Signals and Systems

1. Solution of Difference
equations
DTLTI
H(Z)
X(Z) Y(Z)

2. Properties…
Shifting Property :-
Case 1 : Time Delay



z
k
n 
k 
z

(z)  x(n)z , k  0
then x(n  k)z X
If x(n)  X (z)
n1
-k 
In case x(n) is causal, then x(n - k) 
 z X (z)
z
Proof :-









1
 X (z)
 z x(l)z
Z {x(n  k)}  z x(l)z
k
l1
l
k 
 x(l)z
l0
l
lk
l
k 
 Change the index from l to n = -l
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3. 8
Example 1: Determine the one-sided z-transformof
X1(n) = x(n-2) where x(n) = an
Apply the shifting property for k = 2, we have
Proof :
Z 
{x(n
1
1
1
1
 a 1
z 1
 a  2
1  az
z 2
X 
( z ) 
we obtain
1  az
Since x(-1)  a1
, x (  2 )  a  2
, X ( z) 
- 2)} = z -2
[ X 
( z )  x (1) z  x (2)Z 2
]
 z  2
X 
( z )  x(1) z 1
 x(2)
To obtain x(n-k) (k>0) from x(n), we should shift x(n) by k
samples to the right.

4. Properties…
Shifting Property :-
Case 2 : Time Advance





z
k1
n0
 x(n)zn
, k  0
then x(n  k)

z
 zk
X 
(z)
If x(n) X 
(z)
Proof :-
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

 l
n
 zk
x(l)z
Z {x(n k)}  x(n  k)z
n0 lk
We have changed the index of summation from n to l =
n+k 
lk
 x(l)zl
 x(l)zl
X 
(z)  x(l)zl
 k1
l0 l0





k1
n0
 x(n)zn

X (z)  zk
X 
(z)

5. Properties
… (cntd…)
Example 2: Determine the one-sided z-transformof
X2(n) = x(n + 2) where x(n) = an
Apply the shifting property for k = 2, we have
Proof :
1
z2
2
1
1  az
X 
( z)   z 2
 az
1
) we obtain
Since x(0)  1, and x (1)  a, and X 
( z )  1 (1  az
Z
{x(n  2)} = z2
[ X 
(z)  x(0)  x(1)z]
 z 2
X 
(z)  x(0) z 2
 x(1)z
To obtain x(n+k) (k>0) from x(n), we should shift x(n) by k
samples to the left.
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