Chapt 2 Z-transform

10/7/2009e-TECHNote from IRDC India

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Chapt 02

Z-Transform

Digital Signal Processing

PrePrepared by

IRDC India


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Copyright© with Authors. All right reserved

For education purpose.

Commercialization of this material is

strictly not allowed without permission

from author.


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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


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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.


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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

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


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Solution(2)

∑∞

−∞=

−=n

nznxzX )()(

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

nznxzX )()(

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)( −−− ++−++= zzzzzzXZ=0


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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.

( ) ( )nUanxn=

( ) ( )1−−−= nUanx n

)1()()( −−+= nUbnUanx nn

1

2

3



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|



Solution(2)

Sequence is anti-causal ( ) ( )1−−−= nUanxn

∑∞

−∞=

−=n

nznxzX )()(


∑∞

−∞=

−−−−=n

nn znUa )1(

( )∑−

−∞=

−−=1

1)(n

nazzX

U(-n-1)=0 for n>-1

∑∞

=

+++=0

210 .......n

n aaaaWe know that


∑∞

= −=

0 1

1

n

n


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



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Solution(3)

Sequence is non-causal

∑∞

−∞=

−=n

nznxzX )()(


∑∑−

−∞=

−∞

=

− +=1

0 n

nn

n

nn zbza

∑∑−

−∞=

−∞

=

− +=1

1

0

1 )()()(n

n

n

nbzazzX

∑∞

=

+++=0

210 .......n

n aaaaWe know that


∑∞

= −=

0 1

1

n

n


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|



Problems

Find z- transform for followings:

( ) )()(6)(7)( 21

31 nununx

nn −=

( ) )()sin()( 431 nunnx

n π=

nbnx =)(

1.

2.

3.



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



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



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



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



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Inverse z-transform

•Synthetic Division Method

•Partial Fraction Method

•Cauchy’s Integration Method



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 >



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−−



Problem

Using long division method, determine the z-transform of

22

112

31

1)(

−− +−=

zzzX 1|:|) >zROCa 2

1|:|) <zROCb



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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



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+



Z-transform Pairs



Z-transform Pairs



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Z-transform Pairs



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



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



Different Pole’s Cases

a) Distinct Real Poles

b) Complex Conjugate and Distinct Poles

c) Repeated Poles



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



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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



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



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



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 −+++−=



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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



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



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



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



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



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



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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



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)( +−=



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



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=+=+−+=



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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−−



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 →←−



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


=0

][)(

kn

k

xnx

otherwise

kofmultipleisnif −−−−− )(][ kz

k zXnx →←kR

1


)(][ ***zXnx

z→←

)(][ zXnxz→← RROC = ))((][ zX

dz

dZnnx

z −→←

)(][ zXnxz→← RROC = dz

z

zX

n

nxz

z

∫−→←0

)(][



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



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



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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≥



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



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



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



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



Contd..

)()()( 21 nxnxnx +=

)(5.0))(1()1(5.0)1()( 22

312

1 nunnnunnnx nn −− −+++=∴



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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



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−=

−=

−

−=



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=



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



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



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 =∴



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Contd..

Find out convolution of two sequences given below

}121{)(&}30112{)() −=−=↑↑

nhnxa

}11{)(&}1231{)() == nhnxb



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



Contd..

54321 364352)().( −−−−− −++−−+= zzzzzzzHzX

)}().({)(*)( 1zHzXZnhnx

−=

}3643152{)(*)( −−−=↑

nhnx



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Contd..

Solution (b)

}11{)(

}1231{)(

=

=

nh

nx

321 3231)( −−− −++= zzzzX

11)( −+= zzH

4321321 23231 −−−−−−− −+++−++= zzzzzzz

)1)(3231()().( 1321 −−−− +−++= zzzzzHzX

4321 541 −−−− −+++= zzzz

}11541{)(*)( −=↑

nhnx



End of Chapter 02

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Documents

Chapt 2 Z-transform