H. C. So Page 1 Semester A, 2001

z-Transform

z-transform is a generalization of Fourier transform for discrete-timesignals (DTFT).

Given a discrete-time signal ][nx , the z-transform is defined as

∑==∞

−∞=

−

n

nznxnxZzX ][]}[{)(

where z is a complex variable.

Let ω= jrez ( =r magnitude of z and =ω angle of z ),

}][{][)( n

n

jnn rnxFernxzX −∞

−∞=

ω−− =∑=

⇒ )(zX is the Fourier transform of nrnx −][ .When 1=r or ω= jez ,

)(]}[{)( ω== jeXnxFzX

H. C. So Page 2 Semester A, 2001

The z-Plane

Discrete-time Fouriertransform is equal to thez-transform evaluatedalong the unit circlewith ω varying from 0 toπ2

H. C. So Page 3 Semester A, 2001

Example 32Determine the z-transform of ][][ nuanx n= .Solution:

∑∑∞

=

−∞

−∞=

− ==0

1)(][)(n

n

n

nn azznuazX

)(zX converges if ∞<∑∞

=

−

0

1

n

naz . This requires 11 <−az or az > ,

and

111)( −−

=az

zX

Notice that for another signal ]1[][ −−−= nuanx n ,

( )∑−=∑−=∑ −=∞

=

−∞

=

−−

−∞=

−

1

1

1

1)()(

m

m

m

mm

n

nn zazazazX

H. C. So Page 4 Semester A, 2001

In this case, )(zX converges if 11 <− za or az < , and

111)( −−

=az

zX

ROC of][][ nuanx n=

ROC of]1[][ −−−= nuanx n

H. C. So Page 5 Semester A, 2001

Example 33Determine the z-transform of ][

216][

317][ nununx

nn

−

= .

Solution:

21||,

)21)(

31(

)23(

121,1

31,

211

6

311

7216

317)(

11

11

00

>−−

−=

<<−

−−

=

∑

−∑

=

−−

−−

∞

=

−∞

=

−

zzz

zz

zzzz

zzzXn

nn

n

nn

H. C. So Page 6 Semester A, 2001

Example 34

Determine the z-transform of ][4

sin31][ nunnx

n

π

= .

Solution:

31],

311

1

311

1[21

2

31

31

2

31

31

4sin

31)(

1414

0

1414

0

44

0

>

−

−

−

=

∑

−

=∑

−

=

∑

π

=

−π−−

π

∞

=

−π−−

π

∞

=

−

π−π

∞

=

−

z

zezej

j

zeze

zj

ee

znzX

jj

n

nj

nj

n

n

nj

nj

n

nn

H. C. So Page 7 Semester A, 2001

ROC for z-Transform

Property 1: The ROC of )(zX consists of a ring centered aboutthe origin in the z -plane.

! }][{)( nrnxFzX −= , ω= jrez , and }][{ nrnFx − converges when

∞<∑∞

−∞=

−

n

nrnx ][ which depends only on zr = and not on z∠ .

That is, if the ROC of )(zX contains a point 0z , it must contain theentire circle with 0zz = .

Property 2: The ROC of )(zX contains no poles.

As with the Laplace transform, it is a consequence of the fact that ata pole )(zX is infinite which means that )(zX does not converge.

H. C. So Page 8 Semester A, 2001

Property 3: If ][nx has finite duration, the ROC of )(zX is theentire z-plane, except possibly 0=z and/or ∞=z .

Property 4: If ][nx is right-sided, and if the circle 0rz = is in ROCof )(zX , then all finite values of z for which 0rz > must be in ROC.

Property 5: If ][nx is left-sided, and if the circle 0rz = is in ROC of)(zX , then all values of z for which 00 rz << must be in ROC.

Property 6: If ][nx is two-sided, and if the circle 0rz = is in ROC of)(zX , then the ROC will consist of a ring that includes 0rz = .

ROCSignal typeLaplace transform z-transform

Finite duration Entire s-plane Entire z-planeRight-sided The right of a line Outside a circleLeft-sided The left of a line Inside a circleTwo-sided A strip A ring

H. C. So Page 9 Semester A, 2001

z-Plane and s-PlaneConsider a continuous-time signal )(tx and its sampled version:

∑ −δ=∞

−∞=np nTtnTxtx )()()(

The Laplace transform of )(tx p is,

∑=∑=∞

−∞=

−∞

−∞=

−

n

snT

n

snTp enxenTxsX ][)()(

where )(][ nTxnx = is a discrete-time signal.

Comparing ∑∞

−∞=

−=n

nznxzX ][)( and ∑∞

−∞=

−=n

snTp enxsX ][)( ,

and since )(zX & )(sX p are representing the same spectrum(except for a scaling of time axis), we have a mapping between zand s:

sTez =

H. C. So Page 10 Semester A, 2001

z-transform can be viewed as the Laplace transform of a sampledcontinuous-time signal with the change of variable sTez = .

s-plane z-plane

ωj -axis (e.g., 1ω= js ) unit circle ( 1ω= jez )left-half (e.g, 22 ω+σ= js , 02 >σ ) Inside unit circle ( TjT eez 22 ωσ= )right-half Outside unit circle

1

Re

Im

Re

Im

x

x

x

x

s-plane z-plane

H. C. So Page 11 Semester A, 2001

Property 7: If )(zX is rational, then its ROC is bounded by poles orextends to infinity.

This is because that ROC cannot contain any poles.

Property 8: If )(zX is rational and ][nx is right-sided, then its ROCis the region outside the origin-centered circle containing theoutermost pole. Furthermore, if 0][ =nx for all 0<n , the ROCcontains ∞=z .

Property 9: If )(zX is rational and ][nx is left-sided, then its ROC isthe region inside the origin-centered circle containing the innermostpole. If 0][ =nx for all 0>n , the ROC contains 0=z .

H. C. So Page 12 Semester A, 2001

Example 35Find out all of the possible ROCs for

)21)(311(

1)(11 −− −−

=zz

zX .

Solution:

)(zX has two poles at 31=z and 2=z . According to Properties 4 –

9, its ROCs can be:

1) Inside the circle 31=z , for which ][nx is left-sided.

2) Outside the circle 2=z , for which ][nx is right-sided.

3) The ring between 31=z and 2=z , for which ][nx is two-sided.

H. C. So Page 13 Semester A, 2001

Inverse z-TransformThe formal expression of inverse z-transform is given by

∫π== −− dzzzX

jzXZnx n 11 )(

21)}({][

where ∫ represents integration around a counterclockwise closedcircular contourlocates in the ROC of )(zX . Alternatively,

∫ ωπ

= πωω

2 ))((21][ drereXnx njj

Two useful methods of evaluating inverse z-transform are:

1) partial fraction2) power-series expansion

H. C. So Page 14 Semester A, 2001

Example 36

Find the inverse z-transform of )

311)(

411(

653

)(11

1

−−

−

−−

−=

zz

zzX ,

31>z

Solution:

11

311

2

411

1)(−− −

+−

=zz

zX

For 31>z , both of the two terms above result in right-sided

sequence. That is,

][312][

41][ nununx

nn

+

=

H. C. So Page 15 Semester A, 2001

If the ROC is 41<z , we have two left-sided sequences,

]1[312]1[

41][ −−

−−−

−= nununx

nn

If the ROC is 31

41 << z ,

]1[312][

41][ −−

−

= nununx

nn

H. C. So Page 16 Semester A, 2001

Example 37Find the inverse z-transform of 11

1)( −−=

azzX , az > , by power

series expansion.

Solution:

∑∞

=

−−−− =+++=

−=

0

2211 1

11)(

n

nn zazaazaz

zX " , az > .

By the definition of z-transform,

][0,00,][ nua

nnanx n

n=

<≥=

If the ROC is az < , the converging power series is

∑−

−∞=

−−−− −=−−−=

−=

1221

1 )(1

1)(n

nn zazazaaz

zX "

Then its inverse z-transform is ]1[][ −−−= nuanx n

H. C. So Page 17 Semester A, 2001

Example 38Find the inverse z-transform of )1log()( 1−+= azzX , az > , bypower series expansion.

Solution:Taylor’s series expansion,

∑∞

=

+−=+1

1)1()1log(n

nn

nvv , 1<v

For az > or 11 <−az ,

∑∑∞

=

−+∞

=

−+− −=−=+=

1

1

1

111 )1()()1()1log()(

n

nnn

n

nnz

na

nazazzX

Thus,

]1[)(

1,0

1,)(][ −−−=

<

≥−−= nu

na

n

nna

nxnn

H. C. So Page 18 Semester A, 2001

Properties of z-Transform

H. C. So Page 19 Semester A, 2001

Example 39

Find out the inverse z-transform of )1log()( 1−+= azzX , az > ,using the properties of z-transform.

Solution:

1

22

1 1)(

11)(

−

−−

− +−=−⋅

+=

azazaz

azdzzdX

1

1

1)(-

−

−

+=∴

azaz

dzzdXz

dxxfd

xfdxfd

dxxfd ))((

))(())(log())(log( ⋅=!

By the property of differentiation in z-domain,

1

1

1][ −

−

+→←

azaznnx Z

H. C. So Page 20 Semester A, 2001

Since ][)(}1

1{ 11 nua

azZ n−=

+ −− , by the time-shift property,

]1[)(}1

{

]1[)(}1

1{

11

11

11

11

−−=+

⇒

−−=+⋅

−−

−−

−−

−−

nuaaaz

azZ

nuaaz

zZ

n

n

]1[)(][ 1 −−=∴ − nuaannx n

⇒ ]1[)(][ −−−= nunanx

n

H. C. So Page 21 Semester A, 2001

Common z-transform pairs

H. C. So Page 22 Semester A, 2001

z-Transform and LTI systemsIn z -domain, a discrete-time LTI system is described as

)()()( zXzHzY =

where )(zY and )(zX are the z-transforms of output ][ny and input][nx , respectively, while )(zH is the z-transform of impulse

response ][nh .

)(zH is called system function or transfer function.

For ω= jez , )( ωjeH becomes the frequency response.

H. C. So Page 23 Semester A, 2001

Causality and ROC

Causality condition: 0][ =nh for all 0<n ,

][nh is right-sided

The ROC for )(zH isthe exterior of an origin-centered circle (including ∞=z )

If )(zH is rational, the ROC for )(zH isthe exterior outside the outermost pole.

And, the order of numerator polynomialis not larger than that of the denominator.

H. C. So Page 24 Semester A, 2001

Stability and ROC

Stability condition: ∞<∑∞

−∞=nnh ][

)( ωjeH , i.e., the Fourier transform of ][nh , converges

The ROC for )(zH includes the unit circle 1=z

A discrete-time causal LTI system with rational system function)(zH is stable if and only if all of the poles of )(zH lie inside the

unit circle.

H. C. So Page 25 Semester A, 2001

Example 40Consider a LTI system with 11 21

1

211

1)( −− −+

−=

zzzH , 2>z . Is it

causal? Is it stable? Why?

Solution: The ROC of )(zH is 2>z , outside the outermost pole. Also,

Thus, the system is causal. Its impulse response is given as

][221][ nunh n

n

+

=

125

252

)21)(211(

252

)(2

2

11

1

+−

−=

−−

−=

−−

−

zz

zz

zz

zzH

The numerator’sorder is equal tothe denominator’s

H. C. So Page 26 Semester A, 2001

Since the ROC of )(zH does not include the unit circle, thesystem is not stable. This can also be seen from

∞→+

= ∑∑

∞

=

∞

−∞= 02

21][

n

nn

nnh

For the system to be stable, )(zH should be 221 << z , i.e.,

including the unit circle.

H. C. So Page 27 Semester A, 2001

Example 41Can the following system be causal and stable?

111)( −−

=az

zH

Solution:

The system function 111−− az

has two possible ROCs: az > and

az < .

For the system to be causal, the ROC of )(zH must be the exteriorof some origin-centered circle. Hence, the ROC should be az > .And the impulse response is ][][ nuanh n= .

For the causal system to be stable, its ROC must include the unitcircle 1=z as well. This requires 1<a .

H. C. So Page 28 Semester A, 2001

Example 42Discuss about the causality and stability of a system with

221)cos2(11)( −− +θ−

=zrzr

zH

Solution:)(zH has a pair of conjugate poles θ= jrez1 and θ−= jrez2 .

For system to be causal, the ROC of )(zH should be rz > .For system to be stable, we need 1<r , i.e., pole inside unit circle.

stable unstable

H. C. So Page 29 Semester A, 2001

Example 43Consider a LTI system for which the input ][nx and output ][nysatisfy the linear constant-coefficient difference equation,

]1[31][]1[

21][ −+=−− nxnxnyny

Find the system function and comment on its causality and stability.

Solution:Taking z-transform on both sides,

)(31)()(

21)( 11 zXzzXzYzzY −− +=−

Thus,

1

1

211311

)()()(

−

−

−

+==

z

z

zXzYzH

H. C. So Page 30 Semester A, 2001

For the system to be causal, the ROC of )(zH should be 21>z .

And in this case, the system is stable since the ROC includes theunit circle. Expressing )(zH as,

1

1

1

2113

1

211

1)(−

−

− −⋅+

−=

z

z

zzH

Then, by the property of time-shift, for 21>z

]1[21

31][

21][

1

−

+

=

−

nununhnn

∑∑==

−=−M

kk

N

kk knxbknya

00][][ ⇒

∑

∑

=

−

=

−

= N

k

kk

M

k

kk

za

zbzH

0

0)(

H. C. So Page 31 Semester A, 2001

Block Diagram & Transfer Function

][]1[][ nbxnayny =−+

)()()( tbxtaydt

tdy =+

∫ ττ−τ+= tt daybxtyty

0)]()([)()( 0

H. C. So Page 32 Semester A, 2001

1

1

411

21)(−

−

−

−=z

zzH 1

411

1−− z

121 −− z

using lessdelay units

H. C. So Page 33 Semester A, 2001

2111

81

411

1

)411)(

211(

1)(−−−− −+

=−+

=zzzz

zH

Equivalently, ][]2[81]1[

41][ nxnynyny =−−−+

H. C. So Page 34 Semester A, 2001

Alternatively, 11

411

1

211

1)(−− −

⋅+

=zz

zH

The system is resulted from the cascade of two systems:

][]1[21][ nxnwnw =−+ and ][]1[

41][ nwnyny =−−

w[n]

H. C. So Page 35 Semester A, 2001

Or 11

411

131

211

132)(

−− −⋅+

+⋅=

zzzH

The system is as the parallel of two systems:

][32]1[

21][ 11 nxnvnv =−+ and ][

31]1[

41][ 22 nxnvnv =−−

and ][][][ 21 nvnvny +=

v1[n]

v2[n]