ma6351 transforms and partial differential equations iii … · · 2016-08-18ma6351 transforms...

MA6351 Transforms and Partial Differential Equations III Semester

Department of Electronics and Communication Engineering 1

MA6351 Transforms and Partial Differential Equations

UNIT I

Partial Differential Equations

1. Solve: 3 2 2 3( ) 0D D D DD D z

Solution:

Auxiliary Equation

3 2

2

2

1 1 2 2 3 3

1 2 3

1 0

1, 2 1 0

( 1) 0

1, 1, 1

( ) ( ) ( )

( ) ( ) ( )

m m m

m m m

m

m m m

z f y m x f y m x xf y m x

z f y x f y x xf y x

2. Solve: 2( 1) 0D DD D z .

Solution:

The given differential equation is non-homogeneous.

Here

' '

'

1 2

1, 1, 0, 1

0

1 1 0

( ) ( )x x

a b c d

D aD b D cD d

D D D

z e f y e f y x

3. Solve: ( )( 2 1) 0D D D D z .

Solution:

The given differential equation is non-homogeneous.

Here

' '

' '

1 2

1, 1, 2, 0

0

1 2 0

( ) ( 2 )

a b c d

D aD b D cD d

D D D D

z f y x f y x



4. Find the particular integral of 2 3 4( 4 ) x yD DD z e .

Solution:

5. Find the particular integral of 2 2( 3 2 ) cos( 2 )D DD D z x y .

Solution:

6. Find the complete integral of 0p q .

0................ 1

This is of th type ( , ) 0

The trialsolution is ........ 2

Tofind complete integral

...........(3)

............(3)

substituting 3 in 1 weget

0

................ 4

usin

p q

F p q

z ax by c

zLet p a

x

zq b

y

a b

a b

g (4) in (2) weget

( )

z ax by c

z ax ay c

z a x y c

3 4

2 '

3 4

2

3 4

3 4

1Particular Integral

4

1

(3) 4(3)(4)

1

9 48

1

39

x y

x y

x y

x y

eD DD

e

e

e

22 ' '

2

1Particular integral cos( 2 )

3 2

1cos( 2 )

( 1) 3( 2) 2( 4)

1cos( 2 )

1 6 8

1cos( 2 )

3

x yD DD D

x y

x y

x y



7. Solve: 1p q

Solution:

8. Find the complete solution of z x y

pqpq q p

.

Solution:

Given equation is of the form ,z px qy f p q

Then the solution is ,z ax by f a b

Given z x y

pqpq q p

xp yq pq pqz

pq pq

3

2

,

z px qy pq pq

Put p a q b

z ax by ab

2

1............... 1

This is of the type F(p,q)=0

The trialsolution is z=ax+by+c........ 2

Tofind complete integral

put p=a and q=b

substitute p and q in (1)

1

1

1

p q

a b

b a

z ax a y c



9. Find the complete integral of p q pq where z

px

and z

qy

.

Solution:

.........(1)

( , ) 0

........(2)

(1)

( 1)

(2)1

1

p q pq

F p q

Solution z ax by c

Put p a q bin

a b ab

a ab b

a b a

ab in

a

az ax y c

a

10. Find the complete integral of 2 2p y q x .

Solution:

2 2

1 2

2 2

2 2

Given:

Given equation isof the ( , ) ( , )

Then thesolution is

p y q x

f x p f y q

z pdx qdy

p x q y k

p x k p k x

2 2

2 2

3 3

( ) ( )

3 3

q y k q k y

z k x dx k y dy

x yz kx ky c



11. Find the general solution of px qy z .

Solution:

1

1

1

Theequation is of the form

Theauxilliary equation is

1:

Integrating on both sides

log log log

log log

2 :


l

Pp Qq R

dx dy dz

P Q R

dx dy dz

x y z

case

dx dy

x y

dx dy

x y

x y C

xC

y

xC

y

case

dy dz

y z

dy dz

y z

2

2

2

1, 2

og log log

log log

( ) 0

, 0

y z C

yC

z

yC

z

The solution is C C

x y

y z



12. Solve: tan tan tanp x q y z

Solution: Theequation isof theform

Theauxilliaryequation is

tan tan tan

Pp Qq R

dx dy dz

P Q R

dx dy dz

x y z

1:

tan tan


tan tan

log sin log sin log

sinlog log

sin

sin

sin

2 :

tan tan


tan tan

log sin log sin log

sinlog

sin

GROUP

dx dy

x y

dx dy

x y

x y a

xa

y

xa

y

GROUP

dy dz

y z

dy dz

y z

y z b

y

z

log

sin

sin

Thesolution is ( , ) 0

sin sin, 0

sin sin

b

yb

z

a b

x y

y z



13. Solve: 2 2 2x p y q z

Solution:

2 2 2

Theequationis of the form Pp Qq R

dx dy dzThe auxilliary equationis

P Q R

dx dy dz

x y z

2 2

1:GROUP

dx dy

x y

2 2

1

1

2 2

2 2

2

2

1, 2


1 1

1 1

2 :


1 1

1 1

Thesolution is ( ) 0

1 1 1 1, 0

dx dy

x y

Cx y

Cx y

GROUP

dy dz

y z

dy dz

y z

Cy z

Cy z

C C

x y y z



14. Form the partial differential equation by eliminating the constants a and b from 2 2 2 2( )( )z x a y b .

Solution:

2 2 2 2

2 2

2 2

( )( )...........(1)

( )(2 )

( )(2 )

z x a y b

zp y b x

x

zq x a y

y

2 2

2 2

..........(2)2

..........(3)2

Sub (2)&(3) in (1)

2 2

4

py b

x

qx a

y

p qz

x y

xyz pq

15. Form the partial differential equation from2 2( ) ( )z x a y b .

Solution: 2 2( ) ( ) ...........(1)

2( )

2( )

z x a y b

zp x a

x

zq y b

y

2 2

2 2

2 2

..........(2)2

..........(3)2

(2)& (3) (1)

2 2

4

4

px a

qy b

Sub in

p qz

p qz

z p q



16. Form the partial differential equation by eliminating the arbitrary constants from 2 2z a x ay b .

Solution:

2 2

2

2

22

2

22

2

22

2 2

(2 )

(2 )

(2 )

4

4

4

z a x ay b

zp a

x

zq a y

y

qy

a

qy

a

qy

a

qy

p

y p q

17. Form the partial differential equation by eliminating the arbitrary function from

( )z f xy .

Solution: ( , )

'( )

z f x y

zp f xy y

x

'( )z

q f xy xy

'( ) ............(1)

'( ) ............(2)

(1) '( )

(2) '( )

pf xy

y

qf xy

x

p

f xy y

qf xy

x

p q

y x

px qy



18. Form the partial differential equation by eliminating the arbitrary constants a and

b from the equation 2 2 2 2 2( ) ( ) 1x a y b z .

Solution:

2 2 2 2 2( ) ( ) 1

Differentiating with respect to 'x'

2 2 0

2

2

..........(1)

Differentiating with respect to 'y'

2 2 0

2

2

..........(2)

From (1)&(2)

0

x a y b z

zx z

x

x xp

z z

xz

p

zy z

y

y yq

z z

yz

q

x y

p q

qx py

xq py



19.Form the partial differential equation by eliminating the arbitrary function

from 2 2( )z f x y Solution:

2 2

2 2

2 2

( )

'( )2

'( )2

z f x y

zp f x y x

x

zq f x y y

y

2 2'( ).............(1)2

pf x y

x

2 2'( ).............(2)2

(1)&(2)

2 2

0

qf x y

y

From

p q

x y

py qx



UNIT –II

Fourier Series

Problem 1

State the Dirichlets conditions

Solution:

A function (x)f can be expanded in Fourier series, 0

1 1

(x) cos sin2

n n

n n

af a nx b nx

where0a ,

na ,nb are constant s provided the following conditions are proved

Conditions:

1. (x)f is periodic single valued and finite

2. (x)f has finite number of discontinuities in one period (piecewise continuous)

3. (x)f has finite number of maxima and minima

Problem 2

State the convergence conditions on Fourier series.

Solution:

A Fourier series of (x)f converges to

(i) (x)f , if x is a point of continuity

(ii) (x ) f(x )

,2

fif x

is a point of discontinuity

Problem 3

State Euler’s formula when (x)f is expanded has Fourier series in 2c x c

Solution:

Euler’s formula:

The Fourier series for (x)f in the interval 2c x c is given by

0

1 1

(x) cos sin2

n n

n n

af a nx b nx



Where

2

0

2

2

1(x)dx

1(x)cosnxdx

1(x)sinnxdx

c

c

c

n

c

c

n

c

a f

a f

b f

Problem 4

Does (x) tanf x possess a Fourier expansion

Solution:

No (x) tanf x possess a Fourier expansion in the interval ( , ) . Since it has infinite discontinuity

at 2

x

Problem 5

State parsevals theorem on Fourier series .

Solution:

If (x)y f is expressed as a Fourier series in (a,b) then where a, b are Fourier constant, y is the

R.M.S given by 2

2 1(x)

b

a

y f dxb a

.

Problem 6

Find the constant term in the Fourier series corresponding to 2f(x) cos x express in terms of the

interval ( , )

Solution: We know that constant term 0

2

a

Given

2

2

2

f(x) cos

( x) cos ( )

cos ( )

f(x)

x

f x

x



The Given function is even

Constant term 0 1

2 2

a .

Problem 7

If (x) sinhxf is defined in ( , ) , write the value of 0a

na

Solution:

(x) sinhxf

Put x x

( x) sinh( x)f

sinhx

(x)f

The given function is an odd function

0a 0na

0

0

2

0

0

0

0

0

2(x)dx

2cos xdx

2 1 cos 2( )dx

2

2(1 cos 2 x)dx

1 2

2

10

1

a f

x

sin xx

a



Problem 8

If (x)f is an odd function of x in ( , )l l , what are the values of 0a and

na in the Fourier series of (x)f

Solution:

Given (x)f is an odd function

0a 0na

SECTION: II

HALF RANGE SERIES

Problem 1

Find the RMS value of (x) xf in 0 x l

Solution:

W.K.T 2

2 1(x)

b

a

y f dxb a

Given (x) xf in the interval (0, )l

3

0

1

3

l

x

l

310

3

l

l

32

3

ly

3

ly

Problem 2

If sin sin 2 sin3 sin 4

x 2 ..........1 2 3 4

x x x x

in 0 x then prove that

2

2

1

6n

Solution:

Given series is a half range Fourier sine series

sin sin 2 sin3 sin 4x 2 ..........

1 2 3 4

x x x x



sin sin 2 sin3 sin 4..........

2 1 2 3 4

x x x x x

(x)2

xf

By parseval’s theorem,

2

2

10

2(x) n

n

f dx b

22

10

2(1)

4n

n

xdx b

W.K.T 0

2(x)sinnxdxnb f

0

2sinnxdx

2n

xb

=2

0

1 cos sin(1)

nx nxx

n n

=2

1 cos sin( ) 0 sin 0

n n

n n

1 ( 1)n

n

( 1)n

nbn



SECTION III

COMPLEX FORM

Problem 1

Write down the complex form of the Fourier series for (x)f in (c,c 2 )

Solution:

(x) inx

n

n

f c e

21

(x)2

c

inx

n

c

c f e dx

SECTION IV

HARMONIC ANALYSIS

Problem 1:

What is known as harmonic analysis?

The process of finding the Fourier series for a function (x)y f from tabulated values x and y at equal

intervals of x is called harmonic analysis



UNIT – III

APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS

SECTION 1

ONE DIMENSIONAL WAVE EQUATION

Problem 1:

In the wave equation 2 2

2

2 2

y yc

t x

. What does 2c stands for?

Solution:

2 T Tensionc

M Mass per unit length

Problem 2:

What are the possible solutions of one dimensional wave equation?

Solution:

The possible solutions are:

i). ( , ) ( )( )px px pat paty x t Ae Be Ce De

ii). ( , ) ( cos sin )( cos sin )y x t A px B px C pat D pat

iii). ( , ) ( )( )y x t Ax B Ct D

Problem 3:

A tightly stretched string with fixed end points 0x and x l is initially at rest in equilibrium

position given by 3

0( ,0) sinx

y x yl

. If it is released from rest in the position, write the

boundary conditions.

Solution:

The boundary conditions are:

i). (0, ) 0, 0y t t

ii). ( , ) 0, 0y l t t



iii). ( ,0)

0y x

t

iv). 3

0( ,0) sin ,0x

y x y x ll

Problem 4:

State Fourier law of heat condition.

Solution:

The rate at which heat flows across an area A at a distance x from one end of a bar is given by

x

uQ KA

x

where K Thermal conductivity,

x

u

x

The temperature gradient at x.

SECTION 2

ONE DIMENSIONAL HEAT EQUATION

Problem 1:

In steady state conditions derive the solution of one dimensional heat flow equation.

Solution:

When steady state conditions exist the heat flow equation is independent of time ‘t’.

2

2

0

0

u

t

u

x

Problem 2:

What is the basic difference between the solutions of one dimensional wave equation and one

dimensional heat equation?

Solution:

i). The solution of the one dimensional wave equation is of periodic in nature.

ii). The solution of the one dimensional heat equation is not periodic in nature.

Problem 3:

State any two laws which are assumed to derive one dimensional heat equation.



Solution:

i). Heat flows from a higher to lower temperature.

ii). The amount of heat required to produce a given temperature change in a body is

proportional to the mass of the body and to the temperature change.

iii). The rate at which heat flows across any area is proportional to the area and to the gradient

normal to the curve.

Problem 4:

State one dimensional heat equation with the initial and boundary conditions.

Solution:

The one dimensional heat equation is 2

2

2

u u

t x

.

The boundary conditions are:

i). 0

1(0, )u t K c for all t > 0

ii). 0

2( , )u l t K c for all t > 0

iii). ( ,0) ( )u x f x in (0, )l

Problem 5:

What are the possible solutions of one dimensional heat equation?

Solution:

The possible solutions of one dimensional heat equation are:

i). 2 2

( , ) ( )px px p tu x t Ae Be Ce

ii). 2 2

( , ) ( cos sin ) p tu x t A px B px Ce

iii). ( , )u x t Ax B

Problem 6:

In the heat equation 2

t xxu u , what does 2 stands for?

Solution:

In heat equation 2 = Thermal diffusivity.



Problem 7:

A rod of 50 cm long with insulated sides has its end A and end B kept at 20 C and 70 C

respectively. Find the steady state temperature distribution of the rod?

Solution:

Let l = 50 cm

When steady state conditions prevail the heat flow equation is 2

20

u

x

.

( )u x ax b ---------------- (1)

When steady state conditions prevail the boundary conditions are (0) 20;u ( ) 70u l ----

(2)

Applying (2) in (1) we get

(0) 20u b -------------------------------------- (3)

( ) 20 70u l al

70 20 50al

50a

l --------------------- (4)

Substituting (3) and (4) in (1), we get

50( ) 20 ; 50

xu x l

l

Problem 8:

A tightly stretched string of length 2l is fixed at both ends. The midpoint of the string is

displaced to a distance ‘b’ and released in this position. Write the initial conditions

Solution:

The initial conditions are

i). (0, ) 0,y t t



ii). (2 , ) 0,y l t t

iii). ( ,0) 0y

xt

iv).

,0

( ,0)

(2 ), 2

bxx l

ly x

bl x l x l

l

Problem 9:

Write down the one dimensional heat flow equation in unsteady state.

Solution:

The one dimensional heat flow equation in unsteady state is 2

2

2

u u

t x

.

SECTION 3

CLASSIFICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS

Problem 1:

Classify the partial differential equation 0xx yyu xu

Solution:

Given 0xx yyu xu

Here A =1, B = 0, C = x.

2

2

2

4 0 4(1)( )

4 4 0 0

4 4 0 0

B AC x

B AC x for x

B AC x for x

The equation is elliptic for 0x

The equation is hyperbolic for 0x

Problem 2:

Classify the partial differential equation 3 4 3 2 0xx xy y xu u u u

Solution:



Given 3 4 3 2 0xx xy y xu u u u

2 24 (4) 4(3)(0)

16 0

16 0

B AC

The nature of the given partial differential equation is hyperbolic.

Problem 3:

Classify the partial differential equation 2 22 (1 ) 2 0xx xy yy xx u xyu y u u

Solution:

Given 2 22 (1 ) 2 0xx xy yy xx u xyu y u u

Here A = x2, B = 2xy, C = 1+y

2

2 2 2 2 2

2 2 2 2 2

2

4 4 4 (1 )

4 4 4

4 0

B AC x y x y

x y x x y

x

The nature of the given partial differential equation is elliptic.



Unit-IV

Fourier Transforms

Part-A

1. State the Fourier integral theorem

Fourier integral theorem:

If f x is piece-wise continuously differentiable and absolutely integrable in

( , ) then

0

1( ) ( ) cos ( )f x f t t x dt d

2. Write down the Fourier transform pair.

The fourier transform and the inverse fourier transforms are called fourier

transform pair

1

[ ( )] ( )2

1( ) ( )

2

isx

isx

F s F f x f x e dx

f x F s e ds

3. State the Parseval’s identity theorem on Fourier transform.

If 1

( ) , ( ) cos2

F f x F s then F f x ax F s a F s a

2 2( ) ( )F s ds f x dx

4. State and prove the Modulation theorem.

Modulation theorem:

If [ ( )]F f x F s then 1

[ ( ) cos ] [ ( ) ( )]2

F f x ax F s a F s a



Proof:

Consider L.H.S

[ ( ) cos ] ( )2

iaxiaxe eF f x ax F f x

1

( )( )2

iax iaxF f x e e

1

( ( ) ) ( ( ) )2

iax iaxF f x e F f x e

1

[ ( ) ( )]2

F s a F s a (by property 2)

1

[ ( ) cos ] [ ( ) ( )]2

F f x ax F s a F s a

5. Prove that ( ) ( )iasF f x a e F s .

Proof:

By definition of Fourier transform

1

( )2

isxF s f x e dx

1

[ ( )] ( )2

isxF f x a f x a e dx

Put x a t

x a t

dx dt

when x t

x t

( )1[ ( )] ( )

2

( )2

[ ]

is a t

iasist

ias

F f x a f a t a e dt

ef t e dt

e F s



[ ( )] [ ]iasF f x a e F s

6. State and prove the change of scale of property.

If 1

( ) , ( ) , 0s

F f x F s then F f ax F provided aa a

Proof:

W.K.T 1

( )2

isxF s f x e dx

1[ ( )] ( )

2

isxF f ax f ax e dx

Put ax t

tx

a

dtdx

a

Case (i):

If 0 , :a t to

1

[ ( )] ( )2

tis

a dtF f ax f t e

a

1 1

( )2

tis

af t e dta

[ ( )]F f ax 1

.........................................(1)s

Fa a

Case (ii):

If 0 , :a t to

1

[ ( )] ( )2

si t

a dtF f ax f t e

a



1 1

( )2

si t

af t e dta

[ ( )]F f ax 1

.........................................(2)s

Fa a

From (1) and (2)

1

[ ( )]s

F f ax Fa a

7. Find the Fourier integral representation of f(x) defined as

0 , 0

1( ) , 0

2

, 0x

x

f x x

e x

.

Solution:

By definition of Fourier integral theorem

0

1( ) ( ) cos ( )f x f t t x dt d

0

0 0

0 0

10 cos ( ) cos ( )

1cos( )

t

t

t x dt d e t x dt d

e t x dt d

0 0

0 0 0

2 2

0

2

0

1(cos cos sin sin )

1cos cos cos sin sin

1 1cos sin

1 1

1 cos sin

1

t

t t

e t x t x dt d

x e t dt x x e t dt d

x x d

x xd



8. Find the Fourier sine transform of

, 0 1

( ) 2 , 1 2

0 , 2

x x

f x x x

x

.

Solution:

By definition of Fourier sine transform

0

2( ) sinsF s f x sx dx

1 2

2 2

0 1

2 cos sin cos sin(2 ) ( 1)

sx sx sx sxx x

s s s s

2 2 2

2 cos sin sin 2 cos sin(0 0) 0

s s s s s

s s s s s

2 2 2

2 cos sin sin 2 cos sins s s s s

s s s s s

2 2

2 2sin sin 2s s

s s

sF s2

2 2sin sin 2s s

s

9. Find the Fourier sine transform of 1

( )f xx

.

Solution:


0


0

2 1sin sx dx

x

Put sx xs

ddx

s



0

2sin

s d

s

0

2 sind

2

2

sF s 2

10. If ( ) , ( )n

nn

n

dF f x F s then F x f x i F s

ds

Solution:

By definition of Fourier transform

1

[ ( )] ( )2

isxF s F f x f x e dx

Differentiate w.r.t ' 's on both sides

1

( )2

isxd dF s f x e dx

ds ds

1

( )2

isxf x e dxs

1

( )2

isxf x e ix dx

In general,

n

n

dF s

ds

1( ) ( )

2

isx nf x e ix dx

1

( ) ( ) ( )2

n n isxi x f x e dx



n

n

dF s

ds( ) [ ( )]n ni F x f x

[ ( )]nF x f x ( )n

n

n

di F s

ds

11. If ( ) ( 0),axf x e a find the Fourier sine transform of f(x).

Solution:


0


0

2sinaxe sx dx

sF s2 2

2 s

a s

12. If ( )F s is the Fourier transform of f(x), show that

1

( )cos ( ) ( )2

c c cF f x ax F s a F s a .

Proof:

By definition of Fourier cosine transform

0

2( ) coscF s f x sx dx

Consider L.H.S

0

0

2( )cos ( ) cos cos

2( ) cos cos

cF f x ax f x ax sx dx

f x sx ax dx

0

2 1( ) [cos( ) cos( ) ]

2f x s a x s a x dx



0 0

1 2 2( ) cos( ) ( )cos( )

2f x s a x dx f x s a x dx

1

( ) ( )2

c cF s a F s a

1

( )cos ( ) ( )2

c c cF f x ax F s a F s a

13. Prove that the Fourier cosine transform of ( )f ax is 1

c

sF

a a

.

Proof:

By definition of Fourier cosine transform

0

2( ) coscF s f x sx dx

Consider L.H.S

0

2( ) ( ) coscF f ax f ax sx dx

Put ax t

tx

a

dtdx

a

0 0when x t

x t

0

2( ) ( ) cosc

st dtF f ax f t

a a

0

1 2( ) cos

sf t t dt

a a

1

( )c c

sF f ax F

a a

;a>0



14. Find the Fourier sine transform of sin , 0

( )0 ,

x x af x

x a

Solution:


0


0

2sin sin 0

a

a

x sx dx dx

0

2sin sin

a

x sx dx

0

2 1cos( 1) cos( 1)

2

a

s x s x dx

0

1 sin( 1) sin( 1)

1 12

as x s x

s s

0

1 sin( 1) sin( 1)

1 12

as a s a

s s

[ ]sF s0

1 sin( 1) sin( 1)

1 12

as a s a

s s

15. Find ax bx

s

e eF

x

.

Solution:

1 12 2tan tan

ax bx ax bx

s s s

e e e eF F F

x x x

s s

a b

1 12tan tan

s s

a b



UNIT – V

Z-TRANSFORM

SECTION I

Problem 1:

State and prove initial value theorem in Z-transform?

Solution:

Statement:

If [ ( )] ( )Z f n F z then 0

lim ( ) (0) limf(n)z n

F z f

Proof:

We know that 0

[ ( )] (n) z n

n

Z f n f

0 1 2

2

( ) (0) z (1) (2) ...

1 1( ) (0) (1) (2) ...

F z f f z f z

F z f f fz z

Taking limz

on both sides

2

(1) (2)lim ( ) lim (0) ...

(0)

z z

f fF z f

z z

f

0lim ( ) limf(n)z n

F z

Problem 2:

State final value theorem in Z-transform.

Solution:

Statement:

If [ ( )] ( )Z f n F z then 1

lim ( ) lim( 1)F( )n z

f n z z



Problem 3:

State time reversal property for bilateral Z-transform.

Solution:

Statement:

If [ (k)] ( )Z f F z then 1

z[f( )] Fkz

Proof:

[ ( k)] ( ) k

k

Z f f k z

Put –k = n k n

[ ( k)] (n)

1 1(n)

n

n

n

n

Z f f z

f Fz z

1[f( )] FZ k

z

Problem 4:

Define unit-step function. Write its Z-transform.

Solution:

A discrete unit-step equation u(n) is defined by 1 0

( ) :{1,1,1,...}0 0

for nU n

for n

Its transformation is

0

1 2

1

[ ( )] ( )

1 ...

11

n

n

Z U n U n z

z z

z

11z

z

[ ( )]1

zZ U n

z



Problem 5:

If 2

( )T

zF z

z e

find (0)f and lim ( )

tf t

.

Solution:

We know that initial value theorem

If [ ( )] ( )Z f n F z then 0


F z f

(0) lim (z)z

f F

1

2limz

z

z e

2(0) lim

1 0zf

(0) 2f

We know that final value theorem is

If [ ( )] ( )Z f n F z then 1


F z z z

1

1

lim ( ) lim( 1) ( )

2lim( 1)

2(1 1)

1

t z

Tz

T

f t z F z

zz

z e

e

lim ( ) 0t

f t

Problem 6:

If 5

( )( 2)( 3)

zF z

z z

find f(0) and lim ( )

tf t

.

Solution:

We know that initial value theorem is

If [ ( )] ( )Z f n F z then 0


F z f



(0) lim (z)

5lim

( 2)( 3)

z

z

f F

z

z z

5(0) lim

( 2).1 ( 3).1

5lim

2 3

(0) 0

z

z

fz z

z z

f

We know that final value theorem is

If [ ( )] ( )Z f n F z then 1


F z z z

1

1

lim ( ) lim( 1) ( )

5lim( 1)

( 2)( 3)

5(1 1)

(1 2)(1 3)

lim ( ) 0

t z

z

t

f t z F z

zz

z z

f t

Problem 7:

Find [ cos ]nz r n and [ sin ]nz r n

Solution:

We Know that

[ ]

[( ) ]

n

i n

i

zZ a

z a

zZ re

z re



2

2 2

2

2 2 2 2

[ (cos sin )]

(cos sin )

( ) r

cos sin

2 cos r 2 cos r

in

i i

i i

z z reZ r n i n

z re z re

z zr i

z rz e e

z zr izr

z rz z rz

2

2 2

cos[ cos ]

2 cos r

n z zrZ r n

z rz

2 2

sin[ sin ]

2 cos r

n izrZ r n

z rz

Problem 8:

Find the Z-transform of !

na

n.

Solution:

0

0

0

[ ( )] (n) z

z!

( / )

!

n

n

nn

n

n

n

Z f n f

a

n

a z

n

2 3( / ) ( / ) ( / )1 ...

1! 2! 3!

a z a z a z

/

!

na za

z en

Problem 9:

Find the Z-transform of 1

!n.

Solution:

0

1 1

! !

n

n

Z zn n



2 3(1/ ) (1/ ) (1/ )

1 ...1! 2! 3!

z z z

1/1

!

zZ en

Problem 10:

Find the Z-transform of ( 1)( 2)n n .

Solution:

2[( 1)( 2)] [n 3 2]Z n n Z n

2

2

3 2

33

3

[n ] [3 ] 2 [1]

23 2

( 1) ( 1) 1

22

( 1) 1

Z Z n Z

z z z

z z z

z z

z z

Problem 11:

Find the Z-transform of na .

Solution:

0

0

0

[ ( )] (n) z

[a ] a z

n

n

n n n

n

n

n

Z f n f

Z

a

z

2 3

1 1

1 ...

1

a a a

z z z

a z a

z z

[a ]n zZ

z a



Problem 12:

Find 1

1Z

n

.

Solution:

0

2

2 3

1 1z

1 1

1/ (1/ )1 ...

2 3

1 (1/ ) (1/ )z ...

3 3

log1

n

n

Zn n

z z

z z

z

zz

z

CONVOLUTION

SECTION 2

Problem 1:

State convolution of two functions.

Solution:

0

( ) ( ) ( ) (n )n

r

f n g n f r g r

Problem 2:

State convolution theorem for Z-transform.

Solution:

i). [ ( ) ( )] ( ) ( )Z f n g n F z G z

ii). [ (t) (t)] ( ) ( )Z f g F z G z



Problem 3:

Define convolution of two sequences { ( )}f n and { ( )}g n .

Solution:

0

( ) ( ) ( ) (n )n

r

f n g n f r g r

SECTION – IV

Problems based on formation of difference equation.

Problem 1:

Form a difference equation by elimination the arbitrary constant A from .3n

ny A .

Solution:

1

1

.3

.3 3 .3

3

n

n

n n

n

n

y A

y A A

y

1 3 0n ny y

Problem 2:

Form a difference equation by eliminating the arbitrary constants from .2n

ny A B .

Solution:

1

1

.2

.2

2 .2

n

n

n

n

n

y A B

y A B

A B

2

2 .2 4 .2n n

ny A B A B

Eliminating ‘A’ and ‘B’, we get

1

2

1 1

1 2 0

1 4

n

n

n

y

y

y



1 2 1 2

1 2 1 2

2 1

(4 2) 1(4 2 ) 1( ) 0

2 4 2 0

3 2 0

n n n n n

n n n n n

n n n

y y y y y

y y y y y

y y y

Problem 3:

Find the difference equation from ( )2n

ny A Bn

Solution:

( )2

.2 .2

n

n

n n

y A Bn

A Bn

1 1

1 .2 ( 1).2

2 .2 2 ( 1).2

n n

n

n n

y A B n

A B n

2 2

2 .2 ( 2).2

4 .2 4 ( 2).2

n n

n

n n

y A B n

A B n

Eliminating ‘A’ and ‘B’, we get

1

2

1

2 2( 1) 0

4 4( 2)

n

n

n

y n

y n

y n

1 2 1 2[8( 2) 8( 1)] [4( 2) 2( 1) ] [4 2 ] 0n n n n ny n n n y n y n y y

1 28 [ 2 1] [4 4( 2)] [2( 1) 2 ] 0n n ny n n y n n y n n

1 28 8 [2] 0n n ny y y

1 24 4 0n n ny y y

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