Signal&SystemsLecture#10

22ndMay18

Z-Transform

TheZ-Transform

v  Thez-transformofasequencex[n]is:

v  The z-transform can also be thought of as an operator Z{.} thattransformsasequencetoafunction:

v  Inbothcaseszisacontinuouscomplexvariable.

v  Thez-transformoperationisdenotedas:

v  Where“z”isthecomplexnumber.Therefore,wemaywritezas:

X(z) = x(n)z−nn=−∞

∞

∑

Z x n⎡⎣ ⎤⎦{ }= x n⎡⎣ ⎤⎦z−n

n=−∞

∞

∑ = X z( )

x(n)z

↔X(z)

z = re jω

TheZ-Transform(cont.)

v  Where r and ω belongs to Real number. When r=-1, the z-transformofadiscrete-timesignalbecomes:

v  Therefore,theDTFTisaspecialcaseofthez-transform.

v  Pictorially,wecanviewDTFTasthez-transformevaluatedontheunitcircle:

X e jω( ) = x[n]e− jωnn=−∞

∞

∑

TheZ-Transform(cont.)

v  Whenr≠1,thez-transformisequivalentto:

v  WhichistheDTFTofthesignalr-nx[n].

v  However, we know that the DTFT does not always exist. It exists onlywhenthesignalissquaresummable,orsatisfiestheDirichletconditions.

v  Therefore, X(z) does not always converge. It converges only for somevaluesofr.thisrangeofriscalledtheregionofconvergence(ROC).

X re jω( ) = x[n] re jω( )−n

n=−∞

∞

∑

= r−nx[n]( )e− jωnn=−∞

∞

∑

= F r−nx[n]⎡⎣ ⎤⎦

RegionofConvergence(ROC)

v  The region of convergence are the values of for which the z-transformconverges.

v  Z-transform is an infinite power series which is not alwaysconvergentforallvaluesofz.

v  Therefore,theregionofconvergenceshouldbementionedalongwiththez-transformation.

v  TheRegionofConvergence(ROC)ofthez-transformisthesetofzsuchthatX(z)converges,i.e.,

x[n] r−nn=−∞

∞

∑ <∞

Example#1

v  Z-transformofright-sidedexponentialsequences.

v  Considerthesignalx[n]=anu[n].Mathematicallyitcanbewrittenas:

v  Thissequenceexistsforpositivevaluesofn:

x n( ) =an for n ≥ 00 for n < 0

⎧⎨⎪

⎩⎪

1 3

x(n)

1a

a2a3

a4

0 2 4-2 -1

Fora<1

1 3

x(n)

1a

a2 a3

0 2 4-2 -1

Fora>1

Example#1(cont.)

v  Thez-transformofx[n]isgivenby:

v  Therefore,X(z)convergesif .Fromgeometricseries,weknowthat:

v  When|az-1|<1,orequivalently|z|>|a|.So,

X z( ) = anu n[ ] z−n−∞

∞

∑

= az−1( )n

n=0

∞

∑

az−1( )n

n=0

∞

∑ <∞

az−1( )n

n=0

∞

∑ =1

1− az−1

X z( ) = 11− az−1

Example#1(cont.)

v  WithROCbeingthesetofzsuchthat|z|>|a|.Asshownbelow:

Example#2

v  Considerthesignalx[n]=-anu[-n-1]with0<a<1.Whichistheleftsided exponential sequence and it can be determinedmathematicallyas:

v  Thissequenceexistsonlyfornegativevaluesofn.

x(n) = −anu(−n−1) for n ≤ 00 for n > 0

⎧⎨⎪

⎩⎪

1

-3

x(n)

0

-5 -4 -2 -1

Example#2(cont.)

v  Thez-transformofx[n]is:

v  Therefore,X(z)convergeswhen|a-1z|<1,orequivalently|z|<|a|.Inthiscase:

X z( ) = − anu −n−1[ ] z−nn=−∞

∞

∑

= − anz−nn=−∞

−1

∑

= − a−nznn=1

∞

∑

=1− a−1z( )n

n=0

∞

∑

X z( ) =1− 11− a−1z

=1

1− az−1

Example#2(cont.)

v  WitchROCbeingthesetofzsuchthat|z|<|a|.

v  Note that the z-transform is the same as that of Example 1, theonlydifferenceistheROC.Whichisshownbelowas:

PropertiesofZ-Transform

Linearity

v  Thispropertystatesthatifand,then

v  Wherea1anda2areconstants.

x1(n)z

↔X1(z) x2 (n)z

↔X2 (z)

a1x1(n)+ a2x2 (n)z

↔a1X1(z)+ a2X2 (z)

TimeScaling

v  Thispropertyofz-transformstatesthatif,thenwecanwrite:

v  Wherekisanintegerwhichisshiftintimeinx(n)insamples.

x(n)z

↔X(z)

x(n− k)z

↔z−kX(z)

ScalinginZ-Domain

v  Thispropertystatesthatif:

v  Whereaisaconstant.

x(n)z

↔X(z) ROC : r1 < z < r2

then anx(n)z

↔X za⎛

⎝⎜⎞

⎠⎟ ROC : a r1 < z < a r2

TimeReversal

v  Thispropertystatesthatif:

v  Then,

x(n)z

↔X(z) ROC : r1 < z < r2

x(−n)z

↔X(z−1) ROC : 1r1< z < 1

r2

DifferentiationinZ-Domain

v  Thispropertystatesthatif,then:

x(n)z

↔X(z)

nx(n)z

↔− zd X(z){ }

dz

Convolution

v  Thispropertystatesthatif and then:

x1(n)z

↔X1(z) x2 (n )z

↔X 2 (z )

x1(n)∗ x2 (n)z

↔X1(z)X2 (z)

Example#3

v  Findtheconvolutionofsequences:

v  Solution:

v  Step1:Determinez-transformofindividualsignalsequences:

x1 = 1,−3,2{ } and x2 = 1,2,1{ }

X1(z) = Z x1(n)[ ] = x1(n)z−n

n=0

2

∑ = x1(0)z0 + x1(1)z

−1 + x1(2)z−2

=1z0 −3z−1 + 2z−2 =1−3z−1 + 2z−2

and X2 (z) = Z x2 (n)[ ] = x2 (n)z−n

n=0

2

∑ = x2 (0)z0 + x2 (1)z

−1 + x2 (2)z−2

=1z0 + 2z−1 +1z−2 =1+ 2z−1 +1z−2

Example#3(cont.)

v  Step2:MultiplicationofX1(z)andX2(z):

v  Step3:Letustakeinversez-transformofX(z):

X(z) = X1(z)X2 (z) = 1−3z−1 + 2z−2( ) 1+ 2z−1 +1z−2( )

=1− z−1 −3z−2 + z−3 + 2z−4

x(n) = IZT 1− z−1 −3z−2 + z−3 + 2z−4⎡⎣ ⎤⎦= 1,−1,−3,1, 2{ }

InverseZ-Transform

InverseZ-Transform

v  TheinverseZ-transformisasfollows:

v  MethodstoobtainInverseZ-transform:v  IfX(z) is rational,wecanuseexpanding the rationalalgebraic intoa

linearcombinationoflowerordertermsandthenonemayuse:v  IfROCisoutofpolez=ai:

v  IfROCisinsideofz=ai:

v  DonotforgettoconsiderROCinobtaininginverseofZT.

x n[ ] = 12π j

X z( ) zn−1 dz!∫

X z( ) = Ai1− aiz

−1 → x n[ ] = Aiaiu n[ ]

X z( ) = Ai1− aiz

−1 → x n[ ] = −Aiaiu −n−1[ ]

InverseZ-Transform(cont.)

v  IfX(z) isnon-rational,usePowerseriesexpansionofX(z), thenapplyδ[n+n0]çèzn0

v  IfX(z)isrational,powerseriescanbeobtainedbylongdivision.

v  IfX(z) is a rational functionof z, i.e., a ratioofpolynomials,we canalsousepartialfractionexpansiontoexpressX(z)asasumofsimpleterms for which the inverse transform may be recognized byinspection.

v  TheROCplaysacriticalroleinthisprocess.

Example#4

v  Considerthez-transform:

X z( ) =3− 56z−1

1− 14z−1

⎛

⎝⎜

⎞

⎠⎟ 1−

13z−1

⎛

⎝⎜

⎞

⎠⎟, z > 1

3

Example#5

v  Consider:

v  Expandinapowerseriesbylongdivision.

X z( ) = 11− az−1

, z > a

ThankYou!