX
U(n)=0 for
n<0
∑
∞
=
+++=
0
210
.......
n
n
aaaaWe know that
Series converges iff |a|<1
∑
∞
= −
=
0 1
1
n
n
a
aWe also know 1<afor
Thus, x (z) converges when | a z–1 | < 1
( ) 1
1
1
−
−
=
az
zX
( )
az
z
zX
−
=
for | a z–1 | < 1
for | a /z | < 1
for | z | > |a|
ROC is outside the
circle |z|=|a|

10/7/2009
Solution(2)
Sequence is anti-causal( ) ( )1−−−= nUanx n
∑
∞
−∞=
−
=
n
n
znxzX )()(
∑
∞
−∞=
−
−−−=
n
nn
znUa )1(
( )∑
−
−∞=
−
−=
1
1
)(
n
n
azzX
U(-n-1)=0 for
n>-1
∑
∞
=
+++=
0
210
.......
n
n
aaaaWe know that
∑
∞
= −
=
0 1
1
n
n
a
Thus, x (z) converges when | a–1 z | < 1
( )
za
za
zX 1
1
1 −
−
−
−=
( )
az
z
zX
−
=
for | a–1 z | < 1
for | z /a | < 1
for | z | < |a|
ROC is inside the
circle |z|=|a|
Put n=-m
( ) ( )∑∑ ∞=
−
∞=
−−
−=−=
1
1
1
1
)(
m
m
m
m
zaazzX

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Solution(3)
Sequence is non-causal
∑
∞
−∞=
−
=
n
n
znxzX )()(
∑∑
−
−∞=
−
∞
=
−
+=
1
0 n
nn
n
nn
zbza
∑∑
−
−∞=
−
∞
=
−
+=
1
1
0
1
)()()(
n
n
n
n
bzazzX
∑
∞
=
+++=
0
210
.......
n
n
aaaaWe know that
∑
∞
= −
=
0 1
1
n
n
a
Thus, x (z) converges when | a z–1 | < 1 & |b-1z|<1
( )
zb
zb
az
zX 1
1
1
11
1
−
−
−
−
+
−
=
( )
zb
z
az
z
zX
−
+
−
=
for | a z–1 | < 1 & | b-1 z | < 1
for | a /z | < 1 & |z/b| <1
for | z | > |a| & | z | < |b|
)1()()( −−+= nUbnUanx nn
Put n=-m in second
term
∑∑
∞
=
−
∞
=
−
+=
1
1
0
1
)()()(
m
m
n
n
zbazzX
ROC is |b|>| z | > |a|

10/7/2009
Problems
Find z- transform for followings:
( ) )()(6)(7)( 2
1
3
1 nununx nn
−=
( ) )()sin()( 43
1 nunnx
n π=
n
bnx =)(
1.
2.
3.

10/7/2009
Solution(1)
( ) )()(6)(7)( 2
1
3
1 nununx nn
−=
∑∑
∞
=
−
∞
=
−
−=
0
2
1
0
3
1 )(6)(7)(
n
nn
n
nn
zzzX
∑∑
∞
=
−
∞
=
−
−=
0
1
2
1
0
1
3
1 )(6)(7)(
n
n
n
n
zzzX
1
2
11
3
1 1
6
1
7
−−
−
−
−
=
zz
11
3
1 <−
z 11
2
1 <−
z&
)1)(1(
67
1
2
11
3
1
1
3
61
2
7
−−
−−
−−
+−−
=
zz
zz
))((
))(1(
2
1
3
1
1
3
6
2
72
−−
+−+
=
−
zz
zz
X&/ by
z2
))((
)(
2
1
3
1
2
3
−−
−
=
zz
zz
3/2
3
1>z & 2
1>z
1
1/2
ROC is
outside the
circle |z|=1/2

10/7/2009
Solution(2)
( ) ][)sin(][ 43
1 nunnx
n π=
][
2
)(][
44
3
1 nu
j
ee
nx
njnj
n





 −
=
− ππ
][)(][)( 44
3
1
2
1
3
1
2
1
nuenue njn
j
njn
j
ππ −
−=
1
3
12
1
1
3
12
1
44
1
1
1
1
)( −−−
−
−
−
=
zeze
zX njjnjj ππ
ROC
1
3
11
3
1 44
&1 −−−
< zeze njnj ππ
3
1>z & 3
1>z
3
1>z
1
1/3
ROC is
outside the
circle |z|=1/3

10/7/2009
Solution(3)
n
bnx =)( b>0
)1()()( −−+= −
nubnubnx nn
We know
1
1
1
][ −
−
⇔ bz
n
nub bz >||
11
1
1
]1[ −−
−
−
⇔−− zb
n
nub bz 1|| <and
111
1
1
1
1
)( −−−
−−
−= zbbz
zX bzb 1|| <<
For b>1, there are no values of z that satisfy ROC b1/b
|z|=1
10 << b

10/7/2009
Problem:
)21)(1(
1
)( 11
3
1 −−
−−
=
zz
zX
Show all possible ROC’s and pole-zero
diagram z-transform given below
1/32 1
|z|=1
1/32 1
|z|=1
1/31
|z|=1
If x[n] is left sided signal
i.e. anti-causal signal If x[n] is right sided signal
i.e. causal signal
If x[n] is double sided
signal
i.e. non-causal signal

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Inverse z-transform
•Synthetic Division Method
•Partial Fraction Method
•Cauchy’s Integration Method

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Synthetic division Example(1)
1
1
1
)( −
−
=
az
zX
1
1 −
− az 1
1
1 −
− az
1
1−
az
221 −−
− zaaz
22 −
za
1−
+ az
3322 −−
− zaza
33 −
za
•Since ROC is |z| >a, x[n] is causal
sequence or right sided sequence.
•Quotient series should have –ve
powers of z as z-transform of causal
sequence has –ve powers of z
22 −
+ za
.....}1{][ 432
aaaanx
↑
=
|||| az >

10/7/2009
Synthetic division Example(2)
1
1
1
)( −
−
=
az
zX
11
+− −
az 1
za 1
1 −
−
za 1−
221
zaza −−
−
22
za−
za 1−
−
3322
zaza −−
−
33
za−
•Since ROC is |z| >a, x[n] is anti-causal
sequence or left sided sequence.
•Quotient series should have +ve powers
of z , as z-transform of anti-causal
sequence has +ve powers of z
22
za−
−
.......}0.......{][ 123
↑
−−−
−−−= aaanx
|||| az <
33
za−
−

10/7/2009
Problem
Using long division method, determine the z-transform of
2
2
11
2
31
1
)( −−
+−
=
zz
zX 1|:|) >zROCa 2
1|:|) <zROCb

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Solution (a)
2
2
11
2
31
1
)( −−
+−
=
zz
zX 1|:| >zROC
As |z|>1, x[n] is causal sequence
2
2
11
2
31 −−
+− zz 1
1
2
2
11
2
31 −−
+− zz
2
2
11
2
3 −−
− zz
1
2
3 −
+ z 2
4
7 −
+ z 3
8
15 −
+ z 4
16
31 −
+ z
3
4
32
4
91
2
3 −−−
+− zzz
3
4
32
4
7 −−
− zz
4
8
73
14
212
4
7 −−−
+− zzz
4
8
73
8
15 −−
− zz
......}1{][ 16
31
8
15
4
7
2
3
↑
=nx

10/7/2009
Solution (b)
2
2
11
2
31
1
)( −−
+−
=
zz
zX 2
1|:| <zROC
As |z|<1/2, x[n] is anti-causal sequence
11
2
32
2
1 +− −−
zz 1
2
231 zz +−
2
23 zz −
2
2z+ 3
6z+ 4
14z+
↑
= }0026143062......{][nx
5
30z+
6
62z+

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Z-transform Pairs

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Z-transform Pairs

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Partial Fraction Method
21
1
861
84
)( −−
−
++
+−
=
zz
z
zH
)21)(41(
84
11
1
−−
−
++
+−
=
zz
z
1
2
1
1
2141
)( −−
+
+
+
=
z
A
z
A
zH
12
21
84
)2/1(
6
1
1
1
4
11
−=−=
+
+−
=
−=
−
−
−
z
z
z
A 8
41
84
1
8
1
1
2
2
11
=−=
+
+−
= −
−=
−
−
−
z
z
z
A
11
21
8
41
12
)( −−
+
+
+
−
=∴
zz
zH
Taking IZT ,we get )(])2(8)4(12[][ nunh nn
−+−−=
Partial Fraction
Expansion

10/7/2009
Another Approach
21
1
861
84
)( −−
−
++
+−
=
zz
z
zH
)86(
)84(
22
1
++
+−
= −
−
zzz
zz
24
)( 21
+
+
+
=∴
z
A
z
A
z
zH
12
2
)84(
2
816
4
1 −==
+
−−
= −
+
−=zz
z
A
Taking IZT ,we get
)(])2(8)4(12[][ nunh nn
−+−−=
)86(
)84(
2
++
−−
=
zz
zz
)2)(4(
)84()(
++
−−
=
zz
z
z
zH
Partial Fraction
Expansion
8
4
)84(
2
)16(
2
2 ==
+
−−
= −−
−=zz
z
A
2
8
4
12)(
+
+
+
−
=∴
zzz
zH
2
8
4
12)(
+
+
+
−=∴
z
z
z
z
zH
11
21
1
8
41
1
12)( −−
+
+
+
−=∴
zz
zH

10/7/2009
Different Pole’s Cases
a) Distinct Real Poles
b) Complex Conjugate and Distinct Poles
c) Repeated Poles

10/7/2009
Distinct Real Poles
)(
)(
)(
zD
zN
zH = Where N(z) and D(z) are polynomials of order m
and n respectively
If m<n,
).......())()((
)(
)(
321 npzpzpzpz
zN
zH
+++++
=
)(
....
)()()(
)(
3
3
2
2
1
1
n
n
pz
A
pz
A
pz
A
pz
A
zH
+
++
+
+
+
+
+
=
where
kpzkk zHpzA −=
+= )()(
If m>=n, use division
)(
)(
)(
zD
zN
QzH +=
such that m’<n, if not repeat division
Q- Quotient
N(z)- remainder
D(z)- Divisor

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Problem
21
21
231
3
)( −−
−−
++
++
=
zz
zz
zH
312
++ −−
zz132 12
++ −−
zz
2
1
2
11
2
32
++ −−
zz
2
51
2
1 +− −
zD(z)
N(z)
Q
132
)( 12
2
51
2
1
2
1
++
+−
+= −−
−
zz
z
zH
)(
)(
)(
zD
zN
QzH +=
122 111
2
51
2
1
2
1
+++
+−
+= −−−
−
zzz
z
)1()1(2 111
2
51
2
1
2
1
+++
+−
+= −−−
−
zzz
z
)1)(12( 11
2
51
2
1
2
1
++
+−
+= −−
−
zz
z
)1()12(
)( 1
2
1
1
2
1
+
+
+
+= −−
z
A
z
A
zH

10/7/2009
Contd..
2
11)1( 1
2
51
2
1
1
−=
−
−
−+
+−
=
z
z
z
A
1
)(
2
1
2
5
2
1
2
1
+−
+−−
= 2
11
2
1
2
5
4
1
=
+
=
1
1
2
51
2
1
2
1)12( −=
−
−
−+
+−
=
z
z
z
A
1)1(2
)1( 2
3
2
1
+−
+−−
= 3
1
2
5
2
1
−=
−
+
=
)1(
3
)12(
)( 11
2
11
2
1
+
−
+
+
+= −−
zz
zH
Taking IZT ,we get
)()1(3)()2()(][ 2
11
2
1 nununnh nn
−−−+= δ
)(])1(3)2([)(][ 2
11
2
1 nunnh nn
−−−+= δ
)1(
3
)21( 11
2
11
2
1
−−
+
−
+
+=
zz

10/7/2009
Complex-Conjugate & Distinct Poles
)52)(2(
)2(
)( 2
2
+++
+
=
zzz
zzz
zX
)52)(2(
2)(
2
2
+++
+
=
zzz
zz
z
zX
)12)(12)(2(
22
jzjzz
zz
−++++
+
=
)12()12()2(
)( 321
jz
A
jz
A
z
A
z
zX
−+
+
++
+
+
=
2
2
1
)12)(12(
2
−=
−+++
+
=
z
jzjz
zz
A 0=
)12(
2
2
)12)(2(
2
jz
jzz
zz
A
+−=
−++
+
=
j−= 2
1
)12(
2
3
)12)(2(
2
jz
jzz
zz
A
−−=
+++
+
=
j+= 2
1

10/7/2009
Contd..
)12()12(
0
)( 2
1
2
1
jz
j
jz
j
z
zX
−+
+
+
++
−
+=
)12(
)(
)12(
)(0)( 2
1
2
1
jz
z
j
jz
z
jzX
−+
++
++
−+=
Taking IZT ,we get
)()2)(()()2)((][ 2
1
2
1
nujjnujjnh nn
−+++−=
)(})2)(()2)({(][ 2
1
2
1
nujjjjnh nn
−+++−=

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Repeated Poles
∑∑ == +
+
+
=
r
k k
k
q
k k
k
pz
B
pz
A
zH
11
)(
where q no. of distinct poles and not repetitive
r no. of repetition of repetitive pole
Ak is calculated by method as described earlier
Bk is calculated by following equation
{ } jpz
r
ikr
kr
k zFpz
dz
d
kr
B −=−
−
+
−
= )()(
)!(
1

10/7/2009
Problem
3
2
)2)(1(
)9(
)(
−−
−
=
zz
zz
zH
3
3
2
211
)2()2()2()1(
)(
−
+
−
+
−
+
−
=
z
B
z
B
z
B
z
A
z
zH
1
3
2
1
)2(
9
=
−
−
=
z
z
z
A 8
1
8
)21(
91
3
2
=
−
−
=
−
−
=
2
3
2
3
2
2
1
)2)(1(
9
)2(
)!13(
1
=






−−
−
−
−
=
z
zz
z
z
dz
d
B
2
2
2
2
)1(
9
2
1
=






−
−
=
z
z
z
dz
d

10/7/2009
Contd..
2
2
22
2
2
)1(
9
)1(2
1
=






−
−
−
=
z
zdz
d
z
z
dz
d
2
2
2
)1(
90)1(
)1(
2)1(
2
1
=












−
−−
−





−
−−
=
z
z
z
dz
d
z
zzz
dz
d
2
2
22
)1(
9
)1(
22
2
1
=












−
−
−





−
−−
=
z
zdz
d
z
zzz
dz
d
2
32
2
)1(
1
)2(9
)1(
2
2
1
=












−
−+





−
−
=
z
zz
zz
dz
d

10/7/2009
Contd..
2
34
22
)1(
1
18
)1(
)1(2)2()22()1(
2
1
=












−
+





−
−−−−−
=
z
zz
zzzzz
2
33
2
)1(
1
18
)1(
42)22)(1(
2
1
=












−
+





−
+−−−
=
z
zz
zzzz
2
33
22
)1(
1
18
)1(
422222
2
1
=












−
+





−
+−+−−
=
z
zz
zzzzz
2
33
)1(
18
)1(
2
2
1
=












−
+





−
=
z
zz 2
3
)1(
16
2
1
=












−
−
=
z
z

10/7/2009
Contd..
2
3
)1(
8
=












−
−
=
z
z
8
)12(
8
3
−=











−
−
= 81 −=B
2
3
2
3
2
)2)(1(
9
)2(
)!23(
1
=






−−
−
−
−
=
z
zz
z
z
dz
d
B
2
2
)1(
9
1
=






−
−
=
z
z
z
dz
d
2
2
)1(
1)9(2)1(
1
=






−
−−−
=
z
z
zzz
dz
d
2
2
22
)1(
)9()22(
=






−
−−−
=
z
z
zzz
2
2
2
)1(
92
=






−
+−
=
z
z
zz
1
)12(
944
2
2
=





−
+−
=
=z
92 =B

10/7/2009
Contd..
2
3
2
3
0
0
3
)2)(1(
9
)2(
)!33(
1
=






−−
−
−
−
=
z
zz
z
z
dz
d
B
2
2
)1(
9
1
=






−
−
=
z
z
z
5
)12(
922
−=





−
−
=
53 −=B
32
)2(
5
)2(
9
)2(
8
)1(
8)(
−
−
+
−
+
−
−
−
=
zzzzz
zH
32
)2()2(
9
)2(
8
)1(
8)(
−
−
−
+
−
−
−
=
z
z
z
z
z
z
z
z
zH
Taking IZT , we get………to be seen after z-transform properties

10/7/2009
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10/7/2009
Cauchy’s Integration (Residue) Method
If X(z) is the z-transform of x(n), then
∫
−
= dzzzX
j
nx n 1
)(
2
1
)(
π
∫= dzzG
j
)(
2
1
π
= sum of residues of G(z) corresponding to poles of G(z)
Residue at pole z=a is given by
azaz zGazR == −= )()(
where
1
)()( −
= n
zzXzG
mth order pole at z=a, will have residue as
az
m
m
m
az zG
m
az
dz
d
R
=
−
−
=






−
−
= )(
)!1(
)(
1
1

10/7/2009
Problem
)2)(1(
10
)(
−−
=
zz
z
zX
)2)(1(
10
)2)(1(
10
)()( 11
−−
=
−−
== −−
zz
z
z
zz
z
zzXzG
n
nn
G(z) has two poles ,z =1 & z=2
x(n) = Residue of G(z) at z=1 + Residue of G(z) at z=2
10
21
1.10
)2)(1(
10
)1(
1
1 −=
−
=
−−
−=
=
=
n
z
n
z
zz
z
zR
n
n
z
n
z
zz
z
zR )2.(10
1
)2.(10
)2)(1(
10
)2(
2
2 ==
−−
−=
=
=
n
nx )2(1010)( +−=

10/7/2009
Problem
2
2
)1(
)(
−
+
=
z
zz
zX
1
2
2
)1(
)( −
−
+
= n
z
z
zz
zG
n
z
z
z
2
)1(
1
−
+
= 2
1
)1( −
+
=
+
z
zz nn
G(z) has one 2nd order pole at ,z =1
1
2
1
1
1 )(
)!12(
)1(
=
=






−
−
=
z
z zG
z
dz
d
R
az
m
m
m
az zG
m
az
dz
d
R
=
−
−
=






−
−
= )(
)!1(
)(
1
1
{ }
1
1
=
+
+=
z
nn
zz
dz
d
{ } 1
1
)1(
=
−
++=
z
nn
nzzn
{ } 12)1(1)1( 1
+=++= −
nnn nn
)()12(
12)(
nun
nnx
+=
+= 0≥n

10/7/2009
Problem
3
)1(2
)13(
)(
−
−
=
z
zz
zX
1
3
)1(2
)13(
)( −
−
−
= n
z
z
zz
zG 3
1
)1(2
3
−
−
=
+
z
zz nn
G(z) has one 3rd order pole at ,z =1
1
3
13
2
2
1
)1(2
3
)!13(
)1(
=
+
= 





−
−
−
−
=
z
nn
z
z
zzz
dz
d
R
1
1
2
2
2.2
3
=
+





 −
=
z
nn
zz
dz
d
( )
1
1
.)1.(3
4
1
=
−
−+=
z
nn
znzn
dz
d
( ) 1
21
)1.(.).1.(3
4
1
=
−−
−−+=
z
nn
znnznn
( ))1.().1.(3
4
1
−−+= nnnn ( ) )(]15.0[133
4
nxnnnn
n
=+=+−+=

10/7/2009
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10/7/2009
Quiz
Find IZT for following z-transforms
)5.0)(1(
1
)(
−−
=
zz
zX
))(1(
)1(
)( a
a
ezz
ze
zX −
−
−−
−
=
Ans )(])5.0(1[2 1
nun−
−
)(]1[ nue an−
−

10/7/2009
z-transform properties
Sr.
No
Given Property ROC
1 Linearity
2 Time shifting R
3 Scaling in z-domain
4 Time Reversal
1/R
)(][&
)(][
22
11
zXnx
zXnx
z
z
→←
→←
2
1
RROC
RROC
=
=
)()()()( 2121 zbXzaXnbxnax z
+→←+ 21 RR ∩
)(][ zXnx z
→← RROC = )(][ 0
0 zXznnx nz −
→←−
)(][ zXnx z
→← RROC = )(][ 00 z
zzn
Xnxz →← Rz || 0
)(][ zXnx z
→← RROC =
)(][ 1
z
z
Xnx →←−

10/7/2009
Contd..
Sr.
No
Given Property ROC
5 Scaling in Time domain
6 Conjugation R
7 Differentiation in z-domain/multiplication by n in t-
domain
R
8 Integration in z-domain/division by n in t-domain R
)(][ zXnx z
→← RROC =



=
0
][
)(
k
n
k
x
nx
otherwise
kofmultipleisnif −−−−−
)(][ kz
k zXnx →←
k
R
1
)(][ zXnx z
→← RROC =
)(][ ***
zXnx z
→←
)(][ zXnx z
→← RROC = ))((][ zX
dz
d
Znnx z
−→←
)(][ zXnx z
→← RROC = dz
z
zX
n
nx
z
z
∫−→←
0
)(][

10/7/2009
Contd..
Sr.
No
Given Property ROC
9 Convolution
10 If x[n] =0 , n<0 Initial Value Theorem
11 Final Value Theorem
12
)(][&
)(][
22
11
zXnx
zXnx
z
z
→←
→←
2
1
RROC
RROC
=
=
)().()(*)( 2121 zXzXnxnx z
→←
21 RR ∩
)(lim]0[ zXx
z ∞→
=



 −
=∞
→
)(
1
lim][
1
zX
z
z
x
z

10/7/2009
Problem on Properties 1-8
)(nuan
1) Find z-transform of
We know
1
1
1
)( −
−
→←
z
nu z
)()( a
zzn
Unua →←and Property 3
1
)(1
1
)( −
−
→←∴
a
z
zn
nua
1
1
1
−
−
=
az

10/7/2009
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10/7/2009
Contd..
2) Find IZT of )1log()( 1−
+= azzX |z|>|a|
We know .....)1log( 432
432
+−+−=+ xxx
xx ∑
∞
=
+
−=
1
1
)1(
n
n
n
n
x
∑
∞
=
−
+
−=
1
1
1 )(
)1(
n
n
n
n
az
)1log()( 1−
+= azzX
n
n
n
n
z
n
a −
∞
=
+
∑






−=
1
1
)1(
n
n
n
n
znu
n
a −
∞
−∞=
+
∑






−−= )1()1( 1
Comparing above equation with z-transform definition,
we get
∑
∞
−∞=
−
=
n
n
znxzX )()(
)1()1()( 1
−−= +
nu
n
a
nx
n
n





−
=
+
0
)1(
)(
1
n
a
nx
n
n
otherwise
n 1≥

10/7/2009
Contd..
3) Find IZT of 3
2
)5.0(
3
)(
−
+
=
z
zz
zX
33
2
)5.0(
3
)5.0(
)(
−
+
−
=
z
z
z
z
zX
3131
1
)5.01(
3
)5.01( −−
−
−
+
−
=
z
z
z
z
)()()( 21 zXzXzX +=
Taking IZT ,we get )()()( 21 nxnxnx += -----------(1)
where
31
1
1
)5.01(
)( −
−
−
=
z
z
zX 312
)5.01(
3)( −
−
=
z
z
zX

10/7/2009
Contd..
We know , property of differentiation in z domain
)(][ zXnx z
→← RROC = ))((][ zX
dz
d
Znnx z
−→←If with then
Thus we have
1
1
1
)( −
−
→←
az
nua zn
1
1
1
)( −
−
−→←
azdz
d
znuna zn
)).1.((
)1(
1
)1( 2
21
−
−
−−
−
−−= za
az
z
21
2
)1( −
−
−
=
az
az
z 21
1
)1( −
−
−
=
az
az

10/7/2009
Contd..
To remove z-1 , we use property of time shifting
)(][ zXnx z
→← RROC = )(][ 0
0 zXznnx nz −
→←−If with then
We get
21
1
1
)1(
)1()1( −
−
+
−
→←++
az
az
znuan zn
21
)1( −
−
=
az
a
Further multiplication by n in time domain is required to make power of D(z) as 3






−
−→←++ −
+
21
1
)1(
)1()1(
az
a
dz
d
znuann zn
)(
)1(
2 2
31
−
−
−
−
−= az
az
z 31
12
)1(
2
−
−
−
=
az
za

10/7/2009
Contd..
31
12
1
)1(
2
)1()1( −
−
+
−
→←++
az
za
nuann zn
31
12
1
2
1
)1(
)1()1( −
−
+
−
→←++
az
za
nuann zn
31
1
1
2
1
)1(
)1()1( −
−
−
−
→←++
az
z
nuann zn
Multiply by 1/2
Divide by a2
If a=0.5, we get
31
1
1
2
1
)5.01(
)1(5.0)1( −
−
−
−
→←++
z
z
nunn zn )(1 zX=
------------(2)
Multiply by 3z-1 in eq. (2) and putting a=0.5, we get
31
2
2
2
3
)1(
3
)(5.0))(1( −
−
−
−
→←−
az
z
nunn zn )(2 zX=
=)(1 nx
=)(2 nx

10/7/2009
Contd..
)()()( 21 nxnxnx +=
)(5.0))(1()1(5.0)1()( 2
2
31
2
1 nunnnunnnx nn −−
−+++=∴

10/7/2009
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10/7/2009
Problem on Initial & Final value theorem
Find the initial and final value of x(n) ,if
)5.0)(1(
)2(
)(
−−
−
=
zz
zz
zX
)5.0)(1(
)2(
lim]0[
−−
−
=
∞→ zz
zz
x
z
We know initial value theorem
)(lim]0[ zXx
z ∞→
=
)5.01)(1(
)21(
lim 112
12
−−
−
∞→ −−
−
=
zzz
zz
z
1
)01)(01(
01
]0[ =
−−
−
=x

10/7/2009
Contd..
We know final value theorem





 −
=∞
→
)(
1
lim][
1
zX
z
z
x
z






−−
−−
=
→ )5.0)(1(
)2(1
lim
1 zz
zz
z
z
z
2
5.0
1
)5.01(1
21
−=
−
=
−
−
=

10/7/2009
Contd..
Find the initial and final value of following functions a) u(n) b) r(n)
Solution (a): we know
1
1
1
)( −
−
→←
z
nu z
)(lim]0[ zUu
z ∞→
= 1
01
1
1
1
lim 1
=
−
=
−
= −∞→ zz
1
1
1 1
1
)1(lim)( −
−
→ −
−=∞
z
zu
z
1=

10/7/2009
Contd..
Solution (b): we know
1
1
1
)( −
−
→←
z
nu z
we also know )()( nnunr =






−
−→←∴ −1
1
1
)(
zdz
d
znnu z
2
21
)1(
1 −
−
−
−= z
z
z
21
1
)1(
)( −
−
−
−
→←∴
z
z
nr z
0
)01()1(
lim)0( 2
1
21
1
=
−
−
=
−
−
= ∞
−
−
∞→ z
z
r
z
∞=
−
=
−
−
=
−
−
−=∞ −
−
−
→ 0
1
)11(
1
)1(
)1(lim)( 21
1
1
1 z
z
zr
z

10/7/2009
Problems on convolution
Find x(n) using convolution theorem if 2
)1(
)(
−
=
z
z
zX
2
)1(
)(
−
=
z
z
zX
)1(
1
)1( −−
=
zz
z
)().()( 21 zXzXzX =
)1(
1
)1(
)( 11 −
−
=
−
=
zz
z
zX
Taking IZT ,we get )()(1 nunx =
)1(
1
)(2
−
=
z
zX
1
1
1
)( −
−
→←
z
nu z
We know
1
1
1
1
)1( −
−
−
→←−
z
znu z
1
1
)1(
−
→←−
z
nu z
=)(2 nx )(2 zX=
)1()(2 −= nunx

10/7/2009
Contd..
Now we know convolution property )().()(*)( 2121 zXzXnxnx z
→←
)(*)()( 21 nxnxnx =∴
)1(*)( −= nunu
∑
∞
−∞=
−−=
k
knuku )1()(
u(k)
u(-k-1)
n<0,x(n)=0
n=0 ,x(n)=0
n=1 ,x(n)=1u(1-k-1)
n=2 ,x(n)=2u(2-k-1)
)()( nnunx =∴

10/7/2009
e-TECHNote
IRDC India
www.irdcindia.com

10/7/2009
Contd..
Find out convolution of two sequences given below
}121{)(&}30112{)() −=−=
↑↑
nhnxa
}11{)(&}1231{)() == nhnxb

10/7/2009
Contd..
}121{)(
}30112{)(
−=
−=
↑
↑
nh
nxSolution (a)
421
302)( −−−
++−+= zzzzX
1
2)( −
−+= zzzH
Using convolution property
)}().({)(*)( 1
zHzXZnhnx −
=
)2)(302()().( 1421 −−−−
−+++−+= zzzzzzHzX
5321
42131
2
6224312
−−−−
−−−−−
−+−−
+−+++−+=
zzzz
zzzzzz

10/7/2009
Contd..
54321
364352)().( −−−−−
−++−−+= zzzzzzzHzX
)}().({)(*)( 1
zHzXZnhnx −
=
}3643152{)(*)( −−−=
↑
nhnx

10/7/2009
e-TECHNote
IRDC India
www.irdcindia.com

10/7/2009
Contd..
Solution (b)
}11{)(
}1231{)(
=
=
nh
nx
321
3231)( −−−
−++= zzzzX
1
1)( −
+= zzH
4321321
23231 −−−−−−−
−+++−++= zzzzzzz
)1)(3231()().( 1321 −−−−
+−++= zzzzzHzX
4321
541 −−−−
−+++= zzzz
}11541{)(*)( −=
↑
nhnx

10/7/2009
End of Chapter 02
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Digital Signal Processing Tutorial:Chapt 2 z transform

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