Properties of z transform day 2

UNIT II DISCRETE TIME SYSTEM ANALYSIS 6+6
Z-transform and its properties, inverse z-transforms; difference equation – Solution by z transform,application to discrete systems - Stability analysis, frequency response –Convolution – Discrete Time Fourier transform , magnitude and phase representation.

1. WEBINAR ON DISCRETE TIME SYSTEM
ANALYSIS

Day 2 – 21/7/2020


1
K.Vijay Anand - Associate Professor

Department of Electronics and Instrumentation Engineering

R.M.K Engineering College

2. •Unit step Function
2

3. Ex. 3 Find the z-transform of [n -1]

z
X ( z)   [n 1] z
n
 z
1

1
n
with an ROC consisting of the entire z - plane except z  0 .

Ex. 4 Find the z-transform of  [n +1]

X ( z)   [n 1] z
n
 z
n
with an ROC consisting of the entire z - plane except z  ,
i.e., there is a pole at infinity.

4. Infinite duration sequence

7. Scanned with CamScanner

8. Bilateral Z-Transform
8

10. Scanned with CamScanner

11. Scanned with CamScanner

12. DISCUSSION FORUM

21. Properties of Z transform
•Linearity Property
•Time Shifting Property
•Multiplication by Exponential
Sequence Property
•Time Reversal Property
•Differentiation in Z-Domain OR
Multiplication by n Property
•Convolution Property
•Conjugation property
•Intial Value theorem
•Final Value theorem

22. 




n
n
x(n)z ] x(n)ZT[ (z)X b (z)X a
] (n) x[ZT b ] (n) x[ZT a
]z (n) x[bz (n) xa
]z (n) xb []z (n) xa [
]z (n) xb z (n) xa [
z ] (n) xb (n) xa [
z ] (n) xb (n) xa [ (n)] xb (n) xa ZT[
21
21
n
n
2
n
n
1
n
2
n
1
n
2
n
1
n
21
n
2121







 
 
























n n
n
n
n LINEAR PROPERTY:
Let x
1
(n), x
2
(n) are two discrete sequences and ZT[ x
1
(n) ] = X
1
(z), ZT[ x
2
(n) ] = X
2
(z), then
according to linear property of z transform
ZT[ a x
1
(n) + b x
2
(n) ] = a X
1
(z) + b X
2
(z)
PROOF:
From basic definition of z transform of a sequence x(n)

Replace x(n) by a x
1
(n) + b x
2
(n)


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24. 




n
n
x(n)z ] x(n)ZT[ 

































z
1
X
1/zz of treplacemen with] x(n)ZT[
z
1
x(m)
)(z x(m)
x(m)z
m,n m,n Let ,x(-n)z ] x(-n)ZT[
m
m
m
m1
m
m)(
n
n TIME REVERSAL PROPERTY :
Let x(n) be a discrete time sequence and ZT[ x(n) ] = X(z), then according to time reversal property of z transform
ZT[ x(– n) ] = X(1/z)
PROOF:
From basic definition of z transform of a sequence x(n)


Replace x(n) by x(– n)


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26. 




n
n
x(n)zX(z) ] x(n)ZT[ ] X(z) [
dz
d
] nx(n) ZT[
] nx(n) ZT[
z
1
] X(z) [
dz
d
z ] x(n)n [z
z(-n)z x(n)
(-n)z x(n)
z
dz
d
x(n)] X(z) [
dz
d
n1
1n
1n
n
z
n
n
n
n






















 DERIVATIVE PROPERTY :
Let x(n) be a discrete time sequence and ZT[ x(n) ] = X(z), then according to derivative property of z transform
ZT[ n x(n) ] = - d/dz [ X(z) ]
PROOF:
From basic definition of z transform of a sequence x(n)

Differentiate w.r.t z


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30. X(z)
z
Lt
x(n)
0n
Lt
x(0)



 ...........
x(2)x(1)
x(0)
...........x(2)zx(1)zx(0)
x(n)zX(z)
x(n)z ] x(n)ZT[
2
21
0
n
n













zz
n
n X(z)
z
Lt
x(n)
0n
Lt
)0(
)0(
...............00x(0)
...........
x(2)x(1)
x(0)X(z)
z
Lt
2












x
x INITIAL VALUE THEOREM:
Let x(n) be a discrete time causal sequence and ZT[ x(n) ] = X(z), then according to
initial value theorem of z transform

PROOF:
From basic definition of z transform of a sequence x(n)

Apply as z  

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