One sided z transform

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

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 forforthethe entireentire timetime rangerange ∞<<∞ n -

�� CanCan notnot bebe usedused toto evaluateevaluate thethe outputoutput ofof nonnon--relaxedrelaxedsystemssystems..

�� NonNon--relaxedrelaxed areare systemssystems describeddescribed byby differencedifference�� NonNon--relaxedrelaxed areare systemssystems describeddescribed byby differencedifferenceequationsequations withwith nonzerononzero initialinitial conditionsconditions..

� We’ll Develop the one sided z-transform to solvedifference equations with initial conditions.


�� SinceSince thethe inputinput isis appliedapplied atat aa finitefinite timetime (n(n00),), bothboth thetheinputinput andand outputoutput signalssignals areare specifiedspecified forfor n≥n≥ nn00,, butbut byby oomeansmeans areare zerozero forfor nn << nn00 ..

∑∞

=

−+ =0

)()(n

nznxzX

Definition…Definition…Definition…Definition…Definition…Definition…Definition…Definition…

�� TheThe OneOne sidedsided (Unilateral)(Unilateral) zz--transformtransform ofof aa causalcausalDTDT signalsignal x[n]x[n] isis defineddefined asas ::

�� WeWe cancan alsoalso writewrite :: ZZ++{x(n)}{x(n)} andand )( )( zXnxz

+

+

↔�� 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 initialinitialconditionsconditions..


Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (Definition… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)

�� ItIt doesdoes notnot containcontain informationinformation aboutabout thethe signalsignalx(n)x(n) forfor negativenegative valuesvalues ofof timetime (for(for nn << 00 ))

�� ItIt isis uniqueunique onlyonly forfor causalcausal signals,signals, becausebecause onlyonlythesethese 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 whenwhenwewe dealdeal withwith oneone sidedsided zz--transform,transform, itit isis notnotnecessarynecessary toto referrefer toto theirtheir ROCROC..



(A) 1X1(n) = { 1, 2, 5, 7, 0, 1 }

(B) 2X2(n) = { 1, 2, 3, 0, 8, 1 }

-5-3-2-1

1z7z5z2z1=(z)x ++++

+

-3-2

2z8z3=(z)x ++

+

2z8z3=(z)x ++

(C) 3X3(n) = { 0, 0, 1, 2, 5, 7, 0, 1 }

-7-5-4-3-2

3zz7z5z2z=(z)x ++++

+

(D) 4X4(n) = { 2, 4, 5, 7, 0, 1 }

-3-1

4z7z5=(z)x ++

+



(E) X5(n) = δ (n)

(F) X6(n) = δ (n - k)

1=(z)x 5

+

0 k ,z=(z) -k

6x >+

(G) X7(n) = δ (n + k)

0 k 0,=(z)x 7>

+


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 isisnotnot uniqueunique..

�� ForFor aa causalcausal signal,signal, thethe oneone sidedsided zz--transformtransform isis uniqueunique..


�� ForFor antianti--causalcausal signals,signals, thethe oneone sidedsided zz--transformtransform isisalwaysalways zerozero..

Properties… Properties… Properties… Properties… Properties… Properties… Properties… Properties… Shifting Property :-

Case 1 : Time Delay

0 k ,)()()(

)( )(

1

z

>

−+→←−

→←

∑=

+−

+

+

+

k

n

nkzznxzXzknxthen

zXnxIf

1 =n

)(z k)- then x(ncausal, is x(n)case -kzzXIn

+→←+

Proof :-

+=

+=−

+−

−=

−−

∞

=

−−

−=

−−+

∑

∑∑

)()(

)()( )}({

1

0

1

zXzlxz

zlxzlxzknxZ

k

l

lk

l

l

kl

lk

�� ChangeChange thethe indexindex fromfrom ll toto nn == --ll9/12/2013 Mahesh J. vadhavaniya 7

Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (Properties… (cntdcntdcntdcntdcntdcntdcntdcntd…)…)…)…)…)…)…)…)

Example 1: Determine the one-sided z-transform ofX1(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()(

−−−

−

−+

−

−−

−+−

++−

=

−==−=

−+−+=

azaaz

zzX

obtainweaz

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 kksamplessamples toto thethe rightright..


Properties… Properties… Properties… Properties… Properties… Properties… Properties… Properties… Shifting Property :-

Case 2 : Time Advance

0 k ,)()()(

)( )(

1

0

z

>

−→←+

→←

∑−

=

−+

+

+

+

k

n

nkzznxzXzknxthen

zXnxIf

0 =n

Proof :- ∑∑∞

=

−∞

=

−+ =+=+kl

lk

n

nzlxzzknxknxZ )()( )}({

0

We have changed the index of summation from n to l = n+k

∑∑∑∞

=

−−

=

−∞

=

−+ +==kl

lk

l

l

l

lzlxzlxzlxzX )()()( )(

1

00

−= ∑

−

=

−++1

0

)()()(k

n

nkznxzXzzX



Example 2: Determine the one-sided z-transform ofX2(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

azzaz

zzX

obtainweazzXaxand

zxzxzXz

−−−

=

−===

−−=

−

+

−+

+

2

1

2

2

1

1

22

1)(

)1(1)( and ,)1( ,1 x(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 kksamplessamples toto thethe leftleft..



Final Value Theorem :

Proof :

)()1(lim)()(lim

)( )(

1zn

z

zXzxnxthen

zXnxIf

+

→∞→

+

−=∞=

→←+

∑∞

−∞=∞ )()]([ nzxxZ

∑

∑

∑

∑

∞

=

−+

∞

=

−++

∞

=

−

=

−+=−−

−+=−−

−+=−+

∞=∞

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


Final Value Theorem :

exp

)]()1([)]0()()1[(lim

])]()1([[lim)]0()()1[(lim

01z

0

1

1z1z

l n termanding tilnow

nxnxxzXz

znxnxxzXz

n

n

∞

=

+

→

∞

=

−

→

+

→

−+=−−

−+=−−

∑

∑

ThisThis theoremtheorem enablesenables usus toto findfind thethe steadysteady statestate valuevalue ofofx(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

+

→

+

→

∞→

+

→

−=∞

−∞=−−

−++

+−+−=−−


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 +−=

1TakingTaking zz--transformtransform onon bothboth sidessides )()]1()([

2

1)( 1

zXyzYzzY +−+= −

SubstituteSubstitute y(y(--11)=)=11 andand

3

1)(

3

1)(

−

=

=

z

znuZzX

n

3

15.0)(5.0)(

3

1]1)([

2

1)(

1

1

−

++=

−

++=

−

−

z

zzYzzY

z

zzYzzY



( )

5.0)(

5.013

15.01

5.0)(

3

15.0)()5.01(

11

1

+=

−

−

+−

=

−

+=−

−−

−

zzY

zz

z

z

zzY

z

zzYz

( )

3

1

2

5.0

3

5.0

5.0)(

3

1

2

5.0

3

5.0

5.0)(

5.03

15.0

5.0)(

−

−−

+−

=

−

−−

+−

=

−

−

+−

=

z

z

z

z

z

zzY

zzzz

zY

zz

z

zz

zY



( ) ( ) ( )

( ) ( ) )(]3125.05.3[)(

)(]3125.035.05.0[)(

nuny

nuny

nn

nnn

−=

−+=

TakingTaking inverseinverse zz--transform,transform, wewe getget

( ) ( ) )(]3125.05.3[)( nunynn

−=


The unilateral z transform is well suited to solving difference

equations with initial conditions. For example,

y n + 2[ ] −3

2y n +1[ ] +

1

2y n[ ] = 1 / 4( )

n, for n ≥ 0


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

2z Y z( ) − y 0[ ] +

1

2Y z( ) =

z

z −1 / 4

the initial conditions are called for systematically.


Applying initial conditions and solving,

Y z( ) = z16 / 3

z −1 / 4+

4

z −1 / 2+

2 / 3

z −1

and


and

y n[ ] =16

3

1

4

n

+ 41

2

n

+2

3

u n[ ]

This solution satisfies the difference equation and the initial

conditions.


9/12/2013 Mahesh J. Vadhavaniya 20

Education

One sided z transform