Bài giảng Xử lý tín hiệu số: Chapter 5 - Hà Hoàng Kha

Bài giảng Xử lý tín hiệu số - Chapter 5: z-Transform has contents: z-Transform, properties of the z-transform, causality and stability, inverse z-transform,.... and other contents.

™ The z-transform is a tool for analysis, design and implementation of discrete time signals and LTI systems

discrete-time signals and LTI systems

™ Convolution in time-domain ⇔ multiplication in the z-domain

1 z-transform

2 Properties of the z-transform

3 Causality and Stability

4 Inverse z-transform

1 The z-transform

™ The z-transform of a discrete-time signal x(n) is defined as the power series:

+ +

+ +

− +

1 ( )

0 ( )

1 ( )

2 ( )

( )

( )

(

| C

z n x z

X z

−∞

=

n

™ The z-transform of impulse response h(n) is called the transform

function of the filter:

function of the filter:

z n h z

z-transform and ROC

™ It is possible for two different signal x(n) to have the same

z-transform Such signals can be distinguished in the z-domain by their g g y

™ z-transforms:

and their ROCs:

ROC of a causal signal is the

t i f i l

ROC of an anticausal signal

is the interior of a circle exterior of a circle. is the interior of a circle.

™ Determine the z-transform of the signal

) 1 (

) ( )

(n = a u n +b u −n −

x n n

™ The ROC of two-sided signal is a ring (annular region).

2 Properties of the z-transform

™ Linearity:

1 1

x ← ⎯→z

if

2 2

2 1

2 Properties of the z-transform

™ Time shifting:

) ( )

x ← ⎯→z

if

) ( )

x − ← ⎯→z −D

then

™ The ROC of z−D X (z) is the same as that of X(z) except for z=0 if

™ The ROC of is the same as that of X(z) except for z=0 if D>0 and z=∞ if D<0

)

(z

X z

Example: Determine the z transform of the signal x(n)=2nu(n 1)

Example: Determine the z-transform of the signal x(n)=2nu(n-1)

™ Scaling in the z-domain:

2

1 | | :

ROC )

( )

x ← ⎯→z ≤ ≤

if

2 1

( )

( )

x

for any constant a, real or complex

Example: Determine the z-transform of the signal x(n)=ancos(w n)u(n)Example: Determine the z transform of the signal x(n) a cos(w n)u(n)

2 Properties of the z-transform

( )

(

1 2

|

| r

1 : ROC )

( )

(

r

z z

X n

x − ← ⎯→z − ≤ ≤

Example: Determine the z-transform of the signal x(n)=u(-n)

™ Convolution of two sequence:

if and x1(n) ← ⎯→z X1(z) x2(n) ← ⎯→z X2(z)

) ( )

( )

( )

( )

( )

(n x1 n x2 n X z X1 z X2 z

x = ∗ ← ⎯→z =

then

the ROC is, at least, the intersection of that for X, , 11(z) and X( ) 22(z).( )

Example: Compute the convolution of x=[1 1 3 0 2 1] and h=[1, -2, 1] ?

2 Properties of the z-transform

™ Differentiation in the z-domain

if x(n) ← ⎯→z X (z)

then

) ( )

(

dz

z

dX z n

nx( ) ← ⎯→z − ( )

Example: Determine the z transform of the signal x(n)=nanu(n)

the ROCs of both are the same

Example: Determine the z-transform of the signal x(n)=nanu(n)

3 Causality and stability

+ +

) (n A p u n A p u n

will have z-transform

+ +

) (n A1p1u n A2 p2u n x

|

| max

|

| ROC )

1 1

)

2

2 1

p z

the ROC of causal signals are g outside of the circle

|

| min

|

| ROC )

1 1

)

2

1 1

i

z z

p z

3 Causality and stability

™ Mixed signals have ROCs that are the annular region between two circles

™ It can be shown that a necessary and sufficient condition for the

™ It can be shown that a necessary and sufficient condition for the

stability of a signal x(n) is that its ROC contains the unit circle

4 Inverse z-transform

ROC

), ( )

(n z transform X z

x ⎯ ⎯ − ⎯ ⎯ →

) ( ROC

), (z inversez-transform x n

ROC ),

( )

(n X z

x ← ⎯→z

™ In inverting a z-transform, it is convenient to break it into its partial

ROC ),

( )

(n X z

fraction (PF) expression form, i.e., into a sum of individual pole

terms whose inverse z transforms are known

1

™ Note that with we have

⎧ n ( ) if ROC| |>| |( l i l )

1 -

az - 1

1 )

(anticausa |

a

|

| z

| ROC if

) 1 (

signals) (causal

| a

|

| z

| ROC if

)

( )

(

n u a

n u

a n

n

Partial fraction expression method

™ In general, the z-transform is of the form

N

N z b z

b b

z N z

™ The poles are defined as the solutions of D(z)=0 There will be M poles say at p p p Then we can write

M

M z a z

a z

D

z

+ +

=

0

1 )

(

) (

poles, say at p1, p2,…,pM Then, we can write

) 1

( ) 1

)(

1 ( ) (z = − p1z−1 − p2z−1 − p z−1

™ If N < M and all M poles are single poles

where

Example od

™ Compute all possible inverse z-transform of

Solution:

- Find the poles: 1-0.25z-2 =0 Æ p1=0.5, p2=-0.5

- We have N=1 and M=2, i.e., N < M Thus, we can writeWe have N 1 and M 2, i.e., N M Thus, we can write

where

Example od

Partial fraction expression method

™ If N=M

Where and for i=1,…,M

™ If N> M

Example od

™ Compute all possible inverse z-transform of

Solution:

- Find the poles: 1-0.25z-2 =0 Æ p1=0.5, p2=-0.5

- We have N=2 and M=2, i.e., N = M Thus, we can writeWe have N 2 and M 2, i.e., N M Thus, we can write

where

Example od

Example od

™ Determine the causal inverse z-transform of

Solution:

- We have N=5 and M=2, i.e., N > M Thus, we have to divide the

denominator into the numerator, giving

Partial fraction expression method

complex-valued poles of X(z) must come in complex-conjugate pairs

Considering the causal case, we have

Writing A1 and p1 in their polar form, say, with B11 and R11 > 0, and thus, we have , ,

As a result, the signal in time-domain is

Example od

™ Determine the causal inverse z-transform of

Solution:

Example od

™ Problems: 5.2, 5.3, 5.4, 5.6, 5.6, 5.8, 5.16