Digital Signal Processing

The z-Transform

�A useful tool in the analysis of discrete-time signalsand systems and is the discrete-time counterpart ofthe Laplace transform for continuous-time signals andsystems.

�May also be used to solve constant coefficientdifference equations, evaluate the response of a lineartime-invariant system to a given input and designlinear filters.

Introduction

Definition of the z-Transform

�Recall the discrete-time Fourier transform of asequencex(n), which is equal to the sum

�However, in order for this series to converge, itis necessary that the signal be absolutelysummable. Unfortunately, many signals are notabsolutely summable such as

( ) ( )j jn

n

X e x n eω ω∞

−

=−∞

= ∑

( ) ( ) ( ) ( ) ( ) ( ) 0 0.5 sinn

x n u n x n u n x n nω= = − =

Definition:The z-transform of a discrete-time signal x(n) is

defined by

wherez = rejω is a complex variable. The valuesof z for which the sum converges define aregion in thez-plane referred to as the regionof convergence (ROC).

Definition of the z-Transform

( ) ( ) n

n

X z x n z∞

−

=−∞

= ∑

�Notationally,

�The z-transform may be viewed as the DTFTof an exponentially weighted sequence.Specifically note that withz = rejω

�Furthermore, the ROC is determined by therange of values ofr for which

.

Definition of the z-Transform

( ) ( )Z

x n X z↔

( ) ( ) ( )n n jn

n n

X z x n z r x n e ω∞ ∞

− − −

=−∞ =−∞

= = ∑ ∑

( ) n

n

x n r∞

−

=−∞

< ∞∑

The z-transform is a complex variable in thez-plane

Using the unit circle where |z| = 1, the z-transform evaluated on the unit circlecorresponds to the DTFT

Definition of the z-Transform

( ) ( )Re Im jz z j z re ω= + =

( ) ( ) j

j

z eX e X z ω

ω=

=

The unit circle in the complex z-plane

Many of the signals of interest in digital signalprocessing have z-transforms that are rationalfractions ofz:

Factoring the numerator and denominatorpolynomials, a rationalz-transform may beexpressed as follows:

Definition of the z-Transform

( ) ( )( )

( )

( )0

0

qk

kp

k

k

b k zB z

X zA z

a k z

−

=

−

=

= =∑

∑

( )( )

( )

1

1

1

1

1

1

q

kk

p

kk

zX z C

z

β

α

−

=

−

=

−=

−

∏

∏

�The roots of the numerator polynomial,βk, arereferred to as the zeros ofX(z), and the roots ofthe denominator polynomial,αk, are referred to aspoles.

�The Region of Convergence (ROC) is, in general,an annulus of the form

� If α = 0, the ROC may also include the pointz =0, and ifβ = ∞, the ROC may also be infinity.

Definition of the z-Transform

zα β< <

The Properties of ROC

1. A finite-length sequence has az-transform withan ROC that includes the entirez-plane except,possibly,z = 0 andz = ∞. The pointz = ∞ will beincluded ifx(n) = 0 for n < 0, and the pointz = 0will be included ifx(n) = 0 for n > 0.

2. A right-sided sequence has az-transform with anROC that is the exterior of a circle.

3. A left-sided sequence has az-transform with anROC that is the interior of a circle.

.

ROC: z α>

ROC: z β<

Relationship Between X(z) and X(ω)

Example 1

Find the z-transform of the sequencex(n) =anu(n). Using the definition of thez-transformand the geometric series

with the sum converging if |αz –1| < 1. Thereforethe region of convergence is the exterior of acircle defined by the set of points |z| > |α|.

( ) ( )

( )0

11

0

1

1

n n n

n n

n

n

X z x n z z

zz

α

αα

∞ ∞− −

=−∞ =

∞−

−=

= =

= =−

∑ ∑

∑

Example 1

ExpressingX(z) in terms of positive powers ofz,

It is obvious thatX(z) has a zero atz = 0 and apole at z = α. A pole-zero diagram with theregion of convergence is shown in the nextslide.

( ) zX z

z α=

−

Example

Example 2

Find the z-transform of the sequence

x(n) = –αnu(– n – 1)

From the previous example

( ) ( ) ( )

( )

1 11

0

11 1

1 10

1

1 1

nn n n

n n n

n

n

X z x n z z z

zz z

z z

α α

αα αα α

∞ − ∞ +− − −

=−∞ =−∞ =

−∞− −

− −=

= = − = −

= − = − =− −

∑ ∑ ∑

∑

With the sum converging if |α–1z| < 1 or |z| < |α|.A pole-zero diagram with the region ofconvergence is indicated below

Comparing |α| ≤ 1, the unit circle is not includedwithin the region of convergence, and theDTFT of x(n) does not exist.

Example 2

Example 3

Find thez-transform ofx(n) = (½)nu(n) – 2nu(–n –1), and find another signal that has the samez-transform but a different region of convergence.

The z-transform ofx1(n) = (½)nu(n) is

and thez-transform ofx2(n) = –2nu(–n – 1) is

.

( ) 11 211

2

1

1X z z

z−= >−

( )2 1

1 2

1 2X z z

z−= <−

Example 3

Therefore, thez-transform ofx(n) = x1(n) + x2(n)is

with a region of convergence of ½ < |z| < 2,which is the set of all points that are in theROC of bothX1(z) andX2(z).

( ) ( )( )15

21 1 1 11 1

2 2

21 1

1 1 2 1 1 2

zX z

z z z z

−

− − − −

−= + =− − − −

Example 3

To find another sequence that has the samez-transform,note that becauseX(z) is a sum of twoz-transforms

Each term corresponds to thez-transform of either a right-sided or a left-sided sequence, depending upon theconvergence. Therefore, choosing the right-sidedsequences for both terms, it follows that

has the samez-transform asx(n), except that the region ofconvergence is |z| > 2.

( ) 1 112

1 1

1 1 2X z

z z− −= +− −

( ) ( ) ( ) ( )12 2

n nx n u n u n= +

Common z-Transform Pairs

Properties

LinearityAs with the DTFT, the z-transform is a linear

operator. Therefore, ifx(n) has az-transformX(z)with an ROC ofRx and if y(n) has az-transformY(z) with an ROCRy

And the ROC ofw(n) will include the intersectionof Rx andRy, that is

However, the ROC ofW(z) may be larger.

( ) ( ) ( ) ( ) ( ) ( )Z

w n ax n by n W z aX z bY z= + ↔ = +

contains w x yR R R∩

Properties

Shifting PropertyShifting a sequence (delaying or advancing) multiplies the

z-transform by a power ofz. That is to say, ifx(n) has az-transform X(z)

Because shifting a sequence does not affect its absolutesummability, shifting does not change the region ofconvergence. Therefore, thez-transforms ofx(n) andx(n – n0) have the same region of convergence, with thepossible exception of adding or deleting the pointsz = 0andz = ∞.

( ) ( )00

Znx n n z X z−− ↔

Properties

Time reversal

If x(n) has az-transformX(z) with an ROCRx,that is the annulusα < |z| < β, the z-transformof the time-reversed sequencex(– n) is

and has an ROC 1/β < |z| < 1/α, which is denotedby 1/Rx.

( ) ( )1Z

x n X z−− ↔

Properties

Multiplication by an exponentialIf a sequencex(n) is multiplied by a complex exponentialαn,

This corresponds to a scaling of thez-plane. If the ROC ofX(z)is r– < |z| < r+, which will be denoted byRx, the ROC ofX(α–1z) is |α|r– < |z| < |α|r+ , which is denoted by |α|Rx. Asspecial case, note that ifx(n) is multiplied by a complexexponentialejnω0.

which corresponds to a rotation of thez-plane.

( ) ( )1Z

n x n X zα α −↔

( ) ( )0 0

Zjn je x n X e zω ω−↔

Properties

Convolution TheoremPerhaps the most importantz-transform property is the

convolution theorem, which states that convolution inthe time domain is mapped into multiplication in thefrequency domain, that is,

The ROC of Y(z) includes the intersection of Rx and Ry,

However, the ROC ofY(z) may be larger, if there is apole-zero cancellation in the productX(z)H(z).

( ) ( ) ( ) ( ) ( ) ( )*Z

y n x n h n Y z X z H z= ↔ =

contains w x yR R R∩

Example 4

Consider the two sequences

Thez-transform ofx(n) is

and thez-transform ofh(n) is

However, thez-transform of the convolution ofx(n) withh(n) is

which, due to apole-zero cancellation, has an ROC that isthe entirez-plane.

( ) ( ) ( ) ( ) ( ) 1nx n u n h n n nα δ αδ= = − −

( ) 1

1

1X z z

zα

α −= >−

( ) 11 0X z z zα −= − <

( ) ( ) ( ) ( )11

11 1

1Y z X z H z z

zα

α−

−= = • − =−

Properties

Conjugation

If X(z) is thez-transform ofx(n), thez-transformof the complex conjugate ofx(n) is

As a corollary, note that ifx(n) is real-valued,x(n) = x*(n), then

( ) ( )* * *Z

x n X z↔

( ) ( )* *X z X z=

Properties

Derivative

If the X(z) is the z-transform of x(n), the z-transform ofnx(n) is

Repeated application of this property allows forthe evaluation of thez-transform ofnkx(n) forany integerk.

( ) ( )Z dX znx n z

dz↔ −

Example 5

Find the z-transform of x(n) = nαnu(–n). Hint: use thetime-reversal and derivative properties

Therefore

and, using the time-reversal property,

Finally, using the derivative property, it follows that thez-transform ofnαnu(–n) is

.

( ) 1

1

1

Znu n z

zα α

α −↔ >−

( ) 1 1

1 1 1

1

nZ

u n zzα α α− −

↔ > −

( ) 1

1

1

Znu n z

zα α

α −− ↔ <−

( )1

21 1

1=

1 1

d zz z

dz z z

α αα α

−

− −− − <

− −

Properties

If x(n) is equal to zero forn < 0, the initial value,x(0), may be found fromX(z) as follows:

This property is a consequence of the fact that ifx(n) = 0 for n < 0,

Therefore, if z →∞, each term inX(z) goes tozero except the first.

( ) ( )0 limz

x X z→∞

=

( ) ( ) ( ) ( )1 20 1 2 ...X z x x z x x− −= + + +

Properties of the z-Transform