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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!