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]

zX(z)   [n 1]zn
 z1

1
n
with an ROCconsisting of the entire z -planeexcept z  0.
Ex. 4 Find the z-transform of  [n +1]

X(z)   [n 1]zn
 z
n
with an ROCconsisting of the entire z -planeexcept z  ,
i.e.,thereis a pole at infinity.

4. Infinite duration sequence

7. ScannedwithCamScanner

8. Bilateral Z-Transform
8

10. ScannedwithCamScanner

11. ScannedwithCamScanner

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)Xb(z)Xa
](n)x[ZTb](n)x[ZTa
]z(n)x[bz(n)xa
]z(n)xb[]z(n)xa[
]z(n)xbz(n)xa[
z](n)xb(n)xa[
z](n)xb(n)xa[(n)]xb(n)xaZT[
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 x1(n), x2(n) are two discrete sequences and ZT[ x1(n) ] = X1(z), ZT[ x2(n) ] = X2(z), then
according to linear property of z transform
ZT[ a x1(n) + b x2(n) ] = a X1(z) + b X2(z)
PROOF:
From basic definition of z transform of a sequence x(n)
Replace x(n) by a x1(n) + b x2(n)
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24. 




n
n
x(n)z]x(n)ZT[


































z
1
X
1/zzoftreplacemenwith]x(n)ZT[
z
1
x(m)
)(zx(m)
x(m)z
m,nm,nLet,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)zx(n)
(-n)zx(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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