L04-ztrans

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2013-10-02Dan Ellis 1

ELEN E4810: Digital Signal Processing

Topic 4: The Z Transform

1. The Z Transform

2. Inverse Z Transform

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2013-10-02Dan Ellis 2

1.The Z Transform! Powerful tool for analyzing & designing

DT systems

! Generalization of the DTFT:

!

zis complex...! z= ej!DTFT

! z= rej!

G(z) = Z g n[ ]{ }= g[n]z!n

n=!"

"

# Z Transform

g n[ ]r!n

e!j"n

n#

DTFT ofr-ng[n]

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2013-10-02Dan Ellis 3

Region of Convergence (ROC)! Critical question:

Does summation

converge(to a finite value)?

! In general, depends on the value of z

! !Region of Convergence:

Portion of complexz-plane

for which a particularG(z)

will converge

G(z) = x[n]z!n

n=!"

"

#

z-plane

Re{z}

Im{z}

ROC|z| > "

"

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2013-10-02Dan Ellis 4

ROC Example! e.g.x[n] = n[n]

! converges onlyfor |z-1| < 1i.e. ROC is |z| > ||

! || < 1(e.g. 0.8) - finite energy sequence

! || > 1(e.g. 1.2) - divergent sequence,

infinite energy, DTFT does notexistbut still has ZTwhen |z| > 1.2 (in ROC)

! X(z) = "nz#n

n=0

$

% =1

1# "z#1

(previous slide)

n-1 1 2 3 4-2

M

when|z-1| < 1

closed form

Re{z}

Im{z}

"

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2013-10-02Dan Ellis 5

About ROCs! ROCs always defined in terms of |z|

!circularregions onz-plane

(inside circles/outside circles/rings)

! If ROC includesunit circle(|z| = 1),

!g[n]has a DTFT

(finite energysequence)

z-plane

Re{z}

Im{z}

Unit circlelies in ROC

!DTFT OK

1

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Another ROC example! Anticausal (left-sided) sequence:

! SameZT as n[n], differentsequence?

x n[ ] = !"n!n !1[ ] n-1

1 2 3 4

-2-3-4-5

X z( )=

!"

n

!n !1

[ ]( )z!n

n

#= ! "

nz

!n

n =!$

!1

# = ! "!m

zm

m=1

$

#

=

!"!1

z

1

1! "!1z=

1

1! "z!1

ROC:|"| > |z|

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2013-10-02Dan Ellis 7

ROC is necessary!! A closed-form expression for ZT

mustspecify the ROC:

x[n] = n[n]

ROC|z| > ||

x[n] = -n[-n-1]

ROC|z| < ||! A single G(z)expression can match

several sequences with different ROCs

!X(z) =1

1" #z"1

!X(z) =1

1" #z

"1

n-1

1 2 3 4-2-3-4 z-plane

Re

Im

|"|

n-1 1 2 3 4-2-3-4

DTFTs?

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2013-10-02Dan Ellis 8

Rational Z-transforms! G(z)expression can be any function;

rational polynomialsare importantclass:

!

By convention, expressed in terms ofz-1

matches ZT definition

! (Reminiscent of LCCDE expression...)

G z( ) = P z( )D z( )

=p0 + p1z!1 +

+ pM!1z!(M!1) + pMz!M

d0 + d1z!1++ dN!1z

!(N!1)+ dNz

!N

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2013-10-02Dan Ellis 9

Factored rational ZTs! Numerator, denominator can be

factored:

! {!}are roots of numerator

!G(z) = 0!{!}are thezerosofG(z)

! {!}are roots of denominator

!G(z) =#!{!}are the polesofG(z)

G z( )=

p0!

!=1

M1"#

!z"1( )

d0!

!=1

N 1" $!z"1( )

=

zM

p0!

!=1

Mz "#

!( )

zNd0!

!=1N z " $

!( )

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2013-10-02Dan Ellis 10

Pole-zero diagram! Can plot poles and zeros on

complexz-plane:

! (Value of) expression determinedby roots

z-plane

Re{z}

Im{z}

1! o

o

o

o

!

!

poles !

(cpx conj for real g[n])

zeros !

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2013-10-02Dan Ellis 11

Z-plane surface! G(z): cpx functionof a cpx variable

! Can calculate value over entire z-plane

M

ROCnot

shown!!

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2013-10-02Dan Ellis

1

0.5

0

0.5

1

1

0.5

0

0.5

1

0

2

4

6

8

10

12

ROCs and sidedness! Two sequences have:

! Each ZT pole!region in ROCoutsideor inside||for R/Lsided term in g[n]

! Overall ROC is intersection of each terms

G(z) = 11! "z

!1

z-plane

|"|

ROC |z| >||!g[n] = n[n]

n

ROC |z|

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2013-10-02Dan Ellis 13

ZT is Linear!

! Thus, if

then

G(z) = Z g n[ ]{ }= g[n]z

!n

"n# Z Transform

y[n] =!1"1n

[n]+!2"2n

[n]

Y(z) =!

1

1# "1z#1+

!2

1# "2z#1

y[n] = g[n]+ h[n]

Y(z) = (g[n]+h[n])z-n

=g[n]z-n+ h[n]z-n= G(z)+H(z)

ROC:

|z|>|"1|,|"2|

Linear!

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G(z) = 11! "

1z!1

+ 11! "

2z!1

ROC intersections

! Consider

with |1| < 1, |2| > 1...

!

Two possible sequences for

1term...

! Similarly for 2 ...

!4 possible g[n]seqs and ROCs ...

no ROC specified

1n[n]

n-

1n[-n-1]

nor

n-2n[-n-1]

n

2n[n]or

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2013-10-02Dan Ellis 15

ROC intersections: Case 1

ROC:|z| >|1| and|z| >|2|

ImRe

|"1| |"2|

Im

G(z) =1

1! "1z!1+

1

1! "2z!1

n

g[n] = 1n[n]+ 2

n[n]

both right-sided:

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ImRe

|"1| |"2|

Im

G(z) =1

1! "1z!1+

1

1! "2z!1

n

g[n] = -1n[-n-1]- 2

n[-n-1]

both left-sided:

ROC:|z|

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ROC:|z| >|1| and|z|

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ROC:|z| |2| ?

Re

|"1| |"2|

Im

G(z) =1

1! "1z!1+

1

1! "2z!1

n

g[n] = -1n[-n-1]+ 2

n[n]

two-sided:

no ROC...

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ROC intersections! Note: Two-sided exponential

! Nooverlap in ROCs!ZT does not exist(does not converge for anyz)

g n[ ] =!n

=!

n

n[ ]+!n

"

n"1[ ]

!"< n

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ZT of LCCDEs! LCCDEs have solutions of form:

! Hence ZT

! Each terminin g[n] corresponds to a

poleiof G(z) ... and vice versa

! LCCDE solns are right-sided

ROCs are |z| > |i|

yc[n]=!i"inn[ ]+...

Yc z( ) =!

i

1" #iz"1 +!

(sames)

outsidecircles

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2013-10-02Dan Ellis 21

Z-plane and DTFT! Slice between surfaceand unit cylinder

(|z| = 1z= ej) is G(ej), the DTFT

|G(ej

)|z = ej

0 / rad/samp

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2013-10-02Dan Ellis 22

Some common Z transforms

g[n] G(z) ROC [n] 1 !z

[n] |z|> 1

n[n] |z|> ||

rncos(0n)[n] |z|> r

rnsin(0n)[n] |z|> r

1

1!z!1

11!"z

!1

1!rcos "0( )z

!1

1!2rcos "0( )z

!1+r

2z!2

rsin !0( )z"1

1"2rcos !0( )z

"1+r

2z"2

sum ofrnej0n + rne-j0n

poles atz = rej0!

!

conjugate polepair

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Z Transform properties

g[n] G(z) w/ROCRg

Conjugation g*[n] G*(z*) Rg

Time reversal g[-n]

G(1/z)

1/Rg

Time shift g[n-n0] z-n0G(z) Rg(0/#?)

Exp. scaling n

g[n]

G(z/) ||Rg

Diff. wrtz ng[n] Rg(0/#?)!zdG(z)

dz

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Z Transform properties

g[n] G(z) ROC

Convolution g[n] h[n] G(z)H(z)

Modulation g[n]h[n]

Parseval:

"at least

Rg#Rh

1

2!j G v( )H z v( )v

"1dvC#

g n[ ]h* n[ ]n=!"

"

# = 12$j G v( )H* 1

v( )v!1

dvC%

at least

RgRh

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ZT Example! ; can express as

! Hence,

x n[ ] = rn cos !0n( ) n[ ]

1

2 n[ ] rej

!0( )

n

+ re"j!

0( )n

$%

'( =v n[ ]+ v* n[ ]

v[n] = 1/2[n]n; = rej0

!V(z) = 1/(2(1- rej0z-1))ROC:|z| > r

X z( ) = V z( )+ V* z*( )= 1

21

1!re j"0z

!1+ 1

1!re! j"0z

!1( )=

1!rcos "0( )z

!1

1!2rcos "0( )z

!1+r

2z!2

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2013-10-02Dan Ellis 26

Another ZT example

y n[ ] = n +1( )!n n[ ]= x[n]+ nx[n] wherex[n] = n[n]

X z( ) = 11!"z

!1

! "zdX z( )

dz

="z d

dz

1

1"#z"1

$

%&

'

() =

#z"1

(1"#z"1

)

2

!Y z( ) =1

1"#z"1+

#z"1

(1"#z"1)2=

1

(1"#z"1)2

repeatedroot - IZT

( |z| > || )

ROC |z| > ||

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2.Inverse Z Transform (IZT)! Forwardztransform was defined as:

! 3 approaches to invertingG(z)to g[n]:! Generalization of inverse DTFT! Power series inz(long division)

! Manipulate into recognizablepieces (partial fractions)

G(z) = Z g n[ ]{ }= g[n]z!n

n=!"

"

#

the usefulone

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IZT #1: Generalize IDTFT! If then

!

so

! Any closed contour around origin will do! Cauchy: g[n] = [residues of G(z)zn-1]

z =re j!

IDTFTz = rejd= dz/jz

Counterclockwise

closed contour at |z| = rwithin ROC

Re

Im

g[n]rn = 12

G rej

ejnd

= 12j

CG (z) zn1rndz

G(z) =G(rej) =

g[n]rnejn = DTFTg[n]rn

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IZT #2: Long division! Since if we could express G(z)as a simple

power series G(z) = a+ bz-1+ cz-2...

then can just read off g[n] = {a, b, c, ...}! Typically G(z)is right-sided (causal)

and a rational polynomial

! Can expand as power series throughlong division of polynomials

G(z) = g[n]z!nn=!"

"

#

G z( ) =P(z)

D(z)

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IZT #2: Long division! Procedure:

! Express numerator, denominator in

descending powers ofz(for a causal fn)

!

Find constant to cancel highest term!first term in result

! Subtract & repeat !lower terms in result

!

Just like long division for base-10numbers

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IZT #2: Long division

! e.g.

H z( ) = 1+ 2z!1

1+ 0.4z!1! 0.12z

!2

1+ 0.4z!1!0.12z!2

1+ 2z!1)

1+1.6z!1! 0.52z

!2+ 0.4z

!3...

1+ 0.4z!1! 0.12z

!2

1.6z!1

+ 0.12z!2

1.6z!1

+ 0.64z!2

!0.192z

!3

!0.52z!2+ 0.192z

!3

...

Result

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IZT#3: Partial Fractions! Basic idea: Rearrange G(z)as sumof

terms recognizedas simple ZTs

! especially

or sin/cos forms

! i.e. given products

rearrange to sums

1

1!"z

!1 # "

n

n[ ]

P(z)

1!

"z!1

( )1!

#z!1

( )!

A

1!"z!1+

B

1! #z!1+!

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Partial Fractions!

Note that:

! Can do the reversei.e.

go from to

! if orderof P(z)is less thanD(z)

A

1!"z!1+

B

1! #z!1+

C

1! $z!1=

A 1! #z!1( ) 1! $z!1( )+ B 1!"z!1( ) 1! $z!1( )+C1!"z!1( ) 1! #z!1( )1!"z!1( ) 1! #z!1( ) 1! $z!1( )order 3 polynomial

order 2 polynomial

u + vz-1 + wz-2

P(z)

!!=1N (1" #

!z"1)

!!

1" #!z"

1!=1

N

$else cancelw/ long div.

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2013-10-02Dan Ellis 34

Partial Fractions! Procedure:

! where

i.e. evaluateF(z)at the pole but multipliedby the pole term !dominates = residueof pole

F(z) =P(z)

!!=1N

(1" #!z"1)=

$!

1" #!z"1!=1

N

%

!!= 1" #

!z"1( )F z( ) z=#

!

! f n[ ] = "! #!( )n n[ ]

!=1

N

$

orderN-1

(cancels term indenominator)

no repeatedpoles!

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Partial Fractions Example! Given (again)

factor:

! where:

H z( ) =1+ 2

z

!1

1+ 0.4z!1! 0.12z

!2

=1+ 2z

!1

1+ 0.6z!1( ) 1! 0.2z!1( )=

"1

1+ 0.6z!1+

"2

1!0.2z!1

!1 = 1+ 0.6z

"1

( )H z( ) z="0.6 =1+ 2z

"1

1" 0.2z"1

z="0.6

="1.75

!2= 1+ 2z

"1

1+ 0.6z"1

z=0.2

=2.75

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2013 10 02Dan Ellis 36

Partial Fractions Example

! Hence

! If we know ROC |z| > || i.e. h[n]causal:

= 1.75{ 1 -0.6 0.36 -0.216 ...}

+2.75{ 1 0.2 0.04 0.008 ...}

= {1 1.6 -0.52 0.4 ...}

H z( ) = !1.751+ 0.6z

!1+ 2.75

1! 0.2z!1

! h n[ ] = "1.75( ) "0.6( )n n[ ]+ 2.75( ) 0.2( )

n n[ ]

same aslong division!

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