lecture 34: properties of the z-transform and the inverse z-transform

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ECE 8443 – Pattern Recognition EE 3512 – Signals: Continuous and Discrete Objectives: Modulation, Summation, Convolution Initial Value and Final Value Theorems Inverse z-Transform by Long Division Inverse z-Transform by Partial Fractions Difference Equations Resources: MIT 6.003: Lecture 23 Wiki: Inverse Z-Transform CNX: Inverse Z-Transform Arslan: The Inverse Z-Transform CNX: Properties ISIP: Pole/Zero Demo LECTURE 34: PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM Audio: URL:

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LECTURE 34: PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM. Objectives: Modulation, Summation, Convolution Initial Value and Final Value Theorems Inverse z-Transform by Long Division Inverse z-Transform by Partial Fractions Difference Equations - PowerPoint PPT Presentation

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Page 1: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

ECE 8443 – Pattern RecognitionEE 3512 – Signals: Continuous and Discrete

• Objectives:Modulation, Summation, ConvolutionInitial Value and Final Value TheoremsInverse z-Transform by Long DivisionInverse z-Transform by Partial FractionsDifference Equations

• Resources:MIT 6.003: Lecture 23Wiki: Inverse Z-TransformCNX: Inverse Z-TransformArslan: The Inverse Z-TransformCNX: PropertiesISIP: Pole/Zero Demo

LECTURE 34: PROPERTIES OF THE Z-TRANSFORMAND THE INVERSE Z-TRANSFORM

Audio:URL:

Page 2: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 2

Properties of the z-Transform

• Linearity:

• Time-shift:

• Multiplication by n:

Proof:

• Multiplication by an:

Proof:

• Multiplication by ejn:

• Multiplication by cosn:

• Multiplication by sinn:

• Summation:

][][][][ 2121 zbXzaXnbxnax

][][ 00 zXznnx n

dz

zdXznnx

][][

][][)(

][)(

][

1 nnxznxndz

zdXzznxn

dz

zdX

znxX(z)

n

n

n

n

n

n

Z

a

zXnxa n ][

a

zX

a

znxznxanua

n

n

n

nnn ][])[()(Z

zeXnxe njnj ][

zeXzeXnxn njnj )2/1(][)cos(

zeXzeXjnxn njnj )2/(][)sin(

)(1

1)(][][

10

zXz

zVixnvn

i

Page 3: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 3

Convolution

• Convolution:

Proof:

Change of index on the second sum:

The ROC is at least the intersection of the ROCs of x[n] and h[n], but can be a larger region if there is pole/zero cancellation.

• The system transfer function is completely analogous to the CT case:

• Causality:

Implies the ROC must be the exterior of a circle and include z = .

)()(][][][*][ zHzXknhkxnhnxk

n

nk

n

n kk

zknhkx

zknhkxknhkxnhnx

][][

][][][][][*][ ZZ

knm

)()(

][][][][][*][ ))(

zHzX

zmhzkxzmhkxnhnx m

m

k

k

im

mk

Z

n

nznhzHnh ][)(][

00][ nnh

Page 4: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 4

Initial-Value and Final-Value Theorems (One-Sided ZT)

• Initial Value Theorem:

Proof:

• Final Value Theorem:

• Example:

• Tables 7.2 and 7.3 in the textbook contain a summary of the z-Transform properties and common transform pairs.

)(lim]0[ zXxz

]0[...]1[]0[lim][lim)(lim 1

0

xzxxznxzXz

n

n

zz

)()1(lim][lim1

zXznxzn

105.

5

5.0

423)]()1[(][lim

)5.0)(1(

423

5.05.12

423)(

12

2

1

2

2

23

2

zzn zz

zzzXznx

zzz

zz

zzz

zzzX

Page 5: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 5

Inverse Laplace Transform

• Recall the definition of the inverse Laplace transform via contour integration:

• The inverse z-transform follows from this:

Evaluation of this integral is beyond the scope of this course. Instead, as with the Laplace transform, we will restrict our interest in the inverse transform to rational forms (ratio of polynomials). We will see shortly that this is convenient since linear constant-coefficient difference equations can be converted to polynomials using the z-transform.

• As with the Laplace transform, there are two common approaches:

Long Division

Partial Fractions Expansion

• Expansion by long division Is also known as the power series expansion approach and can be easily demonstrated by an example.

C

ststj

j

dsesXj

dsesXj

tx )(2

1

2

1)(

C

n dzzzXj

nx 1)(2

1][

Page 6: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 6

Long Division

• Consider:

Solution:42

1)(

3

2

zz

zzX

1

1

12

23

43

42

142

z

z

zz

zzz

321

321

32

1

12

23

30

1264

1263

43

42

142

zzz

zzz

zz

z

zz

zzz

4321

432

431

321

32

1

12

23

430

16206

1684

1264

1263

43

42

142

zzzz

zzz

zzz

zzz

zz

z

zz

zzz

...]4[4]3[3]1[1][0][

...430)( 4321

nnnnnx

zzzzzX

Implications of stability?

Page 7: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 7

Inverse z-Transform Using MATLAB

• Consider:

MATLAB:

Syms X x z

X = (8*z^3+2*z^2-5*z)/(z^3-1.75*z+.75);

x = iztrans(X)

x = 2*(1/2)^n+2*(-3/2)^n+4

Evaluate numerically:

num = [8 2 -5 0];

den = [1 0 -1.75 .75];

x = filter(num, den, [1 zeros(1,9)])

Output:

8 2 9 -2.5 14.25 -11.125 26.8125 -30.1563 55.2656

75.75.1

528)(

3

23

zz

zzzzX

Page 8: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 8

Inverse z-Transform Using Partial Fractions

• Rational transforms can be factored using the same partial fractions approach we used for the Laplace transforms.

• The partial fractions approach is preferred if we want a closed-form solution rather than the numerical solution long division provides.

• Example:

In this example, the order of the numerator and denominator are the same. For this case, we can use a trick of factoring X(z)/z:

2

1)(

23

3

zzz

zzX

643.02)(

0825.0429.0866.05.0)(

5.02

1)(

)(

2866.05.0866.05.0

)(

)866.05.0)(866.05.0)(2(2)(

23

866.05.01

00

3110

23

z

jz

z

zz

zXc

jjzz

zXc

zz

zXc

z

c

jz

c

jz

c

z

c

z

zX

jzjzzzzzzA

Page 9: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 9

Inverse z-Transform (Cont.)We can compute the inverse using our table of common transforms:

The exponential terms can be converted to a single cosine using a magnitude/phase conversion:

][2][)866.05.0(][)866.05.0(][][

21866.05.01866.05.01

2866.05.0866.05.0)(

3110

13

11

11

0

3110

nucnujcnujcncnx

z

c

zj

c

zj

cc

z

zc

jz

zc

jz

zcczX

nnn

][)2(643.0)19.03

4cos(874.0][5.0

][)2()cos(2][

][2][)866.05.0(][)866.05.0(][][

)(10.89rad19.0429.0

0825.0tan

437.0)0825.0()429.0(

rad3

4

5.0

866.0tan

1)866.0()5.0(

311110

3110

11

221

11

221

nunn

nuccnppcnc

nucnujcnujcncnx

c

c

p

p

n

n

nnn

Page 10: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 10

Inverse z-Transform (Cont.)This can be verified using MATLAB:

num = [1 0 0 1];

den = [1 -1 -1 -2 0];

[r, p] = residue(num, den)

r = p=

0.6429 2.0000

0.4286 – 0.825i -0.5000 + 0.8660i

0.4286 + 0.825i -0.5000 – 0.8660i

-0.5000 0

The first 20 samples of the output can be computed numerically using:

num = [1 0 0 1];

den = [1 -1 -1 -2 0];

x = filter(num, den, [1 zeros(1,19)]);

• Using MATLAB as a resource for solving homework problems can greatly reduce the time you spend doing busywork.

2

1)(

23

3

zzz

zzX

Page 11: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 11

First-Order Difference Equations

• Consider a first-order difference equation:

• We can apply the time-shift property:

• We can solve for Y(z):

• The response is again a function of two things: the response due to the initial condition and the response due to the input.

• If the initial condition is zero:

• Applying the inverse z-Transform:

• Is this system causal? Why?

• Is this system stable? Why?

• Suppose the input was a sinusoid. How would you compute the output?

][]1[][ nbxnayny

)(]1[)()( 1 zbXyzYzazY

)(11

]1[)(

11zX

az

b

az

ayzY

11 1)(

)()()(

1)(

az

b

zX

zYzHzX

az

bzY

][)(1

][1

1 nuabaz

bnh n

Z

Page 12: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 12

Example of a First-Order System

• Consider the unit-step response of this system:

• Use the (1/z) approach for the inverse transform:

• The output consists of a DC term, an exponential term due to the I.C., and an exponential term due to the input. Under what conditions is the output stable?

)1)((

]1[

1

1

11

]1[)(

11

]1[)(

1

1)(][][

2

11111

1

zaz

bz

az

zay

zaz

b

az

ayzX

az

b

az

ayzY

az

z

zzXnunx

...,2,1,0],1)([1

)](1[

])1()([1

)](1[][

11

]1[

1

)1/()1/(]1[)(

1

)1/()1/(

)1)((

1)(

1

2

naa

baay

aaa

baayny

z

z

az

az

a

b

az

zay

z

zab

az

zaab

az

zayzY

z

ab

az

aab

zaz

bz

zz

zV

nn

nnn

Page 13: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 13

Second-Order Difference Equations

• Consider a second-order difference equation:

• We can apply the time-shift property:

• Assume x[-1] = 0 and solve for Y(z):

• Multiplying z2/z2:

• Assuming the initial conditions are zero:

• Note that the impulse response is of the form:

This can be visualized as a complex pole pair with a center frequency and bandwidth (see Java applet).

]1[][]2[]1[][ 1021 nxbnxbnyanyany

)()(]2[]1[)(]1[)()( 110

122

11 zXzbzXbyyzzYzayzYzazY

)(11

]1[]1[]2[)(

22

11

110

22

11

1212 zX

zaza

zbb

zaza

zyayayazY

212

12

0)(azaz

zbzbzY

22

2

)cos2(

)cos()(][cos][

azaz

zazzHnunanh n

)(

]1[]2[]1[)(

212

12

0

212

22

21 zXazaz

zbzb

azaz

zyazyayazY

Page 14: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 14

Example of a Second-Order System

• Consider the unit-step response of this system:

• We can further simplify this:

• The inverse z-transform gives:

1]2[,2]1[where]1[][]2[5.0]1[5.1][1

1)(][][

1

yynxnxnynynyaz

z

zzXnunx

1

3

5.0

5.05.05.1

5.2)(

2

2

z

z

z

zzz

zzzY

)]1()(:note[5.05.15.05.1

5.3

15.05.15.05.1

)2)(5.0()1)(5.0()2)(5.1(

)(]1[]2[]1[

)(

222

2

2

2

2

2

2

2

212

12

0

212

22

21

zzzzzzz

z

zz

zz

z

z

zz

zz

zz

zz

zXazaz

zbzb

azaz

zyazyayazY

...,2,1,0,)1(3)5.0(5.0][ nny nn

MATLAB:

num = [1 -1 0];

den = [1 1.5 .5];

n = 0:20;

x = ones(1, length(n));

zi = [-1.5*2-0.5*1, -0.5*2];

y = filter(num, den, x, zi);

Page 15: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 15

Nth-Order Difference Equations

• Consider a general difference equation:

• We can apply the time-shift property once again:

• We can again see the important of poles in the stability and overall frequency response of the system. (See Java applet).

• Since the coefficients of the denominator are most often real, the transfer function can be factored into a product of complex conjugate poles, which in turn means the impulse response can be computed as the sum of damped sinusoids. Why?

• The frequency response of the system can be found by setting z = ej.

M

ii

N

ii inxbinyany

11

][][][

NN

MM

N

i

ii

M

i

ii

M

i

ii

N

i

ii

M

i

ii

N

i

ii

zazazaa

zbzbzbb

za

zb

zX

zYzH

zbzXzazY

zXzbzYzazY

...1

...

1)(

)()(

)(1)(

)conditions initial zero (assuming)()()(

22

110

22

110

1

1

11

11

Page 16: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 16

Transfer Functions

• In addition to our normal transfer function components, such as summation and multiplication, we use one important additional component: delay.

• This is often denoted by its z-transform equivalent.

• We can illustrate this with an example (assumeinitial conditions are zero):

][nxz-1

]1[][ nxny

][nxD

]1[][ nxny

)()( 1 zXzzY

Page 17: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 17

Transfer Function Example

• Redraw using z-transform:

• Write equations for the behavior at each of the summation nodes:

• Three equations and three unknowns: solve the first for Q1(z) and substitute into the other two equations.

)()(2)(

)(3)()(

)()()(

21

12

21

zQzQzY

zYzQzzQ

zXzQzzQ

53

12

)(

)()(

)(3)(1

1)(2)(3)(

1

12)(

)(3)(1

1)(

)(3)()()(

)(3)()()(

)()()(

2

122

1122

1

1222

122

22

12

12

12

11

zz

z

zX

zYzH

zYzzXzz

zXzzYzzXzz

zzY

zYzzXzz

zQ

zYzzXzzQzzQ

zYzXzzQzzzQ

zXzzQzzQ

.Simplify..

Page 18: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 18

Basic Interconnections of Transfer Functions

Page 19: LECTURE 34:  PROPERTIES OF THE Z-TRANSFORM AND THE INVERSE Z-TRANSFORM

EE 3512: Lecture 34, Slide 19

Summary

• Introduced additional properties of the z-transform.

• Derived the convolution property for DT LTI systems.

• Introduced two practical ways to compute the inverse z-transform: long division and partial fractions expansion.

• Worked examples of each and demonstrated how to solve these problems using MATLAB.

• Demonstration: Frequency response using a Java applet that allows you to visualize poles and zeros in the complex plane.

• Demonstrated the solution of Nth-order difference equations using thez-transform: general response is an exponential.

• Demonstrated how to develop and decompose signal flow graphs using the z-transform: introduced a component, the delay, which is equivalent to differentiation in the s-plane.