lecture 07 - z-transform

8/6/2019 Lecture 07 - Z-Transform

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ELEN 5346/4304 DSP and Filter Design Fall 2008

1

Lecture 7: Z-transform

Instructor:

Dr. Gleb V. Tcheslavski

Contact:

[email protected]

Office Hours:

Room 2030

Class web site:

http://ee.lamar.edu/gleb/dsp/index.htm


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Definitions

Z-transform converts a discrete-time signal into a complex

frequency-domain representation. It is similar to the Laplace

transform for continuous signals.

( ) nnn

X z x z| If (where) it exists!

(7.2.1)

n is an integer time index;

j

z re[

! is a complex number; [ - angular freq.

When the magnitude r=1,jz e [!

( ) j j nnn

z e x e[ [p ! If it exists!

(7.2.2)


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Region of Convergence (ROC)

j z re[! (7.3.2)Since z is complex:

( ) n j n n j nn nn n

X z x r e x r e[ [ ! !

The Region of convergence (ROC) is the set of points zin the complex plane,

for which the summation is bounded (converges):

n

n

n

x z g (7.3.1)

In general, z-transform exists for r r r

r z r

(7.3.3)

(7.3.4)

(7.3.5)

r- r+ Re

Im


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Examples of ROCs

from Mitra


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Example 7.1: Let xn = an _ a2 1 2..., , , , , ,...n x a a a a !

n

n

n

x z

! gThere are no values ofzsatisfying:

Example 7.2: Let xn = an un a causal sequence

1 10

1( )

1

nn n

n

n n

X z aaz

u z azg

!

! !

!

1 1o af r az z "

a Re

Im

(7.5.1)

ROC

We can modify (7.5.1) as

( )( )

( )

z N zz

z a D z! !

(7.5.2)

( ) 0 0

( ) 0

( )

( )

N z z

D

zero s

pol z z a e s

! !

! !

x

( )n

n

n

X z x z|

roots of numerator: X(z) = 0

roots of denominator: X(z) p g


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Example 7.3: Let xn = -an u-n-1 an anticausal sequence

_ a

1

11

1

1 1

1 11

1 0

1

1( )n

n n n

n

n n

m m

m m

a u z n n a z

az a z

X z

a z z

a z z a

!g

g g

!

!

! ! u e !

! !

!

!

(7.6.1)

a Re

Imx1 1for a z z a

Conclusion 1: z-transform exists only within the ROC!

Conclusion 3: poles cannot exist in the ROC; only on its boundary.

Note: if the ROC contains the unit circle (|z|=

1), the system is stable.

Conclusion 2: z-transform and ROC uniquely specify the signal.

( )n

n

n

X z x z|


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The transfer function

-i n i j n j

i j

a y b x2 1

! !

! Consider an LCCDE:

Time shift:( ) { } ( )n

l m

n m n m lz

n l

x x z z l n m z x z z z p ! ! ! !

and take z-transform

utilizing time shift( ) ( )i j

i j

i j

a z Y z b z z2 1

! !

!

LTI:( )

( )

( ) ( ) ( )n n m mn n n m n m m n mz

n m m n

z

y h x h x z h xY z Hz zz z ! p ! ! ! 1 44 2 4 43

The system transfer function0

0

( )( )

( )

M

j

j

j n

n

i ni

i

b zY z

z h zX z

a z

!

!

! ! !

(7.7.1)

(7.7.2)

(7.7.3)

(7.7.4)

(7.7.5)

( )n

n

n

X z x z|


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Rational z-transform

Frequently, a z-transform can be described as a rational function, i.e. a

ratio of two polynomials in z-1:1 ( 1)

0 1 1

1 ( 1)

0 1 1

...( )( )

( ) ...

M M

M M

N N

N N

n n z n z n zP zz

z d d z d z d z

! !

Here Mand Nare the degrees of the numerators and denominatorspolynomials. An alternative representation is a ratio of two polynomials in z:

( 1)

0 1 1

1

0 1 1

...( )( )

( ) ...

M M

N M M M

N N

N N

n z n z n z nP zz z

z d z d z d z d

! !

(7.8.1)

(7.8.2)

Finally, a rational z-transform can be written in a factorized form:

1

0 0

1 1( )

1

0 0

1 1

(1 ) ( )

( )

(1 ) ( )

M M

j j

j jN M

N N

i i

i i

n z z n z z

z z

d p z d z p

! !

! !

! !

(7.8.3)

zeros: numerator= 0

Poles: denominator= 0

( )n

n

n

X z x z|


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Notes on poles of a system function

Positions of poles of a transfer function are used to evaluate system

stability. Let assume a single real pole at z=E. Therefore:

( ) 1( )( )

Y z z zY z Y z X z

X z zE

E! ! !

The difference equation is:

1 1 1n n n n n n y y x y y xE E ! !

Therefore, the impulse response is:

1 1n n nh hE H !

for 20,1, 2, 3,... 0,1, , ,...nn h E E ! ! Iff |E| < 1, hn decays as npg and the system is BIBO stable; otherwise, hngrows without limits. Therefore, poles of a stable system (and signals

in fact) must be inside the unit circle.

Zeros may be placed anywhere. Zeros at the origin produce a time delay.

(7.9.1)

(7.9.2)

(7.9.3)

(7.9.4)


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The transfer function and the

Frequency response

( )1 jn z en

jh z RO HC e H z [[

! g ! !BIBO:

0

0 1

0

0 1

( )

( )

( )

MM

Mj

jj

j jN N

i N

i i

i i

b z z zb z

H z

a z a z z p

! !

! !

! !

where zj are zeros andpiare poles of the transfer function.

BIBO:

10

0

1

0

( )

1

0

1

0

0

1

1

( 1

( ( ) ( )

)

( )

)

M

M

j M N j

j

jj

N

jj

jj

N

j

i

i

M N

j

j

i

i j j

j i

j i

b e e z

e zb

ea

e pe

a e

e b a M N e z e

p

p

[

[

[

[

[ [

[

[

[ [[

!

!

!

!

! !

!

!

!

(7.10.1)

(7.10.2)

(7.10.3)

( )n

n

n

X z x z|


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The transfer function and the

Frequency response

A good way to evaluate the systems frequency response:

_ a

_ a

0

2

0k

M

mj

Nkp

k

P FT be

P FT a

[T

[ !

!

When the frequency approaches a pole, the frequency response has a local

maximum, a zero forces the response to a local minimum.

For real systems, poles and zeros are symmetrical with respect to the real axis.

(7.11.1)

0 0.1 0.2 0.3 0.4 0.5 0

0. 5

1

1. 5

MagnitudeofH

([)

Fractional f requency, [

pole zero

Zero-padded

( )n

n

n

X z x z|


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The transfer function and the SFG

1 ( ) ( ) ( )

( ) ( ) ( )

n n n

T Tzn n n

S AS bx zS z AS z bX z

y c S dx Y z c S z dX z

! ! p

! !

1

1

1

( ) ( ) ( ) ( )

( ) ( ) ( )

( )

T

T

zI S z b z S z zI

Y z c zI b z d z

H z c zI b d

b z

!

! !

!

Poles ofH(z) correspond to the eigenvalues of the

system matrix.

(7.12.1)

(7.12.2)

(7.12.3)


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More on Transfer function

0

1 ( )0 0 1

00

0 1 1

( ) ( )( )

( )( )

( ) ( )

M MMMm

j mmj M Nm m

N N N

k N

k i k

k i k

b z z z z zb zbP z

z zz a

a z a z z p z p

! ! !

! ! !

! ! ! !

| 1 ( )j

j

j

jz e

e BIB z z e

e[

[

[

[!

! p p

1) N> M: zeros at z= 0 of multiplicity N-M

2) M> N: poles at z= 0 of multiplicity M-N

zeros

poles

( )

( )

( )

1

( ) .

j M N

j M N

j M N

e const magnitudee distortionless

e M N lin phase

[[

[ [

! p

!

(7.13.1)

( ) 0 ( ) 0; ( ) ( )i i i i

H z Y z H p Y pp p p g p g (7.13.2)


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Types of digital filters

0( ) 0

M

nm

m n

m

b n MH z b z h otherwise

!

e e! p !

1. FIR (all-zero) filter:

(7.14.1)

All poles are at z= 0: a nest of poles

ROC: the entire z-plane except of the origin (z= 0).

FIR filters are stable.


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2. IIR (all-pole) filter:

0

0

( )N

k

k

k

bH z

a z

!

!

All zeros are at z= 0: a nest of zeros

(7.15.1)


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3. General IIR (zero-pole) filter:

0

0

( )

M

m

mm

N

k

k

k

b zH z

a z

!

!

!

(7.16.1)


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On test signals

( ) ( ) ( ) n n nz

Y z H z X z y h x! p !

n nhH p

n k n k n

k

x x yH ! p %

Calculate and compare ton k n kk

y x h! ny

%

(7.17.1)

(7.17.2)

(7.17.3)

We dont need any other that a delta function test signals since a unit-pulse

response is a complete systems description.


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Types of sequences and convergence

1. Two-sided:

( )N

n

n M

nG z g z

!

!

Converges everywhere except ofz= 0 and z=g

(7.18.1)


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2. Right-sided:

1

0

( ) n n nn n nn nM Mn

G z g z g z g z g

g

! !!

! ! (7.19.1)

Assume: if converges at z = z0, converges for |z| > | z0|

Blows up at z=g

ROC: r-

< |z| < g - exterior ROC

For a causal sequence: |z| > r- = max|pk| - a max pole ofG(z)

To be causal, a sequence must be right-sided (necessary but not sufficient)


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3. Left-sided:

1

0

( ) n n nn n nn n n

N N

G z g z g z g z

! !g !

g

! ! (7.20.1)

Blows up at z= 0Converges at z0

ROC: 0 < |z| < r+

- interior ROC

When encountering an interior ROC, we need to check convergence at

z= 0. If the sequence blows up at zero its an anti-causal sequence


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Properties

from Mitra


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Common pairs


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Inverse z-transform

11 ( )2

n

n

c

x z z dzjT

! i (7.23.1)

WhereC

is a counterclockwise closed path encircling the origin and is entirely inthe ROC. ContourCmust encircle all the poles ofX(z).

In general, there is no simple way to compute (7.23.1)

A special case:C is the unit circle (can be used when the ROC includes the unit circle).

The inverse z-transform reduces to the IDTFT.

12

j j n

n x X e e d

T[ [

T

[T

! (7.23.2)


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Inverse z-transform

A. Via Cauchy residue theorem

n i

i

x V! For all poles ofX(z)zn-1 inside C (contour of integration)

WhereViare the residues ofX(z)zn-1

1

1

( )1

( 1)! i

k

i

i k

z p

d z

kdz

JV

!

!

for a pole of multiplicity k:

Residue function: 1( ) ( ) kni i

z X z z z pJ !

(7.24.1)

(7.24.2)

(7.24.3)

( )n

n

n

X z x z|


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Inverse z-transform: Example

( ) ;z

z a zz a

!

a Re

Im

xC

1

0

01 1

02 2

nan

n

aC C

nz z x z dz dz

nj z a j z a

V

V VT T

u! ! !

i i

Example:

V0 is a residue ofX(z)z-n-1 at z=0 involves pole of

multiplicity n wnen n < 0.1

1

1

1( )

( 1)!

k nk n n

a k z ak

d z z a z a

k d z z aV

!!

! ! !

multiplicity1 1

0 1 0

1 ( ) 1( 0) ( 1)( 2)...( ( 1))( )

( 1)! ( 1)!

k

k k k n

k k z

d z aLet n k n z k z a a a

k dz z kV

!

! ! ! ! !

0

0 0nn

n n

n

nn

a nx

a aa u

nx

u !

!

!

( )n

n

n

X z x z|


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Inverse z-transform

B. Via recognition (table look-up)

Example:

( )a

zz e!

0 0

1( ) ; 1

!!

na

z

n

nn

n

a a X z e z a

z n

az

n z

g

!

g

!

! ! ! "

!

n

n n

ax u

n!Therefore:

Sometimes, the z-transform can be modified such way that it can be

found in a table

( )n

n

n

X z x z|


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Inverse z-transform

C. Via long division

1. Right-sided z-transform sequences can be expanded into a power series in

z-1. The coefficient multiplying z-n is the nth sample of the inverse z-transform.

Example:2 1

1

2 2( ) ; 1

1

z z X z z

z

! "

Lower powers first:

1 2

1

2 2( ) ; 1

1

z zz z

z

!

and long division:11 |z

2 3 42 ... z z z 1 2

1

2 2

2 2

z z

z

2

2 3

z

z z

{2, 0,1, 1,1,...}nx !

x0x1x2 x3 x4

( )n

n

n

X z x z|


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Inverse z-transform

2. Left-sided z-transform sequences into a power series in z1

Example: 2 1

1

2 2( ) ; 1

1

z z X z z

z

!

Multiply both numerator and denominator by z2

2 2 1 2

2 1 2

2 2 1 2 2( )

1

z z z z zz

z z z z

! !

Long division

1 2 3 4( ) 1 ...X z z z z z z!

x0 x-1 x-2 x-3 x-4x1 Non-causal

( )n

n

n

X z x z|


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Inverse z-transform

Example:

2 1

1 1

2 2 1( ) ; 1

21 2

z zX z z

z z

!

not suitable for

long division!

Example: 1

1

1 1

( ) 1 ; 1

1

( 1)nn n n n

zX z z z

zz

x uH H

!

q q

!

( )n

n

n

X z x z|


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Inverse z-transform

D. Via partial fraction expansion (PFE)

( )( ) ;

( )

N zLet G z r z r

D z

!

If the degree of the numerator is equal or greater than the degree of the

denominator: Mu N, G(z) is an improper polynomial. Then:

1

0

(( )(

( ))

)

( )

M Nl

l

l

N zG z

D

N zc z

zz D

!

! !

A proper fraction: M1 < N

(7.30.1)

(7.30.2)

Then:1

0 1

( )1

M N N

l ll

l l l

cp

G z zz

V

! ! !

Simple poles: multiplicity of 1.

(7.30.3)

( )n

n

n

X z x z|


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Inverse z-transform

Here Vl is a residue

( ) ( )l

n l

l l z pz p G z zV

!!

poles

Therefore:

0 1

:

:

n M N N

l n l

n l n l l nl l l n l l

p u ifexternal ROC p rg c

p u ifinternal ROC p rH V

! !

e!

u

(7.31.1)

(7.31.2)

This method is suitable for complex poles.

Problem: large polynomials are hard to manipulate

??QUESTIONS??

( )n

n

n

X z x z|

lecture 07 - z-transform

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