ma6351 transforms and partial differential equations unit i partial differential...
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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:
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
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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 .
Solution:
0................ 1
This is of th type ( , ) 0
The trialsolution is ........ 2
To find 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
7. Solve: 1p q
Solution:
2
1............... 1
This is of the type F(p,q)=0
The trialsolution is z=ax+by+c........ 2
Tofind completeintegral
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
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
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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
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
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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
Theauxilliaryequation is
1:
Integrating on both sides
log log log
log log
2 :
Integrating on both sides
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
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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
Integrating on both sides
tan tan
log sin log sin log
sinlog log
sin
sin
sin
2 :
tan tan
Integrating on both sides
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
zlog
sin
sin
Thesolution is ( , ) 0
sin sin, 0
sin sin
b
yb
z
a b
x y
y z
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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
Integrating on both sides
1 1
1 1
2 :
Integrating on both sides
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
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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
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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
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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
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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
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
where 0a , 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
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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
.
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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
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
2
2
2
f(x) cos
( x) cos ( )
cos ( )
f(x)
x
f x
x
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
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Put x x
( x) sinh( x)f
sinhx
(x)f
The given function is an odd function
0a 0na
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
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3
ly
3
ly
Problem 2
If sin sin 2 sin 3 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 sin 3 sin 4x 2 ..........
1 2 3 4
x x x x
sin sin 2 sin 3 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
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=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
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UNIT – III
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
PART - A
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
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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
xThe temperature gradient at x.
UNIT – III
APPLICATION OF PARTIAL DIFFERENTIAL EQUATIONS
PART - A
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.
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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?
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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
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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.
UNIT – III
PART - A
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
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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
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
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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
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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
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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
.
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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
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2 2
2 2sin sin 2s s
s s
sF s
2
2 2sin sin 2s s
s
9. Find the Fourier sine transform of 1
( )f xx
.
Solution:
By definition of Fourier sine transform
0
2( ) sinsF s f x sx dx
0
2 1sin sx dx
x
Put sx xs
ddx
s
0
2sin
s d
s
0
2 sind
2
2
sF s2
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
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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:
By definition of Fourier sine transform
0
2( ) sinsF s f x sx dx
0
2sinaxe sx dx
sF s2 2
2 s
a s
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12. If ( )F s is the Fourier transform of f(x), show that
1( )cos ( ) ( )
2c 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
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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:
By definition of Fourier sine transform
0
2( ) sinsF s f x sx dx
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
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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
PART - A
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
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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:
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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
lim ( ) (0) limf(n)z n
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
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If [ ( )] ( )Z f n F z then 1
lim ( ) lim( 1)F( )n z
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
lim ( ) (0) limf(n)z n
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
lim ( ) lim( 1)F( )n z
F z z z
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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:
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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
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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
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SECTION 2:
CONVOLUTION
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
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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
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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