1 Introduction
2 Linear Difference Equations
3 Z-Transforms And Its Properties
4 Inverse Z-Transforms
5 Convolution Theorem
6 Applications Of Z-Transforms To Difference Equations
7 Formation Of Difference Equations

**Z-Transforms AND
DIFFERENCE EQUATIONS**

1 INTRODUCTION

2 LINEAR DIFFERENCE EQUATIONS

3 Z-TRANSFORMS AND ITS PROPERTIES

4 INVERSE Z-TRANSFORMS

5 CONVOLUTION THEOREM

6 APPLICATIONS OF Z-TRANSFORMS TO
DIFFERENCE EQUATIONS

7 FORMATION OF DIFFERENCE EQUATIONS

**1 Introduction**

The Z-transform plays a vital role in the field of communication
Engineering and control Engineering, especially in digital signal processing.
Laplace transform and Fourier transform are the most effective tools in the
study of continuous time signals, where as Z –transform is used in discrete
time signal analysis. The application of Z – transform in discrete analysis is
similar to that of the Laplace transform in continuous systems. Moreover,
Z-transform has many properties similar to those of the Laplace transform. But,
the main difference is Z-transform operates only on sequences of the discrete
integer-valued arguments. This chapter gives concrete ideas about Z-transforms
and their properties. The last section applies Z-transforms to the solution of
difference equations.

**Difference Equations**

Difference
equations arise naturally in all situations in which sequential relation exists
at various discrete values of the independent variables. These equations may be
thought of as the discrete counterparts of the differential equations.
Z-transform is a very useful tool to solve these equations.

A **difference
equation** is a relation between the independent variable, the dependent
variable and the successive differences of the dependent variable.

are difference equations.

The differences Dy_{n}, D^{2}y_{n}, etc can
also be expressed as.

Dy_{n}
= y_{n+1} - y_{n},

D^{2}y_{n} = y_{n+2} - 2y_{n+1} + y_{n}.

D^{3}y_{n} = y_{n+3} - 3y_{n+2} + 3y_{n+1}
- y_{n} and so on.

Substituting these in (i)
and (ii), the equations take the form

Note that the above
equations are free of Ds¢.

If a difference equation
is written in the form free of Ds,¢then the **order** of difference equation is the difference between
the highest and lowest subscripts of y‟s occurring in it. For example, the
order of equation (iii) is 2 and equation (iv) is 1.

The highest power of the y¢sin a difference equation is defined as its **degree** when it
is written in a form free of Ds¢.For example, the degree of the equations y_{n+3} + 5y_{n+2}
+ y_{n} = n^{2} + n + 1 is 3 and y^{3} _{n+3} +
2y_{n+1} y_{n} = 5 is 2.

**2 Linear Difference
Equations**

A **linear difference
equation with constant coefficients** is of the form

a_{0} y_{n+r} + a_{1} y_{n+r -1}
+ a_{2} y_{n+r -2} + . . . . +a_{r}y_{n} = f(n).

i.e., (a_{0}E^{r} + a_{1}E^{r-1}
+ a_{2} E^{r-2} + . . . . + a_{r})y_{n} = f(n) ------(1)

where a_{0,}a_{1},
a_{2}, . . . . . a_{r}
are constants and f(n) are known functions of n.

The
equation (1) can be expressed in symbolic form as

f(E) y_{n}
= f(n)
----------(2)

If f(n) is zero, then equation (2) reduces to

f (E) y_{n}
= 0 ----------(3)

which is
known as the **homogeneous difference equation** corresponding to (2).The
solution of (2) consists of two parts, namely, the complementary function and
the particular integral.

The
solution of equation (3) which involves as many arbitrary constants as the
order of the equation is called the **complementary function**. The **particular
integral** is a particular solution of equation(1) and it is a function of „n‟ without any arbitrary constants.

Thus the complete solution of (1) is given by y_{n} = C.F + P.I.

**Example 1**

Form the difference
equation for the Fibonacci sequence .

The integers
0,1,1,2,3,5,8,13,21, . . . are said to form a Fibonacci sequence.

If y_{n} be the n^{th}
term of this sequence, then

y_{n} = y_{n-1} + y_{n-2} for n > 2

or y_{n+2} - y_{n+1}
- y_{n} = 0 for n > 0

**3 Z - Transforms and its
Properties**

**Definition**

Let {fn} be a sequence defined for n = 0,1,2,…….,then its Z-transform F(z) is defined as

whenever the series
converges and it depends on the sequence {f_{n}}.

The inverse Z-transform of
F(z) is given by Z^{-1}{F(z)} = {f_{n}}.

**Note: **If {f_{n}}
is defined for
n = 0,
± 1, ±
2, …….,

**Properties of Z-Transforms**

1.
The Z-transform is linear.

i.e, if F(z) = Z{f_{n}} and G(z) = Z{g_{n}},
then

Z{af_{n} + bg_{n}} = aF(z) + bG(z).

4. If Z{f_{n}}
= F(z), then

Z{f_{n+k}}=
z^{k}{ F(z) –f_{0} –(f _{1} / z ) - … - ( f_{k-1}
/ z^{k-1}) } (k > 0)

**Proof**

In Particular,

(i) Z{f _{n+1}} = z {F(z) - f_{0}}

(ii) Z{f _{n+2}}= z^{2} { F(z) –f_{0} –(f_{1}/z)
}

**Corollary**

If Z{f_{n}} =
F(z), then Z{f_{n–k}} = z^{-k} F(z).

**(5) Initial value Theorem**

**Proof**

We know that F (z) = f_{0} + f_{1} z^{-1}
+ f_{2}z^{-2} + . . .

Taking
limits as z ® ¥on both sides, we get

**(6) Final
value Theorem**

**SOME STANDARD RESULTS**

1.
Z{a^{n}} = z /
(z-a), for |z| > |a|.

**Proof**

By definition,
we have

In particular, we have

Z{1} = z / (z-1), (taking a = 1).

and Z{(-1)^{n}} = z / (z +1),
(taking a = -1).

2. Z{na^{n}} = az /(z-a)^{2}

**Proof: **By property, we have

Similarly, we can prove

Z{n^{2}a^{n}}
= {az(z+a)}/ (z-a)^{3}

Equating
the Real and Imaginary parts, we get

Equating the Real and Imaginary parts, we get

**Table of Z –Transforms**

**Example 2**

Find the
Z–transform of

1.
n(n-1)

2.
n^{2} + 7n + 4

3.
(1/2)( n+1)(n+2)

**Example 3 **

Find the
Z- transforms of 1/n and 1/n(n+1)

= - log (1 –1/z ) if |1/z|
< 1

= - log (z-1 / z)

=
log (z/z-1), if | z |
>1.

**Example 4**

Find the Z- transforms of

(i) cos np/2

(ii) sin np/2

**Example 5**

Show that
Z{1/ n!} = e^{1/z} and hence find Z{1/ (n+1)!} and Z{1/ (n+2)!}

We know that Z{f_{n+1}}
= z { F(z) –f_{0}}

Therefore,

**Example 6**

Find the
Z- transforms of the following

**4
Inverse Z –Transforms**

The inverse Z –transforms can be obtained by using any one of
the following methods.They are

I.Power
series method

II. Partial
fraction method

III. Inversion
Integral method

IV. Long division method

**I. Power
series method**

This is
the simplest method of finding inverse Z –transform. Here F(z) can the be
expanded in a series of ascending powers
and the coefficient of z ^{–n} will be the of z desired inverse
Z- transform.

**Example 8**

**II. Partial Fraction
Method**

Here, F(z) is resolved into partial fractions and the inverse
transform can be taken directly.

**Example 9**

Find the inverse Z –transform of

**Inversion Integral Method or Residue Method**

The
inverse Z-transform of F (z) is given by the formula

Sum of
residues of F(z).z^{n-1} at the poles of F(z) inside the contour C
which is drawn according to the given Region of convergence.

**Example
12**

Using the
inversion integral method, find the inverse Z-transform of

Its poles
are z = 1,2 which are simple poles.

By
inversion integral method, we have

\Sum of Residues = -3 + 3.2^{n} = 3 (2^{n}-1).

Thus the
required inverse Z-transform is

f_{n}
= 3(2^{n}-1), n =
0, 1, 2, …

**Example 13**

Find the
inverse z-transform of

The pole
of F(z) is z = 1, which is a pole of order 3. By Residue method, we have

**IV. Long Division Method**

If F(z) is expressed as a ratio of two polynomials, namely, F(z)
= g(z^{-1}) / h(z^{-1}), which can not be factorized, then
divide the numerator by the denominator and the inverse transform can be taken
term by term in the quotient.

**Example
14**

Thus F(z) = 1 + 3z^{-1}
+ 3z^{-2} + 3z^{-3} + . . . . . .

Now,
Comparing the quotient with

**Example
15**

Find the
inverse Z-transform of

5 **CONVOLUTION THEOREM**

If Z^{-1}{F (z)} = f_{n} and Z^{-1}{G(z)}
= g_{n}, then

**Example
16**

Use
convolution theorem to evaluate

We know
that Z^{-1} {F(z). G(z)} = f_{n}*g_{n}.

**Example 18**

Use
convolution theorem to find the inverse Z- transform

**6 Application of Z - transform to Difference equations**

As we know, the Laplace transforms method is quite effective in
solving linear differential equations, the Z - transform is useful tool in
solving linear difference equations.

To solve a difference equation, we have to take the Z -
transform of both sides of the difference equation using the property

Z{f_{n+k}}= z^{k}{ F(z) –f_{0} –(f _{1}
/ z ) - … - ( f_{k-1} / z^{k-1}) } (k > 0)

Using the initial
conditions, we get an algebraic equation of the form F(z) = f(z).

By taking
the inverse Z-transform, we get the required solution f_{n} of the
given difference equation.

**Exmaple 19**

Solve the difference equation
y_{n+1} + y_{n} = 1, y_{0} = 0, by Z - transform
method.

Given
equation is y_{n+1} + y_{n} = 1 ---------- (1)

Let Y(z) be the Z
-transform of {y_{n}}.

Taking the Z - transforms
of both sides of (1), we get

Z{y_{n+1}} + Z{y_{n}}
= Z{1}.

ie, z {Y(z) - y_{0}} + Y(z) = z /(z-1).

Using the
given condition, it reduces to

On taking inverse Z-transforms, we obtain

y_{n}
= (1/2){1 - (-1)^{n}}

**Example
20**

Solve y_{n+2} + y_{n} = 1, y_{0} = y_{1}
= 0 , using Z-transforms.

Consider y_{n+2} + y_{n} = 1 ------------- (1)

Taking Z- transforms on both sides, we get

Z{y_{n+2}}+
Z{y_{n}} = Z{1}

Using Inverse Z-transform, we get

y_{n} =(½){1 - cos (np/ 2) - sin (np/ 2)}.

**Example 21**

Solve y_{n+2}
+ 6y_{n+1} + 9y_{n} = 2^{n}, y_{0} = y_{1} = 0, using
Z-transforms.

Consider y_{n+2} + 6y_{n+1} + 9y_{n}
= 2^{n }-------- (1)

Taking
the Z-transform of both sides, we get

Z{y_{n+2}}
+ 6Z{y_{n+1}} + 9Z{y_{n}} = Z {2^{n}}

On taking Inverse
Z-transforms, we get

y_{n} = (1/ 25){ 2^{n}
- (-3)^{n} + (5/3) n (-3)^{n}}.

**Example 22**

Solve the simultaneous
equations

x_{n+1} - y_{n}
= 1; y_{n+1} - x_{n} = 1 with x (0) = 0; y (0) = 0.

The given equations are

x_{n+1} - y_{n}
= 1, x_{0} = 0

y_{n+1} - x_{n} =1,

y_{0}
= 0 -------------- (2)

Taking
Z-transforms, we get

On taking the inverse Z-transform of both sides, we have x_{n}
= n and y_{n} = n , which is the required solution of the simultaneous
difference equations.

**Example
23**

Solve x_{n+1} = 7x_{n} + 10y_{n}
; y_{n+1} = x_{n} + 4y_{n},
with x_{0} = 3, y_{0} = 2

Given

x_{n+1}
= 7x_{n} + 10y_{n}------------- (1)

y_{n+1}
= x_{n} + 4y_{n}------------- (2)

Taking Z- transforms of
equation(1), we get

z { X(z)
- x_{0}} = 7 X(z) + 10 Y(z)

(z - 7)
X(z) –10 Y(z) = 3z ----------(3)

Again taking Z- transforms
of equation(2), we get

z {Y(z) - y_{0}} = X(z) + 4Y(z)

-X(z)
+ (z - 4)Y(z) = 2z ---------- (4)

Eliminating
„x‟
from (3) &
(4), we get

Taking Inverse Z-transforms, we get

y_{n} = 9^{n} + 2^{n}.

From (2),
x_{n} = y_{n+1} - 4y_{n} = 9^{n+1} + 2^{n+1} - 4 (9^{n}
+ 2^{n})

= 9.9^{n} + 2.2^{n} - 4.9^{n} - 4.2^{n}

Therfore,
x_{n} = 5.9^{n} - 2.2^{n}

Hence the
solution is x_{n} = 5.9^{n}
- 2.2^{n} and y_{n} = 9^{n} + 2^{n}.

**Exercises**

Solve the
following difference equations by Z –transform method

1. y_{n+2} + 2y_{n+1} + y_{n} = n, y_{0} = y_{1} = 0

2. y_{n+2} –y_{n} = 2^{n}, y_{0}
= 0, y_{1} = 1

3. u_{n+2} –2cosau_{n+1}+ u_{n}=0, u_{0} = 1, u_{1}
= cosa

4. u_{n+2} = u_{n+1} + u_{n}, u_{0}
= 0, u_{1} = 1

5. y_{n+2} –5y_{n+1}+ 6y_{n} = n (n-1), y_{0}
= 0, y_{1} = 0

6. y_{n+3} –6y_{n+2} + 12y_{n+1} –8y_{n}
= 0, y_{0} = -1, y_{1} = 0, y_{2} = 1

7 **FORMATION OF DIFFERENCE EQUATIONS**

**Example**

Form
the difference equation

16 *y* * _{n}*
-4

-4(
*y* _{n}_{+}_{2}
-4*y _{n}* ) =0

*y _{n} *

**Exercise:**

1.
Derive the difference equation form *y _{n}*
=(

2.
Derive the difference equation form *U*
* _{n}* =

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