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Property6:Ifx[n]istwosided,andifthe︱z︱Property6:Ifx[n]istwosided,andifthe︱z︱isintheROC,thentheROCwillofaringinthez‐planethatincludesthe.Normally,r1︱z︱r2r1︱z︱`m´x[n]=bn,b>Examplenb10<b<..03412¥X(z)=bnHowtogettheclosedformofx[n]=bn,b>Examplenb10<b<..03412¥X(z)=bnHowtogettheclosedformof ¥b‐=z{bz=u[-n-1]}=1b-1z-z<1/bz11`m,=-1-bz-1-b-1z11`m,=-1-bz-1-b-1z-´bb>1,X(z)isnotb<z<1/IftheIftheZtransformX(z)ofx[n]isrational,andx[n]isrightsided,thentheROCistheregioninthez‐planeoutsidetheoutmostpole‐‐‐I.e.,outsidethecircleradiusequaltothemagnitudeofthepolesofX(z).causal,thentheROCalsoincludesz=¥IftheZIftheZtransformX(z)ofx[n]isrational,andifx[n]isleftsided,thentheROCistheregioninthez‐planetheinnermostnonzeroi.e.,insidethecircleofradiusequaltosmallestmagnitudeofthepolesofthananyatz=0andextendinginwardtopossiblyincludingz=0.inparticular.ifx[n]anticausal,thentheROCalsoincludesExample1X(z)3Thereare3Example1X(z)3Thereare3possibleROCsshowninFigure<1zz<z>33rightlefttwoFromX(z)=X(reFromX(z)=X(rejω)=DTFT{r-n\r-nx[n]=IDTFT{X(rejω1jnωdω=x[n]=1X(rejω)(rejω)n2π1z=dωπ1z=dωπn-1X(z)zdz=InverseIntegrationaroundacounterclockwiseclosedcircularcontourcenteredattheoriginandwithradiusr.ThecalculationforInverseIntegrationofcomplexfunctionbyComputedThecalculationforInverseIntegrationofcomplexfunctionbyComputedbyPartialFractionPower‐seriesExpansionExample10.9TheinverseZTisneeded,3-5z->X(z)= ,z(1-1z-1)(1-1z-343(I).PartialFractionExpansioninz־¹,z>1(1-1z-2(1-1z-X(z)+34From(10.9),we31>nx=),z>1(1-1z-2(1-1z-X(z)+34From(10.9),we31>nx=),T‹1144-42)11nTx[n]=)‹,>21333z-1x[n]=x[n]+x[n]=(1)nu[n]+2x[n]=x[n]+x[n]=(1)nu[n]+2(1)n1243(II).PartialFractionExpansioninz>X(z)= ,z3113z-4z(3z-5)X(z)= (z-1)(z-43>3X(z)=1· +2·z,z(z->3X(z)=1· +2·z,z(z-1(z-34Note:Itissameasinz־¹ExampleX(z)issameasinexamplewithachangedROC12113+X(z)<z1(1-z-141(1-z-134 ,1>nx[n]=)u[n](1-1z-1444121nx[ ,1>nx[n]=)u[n](1-1z-1444121nx[n]=2(),<21333z-1x[n]=x[n]+x[n]=(1)nu[n]-2(1)nu[-n-1243ExampleX(z)issameasinexamplewithachangedROCX(z)= + 1zExampleX(z)issameasinexamplewithachangedROCX(z)= + 1z<(1-1z-(1-1z-44311x[n]=x[n]+x[n]= unnn1]- )n1234Normally,PartialFractionofrational X(z)= i=11-az-iLongZd[n]1,0zZn]«z-n0LongZd[n]1,0zZn]«z-n0,0<z<0(0,∞另外考)¥¥Zx[n]=x[m]d[n-m]m=-x[m]z-m=-=X(z)‹¥X(z)=az-ExampleX(z)=4z2+2+3z-1,ExampleX(z)=4z2+2+3z-1,0z<n=n=Fromthepower‐seriesofZTineq.(10.3),wex[n]=x[n]=δδδδ[n+n]z,0<<01ExampleX(z)1-az-z>long1+az-1ExampleX(z)1-az-z>long1+az-+a2z-+11-)1-az-az-1-a2z-a2z-Q-2-=1+az+a+1-az-x[n]=,a,a2=anu[n]z<long-a-1z-a-2z<long-a-1z-a-2z-)1-a-1za-1a-1z-a-2za-2z1=-a-1z-a-2z2-1-az-Readexample10.14byx[n]={a-1,a-2,...,a-m=-anu[-n-ThePropertiesofZ-Transform(10.5、TTThePropertiesofZ-Transform(10.5、TTx‹X(z),11X2(z),x2[n]aX1(z)+bX2(z),Containsax[n]+bx[n]12Note:(1)normally,common(2)R1∩R2maybelargerthanR1or10.5.2TimeTTX(z),‹Rz-x[n-n]X(z),0R(是10.5.2TimeTTX(z),‹Rz-x[n-n]X(z),0R(是否包含0,∞有可能变化Example1zanu[n],>=T‹1-az-z-Wecan1n-1u[n-1]a,=>az-1zaTX(z),‹Rnx[n]X(z/),|z|00ejω0nTX(z),‹Rnx[n]X(z/),|z|00ejω0nw0=X(e-z),RX(-z),(-1)nx[n]fien[n？10.5.4TimeTX(z),‹‹R1X10.5.4TimeTX(z),‹‹R1Xz(ROC方向与R相反¥X(z)n-=¥¥Y(z)=y[n]z- ={x[-n]}z-={x[m]}(z-1=X(z-11anu[n],>T1-az-1,<‹1-11-az-,<n-au[-n-1anu[n],>T1-az-1,<‹1-11-az-,<n-au[-n-‹1,>‹1-,>‹-a- a-nu[-n]10.5.5TimeX(z),‹RTx[n]X10.5.5TimeX(z),‹RTx[n]X(zk),R1/k(k(k>0时,ROC方向相同[n]=n=km,y[n]=(kn„¥¥Y(z)=y[n]z- ¥={x(k)[n]}(zk=X(zk10.5.6TTX(z10.5.6TTX(z),‹RX*(z*),x*[n]isX(z)=X*(z*Zeros,poles共轭成对出10.5.7TheConvolutionX(z),x‹110.5.7TheConvolutionX(z),x‹11x[n]X(z),22TX1(z)X2(z),R1x[n]*x[n]12Note:ifR1∩R2=Ø,X1(z)X2(isnotT‹X(z),xR11累加器的1T‹X(z),xR11累加器的1nx[k]=x[n]*u[n]fiX(z),1-z-一阶差分系统的 x[n]-x[n-1]=x[n]*{d[n]-d[n-‹X(z)[1-z-1],10.5.8Differentiationinthez-X(z),TTdX(z)‹10.5.8Differentiationinthez-X(z),TTdX(z)‹Rd‹X(z),-R¥={-nx[n]}z-n-X(z)=x[n]z-n=-zdX(z)1anu[n],>T1-Wecand1y[1anu[n],>T1-Wecand1y[n]=nanTY(z)=-]‹dz1-az-,=z>(1-az-1)2ExampleTheinverseZTisX(z)=ln(1-1z->fiz22X(z)=-2d1->2z12ExampleTheinverseZTisX(z)=ln(1-1z->fiz22X(z)=-2d1->2z12z1112 )n-1u[nz>1221121nx[n]= 211()nu[n--n10.4.9TheInitial-(andFinal-)ValueX(10.4.9TheInitial-(andFinal-)ValueX(z),x[n]=0,n<