Chapter 5 (Maths 3)

7/28/2019 Chapter 5 (Maths 3)

1/25

1

Chapter 5

Z-Transforms

Introduction:In Communication Engineering, two basic types of signals are encountered.

They are

(1)Continuous time signals.(2)Discrete time signals.Continuous time signals are defined for continuous values of the independent variable,

namely time and are denoted by a function .

Discrete time signals are defined only at discrete set of values of the independent

variable and are denoted by a sequence .

Z-transform plays an important role in analysis of linear discrete time signals.

Definition of z-transform:

If is a sequence defined for .,then is called

the two-sided or bilateral Z-transform of and denoted by or

,where z is a complex variable in general.

If is a casual sequence, i.e if , for n


2/25

2

(i)

(ii)

Proof:

Extending this result, we get

(3)Frequency Shifting:

(i)

(ii)

Proof:

Similarly (ii) can be proved.

Corollary:

If , then


3/25

3

The result follows, if we replace in (ii).

(4)Time Reversal for Bilateral Z-Transform:

If

Proof:

(5) Differentiation in the Z-Domain:

(i)

(ii)

Proof:

(i)

Similarly, (ii) can be proved.

(6) Initial Value Theorem:(i)

(ii)

Proof:


4/25

4

Similarly, (ii) can be proved.

(7) Final value Theorem:

(i)

(ii)

Proof:

Taking limits as z tends to 1,

Similarly, (ii) can be proved, starting with property 2(ii).

(8) Convolution Theorem:

Definitions:

The convolution of the two sequences is defined as

(i) , if the sequence are non causal and

(ii) , if the sequences are causal.

The convolution of two functions is defined as

where T is the sampling period.


5/25

5

Statement of the theorem:

(i)

(ii)

Proof:

(For the bilateral z=transform)

(i)

By changing the order of summation,

, by putting n-r=m

(ii)

(1)

Say, where

(2)

Using (2) in (1), we get


6/25

6

Z-Transforms of some basic functions:(1) is the unit impulse sequence defined by

(2) Where k is a constant and is the unit step sequence

defined by

(i)

Where the region of convergence (ROC) is .

(ii) In particular,

and

(3)

(i) , where the ROC is .

(ii) , where the ROC is .

(iii) .

(iv) .

(v)

(4) .

(i)


7/25

7

(ii) Where the ROC is .

(iii)

(iv)

(v)

(5) .

(i)

(ii)

(6) .

(i) .

(ii) .

(7) .

(i) .

(ii) Putting a=1, we get .

(8) .

(i) .

(ii) .

(iii) .

In particular,


8/25

8

.

(iv) .

In particular,

.

(9) .

(i) .

(ii) .

(iii) .

(iv) .

Problems:

(1)Find the bilateral Z-transforms of(i)

(ii)(iii)

Solution:

(i)

By property 3,


9/25

9

(ii)

By property 3, which is true for bilateral Z-transform also.

s

(iii) by property5.

(2)Find the Z-transforms of(i) and

(ii)

Solution:

(i)

(ii)


10/25

10

(3) Find Z-transforms of

(i) , and

(ii)

Solution:

(i)

[Refer to basic transform (6)]

(ii) ,by partial fractions.



11/25

11

(4) Find the Z-transforms of

(i)

(ii) , and

(iii)

Solution:

(i) Let


(ii) Let


12/25

12

By basic transform (8)

(iii) Let

(5) (i) Use initial value theorem to find , when

(ii) Use final value theorem to find , when

Solution:

(i) By initial value theorem,

(ii) By final value theorem,


13/25

13

(6) Use convolution theorem to find the sum of the first n natural numbers.

Solution:

By convolution theorem,

Taking inverse Z-transforms,

(7) Use convolution theorem to find the inverse Z-transform of

(i) and

(ii)

Solution:

(i)


14/25

14

(ii)

Inverse Z-transforms:

The inverse of Z-transform of has been already defined as

, when .

can be found out by any one of the following methods.

Method 1 (Expansion method)


15/25

15

If can be expanded in a series of ascending powers of , i.e in the

form , by binomial, exponential and logarithmic theorems, the coefficient of

in the expansion gives .

Method 2 (Long division method)

When the usual methods of expansion of fail and if ,

then is divided by in the classical manner and hence the expansion

is obtained in the quotient.

Method 3 (partial fraction Method)

When is a rational function in which the denominator can be factorised,

is resolved in to partial fraction and then is derived as the sum of the

inverse Z-transforms of the partial fractions.

Method 4 (By Cauchys Residue Theorem)

By using the relation between the Z-transform and Fourier transform of asequence, it can be proved that

Where C is a circle whose centre is the origin and radius is sufficiently large to include

all the isolated singularities of .

By Cauchys residue theorem,

x sum of the residues of at the isolated singularities.

Sum of the residues of at theisolated singularities.

Use of Z-transforms to solve Finite Difference equations:

Z-transforms can be used to solve finite difference equation of the form

with given values of y(0) and y(1).

Taking Z-transforms on both sides of the given difference equation and using the values

of y(0) and y(1), we will get . Then

will give .

To express and in terms of .

(i) .

(ii) .


16/25

16

Problems:

(1)Find the inverse Z-transform of , by the long division method.Solution:

Thus

,

(2)Find the inverse Z-transform of , by the long division method.Solution:

Thus ,


17/25

17

(3)Find by the method of partial fractions.Solution:

Let

(4)Find , by using Residue theorem.Solution:

, Where C is the circle whose

centre is the origin and which includes the singularities .

,by Cauchys residue

theorem.are simple poles.

(5)Find , by using Residue theorem


18/25

18

Solution:

By residue theorem,

the residue of at the only triple pole (z=3).

a

(6) Solve the difference equation given that

.

Solution:

Taking Z-transforms on both sides of the given equation, we have

.

.

.

Inverting, we get .

(7) Solve the equation given thatSolution:

Taking Z-transforms of the given equation,

.


19/25

19

Inverting, we get

.


20/25

20

UNIT 5

PART A(1)Form the difference equation from

Ans:

(2)Express in terms of

Ans:

(3)Find the value of when

Ans:

(4)Define bilateral Z-transform.

Ans : If is a sequence defined for .,then is called the

two-sided or bilateral Z-transform of and denoted by or ,where z is

a complex variable in general.

(5)Find the z-transform of

Ans:

(6)Find using z-transform.

Ans:


21/25

21

(7)Define unilateral Z-transform.Ans:

If is a casual sequence, i.e if , for n


22/25

22

(11)Find the Z-transforms of

Ans:

(12)Use convolution theorem to find the inverse Z-transform of

Ans:


23/25

23

(13) Define Inverse Z-transforms:Ans:

The inverse of Z-transform of is defined as ,

When .

(14) Use final value theorem to find , when

Ans:

By final value theorem,

(15) Find the Z-transforms of

Ans: Let


Part B

(1) (a)Prove that .

Find


24/25

24

(b)Find using Convolution theorem.

(2) (a) Prove that

(b) Solve: using

z- transform.

(3)(a) Find the z-transform of

(b)Using Convolution theorem find

(4)(a) Solve the difference equation

(b) Find by the method of partial fractions.

(5)(a) Find by the method of partial fractions.

(b) Solve the difference equation

(6)(a) Prove that

(b) State and prove the second shifting theorem in z-transform.

(7)(a) Using Convolution theorems evaluate inverse z-transform of .

(b) Solve the difference equation

(8)(a) Find the inverse Z-transform of , by the long division method.

(b) Find Z-transforms of

(i) , and


25/25

(ii)

(9) (a) Solve: using

z- transform.

(b) Find

(10)(a) Find the bilateral Z-transforms of

(i)

(ii)

(iii)

(b) Solve the equation given that

(11)(a)Use convolution theorem to find the sum of the first n natural numbers.

(b) Find the inverse Z-transform of , by the long division method.

(12)(i) Use initial value theorem to find , when

(ii) Use final value theorem to find , when

(13) (a) Find , by using Residue theorem.

(b) Find the Z-transforms of

(ii) , and (iii)

Documents

Chapter 5 (Maths 3)