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Thez-Transform and Its Application
Dr. Deepa Kundur
University of Toronto
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application1 / 36 Chapter 3: Thez-Transform and Its ApplicationDiscrete-Time Signals and Systems
Reference:
Sections 3.1 - 3.4 of
John G. Proakis and Dimitris G. Manolakis,Digital Signal Processing: Principles, Algorithms, and Applications, 4th edition, 2007.Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application2 / 36 Chapter 3: Thez-Transform and Its ApplicationThe Directz-Transform
IDirectz-Transform:
X(z) =1X
n=1x(n)zn INotation:
X(z) Zfx(n)g
x(n)Z !X(z)Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application3 / 36 Chapter 3: Thez-Transform and Its ApplicationRegion of Convergence
I the region of convergence(ROC) ofX(z) is the set of all values ofzfor whichX(z) attains a nite value I Thez-Transform is, therefore, uniquely characterized by: 1. exp ressionfo rX(z) 2.ROC of X(z)
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application4 / 36
Chapter 3: Thez-Transform and Its Application
Power Series Convergence
IFor a power series,
f(z) =1X n=0a n(zc)n=a0+a1(zc) +a2(zc)2+ there exists a number 0r 1such that the series I convergences forjzcjImay or may not converge for values onjzcj=r.
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application5 / 36 Chapter 3: Thez-Transform and Its ApplicationPower Series Convergence
IFor a power series,
f(z) =1X n=0a n(zc)n=a0+a1(zc)+a2(zc)2+ there exists a number 0r 1such that the series I convergences forjzcj>r, andIdiverges forjzcj Imay or may not converge for values onjzcj=r.
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application6 / 36 Chapter 3: Thez-Transform and Its Application Region of Convergence
I Consider
X(z) =1X
n=1x(n)zn= 1X n=1x(n)zn+1X n=0x(n)zn= 1X n 0=0x(n0)zn0
|{z} ROC:jzj 1X n=0x(n)zn |{z} ROC:jzj>r2x(0)|{z}
ROC: allzDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application7 / 36 Chapter 3: Thez-Transform and Its Application
Region of Convergence:r1>r2Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application8 / 36
Chapter 3: Thez-Transform and Its Application
Region of Convergence:r1 ROC Families: Finite Duration Signals
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application10 / 36 Chapter 3: Thez-Transform and Its Application ROC Families: Innite Duration Signals
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application11 / 36 Chapter 3: Thez-Transform and Its Application z-Transform Properties Property Time Domainz-Domain ROCNotation:x(n)X(z) ROC:r21(n)X1(z) ROC1 x 2(n)X1(z) ROC2
Linearity:a1x1(n) +a2x2(n)a1X1(z) +a2X2(z) At least ROC1\ROC2 Time shifting:x(nk)zkX(z) ROC, except
z= 0 (ifk>0) andz=1(ifk<0) z-Scaling:anx(n)X(a1z)jajr2Time reversalx(n)X(z1)1r
1 2Conjugation:x(n)X(z) ROC
z-Dierentiation:n x(n)zdX(z)dz r2Dr. Deepa Kundur (University of Toronto)
Imay or may not converge for values onjzcj=r.
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application6 / 36 Chapter 3: Thez-Transform and Its ApplicationRegion of Convergence
IConsider
X(z) =1X
n=1x(n)zn= 1X n=1x(n)zn+1X n=0x(n)zn= 1X n0=0x(n0)zn0
|{z}ROC:jzj 1X n=0x(n)zn |{z} ROC:jzj>r2x(0)|{z}
ROC: allzDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application7 / 36 Chapter 3: Thez-Transform and Its Application
Region of Convergence:r1>r2Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application8 / 36
Chapter 3: Thez-Transform and Its Application
Region of Convergence:r1 ROC Families: Finite Duration Signals
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application10 / 36 Chapter 3: Thez-Transform and Its Application ROC Families: Innite Duration Signals
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application11 / 36 Chapter 3: Thez-Transform and Its Application z-Transform Properties Property Time Domainz-Domain ROCNotation:x(n)X(z) ROC:r21(n)X1(z) ROC1 x 2(n)X1(z) ROC2
Linearity:a1x1(n) +a2x2(n)a1X1(z) +a2X2(z) At least ROC1\ROC2 Time shifting:x(nk)zkX(z) ROC, except
z= 0 (ifk>0) andz=1(ifk<0) z-Scaling:anx(n)X(a1z)jajr2Time reversalx(n)X(z1)1r
ROC:jzj>r2x(0)|{z}
ROC: allzDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application7 / 36 Chapter 3: Thez-Transform and Its Application
Region of Convergence:r1>r2Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application8 / 36
Chapter 3: Thez-Transform and Its Application
Region of Convergence:r1ROC Families: Finite Duration Signals
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application10 / 36 Chapter 3: Thez-Transform and Its Application ROC Families: Innite Duration Signals
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application11 / 36 Chapter 3: Thez-Transform and Its Application z-Transform Properties Property Time Domainz-Domain ROCNotation:x(n)X(z) ROC:r22(n)X1(z) ROC2
Linearity:a1x1(n) +a2x2(n)a1X1(z) +a2X2(z) At least ROC1\ROC2 Time shifting:x(nk)zkX(z) ROC, except
z= 0 (ifk>0) andz=1(ifk<0) z-Scaling:anx(n)X(a1z)jajr2
1 2Conjugation:x(n)X(z) ROC
z-Dierentiation:n x(n)zdX(z)dz r2Dr. Deepa Kundur (University of Toronto)
2Conjugation:x(n)X(z) ROC
z-Dierentiation:n x(n)zdX(z)dz r2The z-Transform and Its Application12 / 36
Chapter 3: Thez-Transform and Its Application
Convolution Property
x(n) =x1(n)x2(n)()X(z) =X1(z)X2(z)Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application13 / 36 Chapter 3: Thez-Transform and Its ApplicationConvolution using thez-Transform
Basic Steps:
1. Compute z-Transform of each of the signals to convolve (time domain!z-domain): X1(z) =Zfx1(n)g
X2(z) =Zfx2(n)g
2.Multiply the t woz-Transforms (inz-domain):
X(z) =X1(z)X2(z)
3. Find the inverse z-Transformof the product (z-domain!time domain): x(n) =Z1fX(z)gDr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application14 / 36 Chapter 3: Thez-Transform and Its ApplicationCommon Transform Pairs
Signal,x(n)z-Transform,X(z) ROC1(n) 1 Allz
2u(n)11z1jzj>1
3anu(n)11az1jzj>jaj
4nanu(n)az1(1az1)2jzj>jaj
5anu(n1)11az1jzj 6nanu(n1)az1(1az1)2jzj 7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application15 / 36 Chapter 3: Thez-Transform and Its Application Common Transform Pairs
Signal,x(n)z-Transform,X(z) ROC1(n)1 All z
2u(n)11z1jzj>1
3anu(n)11az1jzj>jaj
4nanu(n)az1(1az1)2jzj>jaj
5anu(n1)11az1jzj 6nanu(n1)az1(1az1)2jzj 7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application16 / 36
Chapter 3: Thez-Transform and Its Application
Why Rational?
I X(z) is a rationalfunction i it can be represented as the ratio of two polynomials inz1(orz): X(z) =b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzNI
For LTI systems that are represented by
LCCDEs
, the z-Transform of the unit sample responseh(n), denoted H(z) =Zfh(n)g, isrational Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application17 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros
I zerosofX(z): values ofzfor whichX(z) = 0 I polesofX(z): values ofzfor whichX(z) =1 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application18 / 36 Chapter 3: Thez-Transform and Its Application Poles and Zeros of the Rationalz-Transform
Leta0;b06= 0:
X(z) =B(z)A(z)=b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzN=
b0zMa 0zN zM+ (b1=b0)zM1++bM=b0z N+ (a1=a0)zN1++aN=a0=
b0a 0zM+N(zz1)(zz2)(zzM)(zp1)(zp2)(zpN)
=GzNMQ M k=1(zzk)Q N k=1(zpk)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application19 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
X(z) =GzNMQ
M k=1(zzk)Q N k=1(zpk)whereGb0a 0Note: \nite" does not include zero or1.I
X(z) hasMnitezeros at z=z1;z2;:::;zMI
X(z) hasNnitep olesat z=p1;p2;:::;pNI
ForNM6= 0
I ifNM>0, there arejNMjzeroat o rigin,z= 0 IifNM<0, there arejNMjpolesat o rigin,z= 0Total number of zeros = Total number of poles Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application20 / 36
Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
Example:
X(z) =z2z22z+ 116z3+ 6z+ 5= (z0)(z(12
+j12 )(z(12 j12 )(z(14 +j34 )(z(14 j34 )(z(12 )poles:z=14 j34 ;12 zeros:z= 0;12 j12 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application21 / 36 Chapter 3: Thez-Transform and Its Application Pole-Zero Plot
Example: poles:z=14
j34 ;12, zeros:z= 0;12 j120.5 0.5 -0.5 -0.5 unit circle POLE ZERODr. Deepa Kundur (University of Toronto)The z-Transform and Its Application22 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I Graphical interpretation of characteristics ofX(z) on the complex plane I ROC cannot include poles;assuming causality ...
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application23 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I For realtime-domain signals, the coecients ofX(z) are necessarily real I complex poles and zeros must occur in conjugate pairs Inote: real poles and zeros do nothave to be paired up X(z) =z2z22z+1 16z3+6 z+5 =)
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application24 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot Insights
I For causal systems, the ROC will bethe outer region of the smallest (origin-centered) circle encompassing all the poles .I For stable systems, the ROC will includethe unit circle.0.5 0.5 -0.5 -0.5 ROCI Causal?
Y es. I Stable?
Y es.I
For stability of a
causal system, the poles will lieinside the unit circle Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application25 / 36 Chapter 3: Thez-Transform and Its Application The System Function
h(n)Z !H(z) time-domain Z !z-domain
impulse response Z !system functiony(n) =x(n)h(n)Z
!Y(z) =X(z)H(z) Therefore,
H(z) =Y(z)X(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application26 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
y(n) =NX k=1a ky(nk) +MX k=0b kx(nk)Zfy(n)g=ZfNX k=1a ky(nk) +MX k=0b kx(nk)gZfy(n)g=NX k=1a kZfy(nk)g+MX k=0b kZfx(nk)gY(z)=NX k=1a kzkY(z)+ MX k=0b kzkX(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application27 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
Y(z) +NX
k=1a kzkY(z) =MX k=0b kzkX(z)Y(z)" 1 +NX k=1a kzk#=X(z)MX k=0b kzkH(z) =Y(z)X(z)= P M k=0bkzkh 1 +PN k=1akzkiLCCDE !Rational System Function Many signals of practical interest have a rationalz-Transform.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application28 / 36
Chapter 3: Thez-Transform and Its Application
Inversion of thez-Transform
Three popular methods:
I Contour integration:
x(n) =12jI C X(z)zn1dzI
Expansion into a
p owerseries in zorz1: X(z) =1X
k=1x(k)zk and obtainingx(k) for allkby inspectionI Partial-fraction expansion and
table lo okupDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application29 / 36 Chapter 3: Thez-Transform and Its Application
Expansion into Power Series
Example:
X(z) = log(1 +az1);jzj>jaj=
1X n=1(1)n+1anznn =1X n=1(1)n+1ann znBy inspection: x(n) = (1)n+1ann n1 0n0Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application30 / 36 Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
1. Find the distinct p olesof X(z):p1;p2;:::;pKand their corresponding multiplicitiesm1;m2;:::;mK. 2. The pa rtial-fractionexpansion is of the fo rm:
X(z) =KX
k=1 A1kzpk+A2k(zpk)2++Amk(zpk)mk
wherepkis anmkth order pole (i.e., has multiplicitymk). 3. Use an app ropriateapp roacht ocompute fAikg
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application31 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Example: Findx(n) given poles ofX(z) atp1=2 and a double pole atp2=p3= 1; specically, X(z) =1(1 + 2z1)(1z1)2X(z)z=
z2(z+ 2)(z1)2z 2(z+ 2)(z1)2=
A1z+ 2+A2z1+A3(z1)2
Note: we need a strictly proper rational function. DO NOT FORGET TO MULTIPLY BYzIN THE END.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application32 / 36
Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
z 2(z+ 2)(z+ 2)(z1)2=A1(z+ 2)z+ 2+A2(z+ 2)z1+A3(z+ 2)(z1)2z
2(z1)2=A1+A2(z+ 2)z1+A3(z+ 2)(z1)2
z=2A 1= 49
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application33 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2z
2(z+ 2)=
A1(z1)2z+ 2+A2(z1) +A3z=1A
3= 13 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application34 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2
z 2(z+ 2)=A1(z1)2z+ 2+A2(z1) +A3d
dz z2(z+ 2)= ddz A1(z1)2z+ 2+A2(z1) +A3
z=1 A 2=59 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application35 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Therefore,
assuming c ausality , and using the following pairs: a nu(n)Z !11az1 na nu(n)Z !az1(1az1)2 X(z) =49
11 + 2z1+59
11z1+13
z 1(1z1)2x(n)=
49
(2)nu(n)+ 59 u(n)+ 13 nu(n) (2)n+29 +59
+n3 u(n) Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application36 / 36
quotesdbs_dbs7.pdfusesText_13
6nanu(n1)az1(1az1)2jzj 7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application15 / 36 Chapter 3: Thez-Transform and Its Application Common Transform Pairs
Signal,x(n)z-Transform,X(z) ROC1(n)1 All z
2u(n)11z1jzj>1
3anu(n)11az1jzj>jaj
4nanu(n)az1(1az1)2jzj>jaj
5anu(n1)11az1jzj 6nanu(n1)az1(1az1)2jzj 7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application16 / 36
Chapter 3: Thez-Transform and Its Application
Why Rational?
I X(z) is a rationalfunction i it can be represented as the ratio of two polynomials inz1(orz): X(z) =b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzNI
For LTI systems that are represented by
LCCDEs
, the z-Transform of the unit sample responseh(n), denoted H(z) =Zfh(n)g, isrational Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application17 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros
I zerosofX(z): values ofzfor whichX(z) = 0 I polesofX(z): values ofzfor whichX(z) =1 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application18 / 36 Chapter 3: Thez-Transform and Its Application Poles and Zeros of the Rationalz-Transform
Leta0;b06= 0:
X(z) =B(z)A(z)=b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzN=
b0zMa 0zN zM+ (b1=b0)zM1++bM=b0z N+ (a1=a0)zN1++aN=a0=
b0a 0zM+N(zz1)(zz2)(zzM)(zp1)(zp2)(zpN)
=GzNMQ M k=1(zzk)Q N k=1(zpk)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application19 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
X(z) =GzNMQ
M k=1(zzk)Q N k=1(zpk)whereGb0a 0Note: \nite" does not include zero or1.I
X(z) hasMnitezeros at z=z1;z2;:::;zMI
X(z) hasNnitep olesat z=p1;p2;:::;pNI
ForNM6= 0
I ifNM>0, there arejNMjzeroat o rigin,z= 0 IifNM<0, there arejNMjpolesat o rigin,z= 0Total number of zeros = Total number of poles Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application20 / 36
Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
Example:
X(z) =z2z22z+ 116z3+ 6z+ 5= (z0)(z(12
+j12 )(z(12 j12 )(z(14 +j34 )(z(14 j34 )(z(12 )poles:z=14 j34 ;12 zeros:z= 0;12 j12 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application21 / 36 Chapter 3: Thez-Transform and Its Application Pole-Zero Plot
Example: poles:z=14
j34 ;12, zeros:z= 0;12 j120.5 0.5 -0.5 -0.5 unit circle POLE ZERODr. Deepa Kundur (University of Toronto)The z-Transform and Its Application22 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I Graphical interpretation of characteristics ofX(z) on the complex plane I ROC cannot include poles;assuming causality ...
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application23 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I For realtime-domain signals, the coecients ofX(z) are necessarily real I complex poles and zeros must occur in conjugate pairs Inote: real poles and zeros do nothave to be paired up X(z) =z2z22z+1 16z3+6 z+5 =)
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application24 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot Insights
I For causal systems, the ROC will bethe outer region of the smallest (origin-centered) circle encompassing all the poles .I For stable systems, the ROC will includethe unit circle.0.5 0.5 -0.5 -0.5 ROCI Causal?
Y es. I Stable?
Y es.I
For stability of a
causal system, the poles will lieinside the unit circle Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application25 / 36 Chapter 3: Thez-Transform and Its Application The System Function
h(n)Z !H(z) time-domain Z !z-domain
impulse response Z !system functiony(n) =x(n)h(n)Z
!Y(z) =X(z)H(z) Therefore,
H(z) =Y(z)X(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application26 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
y(n) =NX k=1a ky(nk) +MX k=0b kx(nk)Zfy(n)g=ZfNX k=1a ky(nk) +MX k=0b kx(nk)gZfy(n)g=NX k=1a kZfy(nk)g+MX k=0b kZfx(nk)gY(z)=NX k=1a kzkY(z)+ MX k=0b kzkX(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application27 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
Y(z) +NX
k=1a kzkY(z) =MX k=0b kzkX(z)Y(z)" 1 +NX k=1a kzk#=X(z)MX k=0b kzkH(z) =Y(z)X(z)= P M k=0bkzkh 1 +PN k=1akzkiLCCDE !Rational System Function Many signals of practical interest have a rationalz-Transform.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application28 / 36
Chapter 3: Thez-Transform and Its Application
Inversion of thez-Transform
Three popular methods:
I Contour integration:
x(n) =12jI C X(z)zn1dzI
Expansion into a
p owerseries in zorz1: X(z) =1X
k=1x(k)zk and obtainingx(k) for allkby inspectionI Partial-fraction expansion and
table lo okupDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application29 / 36 Chapter 3: Thez-Transform and Its Application
Expansion into Power Series
Example:
X(z) = log(1 +az1);jzj>jaj=
1X n=1(1)n+1anznn =1X n=1(1)n+1ann znBy inspection: x(n) = (1)n+1ann n1 0n0Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application30 / 36 Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
1. Find the distinct p olesof X(z):p1;p2;:::;pKand their corresponding multiplicitiesm1;m2;:::;mK. 2. The pa rtial-fractionexpansion is of the fo rm:
X(z) =KX
k=1 A1kzpk+A2k(zpk)2++Amk(zpk)mk
wherepkis anmkth order pole (i.e., has multiplicitymk). 3. Use an app ropriateapp roacht ocompute fAikg
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application31 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Example: Findx(n) given poles ofX(z) atp1=2 and a double pole atp2=p3= 1; specically, X(z) =1(1 + 2z1)(1z1)2X(z)z=
z2(z+ 2)(z1)2z 2(z+ 2)(z1)2=
A1z+ 2+A2z1+A3(z1)2
Note: we need a strictly proper rational function. DO NOT FORGET TO MULTIPLY BYzIN THE END.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application32 / 36
Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
z 2(z+ 2)(z+ 2)(z1)2=A1(z+ 2)z+ 2+A2(z+ 2)z1+A3(z+ 2)(z1)2z
2(z1)2=A1+A2(z+ 2)z1+A3(z+ 2)(z1)2
z=2A 1= 49
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application33 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2z
2(z+ 2)=
A1(z1)2z+ 2+A2(z1) +A3z=1A
3= 13 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application34 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2
z 2(z+ 2)=A1(z1)2z+ 2+A2(z1) +A3d
dz z2(z+ 2)= ddz A1(z1)2z+ 2+A2(z1) +A3
z=1 A 2=59 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application35 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Therefore,
assuming c ausality , and using the following pairs: a nu(n)Z !11az1 na nu(n)Z !az1(1az1)2 X(z) =49
11 + 2z1+59
11z1+13
z 1(1z1)2x(n)=
49
(2)nu(n)+ 59 u(n)+ 13 nu(n) (2)n+29 +59
+n3 u(n) Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application36 / 36
quotesdbs_dbs7.pdfusesText_13
7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application15 / 36 Chapter 3: Thez-Transform and Its ApplicationCommon Transform Pairs
Signal,x(n)z-Transform,X(z) ROC1(n)1 All z
2u(n)11z1jzj>1
3anu(n)11az1jzj>jaj
4nanu(n)az1(1az1)2jzj>jaj
5anu(n1)11az1jzj 6nanu(n1)az1(1az1)2jzj 7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application16 / 36
Chapter 3: Thez-Transform and Its Application
Why Rational?
I X(z) is a rationalfunction i it can be represented as the ratio of two polynomials inz1(orz): X(z) =b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzNI
For LTI systems that are represented by
LCCDEs
, the z-Transform of the unit sample responseh(n), denoted H(z) =Zfh(n)g, isrational Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application17 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros
I zerosofX(z): values ofzfor whichX(z) = 0 I polesofX(z): values ofzfor whichX(z) =1 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application18 / 36 Chapter 3: Thez-Transform and Its Application Poles and Zeros of the Rationalz-Transform
Leta0;b06= 0:
X(z) =B(z)A(z)=b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzN=
b0zMa 0zN zM+ (b1=b0)zM1++bM=b0z N+ (a1=a0)zN1++aN=a0=
b0a 0zM+N(zz1)(zz2)(zzM)(zp1)(zp2)(zpN)
=GzNMQ M k=1(zzk)Q N k=1(zpk)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application19 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
X(z) =GzNMQ
M k=1(zzk)Q N k=1(zpk)whereGb0a 0Note: \nite" does not include zero or1.I
X(z) hasMnitezeros at z=z1;z2;:::;zMI
X(z) hasNnitep olesat z=p1;p2;:::;pNI
ForNM6= 0
I ifNM>0, there arejNMjzeroat o rigin,z= 0 IifNM<0, there arejNMjpolesat o rigin,z= 0Total number of zeros = Total number of poles Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application20 / 36
Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
Example:
X(z) =z2z22z+ 116z3+ 6z+ 5= (z0)(z(12
+j12 )(z(12 j12 )(z(14 +j34 )(z(14 j34 )(z(12 )poles:z=14 j34 ;12 zeros:z= 0;12 j12 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application21 / 36 Chapter 3: Thez-Transform and Its Application Pole-Zero Plot
Example: poles:z=14
j34 ;12, zeros:z= 0;12 j120.5 0.5 -0.5 -0.5 unit circle POLE ZERODr. Deepa Kundur (University of Toronto)The z-Transform and Its Application22 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I Graphical interpretation of characteristics ofX(z) on the complex plane I ROC cannot include poles;assuming causality ...
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application23 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I For realtime-domain signals, the coecients ofX(z) are necessarily real I complex poles and zeros must occur in conjugate pairs Inote: real poles and zeros do nothave to be paired up X(z) =z2z22z+1 16z3+6 z+5 =)
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application24 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot Insights
I For causal systems, the ROC will bethe outer region of the smallest (origin-centered) circle encompassing all the poles .I For stable systems, the ROC will includethe unit circle.0.5 0.5 -0.5 -0.5 ROCI Causal?
Y es. I Stable?
Y es.I
For stability of a
causal system, the poles will lieinside the unit circle Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application25 / 36 Chapter 3: Thez-Transform and Its Application The System Function
h(n)Z !H(z) time-domain Z !z-domain
impulse response Z !system functiony(n) =x(n)h(n)Z
!Y(z) =X(z)H(z) Therefore,
H(z) =Y(z)X(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application26 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
y(n) =NX k=1a ky(nk) +MX k=0b kx(nk)Zfy(n)g=ZfNX k=1a ky(nk) +MX k=0b kx(nk)gZfy(n)g=NX k=1a kZfy(nk)g+MX k=0b kZfx(nk)gY(z)=NX k=1a kzkY(z)+ MX k=0b kzkX(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application27 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
Y(z) +NX
k=1a kzkY(z) =MX k=0b kzkX(z)Y(z)" 1 +NX k=1a kzk#=X(z)MX k=0b kzkH(z) =Y(z)X(z)= P M k=0bkzkh 1 +PN k=1akzkiLCCDE !Rational System Function Many signals of practical interest have a rationalz-Transform.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application28 / 36
Chapter 3: Thez-Transform and Its Application
Inversion of thez-Transform
Three popular methods:
I Contour integration:
x(n) =12jI C X(z)zn1dzI
Expansion into a
p owerseries in zorz1: X(z) =1X
k=1x(k)zk and obtainingx(k) for allkby inspectionI Partial-fraction expansion and
table lo okupDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application29 / 36 Chapter 3: Thez-Transform and Its Application
Expansion into Power Series
Example:
X(z) = log(1 +az1);jzj>jaj=
1X n=1(1)n+1anznn =1X n=1(1)n+1ann znBy inspection: x(n) = (1)n+1ann n1 0n0Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application30 / 36 Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
1. Find the distinct p olesof X(z):p1;p2;:::;pKand their corresponding multiplicitiesm1;m2;:::;mK. 2. The pa rtial-fractionexpansion is of the fo rm:
X(z) =KX
k=1 A1kzpk+A2k(zpk)2++Amk(zpk)mk
wherepkis anmkth order pole (i.e., has multiplicitymk). 3. Use an app ropriateapp roacht ocompute fAikg
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application31 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Example: Findx(n) given poles ofX(z) atp1=2 and a double pole atp2=p3= 1; specically, X(z) =1(1 + 2z1)(1z1)2X(z)z=
z2(z+ 2)(z1)2z 2(z+ 2)(z1)2=
A1z+ 2+A2z1+A3(z1)2
Note: we need a strictly proper rational function. DO NOT FORGET TO MULTIPLY BYzIN THE END.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application32 / 36
Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
z 2(z+ 2)(z+ 2)(z1)2=A1(z+ 2)z+ 2+A2(z+ 2)z1+A3(z+ 2)(z1)2z
2(z1)2=A1+A2(z+ 2)z1+A3(z+ 2)(z1)2
z=2A 1= 49
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application33 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2z
2(z+ 2)=
A1(z1)2z+ 2+A2(z1) +A3z=1A
3= 13 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application34 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2
z 2(z+ 2)=A1(z1)2z+ 2+A2(z1) +A3d
dz z2(z+ 2)= ddz A1(z1)2z+ 2+A2(z1) +A3
z=1 A 2=59 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application35 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Therefore,
assuming c ausality , and using the following pairs: a nu(n)Z !11az1 na nu(n)Z !az1(1az1)2 X(z) =49
11 + 2z1+59
11z1+13
z 1(1z1)2x(n)=
49
(2)nu(n)+ 59 u(n)+ 13 nu(n) (2)n+29 +59
+n3 u(n) Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application36 / 36
quotesdbs_dbs7.pdfusesText_13
6nanu(n1)az1(1az1)2jzj 7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application16 / 36
Chapter 3: Thez-Transform and Its Application
Why Rational?
I X(z) is a rationalfunction i it can be represented as the ratio of two polynomials inz1(orz): X(z) =b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzNI
For LTI systems that are represented by
LCCDEs
, the z-Transform of the unit sample responseh(n), denoted H(z) =Zfh(n)g, isrational Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application17 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros
I zerosofX(z): values ofzfor whichX(z) = 0 I polesofX(z): values ofzfor whichX(z) =1 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application18 / 36 Chapter 3: Thez-Transform and Its Application Poles and Zeros of the Rationalz-Transform
Leta0;b06= 0:
X(z) =B(z)A(z)=b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzN=
b0zMa 0zN zM+ (b1=b0)zM1++bM=b0z N+ (a1=a0)zN1++aN=a0=
b0a 0zM+N(zz1)(zz2)(zzM)(zp1)(zp2)(zpN)
=GzNMQ M k=1(zzk)Q N k=1(zpk)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application19 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
X(z) =GzNMQ
M k=1(zzk)Q N k=1(zpk)whereGb0a 0Note: \nite" does not include zero or1.I
X(z) hasMnitezeros at z=z1;z2;:::;zMI
X(z) hasNnitep olesat z=p1;p2;:::;pNI
ForNM6= 0
I ifNM>0, there arejNMjzeroat o rigin,z= 0 IifNM<0, there arejNMjpolesat o rigin,z= 0Total number of zeros = Total number of poles Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application20 / 36
Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
Example:
X(z) =z2z22z+ 116z3+ 6z+ 5= (z0)(z(12
+j12 )(z(12 j12 )(z(14 +j34 )(z(14 j34 )(z(12 )poles:z=14 j34 ;12 zeros:z= 0;12 j12 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application21 / 36 Chapter 3: Thez-Transform and Its Application Pole-Zero Plot
Example: poles:z=14
j34 ;12, zeros:z= 0;12 j120.5 0.5 -0.5 -0.5 unit circle POLE ZERODr. Deepa Kundur (University of Toronto)The z-Transform and Its Application22 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I Graphical interpretation of characteristics ofX(z) on the complex plane I ROC cannot include poles;assuming causality ...
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application23 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I For realtime-domain signals, the coecients ofX(z) are necessarily real I complex poles and zeros must occur in conjugate pairs Inote: real poles and zeros do nothave to be paired up X(z) =z2z22z+1 16z3+6 z+5 =)
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application24 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot Insights
I For causal systems, the ROC will bethe outer region of the smallest (origin-centered) circle encompassing all the poles .I For stable systems, the ROC will includethe unit circle.0.5 0.5 -0.5 -0.5 ROCI Causal?
Y es. I Stable?
Y es.I
For stability of a
causal system, the poles will lieinside the unit circle Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application25 / 36 Chapter 3: Thez-Transform and Its Application The System Function
h(n)Z !H(z) time-domain Z !z-domain
impulse response Z !system functiony(n) =x(n)h(n)Z
!Y(z) =X(z)H(z) Therefore,
H(z) =Y(z)X(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application26 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
y(n) =NX k=1a ky(nk) +MX k=0b kx(nk)Zfy(n)g=ZfNX k=1a ky(nk) +MX k=0b kx(nk)gZfy(n)g=NX k=1a kZfy(nk)g+MX k=0b kZfx(nk)gY(z)=NX k=1a kzkY(z)+ MX k=0b kzkX(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application27 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
Y(z) +NX
k=1a kzkY(z) =MX k=0b kzkX(z)Y(z)" 1 +NX k=1a kzk#=X(z)MX k=0b kzkH(z) =Y(z)X(z)= P M k=0bkzkh 1 +PN k=1akzkiLCCDE !Rational System Function Many signals of practical interest have a rationalz-Transform.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application28 / 36
Chapter 3: Thez-Transform and Its Application
Inversion of thez-Transform
Three popular methods:
I Contour integration:
x(n) =12jI C X(z)zn1dzI
Expansion into a
p owerseries in zorz1: X(z) =1X
k=1x(k)zk and obtainingx(k) for allkby inspectionI Partial-fraction expansion and
table lo okupDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application29 / 36 Chapter 3: Thez-Transform and Its Application
Expansion into Power Series
Example:
X(z) = log(1 +az1);jzj>jaj=
1X n=1(1)n+1anznn =1X n=1(1)n+1ann znBy inspection: x(n) = (1)n+1ann n1 0n0Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application30 / 36 Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
1. Find the distinct p olesof X(z):p1;p2;:::;pKand their corresponding multiplicitiesm1;m2;:::;mK. 2. The pa rtial-fractionexpansion is of the fo rm:
X(z) =KX
k=1 A1kzpk+A2k(zpk)2++Amk(zpk)mk
wherepkis anmkth order pole (i.e., has multiplicitymk). 3. Use an app ropriateapp roacht ocompute fAikg
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application31 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Example: Findx(n) given poles ofX(z) atp1=2 and a double pole atp2=p3= 1; specically, X(z) =1(1 + 2z1)(1z1)2X(z)z=
z2(z+ 2)(z1)2z 2(z+ 2)(z1)2=
A1z+ 2+A2z1+A3(z1)2
Note: we need a strictly proper rational function. DO NOT FORGET TO MULTIPLY BYzIN THE END.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application32 / 36
Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
z 2(z+ 2)(z+ 2)(z1)2=A1(z+ 2)z+ 2+A2(z+ 2)z1+A3(z+ 2)(z1)2z
2(z1)2=A1+A2(z+ 2)z1+A3(z+ 2)(z1)2
z=2A 1= 49
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application33 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2z
2(z+ 2)=
A1(z1)2z+ 2+A2(z1) +A3z=1A
3= 13 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application34 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
z 2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2
z 2(z+ 2)=A1(z1)2z+ 2+A2(z1) +A3d
dz z2(z+ 2)= ddz A1(z1)2z+ 2+A2(z1) +A3
z=1 A 2=59 Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application35 / 36 Chapter 3: Thez-Transform and Its Application Partial-Fraction Expansion
Therefore,
assuming c ausality , and using the following pairs: a nu(n)Z !11az1 na nu(n)Z !az1(1az1)2 X(z) =49
11 + 2z1+59
11z1+13
z 1(1z1)2x(n)=
49
(2)nu(n)+ 59 u(n)+ 13 nu(n) (2)n+29 +59
+n3 u(n) Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application36 / 36
quotesdbs_dbs7.pdfusesText_13
7 cos(!0n)u(n)1z1cos!012z1cos!0+z2jzj>1
8 sin(!0n)u(n)z1sin!012z1cos!0+z2jzj>1
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application16 / 36
Chapter 3: Thez-Transform and Its Application
Why Rational?
I X(z) is a rationalfunction i it can be represented as the ratio of two polynomials inz1(orz):X(z) =b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzNI
For LTI systems that are represented by
LCCDEs
, the z-Transform of the unit sample responseh(n), denotedH(z) =Zfh(n)g, isrational Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application17 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros
I zerosofX(z): values ofzfor whichX(z) = 0 I polesofX(z): values ofzfor whichX(z) =1Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application18 / 36 Chapter 3: Thez-Transform and Its ApplicationPoles and Zeros of the Rationalz-Transform
Leta0;b06= 0:
X(z) =B(z)A(z)=b0+b1z1+b2z2++bMzMa
0+a1z1+a2z2++aNzN=
b0zMa 0zN zM+ (b1=b0)zM1++bM=b0zN+ (a1=a0)zN1++aN=a0=
b0a0zM+N(zz1)(zz2)(zzM)(zp1)(zp2)(zpN)
=GzNMQ M k=1(zzk)Q Nk=1(zpk)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application19 / 36 Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
X(z) =GzNMQ
M k=1(zzk)Q N k=1(zpk)whereGb0a0Note: \nite" does not include zero or1.I
X(z) hasMnitezeros at z=z1;z2;:::;zMI
X(z) hasNnitep olesat z=p1;p2;:::;pNI
ForNM6= 0
I ifNM>0, there arejNMjzeroat o rigin,z= 0 IifNM<0, there arejNMjpolesat o rigin,z= 0Total number of zeros = Total number of polesDr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application20 / 36
Chapter 3: Thez-Transform and Its Application
Poles and Zeros of the Rationalz-Transform
Example:
X(z) =z2z22z+ 116z3+ 6z+ 5= (z0)(z(12
+j12 )(z(12 j12 )(z(14 +j34 )(z(14 j34 )(z(12 )poles:z=14 j34 ;12 zeros:z= 0;12 j12Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application21 / 36 Chapter 3: Thez-Transform and Its ApplicationPole-Zero Plot
Example: poles:z=14
j34 ;12, zeros:z= 0;12 j120.5 0.5 -0.5 -0.5 unit circle POLEZERODr. Deepa Kundur (University of Toronto)The z-Transform and Its Application22 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I Graphical interpretation of characteristics ofX(z) on the complex plane IROC cannot include poles;assuming causality ...
0.5 0.5 -0.5 -0.5ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application23 / 36 Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot
I For realtime-domain signals, the coecients ofX(z) are necessarily real I complex poles and zeros must occur in conjugate pairs Inote: real poles and zeros do nothave to be paired upX(z) =z2z22z+1 16z3+6 z+5 =)
0.5 0.5 -0.5 -0.5 ROCDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application24 / 36Chapter 3: Thez-Transform and Its Application
Pole-Zero Plot Insights
I For causal systems, the ROC will bethe outer region of the smallest (origin-centered) circle encompassing all the poles .I For stable systems, the ROC will includethe unit circle.0.5 0.5 -0.5 -0.5 ROCICausal?
Y es. IStable?
Y es.I
For stability of a
causal system, the poles will lieinside the unit circleDr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application25 / 36 Chapter 3: Thez-Transform and Its ApplicationThe System Function
h(n)Z !H(z) time-domainZ !z-domain
impulse responseZ !system functiony(n) =x(n)h(n)Z
!Y(z) =X(z)H(z)Therefore,
H(z) =Y(z)X(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application26 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
y(n) =NX k=1a ky(nk) +MX k=0b kx(nk)Zfy(n)g=ZfNX k=1a ky(nk) +MX k=0b kx(nk)gZfy(n)g=NX k=1a kZfy(nk)g+MX k=0b kZfx(nk)gY(z)=NX k=1a kzkY(z)+ MX k=0bkzkX(z)Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application27 / 36 Chapter 3: Thez-Transform and Its Application
The System Function of LCCDEs
Y(z) +NX
k=1a kzkY(z) =MX k=0b kzkX(z)Y(z)" 1 +NX k=1a kzk#=X(z)MX k=0b kzkH(z) =Y(z)X(z)= P M k=0bkzkh 1 +PN k=1akzkiLCCDE !Rational System FunctionMany signals of practical interest have a rationalz-Transform.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application28 / 36
Chapter 3: Thez-Transform and Its Application
Inversion of thez-Transform
Three popular methods:
IContour integration:
x(n) =12jI CX(z)zn1dzI
Expansion into a
p owerseries in zorz1:X(z) =1X
k=1x(k)zk and obtainingx(k) for allkby inspectionIPartial-fraction expansion and
table lo okupDr. Deepa Kundur (University of Toronto)The z-Transform and Its Application29 / 36 Chapter 3: Thez-Transform and Its Application
Expansion into Power Series
Example:
X(z) = log(1 +az1);jzj>jaj=
1X n=1(1)n+1anznn =1X n=1(1)n+1ann znBy inspection: x(n) = (1)n+1ann n10n0Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application30 / 36 Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
1. Find the distinct p olesof X(z):p1;p2;:::;pKand their corresponding multiplicitiesm1;m2;:::;mK. 2.The pa rtial-fractionexpansion is of the fo rm:
X(z) =KX
k=1A1kzpk+A2k(zpk)2++Amk(zpk)mk
wherepkis anmkth order pole (i.e., has multiplicitymk). 3.Use an app ropriateapp roacht ocompute fAikg
Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application31 / 36 Chapter 3: Thez-Transform and Its ApplicationPartial-Fraction Expansion
Example: Findx(n) given poles ofX(z) atp1=2 and a double pole atp2=p3= 1; specically,X(z) =1(1 + 2z1)(1z1)2X(z)z=
z2(z+ 2)(z1)2z2(z+ 2)(z1)2=
A1z+ 2+A2z1+A3(z1)2
Note: we need a strictly proper rational function.DO NOT FORGET TO MULTIPLY BYzIN THE END.Dr. Deepa Kundur (University of Toronto)The z-Transform and Its Application32 / 36
Chapter 3: Thez-Transform and Its Application
Partial-Fraction Expansion
z2(z+ 2)(z+ 2)(z1)2=A1(z+ 2)z+ 2+A2(z+ 2)z1+A3(z+ 2)(z1)2z
2(z1)2=A1+A2(z+ 2)z1+A3(z+ 2)(z1)2
z=2A 1= 49Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application33 / 36 Chapter 3: Thez-Transform and Its ApplicationPartial-Fraction Expansion
z2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2z
2(z+ 2)=
A1(z1)2z+ 2+A2(z1) +A3z=1A
3= 13Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application34 / 36 Chapter 3: Thez-Transform and Its ApplicationPartial-Fraction Expansion
z2(z1)2(z+ 2)(z1)2=A1(z1)2z+ 2+A2(z1)2z1+A3(z1)2(z1)2
z2(z+ 2)=A1(z1)2z+ 2+A2(z1) +A3d
dz z2(z+ 2)= ddzA1(z1)2z+ 2+A2(z1) +A3
z=1 A 2=59Dr. Deepa Kundur (University of Toronto)
The z-Transform and Its Application35 / 36 Chapter 3: Thez-Transform and Its ApplicationPartial-Fraction Expansion
Therefore,
assuming c ausality , and using the following pairs: a nu(n)Z !11az1 na nu(n)Z !az1(1az1)2X(z) =49
11 + 2z1+59
11z1+13
z1(1z1)2x(n)=
49(2)nu(n)+ 59 u(n)+ 13 nu(n) (2)n+29 +59
+n3 u(n)