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Manolakis, Digital Signal Processing: Principles, Algorithms, and Applications, 4th edition, 2007 Dr Deepa Kundur (University of Toronto) The z-Transform and Its 



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[PDF] The z-Transform and Its Application - University of Toronto

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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 Application

Discrete-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 Application

The Directz-Transform

I

Directz-Transform:

X(z) =1X

n=1x(n)zn I

Notation:

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 Application

Region 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

I

For 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 forjzcjIdiverges forjzcj>r

Imay 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 Application

Power Series Convergence

I

For 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, and

Idiverges 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)

The 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 Application

Convolution using thez-Transform

Basic Steps:

1. Compute z-Transform of each of the signals to convolve (time domain!z-domain): X

1(z) =Zfx1(n)g

X

2(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)g

Dr. Deepa Kundur (University of Toronto)

The z-Transform and Its Application14 / 36 Chapter 3: Thez-Transform and Its Application

Common 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

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