8. Significance of One Sided ZSignificance of One Sided ZSignificance of One Sided ZSignificance of One Sided ZSignificance of One Sided ZSignificance of One Sided ZSignificance of One Sided ZSignificance of One Sided Z--------TransformTransformTransformTransformTransformTransformTransformTransform
ForFor aa nonnon--causalcausal signal,signal, thethe oneone sidedsided zz--transformtransform isis
notnot uniqueunique..
ForFor aa causalcausal signal,signal, thethe oneone sidedsided zz--transformtransform isis uniqueunique..
9/12/2013 Mahesh J. vadhavaniya 18
ForFor antianti--causalcausal signals,signals, thethe oneone sidedsided zz--transformtransform isis
alwaysalways zerozero..
9. Properties…Properties…Properties…Properties…Properties…Properties…Properties…Properties…
Shifting Property :-
Case 1 : Time Delay
0k,)()()(
)()(
1
z
>
−+→←−
→←
∑=
+−
+
+
+
k
n
nkz
znxzXzknxthen
zXnxIf
1 =n
)(zk)-then x(ncausal,isx(n)case -kz
zXIn +
→←
+
Proof :-
+=
+=−
+
−
−=
−−
∞
=
−
−
−=
−−+
∑
∑∑
)()(
)()()}({
1
0
1
zXzlxz
zlxzlxzknxZ
k
l
lk
l
l
kl
lk
ChangeChange thethe indexindex fromfrom ll toto nn == --ll
9/12/2013 Mahesh J. vadhavaniya 7
10. Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)
Example 1: Determine the one-sided z-transform of
X1(n) = x(n-2) where x(n) = an
ApplyApply thethe shiftingshifting propertyproperty forfor kk == 22,, wewe havehaveProof :
12
2-2
)2()1()(
])2()1()([z=2)}-{x(nZ
−+−
++
−+−+=
−+−+
xzxzXz
ZxzxzX
211
1
2
1
1
21
12
1
)(
1
1
)(,)2(,x(-1)Since
)2()1()(
−−−
−
−
+
−
−−
−+−
++
−
=
−
==−=
−+−+=
aza
az
z
zX
obtainwe
az
zXaxa
xzxzXz
ToTo obtainobtain x(nx(n--k)k) (k>(k>00)) fromfrom x(n),x(n), wewe shouldshould shiftshift x(n)x(n) byby kk
samplessamples toto thethe rightright..
9/12/2013 Mahesh J. vadhavaniya 8
11. Properties…Properties…Properties…Properties…Properties…Properties…Properties…Properties…
Shifting Property :-
Case 2 : Time Advance
0k,)()()(
)()(
1
0
z
>
−→←+
→←
∑
−
=
−+
+
+
+
k
n
nkz
znxzXzknxthen
zXnxIf
0 =n
Proof :-
∑∑
∞
=
−
∞
=
−+
=+=+
kl
lk
n
n
zlxzzknxknxZ )()()}({
0
We have changed the index of summation from n to l = n+k
∑∑∑
∞
=
−
−
=
−
∞
=
−+
+==
kl
l
k
l
l
l
l
zlxzlxzlxzX )()()()(
1
00
−= ∑
−
=
−++
1
0
)()()(
k
n
nk
znxzXzzX
9/12/2013 Mahesh J. vadhavaniya 9
12. Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)
Example 2: Determine the one-sided z-transform of
X2(n) = x(n + 2) where x(n) = an
ApplyApply thethe shiftingshifting propertyproperty forfor kk == 22,, wewe havehaveProof :
zxxzX −−+ ++ 2
])1()0()([z=2)}{x(nZ
azz
az
z
zX
obtainweazzXaxand
zxzxzXz
−−
−
=
−===
−−=
−
+
−+
+
2
1
2
2
1
1
22
1
)(
)1(1)(and,)1(,1x(0)Since
)1()0()(
ToTo obtainobtain x(x(n+kn+k)) (k>(k>00)) fromfrom x(n),x(n), wewe shouldshould shiftshift x(n)x(n) byby kk
samplessamples toto thethe leftleft..
9/12/2013 Mahesh J. vadhavaniya 10
13. Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)
Final Value Theorem :
Proof :
)()1(lim)()(lim
)()(
1zn
z
zXzxnxthen
zXnxIf
+
→∞→
+
−=∞=
→←
+
∑
∞
−
∞=∞ )()]([ n
zxxZ
∑
∑
∑
∑
∞
=
−+
∞
=
−++
∞
=
−
=
−+=−−
−+=−−
−+=−+
∞=∞
0
0
0
0
)]()1([)0()()1(
)]()1([)()]0()([
)]()1([)]()1([
)()]([
n
n
n
n
n
n
n
znxnxxzXz
znxnxzXxzzX
znxnxnxnxZ
zxxZ
TakingTaking thethe limitlimit zz 11 onon bothboth sides,sides,
9/12/2013 Mahesh J. vadhavaniya 11
14. Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)
Final Value Theorem :
exp
)]()1([)]0()()1[(lim
])]()1([[lim)]0()()1[(lim
0
1z
0
1
1z1z
l n termanding tilnow
nxnxxzXz
znxnxxzXz
n
n
∞
=
+
→
∞
=
−
→
+
→
−+=−−
−+=−−
∑
∑
ThisThis theoremtheorem enablesenables usus toto findfind thethe steadysteady statestate valuevalue ofof
x(n)x(n) withoutwithout solvingsolving forfor thethe entireentire sequencesequence..
)()1(lim)(therefore
)0()()]0()()1[(lim
)]}()1([...
)]1()2([)]0()1({[lim)]0()()1[(lim
exp
1
1z
n1z
zXzx
xxxzXz
nxnx
xxxxxzXz
l n termanding tilnow
z
+
→
+
→
∞→
+
→
−=∞
−∞=−−
−++
+−+−=−−
9/12/2013 Mahesh J. vadhavaniya 12
15. Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …
Use the one sided z-transformation to determine y(n),
n≥0, if
GivenGiven
Example 1 :
1)1();(
3
1
)();()1(
2
1
)( =−
=+−= ynunxnxnyny
n
Solution :
)()1(
2
1
)( nxnyny +−=
1
TakingTaking zz--transformtransform onon bothboth sidessides )()]1()([
2
1
)( 1
zXyzYzzY +−+= −
SubstituteSubstitute y(y(--11)=)=11 andand
3
1
)(
3
1
)(
−
=
=
z
z
nuZzX
n
3
1
5.0)(5.0)(
3
1
]1)([
2
1
)(
1
1
−
++=
−
++=
−
−
z
z
zYzzY
z
z
zYzzY
9/12/2013 Mahesh J. vadhavaniya 13
16. Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …
( )
5.0)(
5.01
3
15.01
5.0
)(
3
1
5.0)()5.01(
1
1
1
+=
−
−
+
−
=
−
+=−
−
−
−
zzY
zz
z
z
z
zY
z
z
zYz
( )
3
1
2
5.0
3
5.0
5.0
)(
3
1
2
5.0
3
5.0
5.0)(
5.0
3
15.0
5.0)(
−
−
−
+
−
=
−
−
−
+
−
=
−
−
+
−
=
z
z
z
z
z
z
zY
zzzz
zY
zz
z
zz
zY
9/12/2013 Mahesh J. vadhavaniya 14
17. Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …
( ) ( ) ( )
( ) ( ) )(]3125.05.3[)(
)(]3125.035.05.0[)(
nuny
nuny
nn
nnn
−=
−+=
TakingTaking inverseinverse zz--transform,transform, wewe getget
( ) ( ) )(]3125.05.3[)( nuny
nn
−=
9/12/2013 Mahesh J. vadhavaniya 15
18. The unilateral z transform is well suited to solving difference
equations with initial conditions. For example,
y n + 2[ ]−
3
2
y n +1[ ]+
1
2
y n[ ]= 1/ 4( )n
, for n ≥ 0
Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …
Example 2 :
2 2
y 0[ ]= 10 and y 1[ ]= 4
z transforming both sides,
z2
Y z( )− y 0[ ]− z−1
y 1[ ] −
3
2
z Y z( )− y 0[ ] +
1
2
Y z( ) =
z
z −1/ 4
the initial conditions are called for systematically.
9/12/2013 Mahesh J. vadhavaniya 16
19. Applying initial conditions and solving,
Y z( ) = z
16 / 3
z −1/ 4
+
4
z −1/ 2
+
2 / 3
z −1
and
Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …Solution of Difference Equations …
and
y n[ ]=
16
3
1
4
n
+ 4
1
2
n
+
2
3
u n[ ]
This solution satisfies the difference equation and the initial
conditions.
9/12/2013 Mahesh J. vadhavaniya 17
