The document discusses the one-sided z-transform, which is used to solve difference equations with initial conditions. It defines the one-sided z-transform and explains that it contains information only about causal signals that are zero for negative time values. Some key properties are described, including shifting properties for time delay and advance, and the final value theorem. An example shows how to use the one-sided z-transform to solve a simple difference equation.

2. Objectives…Objectives…Objectives…Objectives…Objectives…Objectives…Objectives…Objectives…
SignificanceSignificance ofof OneOne SidedSided (Unilateral)(Unilateral) ZZ –– TransformTransform..
DefinitionDefinition..
PropertiesProperties..
9/12/2013 Mahesh J. vadhavaniya 1
PropertiesProperties..
SolutionSolution ofof DifferenceDifference EquationsEquations..
ShiftingShifting
•• DelayDelay
•• AdvanceAdvance
FinalFinal ValueValue TheoremTheorem

3. 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
TheThe twotwo sidedsided zz--transformtransform –– signalssignals areare specifiedspecified forfor
thethe entireentire timetime rangerange ∞<<∞ n-
CanCan notnot bebe usedused toto evaluateevaluate thethe outputoutput ofof nonnon--relaxedrelaxed
systemssystems..
NonNon--relaxedrelaxed areare systemssystems describeddescribed byby differencedifferenceNonNon--relaxedrelaxed areare systemssystems describeddescribed byby differencedifference
equationsequations withwith nonzerononzero initialinitial conditionsconditions..
We’ll Develop the one sided z-transform to solve
difference equations with initial conditions.
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SinceSince thethe inputinput isis appliedapplied atat aa finitefinite timetime (n(n00),), bothboth thethe
inputinput andand outputoutput signalssignals areare specifiedspecified forfor n≥n≥ nn00,, butbut byby oo
meansmeans areare zerozero forfor nn << nn00 ..

4. ∑
∞
=
−+
=
0
)()(
n
n
znxzX
Definition…Definition…Definition…Definition…Definition…Definition…Definition…Definition…
TheThe OneOne sidedsided (Unilateral)(Unilateral) zz--transformtransform ofof aa causalcausal
DTDT signalsignal x[n]x[n] isis defineddefined asas ::
WeWe cancan alsoalso writewrite :: ZZ++{x(n)}{x(n)} andand )()( zXnx
z
+
+
↔WeWe cancan alsoalso writewrite :: ZZ++{x(n)}{x(n)} andand )()( zXnx +
↔
EquivalentEquivalent toto thethe bilateralbilateral zz--transformtransform ofof x[n]u[n]x[n]u[n]
SinceSince x[n]u[n]x[n]u[n] isis alwaysalways aa rightright sidedsided sequence,sequence,
ROCROC ofof X(z)X(z) isis alwaysalways thethe exteriorexterior ofof aa circlecircle..
UsefulUseful forfor solvingsolving differencedifference equationsequations withwith initialinitial
conditionsconditions..
9/12/2013 Mahesh J. vadhavaniya 3

5. Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)
ItIt doesdoes notnot containcontain informationinformation aboutabout thethe signalsignal
x(n)x(n) forfor negativenegative valuesvalues ofof timetime (for(for nn << 00 ))
ItIt isis uniqueunique onlyonly forfor causalcausal signals,signals, becausebecause onlyonly
thesethese signalssignals areare zerozero forfor nn << 00..thesethese signalssignals areare zerozero forfor nn << 00..
SinceSince x[n]u[n]x[n]u[n] isis alwaysalways aa rightright sidedsided sequence,sequence,
ROCROC ofof X(z)X(z) isis alwaysalways thethe exteriorexterior ofof aa circlecircle.. SoSo whenwhen
wewe dealdeal withwith oneone sidedsided zz--transform,transform, itit isis notnot
necessarynecessary toto referrefer toto theirtheir ROCROC..
9/12/2013 Mahesh J. vadhavaniya 4

6. Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)
(A) 1X1(n) = { 1, 2, 5, 7, 0, 1 }
(B) 2X2(n) = { 1, 2, 3, 0, 8, 1 }
-5-3-2-1
1
z7z5z2z1=(z)x ++++
+
-3-2
2
z8z3=(z)x ++
+
2
z8z3=(z)x ++
(C) 3X3(n) = { 0, 0, 1, 2, 5, 7, 0, 1 }
-7-5-4-3-2
3
zz7z5z2z=(z)x ++++
+
(D) 4X4(n) = { 2, 4, 5, 7, 0, 1 }
-3-1
4
z7z5=(z)x ++
+
9/12/2013 Mahesh J. vadhavaniya 5

7. Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)
(E) X5(n) = δ (n)
(F) X6(n) = δ (n - k)
1=(z)x 5
+
0k,z=(z) -k
6x >
+
(G) X7(n) = δ (n + k)
0k0,=(z)x 7
>
+
9/12/2013 Mahesh J. vadhavaniya 6

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

20. 9/12/2013 Mahesh J. Vadhavaniya 20