dsp-chap3 2010 april7 [¬Û®e¼Ò¦¡]cwlin/courses/dsp/notes/dsp... · 2010-04-07 · 1...

1

z-Transform

清大電機系林嘉文[email protected]

Chapter 3

The Direct z-Transform� A discrete-time version of Laplace Transform� For a given sequence x(n), its (direct) z-transform X(z) is

defined as:

where z = Re(z) + j Im(z) is a complex variable� For convenience, the z-transform of a signal x(n0 is

denoted by

� Whereas the relationship between x(n) and X(z) is indicated by

� The region of convergence (ROC) of X(z) is the set of all values of z of which X(z) attains a finite value

( ) ( ) n

n

X z x n z∞

−

=−∞

≡ ∑

{ }( ) ( )X z Z x n≡

2010/4/7 2Introduction to Digital Signal Processing

( ) ( )zx n X z←→

2

ROC of the z-Transform� Express the complex variable z in polar form as

Where , then X(z) can be expressed as

� In the ROC of X(z), |X(z)| < ∞. But

� Hence |X(z)| is finite if x(n)r−n is absolutely summable� The problem of finding the ROC for X(z) is equivalent to

determining the range of r for which x(n)r−n is absolutely summable

jz re θ=

( ) ( )jn j n

z ren

X z x n r eθθ

∞− −

==−∞

= ∑


and r z zθ= =�

( ) ( )

( ) ( )

n j n

n

n j n n

n n

X z x n r e

x n r e x n r

θ

θ

∞− −

=−∞

∞ ∞− − −

=−∞ =−∞

=

≤ =

∑

∑ ∑

ROC of the z-Transform


1

0

1 0

( ) ( ) ( )

( )( )

n n

n n

nn

n n

X z x n r x n r

x nx n r

r

− ∞− −

=−∞ =

∞ ∞

= =

≤ +

≤ − +

∑ ∑

∑ ∑

3

The Direct z-Transform

� Example – Determine the z-Transform X(z) of the causal sequence x[n] = αn

u(n) and its ROC

� The above power series converges to

� ROC is the annular region |z| > |α|

0

( ) ( )n n n n

n n

X z u n z zα α∞ ∞

− −

=−∞ =

= =∑ ∑

1for ,1

1)( 1

1<

−= −

− zz

zX αα


The Direct z-Transform

� Example – Consider the anti-causal sequence

� Its z-transform is given by

� ROC is the annular region |z| < |α|

1

1

11

10

11

( )

1

1 , for 1

1

n n l l

n l

l l

l

X z z z

zz z

z

zz

α α

αα α

α

αα

− ∞− −

=−∞ =

−∞− −

−=

−−

= − = −

= − = −−

= <−

∑ ∑

∑


0, 0( ) ( 1)

, 1n

n

nx n u n

nα

α≥

= − − − = − ≤ −

4

z-Transform

� Note: the z-Transforms of two sequences αnu(n) and −αnu(−n −1) are identical even though the two parent sequences are different

� Only way a unique sequence can be associated with a z-transform is by specifying its ROC


z-Transform

� Example – the finite energy sequence

has a DTFT given by

which converges in the mean-square sense� However, hLP[n] does not have a z-transform as it is not

absolutely summable for any value of r

sin( ) ,c

LP

nh n n

n

ωπ

= −∞ < < ∞

( )

≤<

≤≤=

πωωωωω

c

cjLP eH

,0

0,1


5

Characteristic Families of Signals with Their Corresponding ROCs


The Inverse z-Transform


� The z-transform X(z) is given by

� Multiply both side by zn−1 and integrate both sides over a closed contour within the ROC of X(z)

� Interchange the order of integration and summation:

� According to the Cauchy integral theorem

( ) ( ) k

k

X z x k z∞

−

=−∞

= ∑

1 1( ) ( )n n k

C Ck

X z z dz x k z dz∞

− − −

=−∞

= ∑∫ ∫� �

1 1( ) ( )n n k

C Ck

X z z dz x k z dz∞

− − −

=−∞

= ∑∫ ∫� �

1 1,1

0,2n k

C

k nz dz

k njπ− − =

= ≠

∫�11

( ) ( )2

n

Cx n X z z dz

jπ−= ∫�

6

The Inverse z-Transform (from IDTFT)


� General Expression : Recall that, for z = rejω, the z-transform X(z) given by

is merely the DTFT of the modified sequence x(n)r−n

� Accordingly, the inverse DTFT is thus given by

� By making a change of variable z = rejω, the previous equation can be converted into a contour integral given by

where C′ is a counterclockwise contour of integration defined by |z| = r

( ) ( ) ( )n n j n

n nX z x n z x n r e ω∞ ∞− − −

=−∞ =−∞= =∑ ∑

( )1( )

2n j j nx n r X re e d

π ω ω

πω

π−

−= ∫

11( ) ( )

2n

Cx n X z z dz

jπ−= ∫�

Properties of the z-Transform

� Linearity - If

and

then

� Example – Determine the z-Transform and the ROC of

Ans:


1 1( ) ( )zx n X z←→

2 2( ) ( )zx n X z←→

1 1 2 2 1 1 2 2( ) ( ) ( ) ( ) ( ) ( )zx n a x n a x n X z a X z a X z= + ←→ = +

( ) ( )( ) 3 2 4 3 ( )n nx n u n = −

1 1 1

1( ) 2 ( ) ( ) , ROC: 2

1 2znx n u n X z z

z−= ←→ = >

−

2 2 1

1( ) 3 ( ) ( ) , ROC: 3

1 3znx n u n X z z

z−= ←→ = >

−

1 1

3 4( ) , ROC: 3

1 2 1 3X z z

z z− −= − >

− −

7


� Example - Consider the two-sided sequence x[n] = αnu[n] − bnu[−n −1]

� Let x[n] = αnu[n] and y[n] = − bnu[−n −1] with X1(z) and X2(z) denoting, respectively, their z-transforms

� Now

and

� Using the linearity property we arrive at

� The ROC of X(z) is given by the overlap regions of |z| > |α| and |z| < |b|

1 1

1( ) ,

1X z z

zα

α −= >

−

2 1

1( ) ,

1X z z b

bz−= <

−

1 2 1 1

1 1( ) ( ) ( )

1 1X z X z X z

z bzα − −= + = +− −




8


� Example - Determine the z-transform and its ROC of the causal sequence

� We can express x[n] = v[n] + v*[n] where

� The z-transform of v[n] is given by

� Using the conjugation property we obtain the z-transform of v*[n] as

01 1

[ ] [ ] [ ]2 2

j n nv n e u n u nω α= =

01 1

1 1 1 1( ) , 1

2 1 2 1 jV z zz e zω α

α − −= ⋅ = ⋅ > =− −

01 1

1 1 1 1*( *) , 1

2 1 * 2 1 jV z z

z e zω αα −− −= ⋅ = ⋅ > =

− −


( )0[ ] cos ( )x n n u nω=


� Using the linearity property we get

or,

� Therefore, we obtain

Similarly

0 01 1

1 1 1( ) ( ) *( *)

2 1 1j jX z V z V z

e z e zω ω−− −

= + = + − −

( )( )

10

1 20

1 cos( ) , 1

1 2cos

zX z z

z z

ωω

−

− −

−= >

− +


( ) ( )( )

10

0 1 20

1 cos( ) cos ( ) ( ) , ROC: 1

1 2 cosz z

x n n u n X z zz z

ωω

ω

−

− −

−= ←→ = >

− +

( ) ( )( )

10

0 1 20

sin( ) sin ( ) ( ) , ROC: 1

1 2 cosz z

x n n u n X z zz z

ωω

ω

−

− −= ←→ = >− +

9



� Time Shifting - If

then

And the ROC remains unchanged except for z = 0 if k > 0 and z = ∞ if k < 0

� Example – Determine the z-Transform of the signal

Ans:

( ) ( )zx n X z←→

( ) ( )z kx n k z X z−− ←→

1, 0 1( )

0, elsewhere

n Nx n

≤ ≤ −=

( ) ( ) ( )x n u n u n N= − −

{ } { } ( ) 1

1( ) ( ) ( ) 1 , ROC: 1

1NX z Z u n Z u n N z z

z−

−= − − = − >−



� Scaling in the z-domain -If

then

Fro any constant a

� Proof

1 2( ) ( ), ROC: zx n X z r z r←→ < <

{ }

( )

( )

1

1

( ) ( )

( )

n n n

n

n

n

Z a x n a x n z

x n a z

X a z

∞−

=−∞

∞ −−

=−∞

−

=

=

=

∑

∑

11 2( ) ( ), ROC: zna x n X a z a r z a r−←→ < <

11 2ROC: r a z r−< < 1 2ROC: a r z a r< <or

10



� Example - Determine the z-transform and its ROC of the causal sequence

and

� Ans:

By applying the scaling property

and similarly

( ) ( )( )

10

0 1 2 20

1 cos( ) cos ( ) ( ) , ROC:

1 2 coszn r z

x n r n u n X z z rr z r z

ωω

ω

−

− −

−= ←→ = >

− +

( )0( ) cos ( )nx n r n u nω=

( )0( ) sin ( )nx n r n u nω=

( ) ( )( )

10

0 1 2 20

sin( ) sin ( ) ( ) , ROC:

1 2 coszn r z

x n r n u n X z z rr z r z

ωω

ω

−

− −= ←→ = >− +



� Time Reversal -If

then

� Proof

� Example -

1 2( ) ( ), ROC: zx n X z r z r←→ < <

{ } ( ) ( )1 1( ) ( ) ( )ln

n l

Z x n x n z x l z X z∞ ∞ −− − −

=−∞ =−∞

− = − = =∑ ∑

1

2 1

1 1( ) ( ), ROC: zx n X z z

r r−− ←→ < <

11 2ROC: r z r−< <

2 1

1 1ROC: z

r r< <or

( ) ( )x n u n= −

1

1( ) , ROC: 1

1zu n z

z−←→ >

−1

( ) , ROC: 11

zu n zz

− ←→ <−

11



� Differentiation in the z-Domain -If

then

ROC remains unchanged

� Example – Find the z-Transform of

� Example – Find the inverse z-transform of

( ) ( )zx n X z←→

( )( ) z dX z

nx n zdz

←→−

( ) ( )nx n na u n=

( )1( ) log 1 , ROC :X z az z a−= + >



� Convolution of Two Sequences -If

then

The ROC is the intersection of that for X1(z) and X2(z)

� Proof

1 1( ) ( )zx n X z←→

1 2 1 2( ) ( ) ( ) ( ) ( ) ( )zx n x n x n X z X z X z= ∗ ←→ =

1 2( ) ( ) ( )k

x n x k x n k∞

=−∞

= −∑

1 2( ) ( ) ( ) ( )n n

n n k

X z x n z x k x n k z∞ ∞ ∞

− −

=−∞ =−∞ =−∞

= = −

∑ ∑ ∑

2 2( ) ( )zx n X z←→and

Intercahnge the order of the summations and apply the time shifting property

1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( ) ( )n k

k n k

X z x k x n k z x k z X z X z X z∞ ∞ ∞

− −

=−∞ =−∞ =−∞

= − = =

∑ ∑ ∑

12



� Correlation of Two Sequences -If

then

The ROC is at least the intersection of that for X1(z) andX2(z-1)

� Proof

� Example -

1 1( ) ( )zx n X z←→

1 2 1 2

11 2 1 2( ) ( ) ( ) ( ) ( ) ( )z

x x x xn

r l x n x n l R z X z X z∞

−

=−∞

= − ←→ =∑

1 2 1 2( ) ( ) ( )x xr l x l x l= ∗ −

{ } { } 11 2 1 2( ) ( ) ( ) ( ) ( )X z Z x l Z x l X z X z−= − =

2 2( ) ( )zx n X z←→and

( ) ( )nx n a u n=

( )1 2 1 1 2

1 1 1 1( ) , ROC:

1 1 1x xR z a z

az az aa z z a− −= = < <

− − − + +



� Multiplication of Two Sequences -If

then

where C is the closed contour that encloses the origina and lies withing the ROC common to both X1(v) and X2(v-1)

� Proof

1 1( ) ( )zx n X z←→

11 2 1 2

1( ) ( ) ( ) ( ) ( )

2z

C

zx n x n x n X z X v X v dv

j vπ− = ←→ =

∫�

1 2( ) ( ) ( ) n

n

X z x n x n z∞

−

=−∞

= ∑

11 1

1( ) ( )

2n

Cx n X v v dv

jπ−= ∫�

2 2( ) ( )zx n X z←→and

11 2

1( ) ( ) ( )

2

n

Cn

zX z X v x n v dv

j vπ

−∞−

=−∞

=

∑∫�

Since

2

zX

v

13



� Parseval’s relation -If

then

� The Initial Value Theorem -

� If x(n) is causal, then

� Proof -

1 1( ) ( )zx n X z←→

* * 11 2 1 2 *

1 1( ) ( ) ( )

2 Cn

x n x n X v X v dvj vπ

∞−

=−∞

=

∑ ∫�

2 2( ) ( )zx n X z←→and

(0) lim ( )z

x X z→∞

=

1 2

0

( ) ( ) (0) (1) (2)n

n

x z x n z x x z x z∞

− − −

=

= = + + +∑ �



14

Common z-Transform Paris


Rational z-Transforms

� In the case of LTI discrete-time systems we are concerned with in this course, all pertinent z-transforms are rational functions of z−1

� That is, they are ratios of two polynomials in z−1

� The degree of the numerator polynomial B(z) is M and the degree of the denominator polynomial A(z) is N

� An alternate representation of a rational z-transform is as a ratio of two polynomials in z:

10 1 0

10 1

0

( )( )

( )

M kMkM k

NN kN kk

b zb b z b zB zX z

A z a a z a z a z

−− −=

− − −

=

+ + += = =

+ + +∑∑

�

�

10 1 0 0

10 1 0 0

( / ) /( )

( / ) /

M M MM

N N NN

b z z b b z b bX z

a z z a a z a a

− −

− −

+ + +=

+ + +

�

�


15

Rational z-Transform

� A rational z-transform can be alternately written in factored form as

� At a root z = zk of the numerator polynomial X(zk) = 0. These values of z are known as the zeros of G(z)

� At a root z = pk of the denominator polynomial X(pk) → ∞. These values of z are known as the poles of G(z)

� There are |N – M| zeros (if N > M) or poles (if N < M) at z = 0

( )( ) ( )( )( ) ( )

( )

( )

1 20

0 1 2

1

1

( ) MM N

N

M

kM N k

N

kk

z z z z z zbX z z

a z p z p z p

z zGz

z p

− +

− + =

=

− − −=

− − −

−=

−

∏

∏

�

�


0

0

bG

a=


� Example – the z-transform

has a zero at z1 = 0 and a pole at p1 = a

( ) 1

1, ROC:

1

zX z z a

az z a−= = >

− −


Pole-zero plot

16


� Example – Determine the pole-zero plot for the signal

Since a > 0 zM = aM has M roots at

The zero z0 = a cancels the pole at z = a. Thus

( ) ( ) ( )( )

111

1 10

1

1

MM MM n

Mn

az z aX z az

az z z a

−−−

− −=

− −= = =

− −∑


Pole-zero plot

, 0 1( )

0, elsewhere

na n Mx n

≤ ≤ −=

0a >

2 / 0,1, , 1j k Mkz ae k Mπ= = −…

( ) ( )( ) ( )1 2 11

MM

z z z z z zX z

z−

−

− − −=

�


� Example – Determine the z-transform and the signal corresponds to the pole-zero plot

� There are 2 zeros (M = 2) at z1 =0 and z2 = rcosω0 and 2 poles (N = 2) at and


Pole-zero plot

( )1

01 2 2

0

1 cos, ROC:

1 2 cos

rzX z G z r

rz r z

ωω

−

− −

−= >

− +

( )0( ) cos ( )nx n G r n u nω=

( ) ( )( )( )( )

( )( )( )0 0

1 2

1 2

0cos, ROC:

j j

z z z zX z G

z p z p

z zG z r

z re z reω ω

ω−

− −=

− −

−= >

− −

01

jp re ω= 02

jp re ω−=

17


� A physical interpretation of the concepts of poles and zeros can be given by plotting the log-magnitude 20log10|X(z)| for

• The magnitude plot exhibits very large peaks around the poles of X(z) (z = 0.4 ± j 0.6928)

• It also exhibits very narrow and deep wells around the location of the zeros (z = 1.2 ± j 1.2)

1 2

1 2

1 2.4 2.88( )

1 0.8 0.64

z zX Z

z z

− −

− −

− +=

− +


Behavior of a Single Real-Pole Causal Signal


1

1( ) ( ) ( ) , ROC:

1znx n a u n X z z a

az−= ←→ = >

−

18

Behavior of a Double Real-Pole Causal Signal


( )21

1( ) ( ) ( ) , ROC:

1

znx n na u n X z z aaz−

= ←→ = >−

Behavior of a Causal Signal with a Pair of Complex-Conjugate Poles


( )0( ) cos ( )nx n r n u nω=

19

Behavior of a Causal Signal with a Double Pair of Complex-Conjugate Poles


( )0( ) cos ( )x n n n u nω=

The System Function of an LTI System

� According to the convolution property, the I/O-relationship can be expressed as

� The system function (impulse/unit-sample response) H(z) is

� If an LTI system is described by the following constant-coefficient difference equation

� Taking the z-Transform at both sides


( ) ( ) ( )Y z H z X z=

( )( )

( )

Y zH z

X z=

( ) ( ) ( )1 0

N M

k kk k

y n a y n k b x n k= =

= − − + −∑ ∑

1 0

( ) ( ) ( )N M

k kk k

k k

Y z a Y z z b X z z− −

= =

= − +∑ ∑

20


� Therefore an LTI system described by a constant-coefficient difference equation has a rational system function

� Two important special forms of pole-zero systems:� If ak = 0 for 1 ≤ k ≤ N

� If bk = 0 for 1 ≤ k ≤ M


1 0

( ) 1 ( )N M

k kk k

k k

Y z a z X z b z− −

= =

+ =

∑ ∑

0

1

( )( )

( ) 1

Mk

kk

Nk

kk

b zY z

H zX z

a z

−

=

−

=

= =+

∑

∑

0 0

1( )

M Mk M k

k kMk k

H z b z b zz

− −

= =

= =∑ ∑

0

1 0

1( )

1

N

N Nk N k

k kk k

b zH z

a z a z− −

= =

= =+∑ ∑

(all-zero system)

(all-pole system)

(pole-zero systems)



� Example – A causal LTI IIR filter is described by a constant coefficient difference equation given by

� y(n) = x(n −1) −1.2 x(n − 2) + x(n − 3) +1.3 y(n −1) −1.04 y(n − 2) + 0.222 y(n − 3)

� Its transfer function is therefore given by

� Alternate forms:

� Note: Poles farthest from z = 0 have a magnitude� ROC:

321

321

222.004.13.11

2.1)( −−−

−−−

−+−+−

=zzz

zzzzH

( )( )( )( )( )7.05.07.05.03.0

8.06.08.06.0222.004.13.1

12.1)(

23

2

jzjzz

jzjzzzz

zzzH

−−+−−−−+−

=

−+−+−

=

0.74

0.74z >

21

MATLAB Functions for Rational z-Transform

� In general, if the rational z-transform has N poles with R distinct magnitudes, then it has R + 1 ROCs, and R + 1 distinct sequences having the same z-transform

� Hence, a rational z-transform with a specified ROC has a unique sequence as its inverse z-transform

� MATLAB [z,p,k] = tf2zp(num,den) determines the zeros, poles, and the gain constant of a rational z-transform with the numerator coefficients specified by num and the denominator coefficients specified by den

� [num,den] = zp2tf(z,p,k) implements the reverse process� The factored form of the z-transform can be obtained using

sos = zp2sos(z,p,k)


MATLAB Functions for Rational z-Transform

� The pole-zero plot is determined using the function zplane� The z-transform can be either described in terms of its

zeros and poles: zplane(zeros,poles) or, in terms of its numerator and denominator coefficients zplane(num,den)

� Example – The pole-zero plot of

obtained using MATLAB

4 3 2

4 3 2

2 16 44 56 32( )

3 3 15 18 12

z z z zX z

z z z z

+ + + +=

+ − + −


22

Inversion of the z-Transform


� The inverse z-transform of X(z) is given by

� There are three methods for the evaluation of the inverse z-transform in practice:1. Direct evaluation by the contour integration2. Power series expansion3. Partial-fraction expansion and table lookup

11( ) ( )

2n

Cx n X z z dz

jπ−= ∫�

Inverse z-Transform by Contour Integration


� Cauchy’s integral theorem . Let f(z) be a function of the complex variable z and C be a closed path in the z-plane. If the derivative df(z)/dz exists on and inside the contour C and if f(z) has no poles at z = z0, then

� More generally, if the (k+1)-order derivative of f(z) exist and f(z) has no poles at z = z0, then

0 0

00

( ), if is inside 1 ( )0, if is outside 2 C

f z z Cf zdz

z Cj z zπ

= −

∫�

( )( )

0

1

01

0

0

1 ( ), if is inside 1 ( ) 1 !

20, if is outside

k

k

k z zC

d f zz Cf z k dzdz

j z zz C

π

−

−

=

−= −

∫�

23



� Suppose the integrand of the contour integral is a proper fraction f(z)/g(z), where f(z) has no poles inside the contour C and g(z) is a polynomial with distinct roots inside C. Then

� In the case of the inverse z-transform, we have

1

1 1

1 ( ) 1

2 ( ) 2

1

2

ni

C Ci i

n ni

iCi ii

Af zdz dz

j g z j z z

Adz A

j z z

π π

π

=

= =

= −

= =−

∑∫ ∫

∑ ∑∫

� �

� ( ) ( )

( ) ii i z z

f zA z z

g z == −

{ }

( )

1

1

all poles inside

1

1( ) ( )

2

residue of ( ) at

( )

i

i

n

C

ni

z C

ni z z

i

x n X z z dzj

X z z z z

z z X z z

π−

−

−

=

=

= =

= −

∫

∑

∑

�



� Example – Evaluate the inverse z-transform of

Using the complex inversion integral� Sol:

(1) If n ≥ 0, f(z) has only zeros and hence no poles inside C. The only pole inside C is z = a. Hence

(2) If n < 0, f(z) has an nth-order poles at z = 0 which is also inside C. If n = 1, we have

If n = 2, we have

x(n) = 0 for n < 0

1

1( ) ,

1X z z a

az−= >

−

0( ) ( ) , 0nx n f z a n= = ≥

1

1

1 1( )

2 1 2

n n

C C

z zx n dz dz

j az j z aπ π

−

−= =

− −∫ ∫� �

0

1 1 1 1( 1) 0

2 ( )Cz z a

x dzj z z a z a zπ = =

− = = + =− −∫�

2 20

1 1 1 1( 2) 0

2 ( )Cz az

dx dz

j z z a dz z a zπ ==

− = = + = − − ∫�

( ) ( )nx n a u n=

24

Inverse z-Transform by Power Series Expansion


� Expand X(z) into a power series as

which converges in the given ROC. Then, by the uniqueness of the z-transform, x(n) = cn for all n.

� Example : Determine the inverse z-transform of

If (a) ROC: |z| > 1, (b) ROC: |z| < 0.5Sol: (a)

( ) nn

n

X z c z∞

−

=−∞

= ∑

1 2

1( )

1 1.5 0.5X z

z z− −=

− +

1 2 3 4

1 2

1 3 7 15 31( ) 1

3 1 2 4 8 1612 2

X z z z z zz z

− − − −

− −= = + + + + +

− +�

3 7 15 31( ) 1, , , , ,

2 4 8 16X n

=

…

Inverse z-Transform by Partial-Fraction Expansion

� A rational X(z) can be expressed as

� X(z) is called improper if M ≥ N. It can be re-expressed as

where the degree of B1(z) is less than N� The rational function B1(z)/A(z) is called a proper fraction� Example – Consider

By long division we arrive at

10 0 1

110

( )( )

( ) 1

M i Mii M

N NiNii

b z b b z b zB zX z

B z a z a za z

− − −=

− −−

=

+ + += = =

+ + +∑∑

�

�

11 10 1

0

( ) ( )( )

( ) ( )

M NM N

M N

B z B zX z c z c c z c z

A z A z

−− − −

−=

= + = + + + +∑ �

�

�

�

1 2 3

1 2

2 0.8 0.5 0.3( )

1 0.8 0.2

z z zX z

z z

− − −

− −

+ + +=

+ +

11

1 2

5.5 2.1( ) 3.5 1.5

1 0.8 0.2

zX z z

z z

−−

− −

+= − + +

+ +2010/4/7 48Introduction to Digital Signal Processing

25


� Simple Poles : In most practical cases, the rational z-transform of interest X(z) is a proper fraction with simple poles

� Let the poles of X(z) be at z = pk 1 ≤ k ≤ N� A partial-fraction expansion of X(z) is then of the form

� The constants in the partial-fraction expansion are called the residues and are given by

1 21 1 1 1

1 1 2

( )1 1 1 1

NN

N

AA A AX z

p z p z p z p z− − − −=

= = + + − − − − ∑ �

� �

�

( )11 ( ) z pA p z X z−== −��



� x(n) = Z-1{X(z)} can be obtained by inverting each term as:

� If x(n) is causal, the ROC is |z| > pmax where pmax = max{|p1|, |p2|, …, |pN|}, and x(n) becomes

� The above approach with a slight modification can also be used to determine the inverse of a rational z-transform of a non-causal sequence

( )

( )

11

if ROC: ( ),

(causal sequence)1

1 if ROC: ( 1)

(anticausal sequence)

n kk

nk kk

z pp u n

Zp z z p

p u n

−−

>

=

− < − − −

( )1 1 2 2( ) ( )n n nN Nx n A p A p A p u n= + + +�


26


� Example – Determine the inverse z-transform of

If (a) ROC: |z| > 1, (b) ROC: |z| < 0.5, and (c) 0.5 < |z| < 1� Ans:

(a) ROC: |z| > 1

(b) ROC: |z| < 0.5

(b) ROC: 0.5 < |z| < 1

( )( )1 2 1 1

1 1( )

1 1.5 0.5 1 1 0.5X z

z z z z− − − −= =

− + − −

1 21 1

( )1 1 0.5

A AX z

z z− −= +

− −


( ) ( )1 11 1 2 0.51 ( ) 2, 1 0.5 ( ) 1z zA z H z A z H z− −

= == − = = − = −

( )( ) 2 0.5 ( )nx n u n= −

( )( ) 2 0.5 ( 1)nx n u n= − + − −

( )( ) 2 ( 1) 0.5 ( )n

x n u n u n= − − − −


� Multiple Poles : If X(z) has multiple poles, the partial-fraction expansion is of slightly different form

� Let the pole at z = ν be of multiplicity L and the remaining N − L poles be simple and at z = λ, 1 ≤ l ≤ N − L

� Then the partial-fraction expansion of G(z) is of the form

where the constants are computed using

� The residues ρl are calculated as before

( )∑ ∑∑−

= =−−

−

=

−

−+

−+=

LN L

ii

iNM

vzzzzG

1 111

0 11)(

� �

�

�

�

�

γλρ

η

( ) ( )( ) ( )[ ] LizGvz

zd

d

viLvz

L

iL

iL

iLi ≤≤−−−

= =−

−−

−

− 1,1)(!

1 11

γ


27


Decomposition of Rational z-Transforms

� Suppose we have a rational z-transform X(z) expressed as

� If M ≥ N (i.e., X(z) is improper), we convert X(z) to a sum of a polynomial and a proper function

� If the roots are distinct, it can be expanded in partial fractions as

( )

( )

1

0 10

11

1

1( )

1 1

M

M k kkk k

N Nkkk k

k

z zb z

X z ba z p z

−−

= =

−−

=

=

−= =

+ −

∏∑∑ ∏

( ) ( ) ( )1 21 1 11 2

1 1 1( )

1 1 1N

N

X z A A Ap z p z p z− − −

= + + +− − −

�

pr0

( ) ( )M N

kk

k

X z c z X z−

−

=

= +∑



� If there some complex conjugate pairs of poles, we can group and combine the pairs as

� The partial fraction expansion becomes

where N = K1 + 2K2

( ) ( )( )

1 10 1

1 1 1 21 21 21 1 11 ( )

A A Ap A p z b b zA A

pz p z a z a zp p z pp z

∗ ∗ ∗ − −∗

− ∗ − − −∗ − ∗ −

+ + − − ++ = =

− − + ++ − − +

1 2 10 1

1 1 20 1 1 1 2

( )1 1

K KM Nk k k k

kk k kk k k

b b b zX z c z

a z a z a z

−−−

− − −= = =

+= + +

+ + +∑ ∑ ∑

28



� An alternative form is obtained by expressing X(z) as a product of simple terms. To avoid complex coefficients in the decomposition, the complex-conjugate poles and zeros should be combined as

� Assuming for simplicity that M = N, X(z) can be expressed as

where N = K1 + 2K2

1 21 1 21 2

0 1 1 21 1 1 2

1 1( )

1 1

K Kk k k

k kk k k

b z b z b zX z b

a z a z a z

− − −

− − −= =

+ + += ⋅ ⋅

+ + +∏ ∏

( )( )( )( )

1 1 1 21 2

1 21 11 2

1 1 1

11 1

k k k k

k kk k

z z z z b z b z

a z a zp z p z

− ∗ − − −

− −− ∗ −

− − + +=

+ +− −


LTI Discrete-Time Systems in the Transform Domain

� An LTI discrete-time system is completely characterized in the time-domain by its impulse response sequence {h(n)}

� Thus, the transform-domain representation of a discrete-time signal can also be equally applied to the transform-domain representation of an LTI discrete-time system

� Besides providing additional insight into the behavior of LTI systems, it is easier to design and implement these systems in the transform-domain for certain applications

� We consider now the use of the DTFT and the z-transform in developing the transform-domain representations of an LTI system

29


LTI Discrete-Time Systems in the Transform Domain

� Consider LTI discrete-time systems characterized by linear constant coefficient difference equations of the form

� Applying the z-transform to both sides of the difference equation and making use of the linearity and the time-invariance properties we arrive at

� A more convenient form of the z-domain representation of the difference equation is given by

( ) ( )0 0

N M

k kk k

a y n k b x n k= =

− = −∑ ∑

0 0

( ) ( )N M

k kk k

k k

a z Y z b z X z− −

= =

=∑ ∑

0 0

( ) ( )N M

k kk k

k k

a z Y z b z X z− −

= =

=

∑ ∑


Frequency Response from Transfer Function

� If the ROC of the transfer function H(z) includes the unit circle, then the frequency response H(ejω) of the LTI digital filter can be obtained simply as follows:

� For a real coefficient transfer function H(z) it can be shown that

� For a stable rational transfer function in the form

� the factored form of the frequency response is given

ωω

jez

j zHeH=

= )()(

( ) ( ) ( ) ( ) ( ) ( ) ( )2

1* j

j j j j j

z eH e H e H e H e H e H z H z ω

ω ω ω ω ω− −

== = =

( )( )

0 1

01( )

M

kN M kN

kk

z zbH z z

a z p

− =

=

−=

−

∏∏

( ) ( ) ( )( )

10

01

M jkj N Mj k

N jkk

e zbH e e

a e p

ωωω

ω

− =

=

−=

−

∏∏

30


Geometric Interpretation of Frequency Response Computation

� It is convenient to visualize the contributions of the zero factor (z − zk) and the pole factor (z − pk) from the factored form of the frequency response

� The magnitude function is given by

which reduces to

� The phase response for a rational transfer function is of the form

( ) ( ) 10

01

M jkj N Mj k

N jkk

e zbH e e

a e p

ωωω

ω

− =

=

−=

−

∏∏

( ) 10

01

M jkj k

N jkk

e zbH e

a e p

ωω

ω

=

=

−=

−

∏∏

( ) ( ) ( ) ( ) ( )0 01 1

arg arg / arg argM N

j j jk k

k k

H e b a N M e z e pω ω ωω= =

= + − + − − −∑ ∑



� The magnitude-squared function of a real-coefficient transfer function can be computed using

� The factored form of the frequency response

is convenient to develop a geometric interpretation of the frequency response computation from the pole-zero plot as ω varies from 0 to 2π on the unit circle

� The geometric interpretation can be used to obtain a sketch of the response as a function of the frequency

( ) ( )( )( )( )

2 *2

10

*0

1

M j jk kj k

N j jk kk

e z e zbH e

a e p e p

ω ωω

ω ω

−

=

−

=

− −=

− −

∏∏

( ) ( ) ( )( )

10

01

M jkj N Mj k

N jkk

e zbH e e

a e p

ωωω

ω

− =

=

−=

−

∏∏

31


� A typical factor in the factored form of the frequency response is given by

(ejω − ρejϕ)where ρejϕ is a zero (pole) if it is zero (pole) factor

� As shown below in the z-plane the factor (ejω − ρejϕ)represents a vector starting at the point z = ρejϕ and ending on the unit circle at z = ejω

� As ω is varied from 0 to 2π, the tip of the vector moves counter-clockwise from the point z = 1 tracing the unit circle and back to the point z = 1




� As indicated by

the magnitude response |H(ejω)|at a specific value of ω is given by the product of the magnitudes of all zero vectors divided by the product of the magnitudes of all pole vectors

� Likewise, from

we observe that the phase response at a specific value of ω is obtained by adding the phase of the term b0/a0 and the linear-phase term ω(N − M) to the sum of the angles of the zero vectors minus the angles of the pole vectors

( ) 10

01

M jkj k

N jkk

e zbH e

a e p

ωω

ω

=

=

−=

−

∏∏

( ) ( ) ( ) ( ) ( )0 0 1 1arg arg / arg arg

M Nj j jk kk k

H e b a N M e z e pω ω ωω= =

= + − + − − −∑ ∑

32

3-D View of Transfer Function & Frequency Response


UNIT CIRCLE

EVALUATE H(z)EVERYWHERE 1− z −2

1 +0.7225z−2

Flying Thru z-Plane


POLES CAUSEPEAKS in H(z)

H(ejωωωω)

H(z)

33


Response of Systems with Rational System function

� Assume that the input signal x(n) has a rational z-transform

� For a initially relaxed system, the output becomes

� The output y(n) can be subdivided into two parts. The first part is a function of {pk}, called the natural response of the system. The second is a function of {qk}, called the forced response

( )( )

( )

N zX z

Q z=

( ) ( )

nr fr

1 1

the natral response ( ) the forced response ( )

( ) ( ) ( )N L

n n

k k k kk k

y n y n

y n A p u n Q q u n= =

= +∑ ∑��

1 11 1

( ) ( )( ) ( ) ( )

( ) ( )

1 1

N Lk k

k kk k

B z N zY z H z X z

A z Q z

A Q

p z q z− −= =

= =

= +− −∑ ∑

Transient & Steady-State Responses

� The zero-state response of a system to a given input can be separate into two components: the natural responseand the forced response

where {pk}, k = 1, 2, …, N are the poles of the system.� If |pk| < 1 for all k, then ynr(n) decays to zero as n

approaches infinity. In such a case, we refer to the natural response of the system as the transient response

� The forced response of the system has the form

� where {pk}, k = 1, 2, …, N are the poles in the forcing function


( )nr1

( ) ( )N

n

k kk

y n A p u n=

=∑

( )fr1

( ) ( )L

n

k kk

y n Q q u n=

=∑

34

Transient & Steady-State Responses

� When the causal input signal is a sinusoid, the forced response is also a sinusoid. In such case the forced output is called steady-state response of the system

� Example – Determine the transient and steady-state response of the following system (assuming initially rest)

where the input is � Sol


( ) 0.5 ( 1) ( )y n y n x n= − +

( ) 10cos( / 4) ( )x n n u nπ=

1

1( )

1 0.5H z

z−=

−( )( )1

1 2

10 1 1/ 2( )

1 2

zX z

z z

−

− −

−=

− +

( )( )( )( )( ) ( ) ( ) ( )

128.7 28.7

1 /4 1 /4 1 1 /4 1 /4 1

10 1 1/ 2 6.3 6.78 6.78( )

1 0.5 1 1 1 0.5 1 1

j j

j j j j

z e eY z

z e z e z z e z e zπ π π π

−−

− − − − − − − −

−= = + +

− − − − − −

nr ( ) 6.3(0.5) ( )ny n u n= ( )fr ( ) 13.56cos / 4 28.7 ( )y n n u nπ= −

Causality and Stability

� As defined previously, a causal LTI system is one whose unit sample response h(n) satisfies the condition

h(n) = 0, n < 0 � The ROC of a causal sequence is the exterior of a circle� A linear LTI system is causal if and only if the ROC of the

system function is the exterior of a circle of radius r < ∞, including the point z = ∞

� Recall that a necessary and sufficient condition for a LTI system to be BIBO stable is

� In turn, this condition implies that H(z) must contain the unit circle within its ROC


( )n

S h n∞

=−∞

= < ∞∑

35


� Since

� When evaluated on the unit circle (i.e., |z| = 1)

� Hence, if the system is BIBO stable, the unit circle is contained in the ROC of H(z), vice versa

� Therefore, a LTI system is BIBO stable if and only if the ROC of the system function includes the unit circle

� Note that causality and stability are different and that one does not imply the other

� A causal LTI system is BIBO stable if and only if all the poles of H(z) are inside the unit circle (why?)


( ) ( ) ( ) ( )n n n

n n n

H z h n z h n z h n z∞ ∞ ∞

− − −

=−∞ =−∞ =−∞

= ≤ =∑ ∑ ∑

( ) ( )n

H z h n∞

=−∞

≤ ∑


� Example – A linear LTI system is characterized by the system function

Specify the ROC of H(z) and determine h(n) for the following conditions:

(a) The system is stable.

(b) The system is causal.

(c) The system is anticausal.


1

1 2 1 1

3 4 1 2( )

1 3.5 1.5 1 0.5 1 3

zH z

z z z z

−

− − − −

−= = +

− + − −

36

Pole-Zero Cancelation

� Pole-zero cancelation: When a z-transform has pole that is the same location as a zero, the pole is canceled by the zero and, consequently, the term containing that pole in the inverse z-transform vanishes

� Pole-zero cancelations can occur either in the system function H(z) itself (a reduced-order system) or in the product of H(z)X(z)

� Properly selecting the position of the zeros of the input signal is possible to suppress one or more modes of the system function, or vice versa

� When the zero is located very near the pole but not exactly at the same location, the term in the response has very small amplitude. The nonexact pole-zero cancellation can occur in practice due to insufficient numerical precision


Pole-Zero Cancelation

� Example – Determine the response of the system

to the input signal Sol:

The system function is

� And the z-transform of the input is


1 2 1 1

1 1( )

5 1 1 11 1 16 6 2 3

H zz z z z− − − −

= = − + − −

1( ) ( 1) ( ) 3 ( 1)

2y n y n x n x n= − + − −

1( ) ( ) ( 1)

3x n n nδ δ= − −

11( ) 1

3X z z−= −

1

1( ) ( ) ( )

11

2

Y z H z X zz−

= =−

1( ) ( )

2

n

y n u n =

37

Stability Condition in Terms of the Pole Locations

� In addition, for a stable and causal digital filter for which h[n] is a right-sided sequence, the ROC will include the unit circle and entire z-plane including the point z = ∞

� An FIR digital filter with bounded impulse response is always stable

� On the other hand, an IIR filter may be unstable if not designed properly

� In addition, an originally stable IIR filter characterized by infinite precision coefficients may become unstable when coefficients get quantized due to implementation


Stability of Second-Order Systems

� Consider a causal two-pole system described by the second-order difference equation,

� The system function is

� This system has two zeros at the origin and poles at

� The system is BIBO stable is the poles lie inside the unit circle, that is, if |p1| < 1 and |p2| < 1

� Thus, if the ROC includes the unit circle |z| = 1, then the system is stable, and vice versa

1 2 0( ) ( 1) ( 2) ( )y n a y n a y n b x n= − − − − +


20 0

1 2 2 11 2 1 2

( )( )

( ) 1

b b zY zH z

X z a z a z z a z a− −= = =

+ + + +

21 1 2

1 2

4,

2 4

a a ap p

−= − ±

38


� The coefficients are related to the roots by

� For stability, the following conditions must be satisfied

� Therefore a two-pole system is stable if and only if the two coefficients satisfy the two conditions.

( )1 1 2 2 1 2, a p p a p p= − + =


2 1 2 1 21, 1a p p a a= < < +



39





40


� Example – Consider the causal LTI IIR transfer function:

� The plot of the impulse response is shown below

� As can be seen from the above plot, the impulse response coefficient h(n) decays rapidly to zero value as n increases

� The absolute summability condition of h(n) is satisfied, ⇒ H(z) is a stable transfer function

21 850586.0845.11

1)( −− +−=

zzzH


Stability of Second-Order Systems� Now, consider the case when the transfer function coef. are

rounded to values with 2 digits after the decimal point:

� A plot of the impulse response of is shown below

� In this case, the impulse response coefficient increases rapidly to a constant value as n increases

� Hence, is an unstable transfer function

21 85.085.11

1)( −−

∧

+−=

zzzH


�[ ]h n

� ( )h n

� ( )h n

41

Stability Condition in Terms of the Pole Locations

� Example : The factored form of

is

which has a real pole at z = 0.902 and a pole at z = 0.943� Since both poles are inside the unit circle H(z) is BIBO stable� Example : The factored form of

is

which has a pole at z = 1 and the other inside the unit circle� Since one pole is not inside the unit circle, H(z) is not BIBO

stable

21 850586.0845.01

1)(

−− +−=

zzzH

( )( )11 943.01902.01

1)( −− −−=

zzzH

( )21 85.085.11

1−−

∧

+−=

zzzH

( ) ( )( )11 85.011

1−−

∧

−−=

zzzH


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