10/7/2009e-TECHNote from IRDC India
Chapt 02
Z-Transform
Digital Signal Processing
PrePrepared by
IRDC India
10/7/2009e-TECHNote from IRDC India
Copyright© with Authors. All right reserved
For education purpose.
Commercialization of this material is
strictly not allowed without permission
from author.
10/7/2009e-TECHNote from IRDC India
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
10/7/2009e-TECHNote from IRDC India
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
nznxzX )()(
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.
10/7/2009e-TECHNote from IRDC India
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
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Solution(1)
∑∞
−∞=
−=n
nznxzX )()(
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
10/7/2009e-TECHNote from IRDC India
Solution(2)
∑∞
−∞=
−=n
nznxzX )()(
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| = ∞.
Because at z = ∞, X(z) = ∞.
Thus ROC is entire z-plane except |z| = ∞.
ROC
321 2123)( zzzzX +++=
32 223)( zzzzX +++=Z= ∞
10/7/2009e-TECHNote from IRDC India
Solution(3)
∑∞
−∞=
−=n
nznxzX )()(
x (n) = { 1 2 1 -2 3 1}
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| = ∞.
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)( −−− ++−++= zzzzzzXZ=0
10/7/2009e-TECHNote from IRDC India
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
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
z-Transform of infinite duration signal
Find the z-transform for following discrete time sequences. Also mention
ROC for all the cases.
( ) ( )nUanxn=
( ) ( )1−−−= nUanx n
)1()()( −−+= nUbnUanx nn
1
2
3
10/7/2009e-TECHNote from IRDC India
Solution(1)
Sequence is causal ( ) ( )nUanxn=
∑∞
−∞=
−=n
nznxzX )()(
Z.T. for the given sequence x (n) is defined as
∑∞
−∞=
−=n
nn znUa )(
∑∞
=
−=0
)(n
nnzazX
( )∑∞
=
−=0
1)(n
nazzX
U(n)=0 for n<0
∑∞
=
+++=0
210 .......n
n aaaaWe know that
Series converges iff |a|<1
∑∞
= −=
0 1
1
n
n
aaWe also know 1<afor
Thus, x (z) converges when | a z–1 | < 1
( )11
1−−
=az
zX
( )az
zzX
−=
for | a z–1 | < 1
for | a /z | < 1
for | z | > |a|
ROC is outside the
circle |z|=|a|
10/7/2009e-TECHNote from IRDC India
Solution(2)
Sequence is anti-causal ( ) ( )1−−−= nUanxn
∑∞
−∞=
−=n
nznxzX )()(
Z.T. for the given sequence x (n) is defined as
∑∞
−∞=
−−−−=n
nn znUa )1(
( )∑−
−∞=
−−=1
1)(n
nazzX
U(-n-1)=0 for n>-1
∑∞
=
+++=0
210 .......n
n aaaaWe know that
Series converges iff |a|<1
∑∞
= −=
0 1
1
n
n
aaWe also know 1<afor
Thus, x (z) converges when | a–1 z | < 1
( )za
zazX
1
1
1 −
−
−−=
( )az
zzX
−=
for | a–1 z | < 1
for | z /a | < 1
for | z | < |a|
ROC is inside the
circle |z|=|a|
Put n=-m
( ) ( )∑∑∞=
−
∞=
−− −=−=1
11
1)(m
m
m
m
zaazzX
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Solution(3)
Sequence is non-causal
∑∞
−∞=
−=n
nznxzX )()(
Z.T. for the given sequence x (n) is defined as
∑∑−
−∞=
−∞
=
− +=1
0 n
nn
n
nn zbza
∑∑−
−∞=
−∞
=
− +=1
1
0
1 )()()(n
n
n
nbzazzX
∑∞
=
+++=0
210 .......n
n aaaaWe know that
Series converges iff |a|<1
∑∞
= −=
0 1
1
n
n
aaWe also know 1<afor
Thus, x (z) converges when | a z–1 | < 1 & |b-1z|<1
( )zb
zb
azzX
1
1
1 11
1−
−
− −+
−=
( )zb
z
az
zzX
−+
−=
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
nzbazzX
ROC is |b|>| z | > |a|
10/7/2009e-TECHNote from IRDC India
Problems
Find z- transform for followings:
( ) )()(6)(7)( 21
31 nununx
nn −=
( ) )()sin()( 431 nunnx
n π=
nbnx =)(
1.
2.
3.
10/7/2009e-TECHNote from IRDC India
Solution(1)
( ) )()(6)(7)( 21
31 nununx
nn −=
∑∑∞
=
−∞
=
− −=0
21
0
31 )(6)(7)(
n
nn
n
nnzzzX
∑∑∞
=
−∞
=
− −=0
12
1
0
13
1 )(6)(7)(n
n
n
nzzzX
12
113
1 1
6
1
7−− −
−−
=zz
113
1 <−z 11
21 <−
z&
)1)(1(
671
211
31
13
612
7
−−
−−
−−
+−−=
zz
zz
))((
))(1(
21
31
13
62
72
−−
+−+=
−
zz
zz
X&/ by z2
))((
)(
21
31
23
−−
−=
zz
zz
3/2
31>z &
21>z
11/2
ROC is outside the circle |z|=1/2
10/7/2009e-TECHNote from IRDC India
Solution(2)
( ) ][)sin(][ 431 nunnx
n π=
][2
)(][44
31 nu
j
eenx
njnj
n
−=
− ππ
][)(][)( 44
31
21
31
21 nuenue
njn
j
njn
j
ππ −−=
13
121
13
121
44 1
1
1
1)(
−−− −−
−=
zezezX
njjnjj ππ
ROC1
311
31 44 &1 −−− < zeze
njnj ππ
31>z &
31>z
31>z
11/3
ROC is outside the circle |z|=1/3
10/7/2009e-TECHNote from IRDC India
Solution(3)
nbnx =)( b>0
)1()()( −−+= − nubnubnx nn
We know
11
1][ −−⇔
bz
n nub bz >||
111
1]1[ −−−
− ⇔−−zb
n nub bz 1|| <and
111 1
1
1
1)( −−− −−−=
zbbzzX bzb 1|| <<
For b>1, there are no values of z that satisfy ROC b1/b
|z|=1
10 << b
10/7/2009e-TECHNote from IRDC India
Problem:
)21)(1(
1)(
113
1 −− −−=
zzzX
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 signali.e. causal signal
If x[n] is double sided
signali.e. non-causal signal
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Inverse z-transform
•Synthetic Division Method
•Partial Fraction Method
•Cauchy’s Integration Method
10/7/2009e-TECHNote from IRDC India
Synthetic division Example(1)
11
1)(
−−=
azzX
11 −− az 11
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 z22 −+ za
.....}1{][ 432aaaanx
↑=
|||| az >
10/7/2009e-TECHNote from IRDC India
Synthetic division Example(2)
11
1)(
−−=
azzX
11 +− −az 1
za11 −−
za1−
221 zaza −− −
22 za−
za1−−
3322zaza
−− −
33za−
•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 z22 za−−
.......}0.......{][ 123
↑
−−− −−−= aaanx
|||| az <
33za−−
10/7/2009e-TECHNote from IRDC India
Problem
Using long division method, determine the z-transform of
22
112
31
1)(
−− +−=
zzzX 1|:|) >zROCa 2
1|:|) <zROCb
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Solution (a)
22
112
31
1)(
−− +−=
zzzX 1|:| >zROC
As |z|>1, x[n] is causal sequence
22
112
31 −− +− zz 1
1
22
112
31 −− +− zz
22
112
3 −− − zz
12
3 −+ z 24
7 −+ z 38
15 −+ z 416
31 −+ z
34
324
912
3 −−− +− zzz
34
324
7 −− − zz4
873
14212
47 −−− +− zzz
48
738
15 −− − zz
......}1{][ 1631
815
47
23
↑=nx
10/7/2009e-TECHNote from IRDC India
Solution (b)
22
112
31
1)(
−− +−=
zzzX 2
1|:| <zROC
As |z|<1/2, x[n] is anti-causal sequence
112
322
1 +− −− zz 12231 zz +−
223 zz −
22z+ 36z+ 4
14z+
↑
= }0026143062......{][nx
530z+662z+
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Partial Fraction Method
21
1
861
84)(
−−
−
++
+−=
zz
zzH
)21)(41(
8411
1
−−
−
++
+−=
zz
z
1
2
1
1
2141)(
−− ++
+=
z
A
z
AzH
1221
84)2/1(
61
1
1
411
−=−=+
+−=
−=
−
−
−z
z
zA 8
41
841
81
1
2
211
=−=+
+−= −
−=
−
−
−z
z
zA
11 21
8
41
12)(
−− ++
+
−=∴
zzzH
Taking IZT ,we get )(])2(8)4(12[][ nunhnn −+−−=
Partial Fraction
Expansion
10/7/2009e-TECHNote from IRDC India
Another Approach
21
1
861
84)(
−−
−
++
+−=
zz
zzH
)86(
)84(22
1
++
+−=
−
−
zzz
zz
24
)( 21
++
+=∴
z
A
z
A
z
zH
122
)84(2816
4
1 −==+
−−= −
+
−=zz
zA
Taking IZT ,we get
)(])2(8)4(12[][ nunhnn −+−−=
)86(
)84(2 ++
−−=
zz
zz
)2)(4(
)84()(
++
−−=
zz
z
z
zH
Partial Fraction
Expansion
84
)84(2
)16(
2
2 ==+
−−= −−
−=zz
zA
2
8
4
12)(
++
+
−=∴
zzz
zH
28
412)(
++
+−=∴
z
z
z
zzH
11 21
18
41
112)(
−− ++
+−=∴
zzzH
10/7/2009e-TECHNote from IRDC India
Different Pole’s Cases
a) Distinct Real Poles
b) Complex Conjugate and Distinct Poles
c) Repeated Poles
10/7/2009e-TECHNote from IRDC India
Distinct Real Poles
)(
)()(
zD
zNzH = Where N(z) and D(z) are polynomials of order m
and n respectively
If m<n,
).......())()((
)()(
321 npzpzpzpz
zNzH
+++++=
)(....
)()()()(
3
3
2
2
1
1
n
n
pz
A
pz
A
pz
A
pz
AzH
+++
++
++
+=
wherekpzkk zHpzA
−=+= )()(
If m>=n, use division
)(
)()(
zD
zNQzH +=
such that m’<n, if not repeat division
Q- Quotient
N(z)- remainder
D(z)- Divisor
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Problem
21
21
231
3)(
−−
−−
++
++=
zz
zzzH
312 ++ −− zz132 12 ++ −− zz
21
211
232 ++ −− zz
251
21 +− −
zD(z)
N(z)
Q
132)(
12
251
21
21
++
+−+=
−−
−
zz
zzH
)(
)()(
zD
zNQzH +=
122 111
251
21
21
+++
+−+=
−−−
−
zzz
z
)1()1(2 111
251
21
21
+++
+−+=
−−−
−
zzz
z
)1)(12( 11
251
21
21
++
+−+=
−−
−
zz
z
)1()12()(
1
2
1
1
21
++
++=
−−z
A
z
AzH
10/7/2009e-TECHNote from IRDC India
Contd..
211)1( 1
251
21
1
−=
−
−
−+
+−=
zz
zA
1
)(
21
25
21
21
+−
+−−= 2
11
21
25
41
=+
=
1
1
251
21
21)12(
−=
−
−
−+
+−=
zz
zA
1)1(2
)1( 23
21
+−
+−−= 3
1
25
21
−=−
+=
)1(
3
)12()(
11
211
21
+
−+
++=
−−zz
zH
Taking IZT ,we get)()1(3)()2()(][ 2
112
1 nununnhnn −−−+= δ
)(])1(3)2([)(][ 211
21 nunnh
nn −−−+= δ
)1(
3
)21( 11
211
21
−− +−
++=
zz
10/7/2009e-TECHNote from IRDC India
Complex-Conjugate & Distinct Poles
)52)(2(
)2()(
2
2
+++
+=
zzz
zzzzX
)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
−=−+++
+=
zjzjz
zzA 0=
)12(
2
2)12)(2(
2
jzjzz
zzA
+−=−++
+=
j−=21
)12(
2
3)12)(2(
2
jzjzz
zzA
−−=+++
+=
j+=21
10/7/2009e-TECHNote from IRDC India
Contd..
)12()12(0
)( 21
21
jz
j
jz
j
z
zX
−+
++
++
−+=
)12()(
)12()(0)(
21
21
jz
zj
jz
zjzX
−+++
++−+=
Taking IZT ,we get
)()2)(()()2)((][21
21 nujjnujjnh nn −+++−=
)(})2)(()2)({(][21
21 nujjjjnh nn −+++−=
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Repeated Poles
∑∑== +
++
=r
k k
k
q
k k
k
pz
B
pz
AzH
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 zFpzdz
d
krB −=−
−
+−
= )()()!(
1
10/7/2009e-TECHNote from IRDC India
Problem
3
2
)2)(1(
)9()(
−−
−=
zz
zzzH
3
3
2
211
)2()2()2()1(
)(
−+
−+
−+
−=
z
B
z
B
z
B
z
A
z
zH
1
3
2
1)2(
9
=−
−=
zz
zA 8
1
8
)21(
913
2
=−
−=
−
−=
2
3
23
2
2
1)2)(1(
9)2(
)!13(
1
=
−−
−−
−=
zzz
zz
dz
dB
2
2
2
2
)1(
9
2
1
=
−
−=
zz
z
dz
d
10/7/2009e-TECHNote from IRDC India
Contd..
2
2
22
2
2
)1(
9
)1(2
1
=
−−
−=
zzdz
d
z
z
dz
d
2
2
2
)1(
90)1(
)1(
2)1(
2
1
=
−
−−−
−
−−=
zz
z
dz
d
z
zzz
dz
d
2
2
22
)1(
9
)1(
22
2
1
=
−
−−
−
−−=
zzdz
d
z
zzz
dz
d
2
32
2
)1(
1)2(9
)1(
2
2
1
=
−−+
−
−=
zzz
zz
dz
d
10/7/2009e-TECHNote from IRDC India
Contd..
2
34
22
)1(
118
)1(
)1(2)2()22()1(
2
1
=
−+
−
−−−−−=
zzz
zzzzz
2
33
2
)1(
118
)1(
42)22)(1(
2
1
=
−+
−
+−−−=
zzz
zzzz
2
33
22
)1(
118
)1(
422222
2
1
=
−+
−
+−+−−=
zzz
zzzzz
2
33)1(
18
)1(
2
2
1
=
−+
−=
zzz
2
3)1(
16
2
1
=
−
−=
zz
10/7/2009e-TECHNote from IRDC India
Contd..
2
3)1(
8
=
−
−=
zz
8)12(
83
−=
−
−= 81 −=B
2
3
23
2)2)(1(
9)2(
)!23(
1
=
−−
−−
−=
zzz
zz
dz
dB
2
2
)1(
91
=
−
−=
zz
z
dz
d
2
2
)1(
1)9(2)1(1
=
−
−−−=
zz
zzz
dz
d
2
2
22
)1(
)9()22(
=
−
−−−=
zz
zzz
2
2
2
)1(
92
=
−
+−=
zz
zz1
)12(
944
2
2=
−
+−=
=z
92 =B
10/7/2009e-TECHNote from IRDC India
Contd..
2
3
23
0
0
3)2)(1(
9)2(
)!33(
1
=
−−
−−
−=
zzz
zz
dz
dB
2
2
)1(
91
=
−
−=
zz
z5
)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
zzH
Taking IZT , we get………to be seen after z-transform properties
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Cauchy’s Integration (Residue) Method
If X(z) is the z-transform of x(n), then
∫−= dzzzX
jnx n 1)(
2
1)(
π
∫= dzzGj
)(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== −= )()(
where1)()( −= nzzXzG
mth order pole at z=a, will have residue as
az
m
m
m
az zGm
az
dz
dR
=
−
−
=
−
−= )(
)!1(
)(1
1
10/7/2009e-TECHNote from IRDC India
Problem
)2)(1(
10)(
−−=
zz
zzX
)2)(1(
10
)2)(1(
10)()( 11
−−=
−−== −−
zz
zz
zz
zzzXzG
nnn
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
1021
1.10
)2)(1(
10)1(
1
1 −=−
=−−
−==
=
n
z
n
zzz
zzR
nn
z
n
zzz
zzR )2.(10
1
)2.(10
)2)(1(
10)2(
2
2 ==−−
−==
=
nnx )2(1010)( +−=
10/7/2009e-TECHNote from IRDC India
Problem
2
2
)1()(
−
+=
z
zzzX
1
2
2
)1()(
−
−
+= n
zz
zzzG
nz
z
z2)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 zGz
dz
dR
az
m
m
m
az zGm
az
dz
dR
=
−
−
=
−
−= )(
)!1(
)(1
1
{ }1
1
=
+ +=z
nn zzdz
d { }1
1)1(=
−++=z
nn nzzn
{ } 12)1(1)1( 1 +=++= − nnn nn
)()12(
12)(
nun
nnx
+=
+= 0≥n
10/7/2009e-TECHNote from IRDC India
Problem
3)1(2
)13()(
−
−=
z
zzzX
1
3)1(2
)13()( −
−
−= nz
z
zzzG
3
1
)1(2
3
−
−=
+
z
zznn
G(z) has one 3rd order pole at ,z =1
1
3
13
2
2
1)1(2
3
)!13(
)1(
=
+
=
−
−
−
−=
z
nn
zz
zzz
dz
dR
1
1
2
2
2.2
3
=
+
−=
z
nnzz
dz
d
( )1
1.)1.(34
1
=
−−+=z
nnznzn
dz
d ( )1
21 )1.(.).1.(34
1
=
−− −−+=z
nnznnznn
( ))1.().1.(34
1−−+= nnnn ( ) )(]15.0[133
4nxnnnn
n=+=+−+=
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Quiz
Find IZT for following z-transforms
)5.0)(1(
1)(
−−=
zzzX
))(1(
)1()(
a
a
ezz
zezX
−
−
−−
−=
Ans )(])5.0(1[21
nun−−
)(]1[ nuean−−
10/7/2009e-TECHNote from IRDC India
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 zbXzaXnbxnaxz +→←+ 21 RR ∩
)(][ zXnxz→← RROC = )(][ 0
0 zXznnxnz −→←−
)(][ zXnxz→← RROC = )(][
00 zzzn
Xnxz →← Rz || 0
)(][ zXnxz→← RROC =
)(][ 1z
zXnx →←−
10/7/2009e-TECHNote from IRDC India
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
)(][ zXnxz→← RROC =
=0
][)(
kn
k
xnx
otherwise
kofmultipleisnif −−−−− )(][ kz
k zXnx →←kR
1
)(][ zXnxz→← RROC =
)(][ ***zXnx
z→←
)(][ zXnxz→← RROC = ))((][ zX
dz
dZnnx
z −→←
)(][ zXnxz→← RROC = dz
z
zX
n
nxz
z
∫−→←0
)(][
10/7/2009e-TECHNote from IRDC India
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 zXzXnxnxz→←
21 RR ∩
)(lim]0[ zXxz ∞→
=
−=∞
→)(
1lim][
1zX
z
zx
z
10/7/2009e-TECHNote from IRDC India
Problem on Properties 1-8
)(nuan1) Find z-transform of
We know11
1)(
−−→←
znu
z
)()( azzn Unua →←and Property 3
1)(1
1)(
−−→←∴
az
zn nua
11
1−−
=az
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Contd..
2) Find IZT of )1log()( 1−+= azzX |z|>|a|
We know .....)1log(432
432
+−+−=+ xxxxx ∑∞
=
+−=1
1)1(n
nn
n
x
∑∞
=
−+−=
1
11 )(
)1(n
nn
n
az
)1log()( 1−+= azzX
n
n
nn
zn
a −∞
=
+∑
−=1
1)1(
n
n
nn
znun
a −∞
−∞=
+∑
−−= )1()1( 1
Comparing above equation with z-transform definition,
we get
∑∞
−∞=
−=n
nznxzX )()(
)1()1()( 1 −−= +nu
n
anx
nn
−
=+
0
)1()(
1
n
a
nx
nn
otherwise
n 1≥
10/7/2009e-TECHNote from IRDC India
Contd..
3) Find IZT of 3
2
)5.0(
3)(
−
+=
z
zzzX
33
2
)5.0(
3
)5.0()(
−+
−=
z
z
z
zzX
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
zzX 312
)5.01(3)(
−−=
z
zzX
10/7/2009e-TECHNote from IRDC India
Contd..
We know , property of differentiation in z domain
)(][ zXnxz→← RROC = ))((][ zX
dz
dZnnx
z −→←If with then
Thus we have 11
1)(
−−→←
aznua
zn
11
1)(
−−−→←
azdz
dznuna
zn
)).1.(()1(
1)1( 2
21
−−
−−−
−−= zaaz
z
21
2
)1( −
−
−=
az
azz
21
1
)1( −
−
−=
az
az
10/7/2009e-TECHNote from IRDC India
Contd..
To remove z-1 , we use property of time shifting
)(][ zXnxz→← RROC = )(][ 0
0 zXznnxnz −→←−If with then
We get
21
11
)1()1()1(
−
−+
−→←++
az
azznuan
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
dznuann
zn
)()1(
2 2
31
−−−
−−= az
azz 31
12
)1(
2−
−
−=
az
za
10/7/2009e-TECHNote from IRDC India
Contd..
31
121
)1(
2)1()1(
−
−+
−→←++
az
zanuann
zn
31
121
21
)1()1()1(
−
−+
−→←++
az
zanuann
zn
31
11
21
)1()1()1(
−
−−
−→←++
az
znuann
zn
Multiply by 1/2
Divide by a2
If a=0.5, we get
31
11
21
)5.01()1(5.0)1(
−
−−
−→←++
z
znunn
zn )(1 zX=
------------(2)
Multiply by 3z-1 in eq. (2) and putting a=0.5, we get
31
22
23
)1(
3)(5.0))(1(
−
−−
−→←−
az
znunn
zn )(2 zX=
=)(1 nx
=)(2 nx
10/7/2009e-TECHNote from IRDC India
Contd..
)()()( 21 nxnxnx +=
)(5.0))(1()1(5.0)1()( 22
312
1 nunnnunnnx nn −− −+++=∴
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Problem on Initial & Final value theorem
Find the initial and final value of x(n) ,if )5.0)(1(
)2()(
−−
−=
zz
zzzX
)5.0)(1(
)2(lim]0[
−−
−=
∞→ zz
zzx
z
We know initial value theorem
)(lim]0[ zXxz ∞→
=
)5.01)(1(
)21(lim
112
12
−−
−
∞→ −−
−=
zzz
zz
z
1)01)(01(
01]0[ =
−−
−=x
10/7/2009e-TECHNote from IRDC India
Contd..
We know final value theorem
−=∞
→)(
1lim][
1zX
z
zx
z
−−
−−=
→ )5.0)(1(
)2(1lim
1 zz
zz
z
z
z
25.0
1
)5.01(1
21−=
−=
−
−=
10/7/2009e-TECHNote from IRDC India
Contd..
Find the initial and final value of following functions a) u(n) b) r(n)
Solution (a): we know11
1)(
−−→←
znu
z
)(lim]0[ zUuz ∞→
= 101
1
1
1lim
1=
−=
−=
−∞→ zz
1
1
1 1
1)1(lim)(
−−
→ −−=∞
zzu
z1=
10/7/2009e-TECHNote from IRDC India
Contd..
Solution (b): we know11
1)(
−−→←
znu
z
we also know )()( nnunr =
−−→←∴
−11
1)(
zdz
dznnu
z
2
21)1(
1 −−−
−= zz
z
21
1
)1()(
−
−
−
−→←∴
z
znr
z
0)01()1(
lim)0(2
1
21
1
=−
−=
−
−= ∞
−
−
∞→ z
zr
z
∞=−
=−
−=
−
−−=∞
−
−−
→ 0
1
)11(
1
)1()1(lim)(
21
11
1 z
zzr
z
10/7/2009e-TECHNote from IRDC India
Problems on convolution
Find x(n) using convolution theorem if 2)1()(
−=
z
zzX
2)1()(
−=
z
zzX
)1(
1
)1( −−=
zz
z
)().()( 21 zXzXzX =
)1(
1
)1()(
11 −−=
−=
zz
zzX
Taking IZT ,we get )()(1 nunx =
)1(
1)(2 −
=z
zX
11
1)(
−−→←
znu
z
We know
1
1
1
1)1(
−−
−→←−
zznu
z
1
1)1(
−→←−
znu
z=)(2 nx )(2 zX=
)1()(2 −= nunx
10/7/2009e-TECHNote from IRDC India
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 =∴
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Contd..
Find out convolution of two sequences given below
}121{)(&}30112{)() −=−=↑↑
nhnxa
}11{)(&}1231{)() == nhnxb
10/7/2009e-TECHNote from IRDC India
Contd..
}121{)(
}30112{)(
−=
−=
↑
↑
nh
nxSolution (a)
421 302)( −−− ++−+= zzzzX
12)( −−+= zzzH
Using convolution property
)}().({)(*)( 1 zHzXZnhnx −=
)2)(302()().( 1421 −−−− −+++−+= zzzzzzHzX
5321
42131
2
6224312
−−−−
−−−−−
−+−−
+−+++−+=
zzzz
zzzzzz
10/7/2009e-TECHNote from IRDC India
Contd..
54321 364352)().( −−−−− −++−−+= zzzzzzzHzX
)}().({)(*)( 1zHzXZnhnx
−=
}3643152{)(*)( −−−=↑
nhnx
10/7/2009e-TECHNote from IRDC India
e-TECHNote
This PPT is sponsored by IRDC India
www.irdcindia.com
10/7/2009e-TECHNote from IRDC India
Contd..
Solution (b)
}11{)(
}1231{)(
=
=
nh
nx
321 3231)( −−− −++= zzzzX
11)( −+= zzH
4321321 23231 −−−−−−− −+++−++= zzzzzzz
)1)(3231()().( 1321 −−−− +−++= zzzzzHzX
4321 541 −−−− −+++= zzzz
}11541{)(*)( −=↑
nhnx
Top Related