chapter 6 - the z-transform

Poularikas, A.D. “The Z-Transform.”The Transforms and Applications Handbook: Second Edition.Ed. Alexander D. PoularikasBoca Raton: CRC Press LLC, 2000

6The Z-Transform

6.1 IntroductionA. One-Sided Z-Transform

6.2 The Z-Transform and Discrete Functions6.3 Properties of the Z-Transform

Linearity • Shifting Property • Time Scaling • Periodic Sequence • Multiplication by n and nT • Convolution • Initial Value • Final Value • Multiplication by (nT)k • Initial Value of f(nT) • Final Value for f(nT) • Complex Conjugate Signal • Transform of Product • Parseval’s Theorem • Correlation • Z-Transforms with Parameters

6.4 Inverse Z-Transform Power Series Method • Partial Fraction Expansion • Inverse Transform by Integration • Simple Poles • Multiple Poles • Simple Poles Not Factorable • F(z) is Irrational Function of z

B. Two-Sided Z-Transform6.5 The Z-Transform6.6 Properties

Linearity • Shifting • Scaling • Time Reversal • Multiplication by nT • Convolution • Correlation • Multiplication by e–anT • Frequency Translation • Product • Parseval’s Theorem • Complex Conjugate Signal

6.7 Inverse Z-Transform Power Series Expansion • Partial Fraction Expansion • Integral Inversion Formula

C. Applications6.8 Solutions of Difference Equations with Constant

Coefficients 6.9 Analysis of Linear Discrete Systems

Transfer Function • Stability • Causality • Frequency Characteristics • Z-Transform and Discrete Fourier Transform (DFT)

6.10 Digital Filters Infinite Impulse Response (IIR) Filters • Finite Impulse Responses (FIR) Filters

6.11 Linear, Time-Invariant, Discrete-Time, Dynamical Systems

6.12 Z-Transform and Random ProcessesPower Spectral Densities • Linear Discrete-Time Filters • Optimum Linear Filtering

6.13 Relationship Between the Laplace and Z-Transform

Alexander D. PoularikasUniversity of Alabama in Huntsville

© 2000 by CRC Press LLC

6.14 Relationship to the Fourier TransformAppendix: Tables 1 to 5

Table 1: Z-Transform Properties of the Positive-Time Sequences • Table 2: Z-Transform Properties for Positive- and Negative-Time Sequences • Table 3: Inverse Transform of the Partial Fractions of F(z) • Table 4: Inverse Transform of the Partial Fractions of Fi(z) • Table 5: Z-Transform Pairs

6.1 Introduction

The Z-transform is a powerful method for solving difference equations and, in general, to representdiscrete systems. Although applications of Z-transforms are relatively new, the essential features of thismathematical technique date back to the early 1730s when DeMoivre introduced the concept of agenerating function that is identical with that for the Z-transform. Recently, the development andextensive applications of the Z-transform are much enhanced as a result of the use of digital computers.

A. One-Sided Z-Transform

6.2 The Z-Transform and Discrete Functions

Let f(t) be defined for t ≥ 0. The Z-transform of the sequence {f(nT)} is given by

(6.2.1)

where T, the sampling time, is a positive number.1

To find the values of z for which the series converges, we use the ratio test or the root test. The ratiotest states that a series of complex numbers

with limit

(6.2.2)

converges absolutely if A < 1 and diverges if A > 1 the series may or may not converge.The root test states that if

(6.2.3)

then the series converges absolutely if A < 1, and diverges if A > 1, and may converge or diverge if A = 1.More generally, the series converges absolutely if

(6.2.4)

where denotes the greatest limit points of �f(nT)�1/n, and diverges if

1The symbol means equal by definition.

Z f nT F z f nT zn

n( ){ } = ( ) = ( )=

∞−∑˙

0

=̇

an

n=

∞

∑0

limn

n

n

a

aA

→∞+ =1

limn n

n a A→∞

=

limn n

n a→∞

<1

lim limn→∞


(6.2.5)

If we apply the root test in (2.1) we obtain the convergence condition

or

(6.2.6)

where R is known as the radius of convergence for the series. Therefore, the series will converge absolutelyfor all points in the z-plane that lie outside the circle of radius R, and is centered at the origin (with thepossible exception of the point at infinity). This region is called the region of convergence (ROC).

Example

The radius of convergence of f(nT) = e–a n Tu(nT), a positive number, is

�z –1e– a T � < 1 or �z � > e – a T

The Z-transform of f(nT) = e –anTu(nT) is

If a = 0

Example

The function f(nT) = anT cos nTω u(nT) has the Z-transform

The ROC is given by the relations

�aTe j T ωz –1 � < 1 or �z � > �aT �

�aTe – j T ωz –1� < 1 or �z � > �aT �

Therefore, the ROC is � z � > �aT �.

limn n

n a→∞

>1

lim limn

nnn

nnf nT z f nT z

→∞

−

→∞

−( ) = ( ) <1 1

z f nT Rn

n> ( ) =→∞

lim_____

F z f nT z e ze z

n

n

aTn

naT( ) = ( ) = ( ) =

−−

=

∞− −

=

∞

− −∑ ∑0

1

01

1

1

F z u nT zz

z

zn

n

( ) = ( ) =−

=−

−

=

∞

−∑0

1

1

1 1

F z ae e

z a e z a e z

a e z a e z

a z T

a

nT

n

jnT jnTn T jT

n

n

T jTn

n

T jT T jT

T

T

( ) = + = ( ) + ( )

=−

+−

= −−

=

∞ −− −

=

∞− −

=

∞

− − −

−

∑ ∑ ∑0

1

0

1

0

1 1

1

2

1

2

1

2

1

2

1

1

1

2

1

1

1

1 2

ω ωω ω

ω ω

ωcos

zz T a zT− −+1 2 2cos.

ω


6.3 Properties of the Z-Transform

Linearity

If there exists transforms of sequences Z{ci fi (nT)} = ci Fi (z), ci are complex constants, with radii ofconvergence Ri > 0 for i = 0, 1, 2, …, �(� finite), then

(6.3.1)

Shifting Property

Z{f (nT – kT)} = z –kF (z), f (–nT) = 0 n = 1, 2, … (6.3.2)

(6.3.3)

(6.3.4)

Z{f (nT + T)} = z[F (z) – f (0)] (6.3.4a)

Example

To find the Z-transform of y(nT) we proceed as follows:

or

Time Scaling

(6.3.5)

Z c f nT c F z z Ri

i

i i

i

i i

= =∑ ∑( )

= ( ) >

0 0

l l

max

Z f nT kT z F z f nT zk

n

kk n−( ){ } = ( ) + −( )−

=

− −( )∑1

Z f nT kT z F z f nT zk

n

k

k n+( ){ } = ( ) − ( )=

−−∑

0

1

d y t

dtx t

y nT y nT T y nT T

Tx nT

Y z z Y z y T z z Y z y T z y T z X z T

2

2 2

1 0 2 1 0 2

2 2

2 2

( )= ( ) ( ) − −( ) + ( )

= ( )

( ) − ( ) + −( )[ ] + ( ) + −( ) + −( ) = ( )− − − − −

,–

,

Y zy T y T z y T X z T

z z( ) =

−( ) − −( ) − ( ) + ( )− +

−

− −

2 2

1 2

1 2

1 2

–

Z a f nT F a z f nT a znT T

n

Tn( ){ } = ( ) = ( )( )−

=

∞− −∑

0


Example

Periodic Sequence

(6.3.6)

N is the number of time units in a period, �z � > R

where R is the radius of convergence of F1(z).

Proof

For finite sequence of K terms

(6.3.6a)

Multiplication by n and nT

R is the radius of convergence of F(z)

(6.3.7)

Z

Z

sinsin

cos,

sinsin

cos

ω ωω

ω ωω

nTu nTz T

z z Tz

e nTu nTe z T

e z e z Tz en

( ){ } =− +

>

( ){ } =− +

>−+

+ +−

2

1

2 2 1

1

2 11

2 1

Z Zf nTz

zf nT

z

zF z f nT

N

N

N

N( ){ } =−

( ){ } =−

( ) ( ) =1 1

1 1 1, first period

Z Z Z Zf nT f nT f nT NT f nT NT

F z z F z z F z

F zz

z

zF z

N N

N

N

N

( ){ } = ( ){ } + −( ){ } + −( ){ } +

= ( ) + ( ) + ( ) +

= ( )−

=−

( )

− −

−

1 1 1

1 12

1

1 1

2

1

1 1

L

L

F z F zz

z

N K

N( ) = ( ) −−

− +( )−1

11

1

Z nf nT zdF z

dz( ){ } = −

( )

Z nT f nT TzdF z

dzz R( ){ } = −

( )>


Proof

Example

Convolution

If Z{f(nT)} = F(z) �z � > R1 and Z{h(nT)} = H(z) �v � > R2 , then

� z � > max(R1, R2) (6.3.8)

Proof

The value of h(nT) for n < 0 is zero.Additional relations of convolution are

Z{ f (nT) ∗ h (nT)} = F (z) H (z) = Z{h (nT) ∗ f (nT)} = F(z) H (z) (6.3.8a)

Z{{ f (nT) + h (nT)} ∗ {g (nT)}} = Z{ f (nT) ∗ g (nT)} + Z{h (nT) ∗ g (nT)}

= F (z) G (z) + H (z) G (z) (6.3.8b)

nT nT z Tz f nTd

dzz Tz

d

dzf nT z

TzdF z

dz

n

n

n

n

n

n

=

∞−

=

∞−

=

∞−∑ ∑ ∑( ) = ( ) −

= − ( )

= −( )

0 0 0

Z Z

Z

u nz

znu n z

d

dz

z

z

z

z

n u n zd

dz

z

z

z z

z

( ){ } =− ( ){ } = −

−

=−( )

( ){ } = −−( )

=−( )

−( )

1 1 1

1

1

1

2

2

2

2

4

, ,

Z Zf nT h nT f mT h nT mT F z H zm

( )∗ ( ){ } = ( ) −( )

= ( ) ( )

=

∞

∑0

Z f nT h nT f mT h nT mT z

f mT h nT mT z

f mT h rT z z

f mT z h rT

mn

n

nm

n

r mm

r m

m

rm

( )∗ ( ){ } = ( ) −( )

= ( ) −( )

= ( ) ( )

= ( ) ( )

=

∞

=

∞−

=

∞

=

∞−

= −

∞

=

∞− −

−

=

∞

=

∞

∑∑

∑∑

∑∑

∑∑

00

00

0

00

zz F z H zr− = ( ) ( ) .


Z{{ f (nT) ∗ h (nT)} ∗ g (nT)} = Z{ f (nT) ∗ {h (nT) ∗ g (nT)}} = F(z) H (z) G (z) (6.3.8c)

Example

The Z-transform of the output of the discrete system y(n) = y(n – 1) + x(n), when the input is the

unit step function u(n) given by Y(z) = H(z)U(z). The Z-transform of the difference equation with adelta function input δ(n) is

Therefore, the output is given by

Example

Find the f(n) if

a, b are constants.

From this equation we obtain

Therefore,

Initial Value

(6.3.9)

The above value is obtained from the definition of the Z-transform. If f(0) = 0, we obtain f(1) as the limit

(6.3.9a)

1

2

1

2

H z z H z H z

z

z

z( ) − ( ) = ( ) =

−=

−

−

−

1

2

1

2

1

2

1

11

2

1

2 1

2

1

1

or

Y zz

z

z

z( ) =

− −1

2 1

2

1

F zz

z e z ea b( ) =

−( ) −( )− −

2

f nz

z ee f n

z

z ee

a

an

b

bn1

12

1( ) =−

= ( ) =−( )

=−

−− −

−−

Z Z,

f n f n f n e e e e

ee

e

am

m

nb n m bn a b m

m

n

bna b n

a b

( ) = ( )∗ ( ) = =

= −

−

−

=

− −( ) − − −( )

=

−− −( ) +( )

− −( )

∑ ∑1 2

0 0

11

1

f F zz

0( ) = ( )→∞

lim

limz

zF z→∞

( )


Final Value

if f(∞) exists (6.3.10)

Proof

By taking the limit as z → 1, the above equation becomes

which is the required result.

Example

If F(z) = 1/[(1 – z –1)(1 – e–1z–1)] with � z � > 1 then

Multiplication by (nT)k

k > 0 and is an integer (6.3.11)

As a corollary to this theorem, we can deduce

(6.3.11a)

lim limn z

f n z F z→∞ →

( ) = −( ) ( )1

1

Z f k f k f k f k z

zF z zf F z z F z zf f k f k z

nk

n

k

n

k

k

n

+( ) − ( ){ } = +( )[ ] − ( )[ ]

( ) − ( ) − ( ) = −( ) ( ) − ( ) = +( )[ ] − ( )[ ]→∞

=

−

→∞

−

=

∑

∑

1 1

0 1 0 1

0

0

lim

lim

lim lim

lim

–

lim

z nk

n

n

n

z F z f f k f k

f f f f

f n f n f n f n

f f n

f

→ →∞=

→∞

→∞

−( ) ( ) − ( ) = +( )[ ] − ( )[ ]= ( ) − ( ) + ( ) − ( ) +{+ ( ) − −( ) + +( ) ( )}

= − ( ) + +( ){ }= −

∑10

1 0 1

1 0 2 1

1 1

0 1

L

00( ) + ∞( )f

f F z

e

f n zz e z

z

z e e

z

n z z

01

11

11

1

11

1 1

1

1

1

1 1 1 1 1

2

1 1

( ) = ( ) =−

∞

−∞

=

( ) = −( )−( ) −( ) =

−( ) =−( )

→∞−

→∞ → − − − → − −

lim

lim lim lim

Z Zn T f nT Tzd

dznT f nTk k k( ){ } = − ( ) ( )

−1

Z n f n zd F z

d zn n n n n k

k k

k

k

k( ) −

−

( )( ){ } =( )

( )= −( ) −( ) − +( )

11 2 1, L


The following relations are also true:

(6.3.11b)

(6.3.11c)

Example

Initial Value of f(nT)

Z{ f (nT)} = f (0T) + f (T)z –1 + f (2T)z –2 + L = F (z)

� z � > R (6.3.12)

Final Value for f(nT)

f(∞T) exists (6.3.13)

Example

For the function

� z � > 1

we obtain

Z −( ) − +( )

=( )( )1 1

k kk

kn f n k z

d F z

dz

Z n n n n k f n zd F z

dz

k k

k

k+( ) +( ) + −( ) ( ){ } = −( ) ( )

1 2 1 1L

Z

Z Z

Z

n zd

dz

z

z

z

z

n zd

dzn z

d

dz

z

z

z z

z

n zd

dz

z z

z

z z z

z

{ } = −−

=−( )

{ } = − { } = −−( )

=+( )

−( )

{ } = −+( )

−( )=

+ +( )−( )

1 1

1

1

1

1

1

4 1

1

2

2

2 3

3

3

2

4

,

,

f T F zz

0( ) = ( )→∞

lim

lim limn z

f nT z F z→∞ →

( ) = −( ) ( )1

1

F zz e zT

( ) =−( ) −( )− − −

1

1 11 1

f T F ze

f nT zz

z

z

e e

z T

n z T T

01

11

1

1

11 1

1

11

( ) = ( ) =−

∞

−∞

=

( ) = −( ) − −=

−

→∞ −

→∞ → − −

lim

lim lim


Complex Conjugate Signal

or

Hence,

Z{ f*(nT)} = F*(z*) � z � > R (6.3.14)

Transform of Product

If

Z{ f (nT)} = F (z) � z � > Rf

Z{ h (nT)} = H (z) � z � > Rh

then

� z � > RjRh (6.3.15)

where C is a simple contour encircling counterclockwise the origin with (see Figure 6.3.1)

(6.3.15a)

ProofThe integration is performed in the positive sense along the circle, inside which lie all the singular pointsof the function F(τ) and outside which lie all the singular points of the function H(z/τ ). From (6.3.15),we write

(6.3.16)

which converges uniformly for some choice of contour C and values of z. From (6.3.16), we must have

F z f nT z z R F z f nT zn

n

n

n( ) = ( ) > ( ) = ( )( )=

∞−

=

∞−∑ ∑

0 0

or * *

F z f nT z f nTn

n* * * *( ) = ( ) = ( ){ }=

∞−∑

0

Z

Z Zg nT f nT h nT

f nT h nT z

jF H

z d

n

n

C

( ){ } = ( ) ( ){ }= ( ) ( )

= ( )

=

∞−∑

∫

˙

0

1

2πτ

ττ

τ

Rz

Rfh

< <τ

G zj

F h nTz d

Cn

n

( ) = ( ) ( )

∫ ∑

=

∞ −1

20

πτ

ττ

τ


(6.3.17)

so that the sum in (6.3.16) converges. Because �z � > Rf and τ takes the place of z, then (6.3.16) implies that

�τ � > Rf (6.3.18)

(6.3.19)

and also

Rf Rh < � z �.

Figure 6.3.1 shows the region of convergence.The integral is solved with the aid of the residue theorem, which yields in this case

(6.3.20)

where K is the number of different poles τi(i = 1, 2, …, K) of the function F(τ)/τ. For the residue at thepole τi of multiplicity m of the function F(τ)/τ , we have

(6.3.21)

Hence, for a simple pole, m = 1, we obtain

FIGURE 6.3.1

Rz z

Rz

Rh hhτ τ

τ

< > <−1

1 or or

Rz

Rfh

< <τ

G zF H z

i

i

K

( ) =( ) ( )

=

=∑resτ τ

τ ττ

1

resτ τ τ τ

τ ττ τ

τ ττ

ττ= →

−

−

( ) ( )

=

−( ) −( )( )

ii

F H z

m

d

d

F Hz

m

m i

m1

1

1

1!lim


(6.3.22)

Example

See Figure 6.3.2 for graphical representation of the complex integration.

Hence,

The contour must have a radius �τ � of the value e –T < �τ � < �z � = 1 and we have from (6.3.22)

From (6.3.11)

and verifies the complex integration approach.

FIGURE 6.3.2

resτ τ τ τ

τ ττ

τ ττ

ττ= →

( ) ( )

= −( )

( )

ii

F H zF H

z

ilim

Z ZnT H zz

zT z e F z

z

z ez enT

T

T{ } = ( ) =−( )

> { } = ( ) =−

>−−

−˙ , ˙1

12

Z nTej

Tz

ez

dnT

CT

−

−

{ } =

−( ) −

∫1

21

2πτ τ

τ

τ .

Z nTe e Tz

e zT

ze

z e

nT

e

T

T

T

TT

−=

−

−

−

−{ } = −( )

−( ) −( )

=

−( )−res

ττ τ

τ τ2 2

Z nTe Tzd

dz e zT

ze

z e

nT

T

T

T

−− −

−

−{ } =

−

=−( )

–1

1 1 2


Parseval’s Theorem

If Z{ f (nT)} = F (z), �z � > Rf and Z{h(nT)} = H (z), �z � > Rh with � z � = 1 > Rf Rh , then

(6.3.23)

where the contour is taken counterclockwise.

ProofFrom (6.3.15) set z = 1 and change the dummy variable τ to z.

Example

f(nT) = e–nT u(nT) has the following Z-transform:

� z � > e–T

From (6.3.23) and with C a unit circle (Rf = e–T < 1)

Correlation

Let the Z-transform of the two consequences Z{f(nT)} = F(z) and Z{h(nT)} = H(z) exist �for � z = 1.Then the cross correlation is given by

But Z{h(mT – nT)} = z – nH(z) and, therefore, (see [6.3.15])

n ≥ 1

(6.3.24)

This relation is the inverse Z-transform of g(nT) and, hence,

for � z � = 1 (6.3.25)

f nT h nTj

F z H zdz

zn

C( ) ( ) = ( ) ( )

=

∞−∑ ∫

0

11

2π

F ze zT( ) =

− − −

1

1 1

f nT f nTj e z e z

dz

z j z e

e

e zdz

j

j

e

e e e

nC

T TC

T

T

T

i

T

T T T

( ) ( ) =− −

=− −

= =−

=−

=

∞

− − − −

− −

∑ ∫ ∫

∑0

1

2

1

2

1

1

1

1

1

2

1

2

2

1

1

π π

ππ

residues

g nT f nT h nT f mT h mT nT f mT h mT nT z

f mT h mT nT

mz

m

m

z

( ) = ( ) ⊗ ( ) = ( ) −( ) = ( ) −( )

= ( ) −( ){ }=

∞

→ +=

∞−

→ +

∑ ∑˙ lim

lim

01

0

1Z

g nTj

Fz

Hz d

jF H dt

z C

n

C

n

( ) = ( )

= ( )

→ +

−

−

∫

∫

lim1

1

1

2

1

2

1

πτ

τ ττ

τ

πτ

ττ

Z Zg nT f nT h nT F z Hz

( ){ } = ( ) ⊗ ( ){ } = ( )

˙1


If f(nT) = h(nT) for n ≥ 0 the autocorrelation sequence is

(6.3.26)

and, hence,

(6.3.27)

If we set n = 0, we obtain the Parseval’s theorem in the same form it was developed above.

Example

The sequence f(t) = e –nT , n ≥ 0, has the Z-transform

� z � > e–T

The autocorrelation is given by (6.3.26) in the form

The function is regular in the region e –T < �z � < eT. Using the residue theorem from (6.3.24), we obtain

(6.3.28)

where τi are all poles of the integrand inside the circle �τ � = 1. Similarly from (6.3.27)

(6.3.29)

where τi are the poles included inside the unit circle.

Example

From the previous example we obtain (only the root inside the unit circle)

g nT f nT h nT( ) = ( ) ⊗ ( )˙

= ( ) −( )

= ( )

=

∞

−

∑

∫

f mT f mT nT

jF F d

m

C

n

0

11

2

1

πτ

ττ τ

G z g nT f nT h nT F z Fz

( ) = ( ){ } = ( ) ⊗ ( ){ } = ( )

Z Z1

Z ez

z enT

T

−−{ } =

−

G z f nT f nTz

z ez

ze

z

z e

e

z eTT

T

T

T( ) = ( ) ⊗ ( ){ } =− −

= −− −−

−˙

–Z

1

1

g nT F Hi

i

K

n( ) = ( )

=

=

−∑resτ τ ττ

τ1

11

g nT F Fi

i

K

n( ) = ( )

=

=

−∑resτ τ ττ

τ1

11

−− −

= −−

=

−∫ −=

−−

1

2 11 1

2

2πj

z

z e

e

z ez dz

ze

z ez

e

ee

TC

T

T

n

z e

T

T

nT

T

T nT–

–res


which is equal to the autocorrelation of f(nT) = e–nT u(nT). Using the summation definitions, we obtain

Z-Transforms with Parameters

(6.3.30)

(6.3.31)

finite integral (6.3.32)

Table 1 in the Appendix contains the Z-transform properties for positive-time sequences.

6.4 Inverse Z-Transform

The inverse Z-transform provides the object function from its given transform. We use the symbolicsolution

f (nT) = Z–1{F (z)} (6.4.1)

To find the inverse transform, we may proceed as follows:

1. Use tables.2. Decompose the expression into simpler partial forms, which are included in the tables.3. If the transform is decomposed into a product of partial sums, the resulting object function is

obtained as the convolution of the partial object function.4. Use the inversion integral.

Power Series Method

When F(z) is analytic for � z � > R (and at z = ∞), the value of f(nT) is obtained as the coefficient of z –n

in the power series expansion (Taylor’s series of F(z) as a function of z–1). For example, if F(z) is theratio of two polynomials in z –1, the coefficients f(0T), …, f(nT) are obtained as follows:

(6.4.2)

e u mT e u mT nT e e

e e e e e e

e e e

ee

ee

e

mT

m

T m n Tn mT

m n

T n nT nT T nT T

nT T T

nT

T

nTT

−

=

∞−( ) −

=

∞

− − − − −

− − −

−−

−

∑ ∑( ) −( ) =

= + + +( )= + +( ) +

=−

=

0

2

2 2 2 2 4

2 22

2

2

2

1

1

1

–

L

L

TT −1

Z∂

∂ ( )

= ∂∂ ( )

af nT a

aF z a, ,

Z lim , lim ,a a a a

f nT a F z a→ →

( )

= ( )0 0

Z f nT a da F z a daa

a

a

a

, ,( )

= ( )∫ ∫

0

1

0

1

F zp p z p z p z

q q z q z q zf T f T z f T zn

n

nn( ) =

+ + + ++ + + +

= ( ) + ( ) + ( ) +− − −

− − −− −0 1

12

2

0 11

22

1 20 2L

LL


where

(6.4.3)

The same can be accomplished by synthetic division.

Example

From (6.4.3): 1 = f(0T)·1 or f(0T) = 1, 1 = f(1T)·1 + 1·2 or f(1T) = –1, 0 = f(2T)·1 + f(1T)·2 + f(0T)·3or f(2T) = +2 – 3 = –1, 0 = f(3T)·1 + f(2T)2 + f(1T)3 + f(0T)·0 or f(3T) = 2 + 3 = 5, and so forth.

Partial Fraction Expansion

If F(z) is a rational function of z and analytic at infinity, it can be expressed as follows:

F (z) = F1(z) + F2(z) + F3(z) + L (6.4.4)

and therefore,

f (nT) = Z–1{F1(z)} + Z–1{F2(z)} + Z–1{F3(z)} + L (6.4.5)

For an expansion of the form

(6.4.6)

the constants Ai are given by

(6.4.7)

p f T q

p f T q f T q

p f nT q f n T q f n T q f T qn n

0 0

1 0 1

0 1 2

0

1 0

1 2 0

= ( )= ( ) + ( )

= ( ) + −( )[ ] + −( )[ ] + + ( )M

L

F zz

z z

z z

z zz z z z( ) = +

+ += +

+ += − − + + >

−

− −− − −1

1 2 3 2 31 5 6

1

1 2

2

2

1 2 3 L

F zF z

z p

A

z p

A

z p

A

z pn

nn( ) =

( )−( )

=−

+−( )

+ +−( )

1 1 22

L

A z p F z

Ad

dzz p F z

Ak

d

dzz p F z

An

d

dzz p F z

n

n

z p

n

n

z p

n k

k

k

n

z p

n

n

n

z p

= −( ) ( )

= −( ) ( )

= −( ) ( )

=−( ) −( ) ( )

=

−=

−=

−

−

=

1

1

1

1

1

1

1

M

M

!

!


Example

Let

Also,

from which we find that

and

Hence,

and, therefore, its inverse transform is f(nT) = δ (nT) + 8u(nT – T) – u(nT – T) with ROC �z � > 1.

Example

(a) If

F zz z

z z

z z

z z

z z z( ) = + +

− += + +

− += + + + >

− −

− −

− −1 2

13

2

1

2

2 1

3

2

1

2

17

2

23

41

1 2

1 2

2

2

1 2 L

F zz

z z

A

z

B

z( ) = +

+

−( ) −

= +−

+−

1

7

2

1

2

11

2

11 1

2

A

z z

z zz

=−( ) +

−( ) −

=

=

17

2

1

2

11

2

8

1

B

z z

z zz

=−

+

−( ) −

= −

=

1

2

7

2

1

2

11

2

9

2

1 2

F zz

z

zz

zz

z

z

( ) = +−

−−

= +−

−−

− −

18

1

9

2

1

1

2

18

1

9

2 1

2

1 1

9

2

1

2

1

−n

F zz

z zA

Bz

z

Cz

zz( ) = +

−( ) −( ) = +−

+−

>2 1

1 2 1 22


then we obtain

and

Hence,

and its inverse is f(nT) = δ(nT) – 2u(nT) + (2)n u(nT).

(b) If

then we obtain

and

Hence,

and

f (nT) = –2u (nT – T) + 3(2)n –1u (nT – T)

with ROC �z � > 2.

A

Bz

z

zz

= +−( ) −( ) =

= +−( ) = −

=

0 1

0 1 0 2

1

2

1 1

22

2

1

,

,

Cz

z

zz

= +−( ) =

=

1 1

1

5

2

2

2

F zz

z

z

z( ) = −

−+

−1

22

1

5

2 2

1

2

5

2

F zz

z z

A

z

B

z( ) = +

−( ) −( ) =−

+−

1

1 2 1 2

Az

zz

= +−( ) = −

=

1

22

1

Bz

zz

= +−( ) =

=

1

13

2

F zz z

( ) = −−( ) +

−( )21

13

1

2


Example

If F(z) = with �z � > 1, then we find

To find B we set any value of z (small for convenience) in the equality. Hence, with say z = 2, we obtain

or B = 1/2. Therefore, F(z) = and its inverse transform is f(nT) = (–1)n–1

u(nT – T) + u(nT – T) + (nT – T)u(nT – T) with ROC � z � > 1.

Example

The function F(z) = z3/(z – 1)2 with �z � > 1 can be expanded as follows: F(z) = z + 2 + or F(z)

= z + 2 + = z + 2 + . Therefore, we obtain B = = 1. Set any value

of z (e.g., z = 2) in the above equality we obtain

Hence,

and its inverse transform is

f (nT) = δ (nT + T) + 2δ (nT) + 3u(nT – T) + (nT – T)u(nT – T)

with ROC �z � > 1.

z

z z

A

z

B

z

C

z

2

2 2

1

1 1 1 1 1

+

+ −=

++

−+

−( )( ) ( )

Az

z

Cz

z

z

z

= +

−( )=

= ++

=

= −

=

2

2

1

2

1

1

1

1

2

1

11

,

.

z

z z zB

z zz

z zz

2

2

22 2

2

2

1

1 1

1

2

1

1

1

1

1

1

+

+( ) −( )=

++

−+

−( )=

= ==

1

2

1

1

1

2

1

1

1

12z z z+

+−

+−( )

1

2

1

2

3 2

12

z

z

−

−( )3 2

12

z

z

−

−( )A

z

B

z−+

−( )1 12

3 2 1

1

2

2

1

z z

zz

− −

−

( )( )( )

=

2 23 2 2

2 12 2

1

2 1

1

2 13

2 2+ + ⋅ −

−( )= + +

−+

−( )=A Aor

F z zz z

( ) = + +−

+−( )

23

1

1

12


Tables 3 and 4 in the Appendix are useful for finding the inverse transforms.

Inverse Transform by Integration

If F(z) is a regular function in the region �z � > R, then there exists a single sequence {f(nT)} for whichZ{ f(nT)} = F(z), namely

(6.4.8)

The contour C encloses all the singularities of F (z) as shown in Figure 6.4.1 and it is taken in a counterclockwise direction.

Simple Poles

If F(z) = H(z)/G(z), then the residue at the singularity z = a is given by

(6.4.9)

Multiple Poles

The residue at the pole zi with multiplicity m of the function F(z)zn–1 is given by

(6.4.10)

Simple Poles Not Factorable

The residue at the singularity am is

FIGURE 6.4.1

f nTj

F z z dz F z z nn

Cz z

i

K

n

i( ) = ( ) = ( ){ } =−

==

−∫ ∑1

20 1 21

1

1

πres , , ,K

lim limz a

n

z a

nz a F z z z aH z

G zz

→

−

→

−−( ) ( ) = −( ) ( )( )

1 1

res z zn

z z

m

m i

m n

ii

F z zm

d

dzz z F z z=

−

→

−

−−( ){ } =

−( ) −( ) ( )

11

1

11

1 !lim


(6.4.11)

F(z) is Irrational Function of z

Let F(z) = [(z + 1)/z]α , where α is a real noninteger. By (6.4.8) we write.

where the closed contour C is that shown in Figure 6.4.2.

It can easily be shown that at the limit as z → 0 the integral around the small circle BCD is zero (setz = re jθ and take the limit r → 0). Also, the integral along EA is also zero. Because along AB z = xe– jπ

and along DE z = xejπ, which implies that x is positive, we obtain

(6.4.12)

But the beta function is given by

(6.4.13)

and, hence,

FIGURE 6.4.2

F z zH z

dG z

dz

zn

z a

n

z a

m

m

( ) =( )( )

−

=

−

=

1 1

f nTj

z

zz dz

C

n( ) = +

∫ −1

2

1 1

π

α

f nTj

xe

xex e dx

xe

xex e dx

jx x e dx x x

j

j

n j nj

j

n j n

n j n

( ) = +

+ +

= − −( ) + −( )

∫ ∫

∫ ∫

−

−− − −

− − − −( )

1

2

1 1

1

21 1

1

01

0

11

0

11

0

1

π

π

π

π

απ

π

π

απ

α α π α α nn j n

n

e dx

nx x dx

− − −( )

− −

=−( )[ ]

−( )∫

1

0

11 1

α π α

α αα π

π

sin

B m km k

m kx x dxm k

,( ) =( ) ( )

+( ) = −( )∫ − −Γ Γ

Γ 0

11 1

1


(6.4.14)

But,

(6.4.15)

and, therefore,

(6.4.16)

The Taylor’s expansion of F(z) is given as follows:

(6.4.17)

But,

Γ(α + 1) = α (α – 1)(α – 2)L(α – n + 1) Γ (α – n + 1), Γ(n + 1) = n! (6.4.18)

and, therefore, (6.4.17) becomes

(6.4.19)

The above equation is a Z-transform expansion and, hence, the function F(nT) is that given in (6.4.16).

Example

To find the inverse of the transform

� z � > 2

we proceed with the following approaches:

1. By fraction expansion

f nTn n

n( ) =

−( )[ ] −( ) +( )+( )

sin α π

πα αΓ Γ

Γ

1

1

Γ Γm mm

( ) −( ) =1ππsin

f nTn

n n n n n( ) =

−( ) +( )+( ) −( ) − +( ) =

+( )+( ) − +( )

Γ Γ

Γ Γ Γ

Γ

Γ Γ

α α

α α

α

α

1

1

1

1

1

1 1

F zz

zz

n

d z

dzz

nn z

n

n

n

z

n

n

n

( ) = +

= +( ) =+( )

( )

= −( ) −( ) − +( )

−

=

∞ −

−

=

−

=

∞−

∑

∑

−

11

1 1

11 2 1

1

0

1

1

0

0

1

αα

α

α α α α

!

!L

F zn n

z n

n

( ) =+( )

+( ) − +( )−

=

∞

∑ Γ

Γ Γ

α

α

1

1 10

F zz

z z

( ) =−( )

+( ) −

1

21

2

z

z z

A

z

B

z

−( )+( ) −

=+

+−

1

21

2

2 1

2

,


2. By integration

3. By power expansion

The multiplier z–1 indicates one time-unit shift and, hence, { f(nT)} = n = 1, 2, … .

Example

1. By expansionBy F(z) has the region of convergence �z � > 5, then

Hence, f(nT) = n5n n = 0, 1, 2, …, which sometimes is difficult to recognize using the expansionmethod.

2. By fraction expansion

Az

z

Bz

z

z

z

=−( )−

= =−( )+( ) = −

= −

=

1

1

2

6

5

1

2

1

5

2

1

2

,

f nTz

z

nn

n

( ) =+

−−

= −( ) −

≥− −−

Z1 1

16

5

1

2

1

5

1

1

2

6

52

1

5

1

21

f nTz

z z

z zz

z z

zzn

z

n

n

( ) = ( ) −

+( ) −

+ −

−

+( ) −

= −( ) −

=−−

=

−

−

res z + 2 res21

1

2

1

1

1

21

2

1

2

1

21

2

6

52

1

5

1

221

1

≥−n

n

z

z z

z z z z z z−

+ −= − + + = − + +

− − − − − −1

3

21

5

2

19

41

5

2

19

42

1 2 3 1 1 2L L

15

2

19

4, , ,−

K

F zz

z

z

z zz z z

z z z z

( ) =−( )

=− +

= + + +

= ⋅ + ⋅ + ⋅ + ⋅ +

− − −

− − − −

5

5

5

10 255 50 375

0 5 1 5 2 5 3 5

2 2

1 2 3

0 0 1 2 2 3 3

L

L

F zz

z

Az

z

Bz

z( ) =

−( )=

−+

−( )5

5 5 52

2

2,


Hence,

and f(nT) = –(5)n + (n + 1)5n = n5n , n ≥ 0.3. By integration

Figure 6.4.3 shows the relation between pole location and type of poles and the behavior of causalsignals; m stands for pole multiplicity. Table 5 (Appendix) gives the Z-transform of a number ofsequences.

B. Two-Sided Z-Transform

6.5 The Z-Transform

If a function f(z) is defined by –∞ < t < ∞, then the Z-transform of its discrete representation f(nT) isgiven by

(6.5.1)

where R+ is the radius of convergence for the positive time of the sequence, and R– is the radius ofconvergence for the negative time of the sequence.

Example

The first sum (negative time) converges if �e–Tz � < 1 or �z � < eT. The second sum (positive time) convergesif �e–Tz–1� < 1 or e–T < � z �. Hence, the region of convergence is R+ = e–T < �z � < R– = eT. The two poles ofF(z) are z = eT and z = e–T.

Bz

z

= ==

51

5

,

5 6

6 1

6

6 5

6

6 51

2

2

2

×

−( )= ×

−+

−( )= −A

Aor .

F zz

z

z

z( ) = −

−+

−( )5 5

2

2

1

2 15

5

55 5 0

2 1

2 1

2

2

1

5

1

5−( ) −( )−( )

= = ≥−

−−

=

−

=!, .

d

dzz

z

zz nz n nn

z

n

z

n

ZI I

n

nf nT F z f nT z R z R( ){ } = ( ) = ( ) > <= −∞

∞−

+ −∑˙

F z e e z e z e z e z

e z e ze z e z

I I

nT nT

n

n nT

n

n nT

n

n nT

n

n

nT

n

n nT

n

n

T T

( ) = { } = + = − +

= − + =−

− +−

−

= −∞

−− −

=

∞−

= −∞

− −

=

∞−

−

=

∞−

=

∞−

− − −

∑ ∑ ∑ ∑

∑ ∑

Z

1

0

0

0

0 01

1

11

11

1

1


FIGURE 6.4.3


Example

The Z-transform of the functions of u(nT) and –u(–nT – T) are

FIGURE 6.4.3 (continued)


Although their Z-transform is identical their ROC is different. Therefore, to find the inverse Z-transformthe region of convergence must also be given.

FIGURE 6.4.3 (continued)

Z

Z

I I

n

n

I I

n

n

n

n

n

n

u nT u nT zz

z

zz

u nT T u nT T z

z

zz

z

zz

( ){ } = ( ) =−

=−

>

− − −( ){ } = − − −( )

= − −

= − = −−

=−

<

=

∞−

−

= −∞

−−

−

= −∞

=

∞

∑

∑

∑

∑

01

1

0

0

1

1 11

1

1 11

1 11


Figure 6.5.1 shows signal characteristics and their corresponding region of convergence.Assuming that the algebraic expression for the Z-transform F(z) is a rational function and that f(nT)

has finite amplitude, except possibly at infinities, the properties of the region of convergence are

1. The ROC is a ring or disc in the z-plane and centered at the origin, and 0 ≤ R+ < �z � < R– ≤ ∞.2. The Fourier transform converges also absolutely if and only if the ROC of the Z-transform of

f(nT) includes the unit circle.3. No poles exist in the ROC.4. The ROC of a finite sequence { f(nT)} is the entire z-plane except possibly for z = 0 or z = ∞.5. If f(nT) is right handed, 0 ≤ n < ∞, the ROC extends outward from the outermost pole of F(z)

to infinity.6. If f(nT) is left handed, –∞ < n < 0, the ROC extends inward from the innermost pole of F(z) to zero.7. An infinite-duration two-sided sequence {f (nT)} has a ring as its ROC, bounded on the interior

and exterior by a pole., The ring contains no poles.8. The ROC must be a connected region.

6.6 Properties

Linearity

The proof is similar to the one-sided Z-transform.

Shifting

ZI I{ f (nT ± kT)} = z ± kF (z) (6.6.1)

Proof

The last step results from setting m = n – k. Proceed similarly for the positive sign. The ROC of theshifted functions is the same as that of the unfinished function except at z = 0 for k > 0 and z = ∞ for k < 0.

Example

To find the transfer function of the system y(nT) – y(nT – T) + 2y(nT – 2T) = x(nT) + 4x(nT – T), wetake the Z-transform of both sides of the equation. Hence, we find

Y (z) – z–1Y (z) + 2z –2Y (z) = X (z) + 4z –1X (z)

or

Example

Consider the Z-transform

ZI I

n

n k

m

mf nT kT f nT kT z z f mT z−( ){ } = −( ) = ( )= −∞

∞− −

= −∞

∞−∑ ∑

H zY z

X z

z

z z( ) =

( )( ) = +

− +

−

− −

1 4

1 2

1

1 2


FIGURE 6.5.1


Because the pole is inside the ROC, it implies that the function is causal. We next write the function inthe form

which indicates that it is a shifted function (because of the multiplier z –1). Hence, the inverse transform

is f(n) = u(n – 1) because the inverse transform of is equal to .

Scaling

If

ZI I{ f (nT)} = F (z) R+ < � z � < R–

then

ZII{a n T f (nT)} = F (a–T z) � aT �R+ < �z � < �aT �R– (6.6.2)

Proof

Because the ROC of F(z) is R+ < �z � < R–, the ROC of F(a–T z) is

R+ < �a–T z � < R– or R+�aT � < �z � < �aT �R–

Example

If the Z-transform of f(nT) = exp(–�nT �) is

e –T < � z � < eT

then the Z-transform of g(nT) = anT f(nT) is

aTe –T < � z � < eTaT

F z

z( ) =

−

1

1

2

z > 1

2

F z zz

z

z

z( ) =

−=

−

− −

−

1 1

11

2

1

11

2

z > 1

2

1

2

1

−n

1 11

2

1− −

z1

2

n

ZI InT nT

n

n

n

Tn

Ta f nT a f nT z f nT a z F a z( ){ } = ( ) = ( )( ) = ( )= −∞

∞−

= −∞

∞− − −∑ ∑

F ze z e znT nT( ) =

−+

−−

− − −

1

1

1

11

1

G ze a z e a znT T nT T( ) =

−+

−−

− − −

1

1

1

11

1–


Time Reversal

If

ZI I{ f (nT)} = F (z) R+ < �z� < R–

then

(6.6.3)

Proof

and

The above means that if z0 belongs to the ROC of F(z) then 1/z0 is in the ROC of F(z–1). The reflectionin the time domain corresponds to inversion in the z-domain.

Example

The Z-transform of f (n) = u(n) is z/(z – 1) for �z � > 1. Therefore, the Z-transform of f(–n) = u(–n) is

Also, from the definition of the Z-transform, we write

Multiplication by nT

If

ZI I{ f (nT)} = F (z) R+ < � z � < R–

then

(6.6.4)

ZI I f nT F zR

zR

−( ){ } = ( ) < <−

− +

1 1 1| |

ZI I

n

n

n

n

f nT f nT z f nT z F z−( ){ } = ( ) = ( )( ) = ( )= −∞

∞−

= −∞

∞− − −∑ ∑ 1 1

R z R zR

zR+

−

+

< < > <1 1 1–

–

or and

1

11

1

1z

z

z−=

−

Z u n z zz

n

n

n

n

−( ){ } = = =−

−

= −∞ =

∞

∑ ∑0

0

1

1

ZI I nTf nT zTdF z

dzR z R( ){ } = −

( )< <+ −


ProofA Laurent series can be differentiated term-by-term in its ROC and the resulting series has the sameROC. Therefore, we have

Multiply both sides by –zT

Example

If F(z) = log(1 + az–1) �z � > �a �, then

The z –1 implies a time shift, and the inverse transform of the fraction is (–a)n. Hence, the inverse transformis a(–a)n–1 u(n – 1). From the differentiation property (with T = 1), we obtain

Example

If f(nT) = au(nT) then its Z-transform is F(z) = a/(1 – z –1) for �z � > 1. Therefore,

Convolution

If

ZI I { f1(nT)} = F1(z) and ZI I{ f2(nT)} = F2(z)

then

F (z) = ZII { f1(nT) ∗ f2(nT)} = F1(z) F2(z) (6.6.5)

The ROC of F(z) is, at least, the intersection of that for F1(z) and F2(z).

dF z

dz

d

dzf nT z nf nT z R z Rn

n

n

n

( )= ( ) = − ( ) < <−

= −∞

∞− −

= −∞

∞

+∑ ∑ 1 for –

−( )

= ( ) = ( ){ } < <−

= −∞

∞

+∑zTdF z

dznTf nT z nTf nT R z Rn

n

Z for –

dF z

dz

az

azz

dF z

dzaz

a zz a

( )= −

+−

( )=

− −( ) >−

−−

−

2

1

1

11

1

1or

nf n a a u n f na

nu n

n n n

( ) = −( ) −( ) ( ) = −( ) −( )− −1 11 1 1or

Z nTau nT zTadF z

dzaT

z

zz( ){ } = −

( )=

−( )>

11

2


Proof

where the shifting property was invoked.

Example

The Z-transform of the convolution of e– nu(n) and u(n) is

Also, from the convolution, definition we find

which verifies the convolution property. The ROC for e–nu(n) is �z � > e –1 and the ROC of u(n) is �z � >1. The ROC of e–nu(n) ∗ u(n) is the intersection of these two ROCs and, hence, the ROC is �z � > 1.

Example

The convolution of f1(n) = {2, 1, –3} for n = 0, 1, and 2, and f2(n) = {1, 1, 1, 1} for n = 0, 1, 2, and 3 is

G (z) = F1(z) F2(z) = (2 + z –1 – 3z –2)(1 + z –1 + z –2 + z –3) = 2 + 3z –1 – 2z –4 – 3z –5

which indicates that the output is g(n) = {2, 3, 0, 0, –2, –3} which can easily be found by simply convolutingf1(n) and f2(n).

F z f nT z f mT f nT mT z

f mT f nT mT z

f mT z F z F z F

n

n

mn

n

m n

n

m

m

( ) = ( ) = ( ) −( )

= ( ) −( )

= ( ) ( ) = ( )

= −∞

∞−

= −∞

∞

= −∞

∞−

= −∞

∞

= −∞

∞−

= −∞

∞−

∑ ∑∑

∑ ∑

∑

1 2

1 2

1 2 1 22 z( )

Z Z Z ZI In m

m

n

ne u n u n e u n m e u nz

z e

z

z− −

=

−−( )( )∗ ( ){ } = −( )

= { } ( ){ } =

− −∑0

1 1

Z Z

Z

e u n me

e

e

e

ee

e

z

ze

z

z e

z

z z e

m

m

n n

n

−

=

− −

−

−

−

−−

−−

−

∑ −( )

= −

= −

=−

−−

=−( ) −

0

1

1

1

1

1

1

1

1

2

1

1

1

1 1

1

1 1

1

–

– –

–

−−( )1


Correlation

If

ZI I { f1(nT)} = F1(z) and ZI I{ f2(nT)} = F2(z)

then

(6.6.6)

The ROC of Rf1 f2(z) is at least the intersection of that for F1(z) and F1(z –1).

ProofBut rf1 f 2

(�T) = f 1(�T) ∗ f2(–T�) and, hence, from the convolution property and the time-reversal propertyRf1 f 2

(z) = F1(z)F2(z –1).

Example

The transform of the autocorrelation sequencing f(nT) = an T u(n), –1 < a <1 is

But,

causal signal

and

anticausal signal

Hence,

Because the ROC of Rff(z) is a ring, it implies that rf f(�T) is a two-sided signal.We proceed to find the autocorrelation first

Z Z ZI I f f I I I I

n

f f

r T f nT f nT f nT f nT T

R z F z F z

1 2

1 2

1 2 1 2

1 21

l l( ){ } = ( ) ⊗ ( ){ } = ( ) −( )

= ( ) = ( ) ( )= −∞

∞

−

∑˙

R z r T F z F zf f I I f f( ) = ( ){ } = ( ) ( )−˙ Z l 1

F za z

z aT

T( ) =−

>−

1

1 1

F za z

za

T T

−( ) =−

<1 1

1

1

R za z z a

a za

f f T T

T

T( ) =− +( ) +

< <−

1

1

11 2

ROC

r nT a a a a a a

aa

aa

a

a

an

r nT a a aa

f fmT

m n

m n T nT T m

m

nT T m

m

n

nT

T

nTT n

T

nT

T

f fmT

m

m n T nT

( ) = = −

=−

− −−

=−

≥

( ) = =−

=

∞−( ) −

=

∞−

=

−

− −

=

∞−( ) −

∑ ∑ ∑

∑

2

0

2

0

1

2

2

2 2

02

1

1

1

1 10

1

1 TTn ≤ 0


and then compare by inverting the function F(z)F(z–1).

Multiplication by e –anT

If

ZII{ f (nT)} = F(z) R+ < � z � < R–

then

ZII{ e– a n Tf (nT)} = F(e – a T z) �e – a T �R+ < �z � < �e – a T �R– (6.6.7)

Proof

Frequency Translation

If the region of convergence of F(z) includes the unit circle and g(nT) = e jω 0nT f(nT), then

G(ω) = F(ω – ω 0) (6.6.8)

ProofFrom (6.6.7) G(z) = F(e– jω 0T z) and has the same region of convergence as F(z) because �exp(jω0T) � =1. Therefore,

G(ω) = G(z)� z = e jωT = F(e j (ω – ω 0)T) = F(ω – ω0)

Product

If

ZI I { f (nT)} = F(z) R+f < � z � < R–f (6.6.9)

ZI I {h (nT)} = H(z) R+h < � z � < R–h (6.6.10)

g (nT) = f (nT) h (nT)

then

(6.6.11)

where C is any simple closed curve encircling the origin counterclockwise with

(6.6.12)

ZI IanT

n

aTn

aT aTe f nT f nT e z F e z R e z R−

=−∞

∞−

+( ){ } = ( )( ) = ( ) < <∑ –

ZI I

n

n

Cf h f h

f nT h nT G z f nT h nT z

jF H

z dR R z R R

( ) ( ){ } = ( ) = ( ) ( )

= ( )

< <

= −∞

∞−

+ + − −

∑

∫

˙

1

2πτ

ττ

τ

max , min ,Rz

RR

z

Rfh

fh

+−

−+

< <

τ


ProofThe series in (6.6.11) will converge to an analytic function G(z) for R+g < �z � < R–g. Using the root test(see Section 6.2), we obtain

(6.6.13)

for positive n. However,

(6.6.14)

and this series converges if

(6.6.15)

Hence,

(6.6.16)

Replacing f(nT) in the summation of (6.6.11) by its inversion formula (6.4.8), we find

(6.6.17)

The interchange of the sum and integral is justified if the integrand converges uniformly for some choiceof C and z. The contour must be chosen so that

R+f < �τ � < R– f (6.6.18)

If

(6.6.19)

R f nT h nT

f nT h nT R R

g n

n

n

n

n

n

f h

+ →∞

→∞ →∞ + +

= ( ) ( )( )≤ ( )( ) ( )( ) =

lim

lim lim

1

1 1

F z f nT z f nT zn

n

n

n( ) = ( ) = −( )= −∞

−

=

∞

∑ ∑0

0

z

f nT

R

n

n f<−( )( )

=

→∞

−1

1

lim

R

f nT h nT

f nT h nT

R R

g

n

n

n

n

n

n

f h

−

→∞

→∞ →∞

− −

=−( ) −( )( )

≥−( )( ) −( )( )

≥

1

1

1

1 1

lim

lim lim

G zj

Fd

h nT zj

F h nTz d

nC

n n

Cn

n

( ) = ( ) ( ) = ( ) ( )

= − ∞

∞−

= − ∞

∞ −

∑ ∫ ∫ ∑1

2

1

2πτ τ τ

τ πτ

ττ

τ

Rz

Rz

R

z

Rh hh h

+ −− +

< < < <τ

τor


the series in the integrand of (6.17) will converge uniformly to H(z/τ), and otherwise will diverge. Figure6.6.1 shows the region of convergence for F(τ) and H(z/τ). From (6.6.18) and (6.6.19) we obtain

or equivalently

R+ f R+ h < � z � < R– f R– h (6.6.20)

When z satisfies the above equation, the intersection of the domain identified by (6.6.18) and (6.6.19) is

(6.6.21)

The contour must be located inside the intersection.When signals are causal, R– f = R–h = ∞ and the conditions (6.6.20) and (6.6.21) reduce to

R+ f R+ h < �z � (6.6.22)

(6.6.23)

Hence, all of the poles of F(τ) lie inside the contour and all the poles of H(z/τ ) lie outside the contour.

FIGURE 6.6.1

z

RR z R R

z

RR z R R

hf f h

hf f h

−− − −

++ + +

< <

> >

or

or

R Rz

R

z

RR

z

RR

z

Rf fh h

fh

fh

+ −− +

+−

−+

< <( )∩ < <

=

< <

τ τ τmax , min ,

Rz

Rfh

++

< <τ


Example

The Z-transform of u(nT) is

and the Z-transform of h(nT) = exp(–�nT �) is

But R– f = ∞ and, hence, from (6.11) 1· exp(–T) < �z � < ∞. The contour must lie in the region max(1, �z �e – T) < �τ � < min(–∞, �z �e –T) as given by (6.6.21). The pole-zero configuration and the contourare shown in Figure 6.6.2.

If we choose �z � > eT, then the contour is that shown in the figure. Therefore, (6.6.11) becomes

The poles of H(z/τ) are at τ = z exp(–T) and τ = z exp(T). Hence, the contour encloses the poles τ =1 and τ = z exp(–T). Applying the residue theorem next we obtain

which has the inverse function g(nT) = e–nTu(nT), as expected.

FIGURE 6.6.2

F zz

z R Rf f( ) =−

> = = ∞− + −

1

11

1,

H ze

e z e zR e z e R

T

T T hT T

h( ) = −

( ) −( ) = < < =−

− − − +−

−1

1 1

2

1–

ZI I

T

T TC

u nT h nT G zj

e

ez

ez

d( ) ( ){ } = ( ) =−

−

−

−

− −∫˙

– –

1

2

1

1

1

1 11

2

π τ ττ

ττ

G ze z

z eT

T( ) =−

>− −

−1

1 1


Parseval’s Theorem

If

ZII{f(nT)} = F(z) R+f < �z � < R–f

ZII{h(nT)} = H(z) R+h < �z � < R–h (6.6.24)

with

R+f R+h < �z � = 1 < R–f R–h (6.6.25)

then we have

(6.6.26)

where the contour encircles the origin with

(6.6.27)

ProofIn (6.6.11) and (6.6.12) set z = 1 and replace the dummy variable τ and z to obtain (6.6.26) and (6.6.27).

For complex signals Parseval’s relation (6.6.26) is modified as follows:

(6.6.28)

If f(nT) and h(nT) converge on the unit circle, we can use the unit circle as the contour. We then obtain

(6.6.29)

where we set z = e jω T. If f(nT) = h(nT) then

(6.6.30)

Example

The Z-transform of f(nT) = exp(–nT)u(nT) is F(z) = 1/(1 – e –Tz –1) for �z � > e –T. From (6.6.26) we obtain

f nT h nTj

F z H zdz

zn

C( ) ( ) = ( ) ( )

= −∞

∞−∑ ∫1

21

π

max , min ,RR

z RRf

hf

h+

−−

+

< <

1 1

f nT h nTj

F z Hz

dz

zn

C( ) ( ) = ( )

= −∞

∞

∑ ∫* **

1

2

1

π

f nT h nT F e H e dT

n s

j T j Ts

s

s( ) ( ) = ( ) ( ) == −∞

∞

−∑ ∫* *

1 2

2

2

ωω ω πω

ω

ωω

f nT F e dn s

j T

s

s( ) = ( )= −∞

∞

−∑ ∫

2 2

2

21

ωωω

ω

ω

f nT f nTj e z e z

dz

zn n

T TC

2 2

01

1

2

1

1

1

1= −∞

∞

=

∞

− − −∑ ∑ ∫( ) = ( ) =−π –


From (6.6.27) we see that max(e –T, 0) < �z � < min(∞, eT ). The contour encircles the pole at z = e –T so that

Also we find directly

Complex Conjugate Signal

If

ZI I { f (nT)} = F (z) R+f < �z � < R–f

then

ZI I{ f *(nT)} = F *(z*) R+ f < �z � < R– f (6.6.31)

ProofBy definition we have

Replacing z with z* and taking the conjugate of both sides of the above equation, we obtain (6.6.31).

6.7 Inverse Z-Transform

Power Series Expansion

The inverse Z-transform in operational form is given by

If F(z) corresponds to a causal signal, then the signal can be found by dividing the denominator intothe numerator to generate a power series in z–1 and recognizing that f(nT) is the coefficient of z–n.Similarly, if it is known that f(nT) is zero for positive time (n positive), the value of f(nT) can be foundby dividing the denominator into the numerator to generate a power series in z.

Example

If F(z) = [z(z + 1)]/(z2 – 2z + 1) = (1 + z –1)/(1 – 2z –1 + z –2) and the ROC is �z � > 1, then

f nTz e

z e e z en

T

T T

z e

T

T

2

02

1

1

1=

∞

− −

=

−∑ ( ) =−( ) −( )

=

−−

res- –

e e e e ee

nT

n

nT nT

n

T T

T

−

=

∞− −

=

∞− −

−∑ ∑= = + +( ) +

=−

0

2

0

2 22

21

1

1L

F z f nT zn

n( ) = ( )= −∞

∞−∑

f nT F zI I( ) = ( ){ }−Z

1


and by continuing the division we recognize that

If f(nT) is known to be zero for positive n, that the ROC is �z � < 1, then

This series is recognized as

Example

If F(z) = log(1 + 2z–1), � z � > 2, then using power series expansion for log(1 + x), with �x � < 1, we obtain

which indicates that

In general, any improper rational function (M ≥ N) can be expressed as

1 2 1

1 2

3

3 6 3

5 3

1 2 1

1 2 3

1 2

1 2

1 2 3

2 3

1 3 5 7

− + +

− +−

− +−

− − −

−

− −

− −

− − −

− −

+ + + +

z z z

z z

z z

z z z

z z

z z z– – L

L

f nTn

n n( ) =

<+( ) ≥

0 0

2 1 0

z z z

z z

z

z z

z z

z z z− − −

−

− + +

− +−

− +−

+ + +2 1 1

2 3

1

2

2

2 1 1

2

3

3 6 3

5 3

3 5 L

L

f nTn n

n( ) = − +( ) <

≥

2 1 0

0 0

F zz

n

n n n

n

( ) =−( ) + −

=

∞

∑ 1 21

1

f nT nn

n

n n

( ) = −( ) ≥

≤

+1

20

0 0

1


(6.7.1)

where the inverse Z-transform of the polynomial can easily be found by inspection.A proper function (M < N) is of the form

or

(6.7.2)

Because N > M , the function

(6.7.3)

is always a proper function.

Partial Fraction Expansion

Distinct PolesIf the poles p1, p2, …, pN of a proper function F(z) are all different, then we expand it in the form

(6.7.4)

where all Ai are unknown constants to be determined.The inverse Z-transform of the kth term of (6.7.4) is given by

(6.7.5)

If the signal is causal, the ROC is �z � > pmax, where pmax = max{�p1 �, �p2 �, …, �pN �}. In this case, all termsin (6.7.4) result in causal signal components.

Example

(a) If F(z) = z(z + 3)/(z2 – 3z + 2) with �z � > 2 then

F zN z

D z

b b z b z

a z a z

c c z c zN z

D z

MM

NN

M N

M N

( ) =( )( ) =

+ + ++ + +

= + + + +( )( )

− −

− −

−−

− −( )

0 11

11

0 11 1

1

L

L

L

F zN z

D z

b b z b z

a z a za M NM

M

NN N( ) =

( )( ) =

+ + ++ + +

≠ <− −

− −0 1

1

111

0L

L,

F zN z

D z

b z b z b z

z a z a

N NM

N M

N NN

( ) =( )( ) =

+ + ++ + +

− −

−0 1

1

11

L

L

F z

z

b z b z b z

z a z a

N NM

N M

N NN

( )=

+ + ++ + +

− − − −

−0

11

2 1

11

L

L

F z

z

A

z p

A

z p

A

z pN

N

( )=

−+

−+ +

−1

1

2

2

L

Z−

−−

=

( ) ( ) > ( )( ) − −( ) < ( )

1

1

1

1 p z

p u nT z p

u nT T z pk

k

n

k

n

k

if ROC : causal signal

– p if ROC : anticausal signalk


Therefore,

(b) If F(z) = z(z + 3)/(z2 – 3z + 2) with 1 < �z � > 2, then following exactly the same procedure

However, the pole at z = 2 belongs to the negative-time sequence and the pole at z = 1 belongsto the positive-time sequence. Hence,

Example

To detrmine the inverse Z-transform of F(z) = 1/(1 – 1.5z–1 + 0.5z–2) if (a) ROC: �z � > 1, (b) ROC: �z �< 0.5, and (c) ROC: 0.5 < �z � < 1, we proceed as follows:

or

(a) f(nT) = 2(1)n – (1/2)n, n ≥ 0 because both poles are outside the region of convergence �z � > 1(inside the unit circle).

(b) f(nT) = –2(1)n u(–nT –T) + (1/2)n u(–nT –T), n ≤ –1 because both poles are outside the regionof convergence (outside the circle �z � = 0.5).

(c) Pole at 1/2 provides the causal part and the pole at 1 provides the anticausal. Hence,

F z

z

z

z z

A

z

A

z

( )= +

−( ) −( ) =−

+−

3

2 1 2 11 2

Az z

z zA

z z

z zz z

1

2

2

1

3 2

2 15

3 1

2 14=

+( ) −( )−( ) −( ) = =

+( ) −( )−( ) −( ) = −

= =

,

F zz

z

z

zf nT n

n n( ) =−

−− ( ) = ( ) − ( ) ≥5

24

15 2 4 1 0or

F zz

z

z

z( ) =

−−

−5

24

1

f nTn

n

n

n( ) =− ( ) ≥

− ( ) ≤ −

4 1 0

5 2 1

F zz

z z

z

z z

ABz

z

Cz

z( ) =

− +=

−( ) −

= +−

+−

2

2

2

1 5 0 51

1

2

1 1

2. .

F zz

z

z

z( ) =

−−

−2

1 1

2

f nT u nT T u nT nn

n

( ) = − ( ) − −( )

( ) − ∞ < < ∞2 11

2–


Multiple PolesIf F(z) has repeated poles, we must modify the form of the expansion. Suppose F(z) has a pole ofmultiplicity m at z = pi . Then one form of expansion is of the form

(6.7.6)

The following example shows how to find Ai’s.

Example

Let the transfer function of each of two cascade systems be 1/(1 – (1/2)z –1). If the input to this systemis the unit step function 1/(1 – z–1), then its output is

If we set z = 0 in both sides, we find that A0 = 0. Next we find A3 by multiplying both sides by (z – 1/2)2

and setting z = 1/2. Hence,

and then we write

Az

z pA

z

z pA

z

z pii

m

m

i

m1 2

2

2−+

−( )+ +

−( )L

F zz

z

z

z z

AA z

z

A z

z

A z

z

z

( ) =−

−

=

−( ) −

= +−

+−

+

−

>

−−

1

1

1

11

21

1

2

1 1

21

2

1

1

1

2

3

2

01 2 3

2

2

A

z z

z z zz

3

3

2

2

2

1

2

1

2

11

2

1

21

21

1=−

−( ) −

=−

= −

=

z

z z

A z

z

A z

z

z

z

A z z A z z z z z

z z

A A z

3

21 2

2

2

1

2

22

2

1 23

11

2

1 1

21

2

1

21

1

21

11

2

1

−( ) −

=−

+−

−

−

=−

+ −( ) −

− −( )

−( ) −

=+ −( ) ++ − −

+ +

−( ) −

13

2

1

4

1

2

11

2

2 12

1 2

2

A A z A A z

z z


Equating coefficients of equal powers, we obtain the system

Hence,

and the output is

Another form of expansion of a proper function (the degree of the denominator is one less than thenumerator) is of the form

(6.7.7)

and the following example explains its use (see Table 4 in the Appendix).

Example

Using the previous example for F(z) with �z � > 1, we obtain

Hence,

A A A A A A1 2 1 2 1 21 1 13

20 4 2+ − = − = = = −, – , , and

z

z z

z

z

z

z

z

z

3

2

2

2

11

2

41

21

21

2−( ) −

=−

−−

−

−

f nT n nn

n n

( ) = ( ) −

− +( )

≥4 1 21

21

1

20

A

z p

A z

z p

A z z p

z pii

i

i

1 22

3

3−+

−( )+

+( )−( )

F zz

z z

z z

z z

A

z

A

z

A z

z

( ) =

−( ) −

= +− +

−( ) −

= +−

+−

+

−

3

2

2

2

1 2 32

11

2

12

5

4

1

4

11

2

11 1

21

2

A

z z z

z zz

1

2

2

1

25

4

1

41

11

2

4=− +

−( )

−( ) −

=

=

,


where A2 was found by setting an arbitrary value of z, that is, z = –1, in both sides of the equation.Therefore, the inverse Z-transform is given by

Example

Now let us assume the same example but with �z � < 1/2. This indicates that the output signal is anticausal.Hence, from

and Table 3 (Appendix), we obtain

Similarly from

and Table 4 (Appendix), we obtain

Az

z z z

z z

A

z

3

2

2

2

1

2

2

12

5

4

1

4

1

2

11

2

1

2

3

2

=− +

−

−( ) −

= −

= −

=

,

f nT

n n

n nn

n n( ) =( ) =

( ) −

−

≥

−− −

δ 0

4 13

2

1

2

1

2

1

21

11 1

F zz

z

z

z

z

z

( ) =−

−−

−

−

41

21

21

2

2

2

f nT n nn

n n

( ) = − ( ) +

+ +( )

≤ −4 1 21

21

1

21

F zz

z

z

z

( ) = +−

−−

−

−

1 41

1

3

2

1

1

2

1

2 1

2

2

f nT

n n

n nn

n n( ) =( ) =

− ( ) +

+

≤ −

−− −

δ 0

4 13

2

1

2

1

2

1

21

11 1


Integral inversion formula

Theorem 7.1

If

(6.7.8)

converges to an analytic function in the annular domain R+ < �z � < R– , then

(6.7.9)

where C is any simple closed curve separating �z � = R+ from �z � = R– and it is traced in the counterclockwisedirection.

ProofMultiply (6.7.8) by zn–1 and integrate around C. Then

(6.7.10)

Set z = Re jθ with R+ < R < R– to obtain

(6.7.11)

Hence, the summation on the right-hand side of (6.7.10) reduces to f(nT).Let {ak} be the set of poles of F(z)zn–1 inside the contour C and {bk} be the set of poles of F(z)zn–1

outside C in a finite region of the z-plane. By Cauchy’s residue theorem

(6.7.12)

(6.7.13)

Example

Let

F z f mT zm

m( ) = ( )= −∞

∞−∑

f nTj

F z zdz

zC

n( ) = ( )∫1

2π

1

2

1

2π πjF z z

dz

zf mT

jz

dz

zC

n

m

n m

C( ) = ( )∫ ∑ ∫

= −∞

∞−

1

2

1

2

1

2

1 0

0

0

21 1

0

2

π πθ

πθ

π θ θ

πθ

jz

dz

z jR e Rje d

R e d

k

n m

C

n m j n m j

k j k

− − − − −( )∫ ∫

∫

=

=

==

elsewhere

f nT F z z a nk

nk( ) = ( ){ } ≥∑ −Re ,s 1 0

f nT F z z b nk

nk( ) = − ( ){ } <∑ −Re ,s 1 0

F zz a z

a zT

( ) =−( ) −( ) < >

− −

1

1 11 1

1 1,


The function F(z)zn–1 = zn+1/(z – 1)(z – aT) has two poles enclosed by C for n ≥ 0. Hence,

Example

Let

For n ≥ 0 the contour C encloses only the pole z = 0.8 of the function F(z)zn–1. Therefore,

For n < 0 only the pole z = 1/0.8 is outside C. Hence,

The residue for a multiple pole of order k at z0 is given by

(6.7.14)

C. Applications

6.8 Solutions of Difference Equations with Constant Coefficients

Based on the relation

(6.8.1)

where Z{ f(n)} = F(z), we can solve a difference equation of the form

f nT F z z F z z a

a

a

an

n n

T

n T

T

( ) = ( ){ } + ( ){ }=

−+

−≥

− −

+( )

Re , Re ,s s1 1

1

1

1

1 10

F zz z

z( ) = −−( ) −( ) < <

−−1 0 8

1 0 8 1 0 80 8 0 8

2

1

1.

. .. .

f nT F z zz z

z znn

z

n

z

n( ) = ( ){ } =−( ) −( )−( ) −( ) = ≥−

==

Re. .

. ..

.

.

s 1

0 8

2

0 8

1 0 8 0 8

1 0 8 0 80 8 0

f nT F z z

z z

zn

n

z

n

z

n

( ) = − ( ){ }

= −−( ) −( )

− −( ) −( ) = ≤ −

−

=

− −

−

=

−

−

Re

. . .

. ..

.

.

s 1

1 0 8

2 1 1

1

0 8

1 0 8 0 8 0 8

1 0 8 0 80 8 1

1

Re lim!

s F z zk

d

dzz z F z zn

z z z z

k

k

k n( ){ } =−( ) −( ) ( )

−

= →

−

−−1

1

1 01

0 0

1

1

Z f n m f z z F zm

m m−( ){ } = ( ) = ( )= −

−− +( ) −∑ l

l

l1


(6.8.2)

using the Z-transform approach.

Example

To find the solution to y(n) = y(n –1) + 2y(n – 2) with initial conditions y(0) = 1 and y (1) = 2, weproceed as follows:

From the difference equation

y (0) = y (–1) + 2y (–2) = 1

y (1) = y (0) + 2y (–1) = 2

Hence, y(–1) = and y(–2) = . The Z-transform of the difference equation is given by

Hence,

and

Example

The solution of the difference equation y(n) – ay(n – 1) = u(n) with initial condition y(–1) = 2 and �a �< 1 proceeds as follows:

a y n k b f n kk

k

N

k

k

L

= =∑ ∑−( ) = −( )

0 0

1

2

1

4

Y z y z z Y z y z z Y z

y z Y z y y z z Y z

z Y z

( ) = ( ) + ( ) + ( ) + ( )

= −( ) + ( ) + −( ) + −( ) + ( )( )= + (

= −

−− +( ) −

= −

−− +( ) −

− − −

−

∑ ∑l

l

l

ll l

1

11 1

2

12 2

1 1 2

1

2

1 2 2 1

1

2)) + + + ( ) = + + ( ) + ( )− − − − −1

22 1 21 2 1 1 2z z Y z z z Y z z Y z

Y zz z

z

z z

z

z z

z

z z( ) =

− −+

− −=

− −+

− −− −

−

− −

1

1 2 1 2 2 21 2

1

1 2

2

2 2

Z Z Z− − −( ){ } = ( ) =

− −

+

− −

1 12

2

1

22 2Y z y n

z

z z

z

z z˙

Y z ay az Y zz

z

Y za

az

z

z az

a

az

z

z z a

a

az a z

a

a az

( ) − −( ) − ( ) =−

( ) =−

+− −

=−

+−( ) −( )

=−

+− −

+− −

−

− − −

− − −

11

2

1 1

1

1

2

1 1

2

1

1

1

1

1 1

1

1

1

1 1 1

2

1 1 1


Hence, the inverse Z-transform gives

6.9 Analysis of Linear Discrete Systems

Transfer Function

From (6.8.2) we obtain the transfer function by ignoring initial conditions. The result is

(6.9.1)

where H(z) is the transform of the impulse response of a discrete system.

Stability

Using the convolution relation between input and output of a discrete systems, we obtain

(6.9.2)

where M is the maximum value of f(n). The above inequality specifies that a discrete system is stable ifto a finite input the absolute sum of its impulse response is finite. From the properties of the Z-transform,the ROC of the impulse response satisfying (6.9.2) is �z � > 1. Hence, all the poles of H(z) of a stablesystem lie inside the unit circle.

The modified Schur-Cohn criterion establishes if the zeros of the denominator of the rational transferfunction H(z) = N(z)/D(z) are inside or outside the unit circle.

The first step is to form the polynomial

Dr p(z) = z ND(z –1) = d0z N + L + dN–1 z + dN

where D(z –1) = d0 + L + dN–1z N–1 + dNzN. This Drp(z) is called the reciprocal polynomial associatedwith D(z). The roots of Dr p(z) are the reciprocals of the roots of D(z) and �Drp(z)� = �D(z)� on the unitcircle. Next, we must divide Dr p(z) by D(z) starting at the high power and obtain the quotient α0 = d0 /dN

and the remainder D1r p(z) of degree N – 1 or less, so that

The division is repeated with D1r p(z) and its reciprocal polynomial D1(z) and the sequence α 0, α 1, …,αN–2 is generated according to the rule

y n a aa

u na

aa

au n

a

aa nn n n( ) = ⋅ +

− ( ) +−

=− ( ) + −

−≥+2

1

1 1

1

1

2 1

101

zero inputzero state steady state transient

1231 2444 3444 1 24 34 1 24 34

H zY z

F z

b z

a z

kL

kk

kN

kk

( ) =( )( ) = =

=−

=−

∑∑

0

0

transfer function

y n h k f n k M h kk

n

k

( ) = ( ) −( ) ≤ ( ) < ∞= =

∞

∑ ∑0 0

D z

D z

D z

D z

r p r p( )( ) = +

( )( )α 0

1


The zeros of D(z) are all inside the unit circle (stable system) if and only if the following three conditionsare satisfied:

1. D(1) > 0

2. D(–1)

3. �αk � < 1 for k = 0, 1, …, N – 2

Check conditions (1) and (2) before proceeding to (3). If they are not satisfied, the system is unstable.

Example

D(z) = z3 – 0.2 z 2 + z – 0.2 , Dr p(z) = –0.2 z 3 + z 2 – 0.2z + 1

Because �α1� = 1, condition (3) is not satisfied and the system is unstable.The transfer function of a feedback system with forward (open-loop) gain D(z)G(z) and unit feedback

gain is given by

Assuming that all the individual systems are causal and have rational transfer function, the open-loopgain D(z)G(z) can be written as

where

A(z) = a Lz L + L + a 0 , B (z) = z M + b M – 1 z M – 1 + L + b0, L ≤ M

Hence, the total transfer function becomes

which indicates that the system will be stable if B(z) + A(z) or 1 + D(z)G(z) has zeros inside the unit circle.

D z

D z

D z

D zk N

k r p

k

k

k r p

k

( )( ) = +

( )( ) = −

+( )α1

0 1 2 2for , , , ,K

<

>

0

0

N

N

odd

even

α α0

3 2

3 2

2

1

2

2

0 2 0 2 1

0 2 0 20 2

0 8 0 96 0 96 0 96

0 96 0 961= + +

− + −= − + +

( ) = ++

=– . – .

. ..

. .,

. .

. .

z z z

z z z

z

D z

z

z

H zD z G z

D z G z( ) =

( ) ( )+ ( ) ( )1

D z G zA z

B z( ) ( ) =

( )( )

H zA z

B z A z( ) =

( )( ) + ( )


Causality

A system is causal if h(n) = 0 for n < 0. From the properties of the Z-transform, H(z) is regular in theROC and at the infinity point. For rational functions the numerator polynomial has to be at most of thesame degree as the polynomial in the denominator.

The Paley-Wiener theorem provides the necessary and sufficient conditions that a frequency responsecharacteristic H(ω) must satisfy in order for the resulting filter to be causal.

Paley-Wiener Theorem

If h(n) has finite energy and h(n) = 0 for n < 0, then

Conversely, if �H(ω)� is square integrable and if the above integral is finite, then we can associate with�H(ω)� a phase response with ϕ(ω) so that the resulting filter with frequency response

H(ω ) = � H(ω )�e j ϕ (ω )

is causal.The relationship between the real and imaginary parts of an absolutely summable, causal, and real

sequence is given by the relation

which is known as the discrete Hilbert transform.

Summary of Causality

1. H(ω) cannot be zero except at a finite set of points.2. �H(ω)� cannot be constant in any finite range of frequencies.3. The transition from pass band to stop band cannot be infinitely sharp.4. The real and imaginary parts of H(ω) are independent and are related by the discrete Hilbert

transform.5. �H(ω)� and ϕ(ω) cannot be chosen arbitrarily.

Frequency Characteristics

With input f(n) = e j ω n, the output is

(6.9.3)

where

H (e j ω) = H (z)� z = e jω = Hr (e j ω) + j Hi (e j ω) = A(ω) e j ϕ (ω) (6.9.4)

A(ω) = [H r2 (e j ω) + H i

2 (e j ω)]1/2 = amplitude response (6.9.5)

ϕ (ω) = tan–1[Hi (e j ω )/H r(e j ω )] = phase response (6.9.6)

−∫ ( ) < ∞π

πω ωln H d

H H di rωπ

λ ω λ λπ

π

( ) = − ( ) −−∫

1

2 2cot

y n h k e e h k e e H ek

j n k j n

k

j k j n j( ) = ( ) = ( ) = ( )=

∞−( )

=

∞−∑ ∑

0 0

ω ω ω ω ω


(6.9.7)

Because H(e jω) = H(e j(ω +2π k)) it implies that the frequency characteristics of discrete systems areperiodic with period 2π.

Z-Transform and Discrete Fourier Transform (DFT)

If x(n) has a finite duration of length N or less, the sequence can be recovered from its N-point DFT.Hence, its Z-transform is uniquely determined by its N-point DFT. Hence, we find

(6.9.8)

Set z = e j ω (evaluated on the unit circle) to find

(6.9.9)

X(ω ) is the Fourier transform of the finite-duration sequence in terms of its DFT.

6.10 Digital Filters

Infinite Impulse Response (IIR) Filters

A discrete, linear, and time invariant system can be described by a higher-order difference equation ofthe form

(6.10.1)

Taking the Z-transform of the above equation and solving for the ratio Y(z)/X(z), we obtain

(6.10.2)

The block diagram representation of (6.10.1), in the form of the following pair of equations:

(6.10.3)

τ ωϕ ω

ω ω

( ) = −( )

= − ( )

==

d

dz

d

dznH z

z ej

Re lgroup delay

characteristic

X z x n zN

X k e z

NX k e z

z

N

X k

e z

n

N

n

k

Nj kn N n

n

N

k

Nj k N

n N

n

N

j k Nk

N

( ) = ( ) = ( )

= ( ) ( ) = − ( )−

=

−−

=

−−

=

−

=

−−

−

=

−

−=

−

∑ ∑∑

∑ ∑

0

1

0

12

0

1

0

12 1

0

1

2 10

1

1 1

1

π

ππ

11

∑

X e Xe

N

X k

e

jj N

j k Nk

Nω

ω

ω πω( ) = ( ) = − ( )

−

−

− −( )=

−

∑˙1

12

0

1

y n a y n k b x n kk

k

N

k

k

N

( ) − −( ) = −( )= =

∑ ∑1 0

H zY z

X z

b z

a z

kk

k

M

kk

k

N( ) =( )( ) =

−

−

=

−

=

∑

∑0

1

1

υ n b x n kk

k

M

( ) = −( )=

∑0


(6.10.4)

is shown in Figure 6.10.1. Each appropriate rearrangement of the block diagram represents a differentcomputational algorithm for implementing the same system.

Figure 6.10.1 can be viewed as an implementation of H(z) through the decomposition

(6.10.5)

or through the pair of equations

(6.10.6)

FIGURE 6.10.1

y n a y n k nk

k

N

( ) = −( ) + ( )=

∑1

υ

H z H z H z

a z

b z

kk

k

N k

k

M

k( ) = ( ) ( ) =

−

−

=

=

−

∑∑2 1

1

0

1

1

V z H z X z b z X zk

k

M

k( ) = ( ) ( ) =

( )

=

−∑1

0


(6.10.7)

If we arrange (6.10.5), we can create the following two equations:

(6.10.8)

(6.10.9)

The last two equations are presented graphically in Figure 6.10.2 (M = N).The time domain of Figure 6.10.2 is the pair of equations

(6.10.10)

(6.10.11)

FIGURE 6.10.2

Y z H z V z

a z

V z

kk

k

N( ) = ( ) ( ) =

−

( )−

=∑

2

1

1

1

W z H z X z

a z

X z

kk

k

N( ) = ( ) ( ) =

−

( )−

=∑

2

1

1

1

Y z H z W z b z W zk

k

M

k( ) = ( ) ( ) =

( )

=

−∑1

1

w n a w n k x nk

k

N

( ) = −( ) + ( )=

∑1

y n b w n kk

k

M

( ) = −( )=

∑0


Because the two internal branches of Figure 6.10.2 are identical, they can be combined in one branch sothat Figure 6.10.3. Figure 6.10.1 represents the direct form I of the general Nth-order system and Figure6.10.3 is often referred to as the direct form II or canonical direct form implementation.

Finite Impulse Responses (FIR) Filters

For causal FIR systems, the difference equation describing such a system is given by

(6.10.12)

which is recognized as the discrete convolution of x(n) with the impulse response

(6.10.13)

The direct form I and direct form II structures are shown in Figures 6.10.4 and 6.10.5. Because of thechain of delay elements across the top of the diagram, this structure is also referred to as a tapped delayline structure or a transversal filter structure.

6.11 Linear, Time-Invariant, Discrete-Time, Dynamical Systems

The mathematical models describing dynamical systems are almost always of finite-order differenceequations. If we know the initial conditions at t = t0 , their behavior can be uniquely determined for t ≥t0 . To see how to develop a dynamic, let us consider the example below.

FIGURE 6.10.3

y n b x n kk

k

M

( ) = −( )=

∑0

h nb n Mn( ) =

=

0 1

0

, , ,K

otherwise


Example

Let a discrete system with input ν(n) and output y(n) be described by the difference equation

y (n) + 2y (n – 1) + y (n – 2) = υ (n) (6.11.1)

If y(n0 – 1) and y(n0 – 2) are the initial conditions for n > n0, then y(n) can be found recursively from(6.11.1). Let us take the pair y(n – 1) and y(n – 2) as the state of the system at time n. Let us call the vector

(6.11.2)

the state vector for the system. From the definition above, we obtain

x1(n + 1) = y (n + 1 – 2) = y (n – 1) (6.11.3)

and

x2(n + 1) = y (n) = υ (n) – y (n – 2) – 2y (n – 1) (6.11.4)

or

x2(n + 1) = υ (n) – x1(n) – 2x2(n) (6.11.5)

Equations (6.11.3) and (6.11.5) can be written in the form

(6.11.6)

FIGURE 6.10.4

FIGURE 6.10.5

x nx n

x n

y n

y n( ) =

( )( )

=−( )−( )

1

2

2

1

x n

x n

x n

x nn1

2

1

2

1

1

0 1

1 2

0

1

+( )+( )

=− −

( )( )

+

( )υ


or

(n + 1) = (n) + υ (n) (6.11.7)

But (11.4) can be written in the form

or

y (n) = + υ (n) (6.11.8)

Hence, the system can be described by vector-matrix difference equation (6.11.7) and an output equation(6.11.8) rather than by the second-order difference equation (6.11.1).

A time-invariant, linear, and discrete dynamic system is described by the state equation

(nT + T) = (nT) + υ (nT) (6.11.9)

and the output equation is of the form

(nT) = (nT) + υ (nT) (6.11.10)

where

When the input is identically zero, (6.11.9) reduces to

(nT + T) = (nT) (6.11.11)

so that

(nT + 2T) = (nT + T) = (nT) = (nT)

and so on. In general we have

(nT + kT) = (nT) (6.11.12)

The state transition matrix from n1T to n2T (n2 > n1) is given by

x A x B

y n n x n x nx n

x nn( ) = ( ) − ( ) − ( ) = − −[ ] ( )

( )

+ ( )υ υ1 21

2

2 1 2

C x

x A x B

y C x D

x nT N

nT M

y nT R

A N N

B N M

C R N

D R M

( ) =

( ) =

( ) =

= ×

= ×

= ×

= ×

-dimensional column vector



nonsingular matrix

matrix

matrix

matrix

υ

x A x

x A x A A x A x2

x A xk


(n2T, n1T) = (6.11.13)

This is a function only of the time difference n2T – n1T. Therefore, it is customary to name the matrix

(nT) = (6.11.14)

the state transition matrix with the understanding that n = n2 – n1. It follows that the system states attwo times, n2T and n1T, are related by the relation

(n2T) = (n2T, n1T) (n1T) (6.11.15)

when the input is zero. From (6.11.13) we obtain the following relationships:

(a) (nT, nT) = = identity matrix (6.11.16)

(b) (n2T, n1T) = (n1T, n2T) (6.11.17)

(c) (n3T, n2T) (n2T, n1T) = (n3T, n1T) (6.11.18)

If the input is not identically zero and x(nT) is known, then the progress (later states) of the systemcan be found recursively from (6.11.9). Proceeding with the recursion, we obtain

In general, for k > 0 we have the solution

(6.11.19)

From (6.11.15), when the input is zero, we obtain the relation

(6.11.20)

According to (6.11.19), the solution to the dynamic system when the input is not zero is given by

(6.11.21)

or

(6.11.22)

ϕ An n2 1–

ϕ An

x ϕ x

ϕ I

ϕ ϕ –1

ϕ ϕ ϕ

x nT T A x nT T B nT T

A A x nT A B nT B nT T

nT T nT x nT nT T nT T B nT B nT T

+( ) = +( ) + +( )= ( ) + ( ) + +( )= +( ) ( ) + + +( ) ( ) + +( )

2

2 2

υ

υ υ

ϕ ϕ υ υ, ,

x nT kT nT kT nT x nT nT kT iT T B iTi n

n k

+( ) = +( ) ( ) + + +( ) ( )=

+ −

∑ϕ ϕ υ, ,1

x n T n T n T x n T A x n Tn n

2 2 1 1 12 1( ) = −( ) ( ) = ( )−ϕ

x nT kT nT kT nT x nT n k i T B iTi n

n k

+( ) = + −( ) ( ) + + − −( ) )[ ] ( )=

+ −

∑ϕ ϕ υ1

1

x nT kT kT x nT n k i T B iT ki n

n k

+( ) = ( ) ( ) + + − −( ) )[ ] ( ) >=

+ −

∑ϕ ϕ υ1

1 0


To find the solution using the Z-transform method, we define the one-sided Z-transform of an R × Smatrix function as the R × S matrix

(6.11.23)

The elements of (z) are the transforms of the corresponding elements of . Taking the Z-trans-form of both sides of the state equation (6.11.9), we find

z (z) – z (0) = (z) + (z)

or

(z) = (z – )–1 z (0) + (z – )–1 (z) (6.11.24)

From the output equation (6.11.10), we see that

(z) = (z) + (z) (6.11.25)

The state of the system (nT) and its output can be found for n ≥ 0 by taking the inverseZ-transform of (6.11.24) and (6.11.25).

For a zero input, (6.11.24) becomes

(z) = (z – )–1 z (0) (6.11.26)

so that

(nT) = Z–1{(z – )–1z} (0) (6.11.27)

If we let n1 = 0 and n2 = n, then (6.11.20) becomes

(nT) = (nT) (0) = (0) (6.11.28)

Comparing (6.11.27) and (6.11.28) we observe that

(nT) = = Z–1{(z – )–1z} n ≥ 0 (6.11.29)

or equivalently,

(z) = Z{ } = (z – )–1z (6.11.30)

The Z-transform provides a straightforward method for calculating the state transition matrix.Next combine (6.11.30) and (6.11.24) to find

(z) = (z) (0) + (z)z–1 (z) (6.11.31)

By applying the convolution theorem and the fact that

Z–1{ (z)z–1} = (nT – T)u(nT – T) (6.11.32)

the inverse Z-transform of (6.11.31) is given by

f nT( )

F z f nT zn

n( ) = ( )=

∞−∑

0

F f nT( )

X x A X B V

X I A x I A B V

Y C X D V

x y nT( )

X I A x

x I A x

x ϕ x A xn

ϕ An I A

Φ An I A

X Φ x Φ B V

Φ ϕ


(6.11.33)

The above equation is identical to (6.11.22) with n = 0.The behavior of the system with zero input depends on the location of the poles of

(z) = (z – )–1z (6.11.34)

Because

(6.11.35)

where adj(·) denotes the regular adjoint in matrix theory, these poles can only occur at the roots of thepolynomial

D(z) = det(z – ) (6.11.36)

D(z) is known as the characteristic polynomial for (for the system) and its roots are known as thecharacteristic values or eigenvalues of . If all roots are inside the unit circle, the system is stable. Ifeven one root is outside the unit circle, the system is unstable.

Example

Consider the system

For this system we have

The characteristic polynomial is

Hence, we obtain (see [6.11.34])

x kT kT x k i T B iTi

k

( ) = ( ) ( ) + − −( ) )[ ] ( )=

−

∑ϕ ϕ υ0 10

1

Φ I A

zI AzI A

zI A–

–

det –( ) =

( )( )

−1 adj

I A

AA

x nT T

x nT T

x nT

x nTnT

y nTx nT

x nTnT

1

2

1

2

1

2

0 2

0 22 2

0

1

0 22 2

+( )+( )

=

( )( )

+

( )

( ) = [ ] ( )( )

+ ( )

.

.

υ

υ

A B C D=

=

= [ ] = [ ]0 2

0 22 2

0

10 22 2 1

., , . ,

D z zI Az

z

z

z

z z z z z z

( ) = −( ) =

−

=−

− −

= −( ) − = − − = −( ) +( )

det det.

det.

. . . .

0

0

0 2

0 22 2

2

0 22 2

2 0 44 2 0 44 2 2 0 22


Because D(z) has a root outside the unit circle at 2.2, the system is unstable. Taking the inverse transformwe find that

To check, set n = 0 to find (0) = and (T) = .Let (0) = and the input be, the unit impulse υ(nT) = δ(nT) so that V(z) = 1. Hence, according

to (6.11.31)

The inverse Z-transform gives

and the output is given by

6.12 Z-Transform and Random Processes

Power Spectral Densities

The Z-transform of the autocorrelation function Rxx(τ) = E{x(t + τ)x(t)} sampled uniformly at nT timesis given by

(6.12.1)

Φ zz

z z

z

z

z z

z z

z

z z

z

z z

z

z z

( ) =−( ) +( )

−

=

−( )−( ) +( ) −( ) +( )

−( ) +( ) −( ) +( )

2 2 0 2

2 2

0 22

2

2 2 0 2

2

2 2 0 2

0 22

2 2 0 2 2 2 0 2

2. . .

. . . .

.

. . . .

ϕ nT n

n n n n

n n n n( ) =

( ) + −( ) ( ) − −( )( ) − −( ) ( ) + −( )

≥

1

122 2

11

120 2

5

62 2

5

60 2

11

1202 2

11

1200 2

11

122 2

1

120 2

0. . . .

. . . .

ϕ I ϕ Ax 0

X z z z B V zz z

z

z

z z z

( ) = ( ) ( ) =−( ) +( )

−

=−( ) +( )

−Φ 1 1

2 2 0 2

2 2

0 22

0

1

1

2 2 0 2

2

. . .

. .

x nT n

n n

n n( ) =( ) − −( )

( ) − −( )

>

− −

5

6

2 2 0 2

1

22 2

1

20 2

0

1 1. .

. .

y nT C x nT D nT

n

nn n

( ) = ( ) + ( )

==

( ) − −( ) >

+ +

υ

1 05

122 2

5

120 2 0

1 1. .

S z R nT zxx xx

n

n( ) = ( )= −∞

∞−∑


where the Fourier transform of Rx x(τ) is designated by Sx x(ω). The sampled power spectral density forx(nT) is defined to be

(6.12.2)

However, from the sampling theorem we have

(6.12.3)

Because Sx x(ω) is real, nonnegative, and even, it follows from (6.12.3) that Sxx(e j ω T) is also real, nonne-gative, and even. If the envelope of Rxx(τ) decays exponentially for �τ � > 0, then the region of convergencefor Sx x(z) includes the unit circle. If Rx x(τ) has undamped periodic components the series in (6.12.2)converges in the distribution sense that contains impulse function.

The average power in x(nT) is

(6.12.4)

where C is a simple, closed contour lying in the region of convergence and the integration is taken incounterclockwise sense. If C is the unit circle, then

(6.12.5)

(6.12.6)

Sx y (z) is called the cross power spectral density for two jointly wide-sense stationary processes x(t) andy(t). It is defined by the relation

(6.12.7)

Because Rx y (nT) = Ry x(–nT) it follows that

Sx y (z) = Sy x (z –1) , Sx x(z) = Sx x(z –1) (6.12.8)

Equivalently, we have

Sx x(e j ω T) = Sx x(e – j ω T) (6.12.9)

If Sx x(z) is a rational polynomial, it can be factored in the form

S e S z R nT exxj T

xxz e

xx

n

j nT

j T

ω ωω( ) = ( ) = ( )

== −∞

∞−∑

S eT

S n Txxj T

xx

n

s sω ω ω ω π( ) = −( ) =

= −∞

∞

∑12,

E x nT Rj

S zdz

zxx xxC

2 01

2( ){ } = ( ) = ( )∫π

R S e dTxx

sxx

j Ts

s

s

01 2

2

2

( ) = ( ) =−∫ω

ω ω πω

ωω

S ed

dxxj T

s

ω ωω

ω( ) = average power in

S z R nT zx y x y

n

n( ) = ( )= −∞

∞−∑


(6.12.10)

where

Linear Discrete-Time Filters

Let Rxx(nT), Ryy(nT), and Rx y(nT) be known. Let two systems have transfer functions H1(z) and H2(z) ,respectively. The output of these filters, when the inputs are x(nT) and y (nT) (see Figure 6.12.1), are

(6.12.11)

(6.12.12)

Let n = n + m in (6.12.11), multiply by y (nT), and take the ensemble average to find

(6.12.13)

Hence, by taking the Z-transform we obtain

Sυ y(z) = H1(z)Sx y(z) (6.12.14)

FIGURE 6.12.1

S zN z

D zG z G zx x ( ) =

( )( ) = ( ) ( )−γ 2 1

G z

z

z

a z

b z

b a b

k

k

L

k

k

M

kk

k

L

kk

k

M

k k k k

( ) =

−( )

−( )=

> < <

−

=

−

=

−

=

−

=

∏

∏

∑

∑

1

1

0 1 1

1

1

1

1

0

0

2

α

β

γ α, , , and are real

υ nT h kT x nT kTk

( ) = ( ) −( )= −∞

∞

∑ 1

w nT h kT y nT kTk

( ) = ( ) −( )= −∞

∞

∑ 2

R mT h kT E x mT nT kT y nT

h kT R mT kT

y

k

k

xy

υ ( ) = ( ) + −( ) ( ){ }

= ( ) −( )

= −∞

∞

= −∞

∞

∑

∑

1

1


Similarly from (6.12.12) we obtain

(6.12.15)

and

Sυ w(z) = H2(z –1)Sυ y (z) (6.12.16)

From (6.12.14) and (6.12.16), we obtain

Sυ w(z) = H1(z)H2(z –1)Sx y (z) (6.12.17)

Also, for x(nT) = y(nT) and h1(nT) = h2(nT) = h(nT), (6.12.17) becomes

Sυ υ(z) = H (z) H(z –1)Sx x(z) (6.12.18)

and

(6.12.19)

Optimum Linear Filtering

Let y(nT) be an observed wide-sense stationary process and x(nT) be a desired wide-sense stationaryprocess. The process y(nT) could be the result of the desired signal x(nT) and a noise signal υ(nT). Itis desired to find a system with transfer function H(z) such that the error e(nT) = x(nT) – (nT) =x(nT) – Z–1{Y(z)H(z)} is minimized. Referring to Figure 6.12.2 and to (6.12.18), we can write

(6.12.20)

where a(nT) is taken as white noise (uncorrelated process). We, therefore, can write

Ra a (mT) = γ 2 δ (mT) (6.12.21)

The signal a(nT) is known as the innovation process associated with y(nT). From Figure 6.12.2, we obtain

(6.12.22)

FIGURE 6.12.2

R mT h kT R mT kTw

k

yυ υ( ) = ( ) +( )= −∞

∞

∑ 2

S e H e H e S e

H e S e

j T j T j Txx

j T

j Txx

j T

υυω ω ω ω

ω ω

( ) = ( ) ( ) ( )= ( ) ( )

−

2

x̂

S zH z H z

S zaa y y( ) = ( ) ( ) ( ) =−

1

1 11

2γ

x̂ nT g kT a nT kTk

( ) = ( ) −( )= −∞

∞

∑


The mean square error is given by

To minimize the error we must set the quantity in the brackets equal to zero. Hence,

and its Z-transform is

but from (6.12.17) (because υ(nT) = x(nT) implies that H1(z) = 1) we have

(6.12.23)

(6.12.24)

From Figure 6.12.2, the optimum filter is given by (see also [6.12.20])

(6.12.25)

E e nT E x nT g kT a nT kT

E x nT E g kT x nT a nT kT

k

k

2

2

2 2

( ){ } = ( ) − ( ) −( )

= ( ){ } − ( ) ( ) −( )

= −∞

∞

= −∞

∞

∑

∑

+ ( ) −( )

= ( ) − ( ) ( ) + ( )

= ( ) + ( ) −( )

−

= −∞

∞

= −∞

∞

= −∞

∞

= −∞

∞

∑

∑ ∑

∑

E g kT a nT kT

R g kT R kT g kT

R g kTR kT

R

k

x x

k

x a

k

x x

x a

k

2

2 2

2

2

0 2

01

γ

γγ γ xx a

k

kT2 ( )= −∞

∞

∑

g nT R nT nx a( ) = ( ) − ∞ < < ∞12γ

G z S zxa( ) = ( )12γ

S z H z S z S zS z

H zx y x a x a

x y( ) = ( ) ( ) ( ) =( )

( )−11

11

– or

G zS z

H z

x y( ) =( )

( )−

12

11γ

H zH z

G zS z

H z H z

S z

S z

x y x y

y y

( ) = ( ) ( ) =( )

( ) ( ) =( )( )−

1

12

1 11γ


The mean square error for an optimum filter is

(6.12.26)

Applying Parseval’s theorem in the above equation, we obtain

(6.12.27)

where C can be the unit circle.

6.13 Relationship Between the Laplace and Z-Transform

The one-sided Laplace transform and its inverse are given by the following two equations:

(6.13.1)

(6.13.2)

where σc is the abscissa of convergence.The Laplace transform of a sampled function

(6.13.3)

is given by

(6.13.4)

because

(6.13.5)

E e nT R R kTx x x a

k

2

2

201( ){ } = ( ) − ( )

= −∞

∞

∑γ

E e nTj

S z S z S zdz

z

jS z

S z S z

S z

dz

z

jS z H z S z

x x x a x aC

x x

x y x y

y yC

x x x y

2

2

1

1

1

1

2

1

1

2

1

2

( ){ } = ( ) − ( ) ( )

= ( ) −( ) ( )

( )

= ( ) − ( ) ( )[ ]

−

−

−

∫

∫

π γ

π

πdzdz

zC∫

F s f t f t e dt ss tc( ) = ( ){ } = ( ) { } >

∞−∫˙ Re�

0

σ

f t F sj

F s e ds cc j

c js t

c( ) = ( ){ } = ( ) >− ∞

+ ∞

∫�–1 1

2πσ

f t f t t nT f t t f nT t nTs

k

T

k

( ) = ( ) ( ) = ( ) ( ) = ( ) ( )= −∞

∞

= −∞

∞

∑ ∑δ δ– ˙ –comb

F s f t f nT es s

k

nT s( ) = ( ){ } = ( )= −∞

∞−∑˙ �

� δ δt nT t nT e dt es t s nT−( ){ } = −( ) =−∞

∞− −∫


From (6.13.4) we obtain

F(z) = Fs(s)� s = T–1� n z (6.13.6)

and, hence,

(6.13.7)

If the region of convergence for F(z) includes the unit circle, �z � = 1, then

(6.13.8)

(6.13.9)

The knowledge of Fs (s) in the strip –ωs /2 < ω ≤ ωs /2 determines Fs(s) for all s. The transformation z =es T maps this strip uniquely onto the complex z-plane. Therefore, F(z) contains all the information inFs (s) without redundancy. Letting σ = s + jω , then

z = eσ Te j ω T (6.13.10)

Because �z � = eσ T , we obtain

(6.13.11)

Therefore, we have the following correspondence between the s- and z-planes:

1. Points in the left half of the s-plane are mapped inside the unit circle in the z-plane.2. Points on the jω-axis are mapped onto the unit circle.3. Points in the right half of the s-plane are mapped outside the unit circle.4. Lines parallel to the jω-axis are mapped into circles with radius �z � = eσ T.5. Lines parallel to the σ-axis are mapped into rays of the form arg z = ω T radians from z = 0.6. The origin of the s-plane corresponds to z = 1.7. The σ-axis corresponds to the positive u = Re z-axis.8. As ω varies between –ωs /2 and ωs /2, arg z = ω T varies between –π and π radians.

Let f (t) and g(t) be causal functions with Laplace transforms F(s) and G(s) that converge absolutelyfor Re s > σ f and Re s > σ g , respectively; then

(6.13.12)

The contour is parallel to the imaginary axis in the complex p-plane with

σ = Re s > σ f + σ g and σ f < c < σ – σ g (6.13.13)

F z F s f t f t tz e

s s TT s( ) = ( ) = ( ){ } = ( ) ( ){ }=

˙ � � comb

F F z f nT esz e

n

j nT

j Tω

ω

ω( ) = ( ) = ( )=

= −∞

∞−∑

F s j F sTs s s s+( ) = ( ) = =ω ω π

periodic2

z =< <= => >

1 0

1 0

1 0

σσσ

� f t g tj

F p G s p dpc j

c j

( ) ( ){ } = ( ) −( )− ∞

+ ∞

∫1

2π


With this choice the poles G(s – p) lie to the right of the integration path.For causal f (t), its sampling form is given by

(6.13.14)

If

(6.13.15)

then its Laplace transform is

(6.13.16)

Because σg = 0, then (6.13.12) becomes

(6.13.17)

The distance p in Figure 6.13.1 is given by

p = c + Re j θ π /2 ≤ θ ≤ 3π /2 (6.13.18)

If the function F(p) is analytic for some �p � greater than a finite number R0 and has a zero at infinity,then in the limit as R → ∞ the integral along the path BDA is identically zero and the integral along thepath AEB averages to Fs (s). The contour C1 + C2 encloses all the poles of F (p). Because of theseassumptions, F(p) must have a Laurent series expansion of the form

FIGURE 6.13.1

f t f t t nT f t t f nT t nTs

n

T

n

( ) = ( ) −( ) = ( ) ( ) = ( ) −( )=

∞

=

∞

∑ ∑δ δ0 0

˙ comb

g t t t nTT

n

( ) = ( ) = −( )=

∞

∑comb ˙ δ0

G s g t ee

snT s

nT s( ) = ( ){ } = =

−>−

=

∞

−∑�0

1

10Re

F sj

F p

edp cs s p T f f

c j

c j

( ) =( )

−> < <

− −( )− ∞

+ ∞

∫1

2 1πσ σ σ σ,


(6.13.19)

Q(p) is analytic in this domain and

� Q(p) � < M < ∞ � p � > R0 (6.13.20)

Therefore, from (6.13.19)

(6.13.21)

From the initial value theorem

a–1 = f (0+) (6.13.22)

Applying Cauchy’s residue theorem to (6.13.17), we obtain

(6.13.23)

where {pk} are the poles of F(p) and σ = Re{s} > σ f .Introducing (6.13.22) and (6.13.19) into the above equation, it can be shown (see Jury, 1973)

(6.13.24)

By letting z = es T , the above equation becomes

(6.13.25)

Example

The Laplace transform of f (t) = tu(t) is 1/s2. The integrand �te–σ t e– j ω t � < ∞ for σ > 0 implies that theregion of convergence is Re{s} > 0. Because f (t) has a double pole at s = 0, (6.13.25) becomes

F pa

p

a

p

a

p

Q p

pp R( ) = + + = +

( )>− − −1 2 1

02 2L

a pF pp− →∞

= ( )1 lim

F sF p

e e j

F p

e edps pT sT

p p

R CpT sT

k

( ) =( )

−

−

( )−∑ ∫−

=→∞ −

Re limsk

1

1

2 12π

F sF p

e e

fs pT sT

p p k

( ) =( )

−

−

+( )∑ −

=

Resk

1

0

2

F z F sF p

e z

fz es s

Tn z pT

p p

T

k

f( ) = ( ) =( )

−

−

+( )>

= −

=

∑1 11

0

2lRe ,s

k

σ

F zp e z

d

dp

p

p e z

Tz

z

pT

p

pT

p

( ) =−( )

−

=−( ) =

−( )

−

=

−

=

−

−

Res1

1

0

2

1 1

2 1

0

2

2 1

0

1

12


Example

The Laplace transform of f(t) = e –a t u(t) is 1/(s + a). The ROC is Res > – a and from (6.13.25) we obtain

The inverse transform is

If we had proceeded to find the Z-transform from f(nT) = exp(–anT)u(nT), we would have found F(z)= 1/(1 – e –a T – z–1). Hence, to make a causal signal f (t) consistent with F (s) and the inversion formula,f (0) should be assigned the value f (0+)/2.

It is conventional in calculating with the Z-transform of causal signals to assign the value of f (0+) tof(0). With this convention the formula for calculating F(z) from F(s) reduces to

(6.13.26)

6.14 Relationship to the Fourier Transform

The sampled signal can be represented by

(6.14.1)

with corresponding Laplace and Fourier transforms

(6.14.2)

(6.14.3)

If we set z = esT in the definition of the Z-transform, we see that

(6.14.4)

If the region of convergence for F(z) includes the unit circle, �z � = 1, then

(6.14.5)

F zp a e z e zpT

p a

aT( ) =+( ) −( )

− =

−−

−

= −

− −Res

1

1

1

2

1

1

1

21 1

f nT n e u nTanT( ) = − ( ) + ( )−1

2δ

F zF p

e zz e

pT

p p

T

k

f( ) =( )

−

>∑ −

=

Re ,sk

1 1

σ

f t f nT t nTs

n

( ) = ( ) −( )= −∞

∞

∑ δ

F s f nT es

n

s nT( ) = ( )= −∞

∞−∑

F f nT es

n

j nTω ω( ) = ( )= −∞

∞−∑

F s F zsz es T( ) = ( )

=

F F zsz e j T

ωω( ) = ( )

=


Because Fs(s) is periodic with period ωs = 2π /T, we need only consider the strip –ωs /2 < ω ≤ ωs /2,which uniquely determines Fs(s) for all s. The transformation z = exp(sT) maps this strip uniquely ontothe complex z-plane so that F(z) contains all the information in Fs(s) without the redundancy.

References

R. A. Gabel and R. A. Roberts, Signals and Linear Systems, John Wiley & Sons, New York, 1980.H. Freeman, Discrete-Time Systems, John Wiley & Sons, New York, 1965.E. I. Jury, Theory and Application of the Z-Transform Method, Krieger Publishing Co., Melbourne, FL, 1973.A. D. Poularikas and S. Seeley, Signals and Systems, reprinted second edition, Krieger Publishing Co.,

Melbourne, FL, 1994.S. A. Tretter, Introduction to Discrete-Time Signal Processing, John Wiley & Sons, New York, 1976.R. Vich, Z-Transform Theory and Applications, D. Reidel Publishing Co., Boston, 1987.

Appendix: Tables

TABLE 1 Z-Transform Properties for Positive-Time Sequences


TABLE 1 Z-Transform Properties for Positive-Time Sequences (continued)


TABLE 1 Z-Transform Properties for Positive-Time Sequences (continued)

TABLE 2 Z-Transform Properties for Positive- and Negative-Time Sequences


TABLE 2 Z-Transform Properties for Positive- and Negative-Time Sequences (continued)


TABLE 3 Inverse Transforms of the Partial Fractions of F(z)

TABLE 4 Inverse Transforms of the Partial Fractions of Fi (z)a


TABLE 5 Z-Transform Pairsa


TABLE 5 Z-Transform Pairsa (continued)


chapter 6 - the z-transform

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