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MA6351-TRANSFORMS AND PARTIAL

DIFFERENTIAL EQUATIONS

SUBJECT NOTES

Department of Mathematics

FATIMA MICHAEL

COLLEGE OF ENGINEERING &

TECHNOLOGY

MADURAI – 625 020, Tamilnadu, India

Department of Mathematics – FMCET – MADURAI

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Basic Formulae

DIFFERENTIATION &INTEGRATION FORMULAE

0 Function

( )y f x

Differentiation

dy

dx

1 nx 1nnx

2 log x 1

x

3 sin x cos x

4 cos x sin x

5 axe a xe

6 C (constant) 0

7 tan x 2sec x

8 sec x sec tanx x

9 cot x 2cos ec x

10 cos ecx cos cotecx x

11 x 1

2 x

12 1sin x 2

1

1 x

13 1cos x 2

1

1 x

14 1tan x 2

1

1 x

15 1sec x 2

1

1x x

16 1cosec x 2

1

1x x

17 1cot x 2

1

1 x

18 xa logxa a


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19. If y uv , then dy du dv

v udx dx dx

20. If u

yv

, then2

du dvv u

dy dx dx

dx v


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1

2

1

2 2

2 2

1.1

2. , &

cos3. sin cos & sin

sin4. cos sin & cos

5. tan log sec log cos

6. sec tan

17. tan

18. log

2

nn

ax axx x ax ax

xx dx

n

e ee dx e e dx e dx

a a

axxdx x axdx

a

axxdx x axdx

a

xdx x x

xdx x

dx xdx

x a a a

dx x adx

x a a x

1

2 2

1

2 2

1

2 2

22 2 2 2 1

22 2 2 2 1

22 2 2 2 1

2 2

2 2

9. sin

10. sinh

11. cosh

12. sin2 2

13. sinh2 2

14. cosh2 2

15. log

216. log

a

dx xdx

aa x

dx xdx

aa x

dx xdx

ax a

x a xa x dx a x

a

x a xa x dx a x

a

x a xx a dx x a

a

dxx

x

xdxx a

x a


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3

2

3

2

17. log log

18.3

19.3

120. 2

xdx x x x

a xa x dx

a xa x dx

dx xx

21. 2 2

cos cos sinax

ax ee bxdx a bx b bx

a b

22.2 2

sin sin cosax

ax ee bxdx a bx b bx

a b

23.1 2 3........udv uv u v u v u v´ ´´ ´´´

24.0

( ) 2 ( )

a a

a

f x dx f x dx when f(x) is even

25. ( ) 0

a

a

f x dx when f(x) is odd

26. 2 2

0

cosax ae bxdx

a b

27. 2 2

0

sinax be bxdx

a b

TRIGNOMETRY FORMULA

1. sin2 2sin cosA A A

2 2

2

2

2.cos 2 cos sin

1 2sin

2cos 1

A A A

A

A


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3. 2 1 cos 2cos

2

xx & 2 1 cos 2

sin2

xx

4. sin( ) sin cos cos sin

sin( ) sin cos cos sin

cos( ) cos cos sin sin

cos( ) cos cos sin sin

A B A B A B

A B A B A B

A B A B A B

A B A B A B

15.sin cos sin( ) sin( )

2

1cos sin sin( ) sin( )

2

1cos cos cos( ) cos( )

2

1sin sin cos( ) cos( )

2

A B A B A B

A B A B A B

A B A B A B

A B A B A B

3

3

16. sin 3sin sin 3

4

1cos 3cos cos3

4

A A A

A A A

2 2

2 2

7.sin 2sin cos2 2

cos cos sin2 2

1 2sin 1 cos 2sin2 2

A AA

A AA

A AA

LOGRATHEMIC FORMULA


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log

log log log

log log log

log log

log 1 0

log 0

log 1

n

a

a

a

x

mn m n

mm n

n

m n m

a

e x

UNIT - 1

PARTIAL DIFFRENTIAL EQUATIONS

PARTIAL DIFFERENTIAL EQUTIONS

Notations

zp

x

zq

y

2

2

zr

x

2 zs

x y

2

2

zt

y

Formation PDE by Eliminating arbitrary functions

Suppose we are given f(u,v) = 0

Then it can be written as u = g(v) or v = g(u)

LAGRANGE’S LINEAR EQUATION

(Method of Multipliers)

General form

Pp + Qq = R

Subsidiary Equation

dx dy dz

P Q R


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dx dy dz x my nz

P Q R P mQ nR

Where ( , m ,n) are the Lagrangian Multipliers

Choose , m, n such that P + mQ + nR = 0

Then dx + m dy + n dz = 0

On Integration we get a solution u = a

Similarly, We can find another solution v = a for another multiplier

The solution is (u, v) = 0

TYPE –2 (Clairut’s form)

General form

Z = px + qy + f(p,q) (1)

Complete integral

Put p = a & q = b in (1), We get (2) Which is the Complete integral

Singular Integral

Diff (2) Partially w.r.t a We get (3)

Diff (2) Partially w.r.t b We get (4)

Using (3) & (4) Find a & b and sub in (2) we get Singular Integral

REDUCIBLE FORM

F(xm

p ,ynq) = 0 (1) (or)

F( xm

p, ynq, z)=0 (1)

F( zkp, z

kq)=0

If 1& 1m n then

X = x1-m

& Y = y1-n

xm

p = P(1-m) & yn q = Q(1-n)

Using the above in (1)we get

F(P,Q) = 0 (or) F(P,Q,z) = 0

If 1k then Z = zk+1

1

k Qz q

k

Using the above in (1) We get

F(P,Q) = 0


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If m=1 & n=1 then

X= logx & Y= logy

xp = P & yq = Q

Using the above in (1) we get

F(P,Q) (or) F(P, Q, z) = 0

If k =-1 then Z = log z

p

Pz

& q

Qz

Using the above in (1)

we get

F(P,Q) = 0


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STANDARD TYPES

TYPE –1

TYPE –3(a)

TYPE –3(b)

TYPE –3(c)

TYPE –4

General form

F(p,q) = 0 (1)

General form

F(x,p,q) = 0 (1)

General form

F(y,p,q) = 0

(1)

General form

F(z,p,q) = 0

(1)

General form

F(x,y,p,q) = 0

(1)

Complete

Integral Put p = a and q =

b in (1)

Find b in terms

of a

Then sub b in

z = ax + by + c

we get (2)

which is the

Complete

Integral

Complete

Integral

Put q = a in (1)

Then, find p and

sub in

dz = p dx + q dy

Integrating ,

We get (2)

which is the

Complete

integral

Complete

Integral

Put p = a in (1)

Then, find q and

sub in dz = p dx

+ q dy

Integrating ,

We get (2)

which is the

Complete

integral

Complete

Integral

Put q = ap in (1)

Then, find p and

sub in

dz = p dx + q dy

Integrating,

We get (2) which

is the

Complete

integral

Complete

Integral

(1) Can be written

as,

f(x,p) =f(y,q) = a

Then, find p and q

sub in

dz = p dx + qD y

Integrating,

We get (2) which

is the

Complete integral

Singular

Integral

Diff (2) partially

w.r.t cWe get,0

=1 (absurdThere

is no Singular

Integral

Singular

Integral

Diff (2) partially

w.r.t cWe get,0

=1 (absurdThere

is no Singular

Integral

Singular

Integral

Diff (2) partially

w.r.t cWe get,0

=1 (absurdThere

is no Singular

Integral

Singular

Integral

Diff (2) partially

w.r.t cWe get,0

=1 (absurdThere

is no Singular

Integral

Singular Integral

Diff (2) partially

w.r.t cWe get,0

=1 (absurd

There is no

Singular Integral

General

Integral

Put c = (a) in

(2)We get

(3)Diff (3)

partially w.r.t

aWe get

(4)Eliminating a

from (3) and (4)

we get General

Integral

General

Integral

Put c = (a) in

(2)We get

(3)Diff (3)

partially w.r.t

aWe get

(4)Eliminating a

rom (3) and (4)

we get General

Integral

General

Integral

Put c = (a) in

(2)We get (3)

Diff (3) partially

w.r.t aWe get (4)

Eliminating a

from (3) and (4)

we get General

Integral

General

Integral

Put c = (a) in

(2)We get

(3)Diff (3)

partially w.r.t

aWe get

(4)Eliminating a

from (3) and (4)

we get General

Integral

General Integral

Put c = (a) in

(2)We get (3)

Diff (3) partially

w.r.t aWe get

(4)Eliminating a

from (3) and (4)

we get General

Integral


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HOMOGENEOUS LINEAR EQUATION

General form

33 2 2( ) ( , )aD bD D cDD dD z f x y (1)

To Find Complementary Function

Auxiliary Equation

Put D = m & D = 1 in (1)

Solving we get the roots m1

, m2

, m3

Case (1)

If the roots are distinct then

C.F. = 1 1 2 2 3 3( ) ( ) ( )y m x y m x y m x

Case (2)

If the roots are same then

C.F. = 2

1 2 3( ) ( ) ( )y mx x y mx x y mx

Case (3)

If the two roots are same and one is distinct, then

C.F = 1 2 3 3( ) ( ) ( )y mx x y mx y m x

Function PI =

1

1( , )

( , )F x y

F D D

F(x,y) = eax+by

Put D = a & D1

= b

F(x,y)= sin(ax+by)(or)

Cos (ax+by)

Put

22 2 2( ), ( )& ( )D a DD ab D b

F(x,y) = xr

ys

PI=

1

( , ) r sF D D x y

Expand and operating D & D

on xr

ys

F(x,y) = eax+by

f(x,y) Put D = D+a & D = D +b


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Particular Integral

F(x,y)=ex+y

cosh(x+y) F(x ,y)=2 21

2

x ye e

F(x,y)=ex+y

sinh(x+y) F(x, y) =2 21

2

x ye e

F(x,y)=sin x cos y

1( , ) sin( ) sin( )

2F x y x y x y

F(x,y)= cos x sin y

1( , ) sin( ) sin( )

2F x y x y x y

F(x,y)= cos x cos y

1( , ) s( ) s( )

2F x y co x y co x y

F(x,y)= sin x sin y

1( , ) cos( ) cos( )

2F x y x y x y

Note:

D represents differentiation with respect to ‘x ‘

D represents differentiation with respect to ‘y ‘

1

D

represents integration with respect to ‘x ‘

1

D

represents integration with respect to ‘y ‘


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PARTIAL DIFFRENTIAL EQUATIONS

1. Eliminate the arbitrary constants a & b from

z = (x2

+ a)(y2

+ b)

Answer:

z = (x2

+ a)(y2

+ b)

Diff partially w.r.to x & y here &z z

p qx y

p = 2x(y2

+ b) ; q = (x2

+ a) 2y

(y2

+ b) = p/2x ; (x2

+ a) = q/2y

z = (p/2x)(q/2y)

4xyz = pq

2. Form the PDE by eliminating the arbitrary function from z = f(xy)

Answer:

z = f(xy)

Diff partially w.r.to x & y here &z z

p qx y

p = ( ).f xy y q = ( ).f xy x

p/q = y/x px – qy = 0

3. Form the PDE by eliminating the constants a and b from z = axn

+ byn

Answer:

z = axn

+ byn

Diff. w .r. t. x and y here &z z

p qx y

p = naxn-1

; q = nbyn-1


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1 1

1 1

;n n

n n

n n

p qa b

nx ny

p qz x y

nx ny

nz px qy

4. Eliminate the arbitrary function f from xy

z fz

and form the PDE

Answer:

xy

z fz

Diff. w .r. t. and y here &z z

p qx y

2

2

.

.

.

xy z xpp f y

z z

xy z yqq f x

z z

p y z xp

q x z yq

0

pxz pqxy qyz pqxy

px qy

5. Find the complete integral of p + q =pq

Answer:

Put p = a, q = b

p + q =pq a+b=ab

b – ab = -a

1 1

a ab

a a

The complete integral is z= ax+

1

a

ay +c


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6. Find the solution of 1p q

Answer:

z = ax+by+c ----(1) is the required solution

given 1p q -----(2)

put p=a, q = b in (2)

2

2

1 1 (1 )

(1 )

a b b a b a

z a x a y c

7. Find the General solution of p tanx + q tany = tanz.

Answer:

1 2

1 2

tan tan tan

cot cot cot

cot cot cot cot

logsin logsin log logsin logsin log

sin sin

sin sin

sin sin, 0

sin sin

dx dy dz

x y z

x dx y dy z dz

take x dx y dy y dy zdz

x y c y z c

x yc c

y z

x y

y z

8. Eliminate the arbitrary function f from 2 2z f x y and form the PDE.

Answer:

2 2z f x y

2 2 2 22 ; ( 2 )

20

2

p f x y x q f x y y

p xpy qx

q y

9. Find the equation of the plane whose centre lie on the z-axis

Answer:

General form of the sphere equation is


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22 2 2x y z c r (1)

Where ‘r’ is a constant. From (1)

2x+2(z-c) p=0 (2)

2y +2(z-c) q = 0 (3)

From (2) and (3)

x y

p q

That is py -qx =0 which is a required PDE.

10. Eliminate the arbitrary constants 2 2z ax by a b and form the PDE.

Answer:

2 2z ax by a b

2 2

;p a q b

z px qy p q

11. Find the singular integral of z px qy pq

Answer:

The complete solution is z ax by ab

0 ; 0

;

( ) ( ) ( . )

0

z zx b y a

a b

b x a y

z y x x y y x

xy xy xy

xy

xy z

12. Find the general solution of px+qy=z

Answer:

The auxiliary equation is

dx dy dz

x y z


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From

dx dy

x y Integrating we get log x = log y + log c

on simplifying 1

xc

y.

2

dy dz yc

y z z

Therefore , 0

x y

y z is general solution.

13. Find the general solution of px2

+qy2

=z2

Answer:

The auxiliary equation is 2 2 2

dx dy dz

x y z

From 2 2

dx dy

x y Integrating we get 1

1 1c

y x

Also 2 2

dy dz

y z Integrating we get 2

1 1c

z y

Therefore 1 1 1 1

, 0y x z y

is general solution.

14. Solve 2 22 3 0D DD D z

Answer:

Auxiliary equation is

2 2 3 0m m

3 1 0m m

1, 3m m

The solution is 1 2 3z f y x f y x


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15. Solve 2 24 3 x yD DD D z e

Answer:

Auxiliary equation is 2 4 3 0m m

3 1 0m m

1, 3m m

The CF is 1 2 3CF f y x f y x

2 2

1

4 3

x yPI eD DD D

Put 1, 1D D Denominator =0.

2 4

x yxPI e

D D

2

x yxe

Z=CF + PI

1 2 3z f y x f y x

2

x yxe

16. Solve. 2 23 4 x yD DD D z e

Answer:


4 1 0m m

C.F is = f1

(y + 4x) + f2

(y - x)

2 2

1 1 1

3 4 1 3 4 6

x y x y x yPI e e eD DD D

17. Find P.I 2 2 24 4 x yD DD D z e

Answer:

2

2 2

1

4 4

x yPI eD DD D

Put 2, 1D D

2

2

1

2

x yPI eD D

2

2

1

2 2

x ye

2

16

x ye


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18. Find P.I 2 2 26D DD D z x y

Answer:

2

22

2

1

61

PI x yD D

DD D

2

2

11

Dx y

D D

32

2

1

3

xx y

D

4 5

12 60

x y x

19. Find P.I

2 2

2sin

z zx y

x x y

Answer:

2

1PI Sin x y

D DD Put

2 1, (1)( 1) 1D DD

1

1 1Sin x y

2

Sin x y

20. Solve 3 33 2 0D DD D Z

Answer:


21 2 0m m m

1 2 1 0m m m

1,1 2m m

The Solution is 1 2 3 2CF f y x x f y x f y x

FOR PRACTICE:

1. Eliminating arbitrary constants

2 2 2

2 2 21

x y z

a b c

2. Solve

2

2sin

zy

x

3. Find the complete the solution of p. d .e 2 2 4 0p q pq

4. Form p.d.e eliminating arbitrary function from 2 ,

2

xz xy


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5. Find the singular soln of 2 2 1z px qy p q

1. (i) Solve 2 2 2x y z p y z x q z x y

(ii) Solve 2 2 2 2 2 2x z y p y x z q z y x

2. (i) Solve z z

mz ny nx lz ly mxx y

(ii) Solve 3 4 4 2 2 3z y p x z q y x

3. (i) Solve 2 2 2 2 2x y z p xyq xz

(ii) Solve 2 2 2 2 2 0y z x p xyq zx

4. (i) Solve y z p z x q x y

(ii) Solve y z p z x q x y

5. Solve 2 2 3 23 2 sin(3 2 )x yD DD D e x y

6. Solve 2 2

2cos cos 2

z zx y

x x y

7. Solve 2 26 cosD DD D z y x

8. Solve 2 2 630 x yD DD D z xy e

9. Solve 2 26 5 sinhxD DD D z e y xy

10. Solve 2 2 24 4 x yD DD D z e

11. Solve 3 2 2 3 2 cos( )x yD D D DD D z e x y

12. Solve (i) 2 21z px qy p q

(ii) 2 2z px qy p q

13. Solve 2 2 21z p q


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14. Solve 2 2 2 2( ) 1z p x q

15. Solve (i) 2 2 2 2( )z p q x y

(ii) 2 2 2 2 2( )z p q x y

UNIT - 2

FOURIER SERIES

0

1

( ) cos sin2

n n

n

af x a nx b nx

(0,2 ) ( - , )

Even (or) Half range

Fourier co sine series

Odd (or) Half range

Fourier sine series

Neither even nor odd


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2

0

0

1( )a f x dx

0

0

2( )a f x dx

0 0a 0

1( )a f x dx

2

0

1( )cosna f x nxdx

0

2( )cosna f x nxdx

0na 1( )cosna f x nxdx

2

0

1( )snb f x innxdx

bn=0

0

2( )snb f x innxdx

1( )snb f x innxdx

0

1

( ) cos sin2

n n

n

a n x n xf x a b

(0,2 ) ( - , )

Even (or) Half range

Fourier cosine series

Odd (or) Half range

Fourier sine series

Neither even nor odd

2

0

0

1( )a f x dx

0

0

2( )a f x dx

0 0a 0

1( )a f x dx

2

0

1( )cosn

n xa f x dx

0

2( )cosn

n xa f x dx

0na 1( )cosn

n xa f x dx

2

0

1( )sn

n xb f x in dx

bn=0

0

2( )sn

n xb f x in dx

1( )sn

n xb f x in dx

Even and odd function:

Even function:

f(-x)=f(x)

eg : cosx,x2

, , , sin , cosx x x are even functions

Odd function:


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f(-x)=-f(x)

eg : sinx,x3 ,sinhx, tanx are odd functions

For deduction

In the interval (0,2 ) if x = 0 or x = 2 then

f(0) = f(2 ) = (0) (2 )

2

f f

In the interval (0,2 ) if x = 0 or x = 2 then

f(0) = f(2 ) = (0) (2 )

2

f f

In the interval (- , ) if x = - or x = then

f(- ) = f( ) = ( ) ( )

2

f f

In the interval (- , ) if x = - or x = then

f(- ) = f( ) = ( ) ( )

2

f f

HARMONIC ANALYSIS

f(x)= 0

2

a + a1 cosx +b1sinx + a2cos2x + b2sin2x ……… for form

0 2y

an

1

cos2

y xa

n, 2

cos 22

y xa

n 1

sin2

y xb

n, 2

sin 22

y xb

n

f(x)= 0

2

a + a1 cos

x

+b1

xsin

+ a2

2cos

x

+ b2

2 xsin

………( form)


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0 2y

an

1

cos

2

xy

an

2

2cos

2

xy

an

1

sin

2

xy

bn

,2

2sin

2

xy

bn

1. Define R.M.S value.

If let f(x) be a function defined in the interval (a, b), then the R.M.S value of

f(x) is defined by

21( )

b

a

y f x dxb a

2. State Parseval’s Theorem.

Let f(x) be periodic function with period 2l defined in the interval (c, c+2l).

2 22 2 2

1

1 1( )

2 4 2

c l

on n

nc

af x dx a b

lWhere , &o n na a b are Fourier constants

3. Define periodic function with example.

If a function f(x) satisfies the condition that f(x + T) = f(x), then we say f(x) is a periodic

function with the period T.

Example:-

i) Sinx, cosx are periodic function with period 2

ii) tanx is are periodic function with period

4. State Dirichlets condition.

(i) f(x) is single valued periodic and well defined except possibly at a

Finite number of points.

(ii) f (x) has at most a finite number of finite discontinuous and no infinite

Discontinuous.

(iii) f (x) has at most a finite number of maxima and minima.

5. State Euler’s formula.


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Answer:

( , 2 )

cos sin2

on n

In c c l

af x a nx b nx

2

2

2

1( )

1( )cos

1( )sin

c l

o

c

c l

n

c

c l

n

c

where a f x dx

a f x nxdx

b f x nxdx

6. Write Fourier constant formula for f(x) in the interval (0,2 )

Answer:

2

0

2

0

2

0

1( )

1( )cos

1( )sin

o

n

n

a f x dx

a f x nxdx

b f x nxdx

7. In the Fourier expansion of

f(x) =

21 , 0

21 ,0

xx

xx

in (-π , π ), find the value of nb

Since f(-x)=f(x) then f(x) is an even function. Hence nb = 0

8. If f(x) = x3

in –π < x < π, find the constant term of its Fourier series.

Answer:

Given f(x) = x3

f(-x) = (- x)3

= - x3

= - f(x)

Hence f(x) is an odd function

The required constant term of the Fourier series = ao

= 0


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e26

9. What are the constant term a0

and the coefficient of cosnx in the Fourier

Expansion f(x) = x – x3

in –π < x < π

Answer:

Given f(x) = x – x3

f(-x) = -x - (- x)3

= - [x - x3

] = - f(x)

Hence f(x) is an odd function

The required constant term of the Fourier series = 0a = 0

10. Find the value of a0

for f(x) = 1+x+x2

in ( 0 ,2 )

Answer:

2

0

1( )oa f x dx

22 2 32

0 0

2 3 2

1 1(1 )

2 3

1 4 8 82 2 2

2 3 3

x xx x dx x

11. (i)Find bn

in the expansion of x2

as a Fourier series in ( , )

(ii)Find bn

in the expansion of xsinx a Fourier series in ( , )

Answer:

(i) Given f(x) = x2

f(-x) = x2

= f(x)

Hence f(x) is an even function

In the Fourier series nb = 0

(ii) Given f(x) = xsinx f(-x) = (-x)sin(-x)

= xsinx = f(x)


In the Fourier series nb = 0


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e27

12. Obtain the sine series for

0 / 2

/ 2

x x lf x

l x l x l

Given

0 / 2

/ 2

x x lf x

l x l x l

Answer:

Given

0 / 2

/ 2

x x lf x

l x l x l

Fourier sine series is sinn

nxf x b

l

0

2

0 2

2

2 2

2 2

0 2

2( )sin

2sin ( )sin

cos sin cos sin2

(1) ( ) ( 1)

l

n

l l

l

l l

l

nxb f x dx

l l

nx nxx dx l x dx

l l l

nx nx nx nx

l l l llx l l l x ll n n n n

2 2 2 2

2 2 2 2

2

2 2 2 2

2 cos 2 sin 2 cos 2 sin 2

2 2

2 2 sin 2 4 sin 2

l n l n l nl l n

l n n n n

l n l n

l n n

Fourier series is 2 21

4 sin 2sin

n

l n n xf x

n l

13. If f(x) is odd function in ,l l . What are the value of a0

&an

Answer:

If f(x) is an odd function, ao

= 0, an

= 0


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e28

14. In the Expansion f(x) = |x| as a Fourier series in (- . ) find the value of a0

Answer:

Given f(x) = |x|

f(-x) = |-x|

= |x| = f(x)


2 2

0 0

2 2 2

2 2o

xa xdx

15. Find half range cosine series of f(x) = x, in 0 x

Answer:

2 2

0 0

2 2 2

2 2o

xa xdx

2

0 0

1

2 1 cos sinsin (1)

1 11 cos0 0

n

n n

nx nxa x nxdx x

n n

n

n n n

Fourier series is

0

1

0

cos2

1cos

2

on

n

n

n

af x a nx

nxn

16. Find the RMS value of f(x) = x2

, 0 1x

Answer:

Given f(x) = x2


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e29

R.M.S value

2 122 2

0 0

15

0

1 1( )

1 2

22

5 5

l

y f x dx x dxl

x

17. Find the half range sine series of ( )f x x in (0, )

Answer:

0

2( )sinnb f x nxdx

2

0 0

1

2 2 cos sinsin (1)

2 ( 1) 2( 1)n n

nx nxx nxdx x

n n

n n

Half range Fourier sine series is

1

0

2( 1)sin

n

n

f x nxn

18. Find the value of a0

in the cosine series of ( )f x x in (0, 5)

Answer:

55 2 2

0 0

2 2 2 55

5 5 2 5 2o

xa xdx

19. Define odd and even function with example.

Answer:

(i) If ( ) ( )f x f x then the function is an even function.

eg : cosx ,x2 , , sin , cosx x x are even functions

(ii) If ( ) ( )f x f x then the function is an odd function.

eg : sinx,x3 ,sinhx, tanx are odd functions


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e30

20. Write the first two harmonic.

Answer:

1 1 2 2cos sin cos 2 sin 22

o

The first twoharmonics are

af x a x b x a x b x

FOURIER SERIES

1. Expand

(0, )( )

2 ( ,2 )

xf x

x as Fourier series

and hence deduce that

2

2 2 2

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

1 3 5 8

2. Find the Fourier series for f(x) = x2

in (- . ) and also prove that

(i)

2

2 2 2

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

1 2 3 6 (ii)

2

2 2 2

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

1 2 3 12

3. (i) Expand f(x) = | cosx | as Fourier series in (- . ).

(ii) Find cosine series for f(x) = x in (0, ) use Parsevals identity to

Show that

4

4 4 4

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

1 2 3 90

4. (i) Expand f(x) = xsinx as a Fourier series in (0,2 )

(ii) Expand f(x) = |x| as a Fourier series in (- . ) and deduce to

2

2 2 2

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

1 3 5 8

5. If

0 , ( ,0)( )

sin , (0, )f x

x Find the Fourier series and hence deduce that

1 1 1 2

.........1.3 3.5 5.7 4

6. (i) Find the Fourier series up to second harmonic

X 0 1 2 3 4 5

Y 9 18 24 28 26 20

(ii)Find the Fourier series up to third harmonic


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e31

X 0 π/3 2π/3 π 4π/3 5π/3 2π

F(x) 10 14 19 17 15 12 10

7. (i) Find the Fourier expansion of 2( ) ( )f x x in (0,2 ) and

Hence deduce that

2

2 2 2

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

1 2 3 6

(ii). Find a Fourier series to represent

2( ) 2f x x x with period 3

in the range (0,3)

(ii) Find the Fourier series of xf x e in ( , ) .

(ii) Find the Fourier series for

1 (0, )

2 ( ,2 )

inf x

in

and hence show that

2

2 2 2

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

1 3 5 8

8. (i) Find the the half range sine series for f x x x in the interval (0, ) and deduce

that 3 3 3

1 1 1....

1 3 5

(ii) Obtain the half range cosine series for 2

1f x x in (0,1)

and also deduce that

2

2 2 2

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

1 2 3 6

9. (i) Find the Fourier series for f(x) = x2

in (- . ) and also prove that

4

4 4

1 11 .........

1 2 90 (use P.I)

(ii) Find the Fourier series for f(x) = x in (- . ) and also prove that

4

4 4

1 11 .........

1 3 96 (use P.I)


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e32

10(i)Obtain the sine series for

,02

,2

lcx x

f xl

c l x x l

(ii). Find the Fourier series for the function

,02

2 ,2

lkx x

f xl

k l x x l

11.(i).Find the Fourier series for the function 21 ( , )f x x x in and also

deduce that

2

2 2 2

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

1 2 3 6

(ii) Find the Fourier expansion of

f(x) =

21 , 0

21 ,0

xx

xx

in (-π , π ), and also deduce that 2

2 2 2

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

1 3 5 8


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e33

UNIT - 3

APPLICATIONS OF P.D.E

S.

N

O

ONE DIMENSIONAL WAVE EQUATION

VELOCITY MODEL INITIAL POSITION MODEL

1 STEP-1

One Dimensional wave equation

is

2 22

2 21

y ya

t x

STEP-1

One Dimensional wave equation

is

2 22

2 21

y ya

t x

2 STEP-2

Boundary conditions

1. y(0,t) = 0 for 0t

2. y( , t) = 0 for 0t

3. y(x,0) = 0 for 0 < x <

4. 0t

y

t= f(x) for 0 < x <

STEP-2

Boundary conditions

1. y(0,t) = 0 for 0t

2. y( , t) = 0 for 0t

3. 0t

y

t = 0 for 0 < x <

4. y(x,0) = f(x) for 0 < x <


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e34

3 STEP-3

The possible solutions are

y(x,t) = (A ex + B e

- x) (C e

at + D e

- at)

y(x,t) = (A cos x + B sin x )( C cos at + D sin

at)

y(x,t) = (Ax + B) ( Ct + D)

STEP-3


y(x,t) = (A ex + B e

- x) (C e

at + D e

- at)

y(x,t) = (A cos x + B sin x )( C cos at + D

sin at)

y(x,t) = (Ax + B) ( Ct + D)

4 STEP-4

The suitable solution for the given

boundary condition is

y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)

(2)

STEP-4

The suitable solution for the given

boundary condition is

y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)

(2)

5 STEP-5

Using Boundary condition 1

y(0,t) = 0

Then (2) becomes,

y(0,t) = (A cos 0 +B sin 0 ) ( C cos at + Dsin at) =0

(A) ( C cos at + D sin at)=0

A = 0

Using A = 0 in (2)

y(x,t) = ( B sin x) ( C cos at + D sin at) (3)

STEP-5


y(0,t) = 0

Then (2) becomes,

y(0,t) = (A cos 0 +B sin 0 ) ( C cos at + D sin at)

=0

(A) ( C cos at + D sin at)=0

A = 0

Using A = 0 in (2)

y(x,t) = ( B sin x) ( C cos at + D sin at) (3)

6

STEP-6


y( ,t) = 0

Then (3) becomes,

y( ,t) = (B sin ) ( C cos at + D sin at)=0

(B sin ) ( C cos at + D sin at)=0

= n

Then (3) becomes,

( , ) sin( ) cos( ) sin( )n x n at n at

y x t B C D

(4)

STEP-6


y( ,t) = 0

Then (3) becomes,

y( ,t) = (B sin ) ( C cos at + D sin at)=0

(B sin ) ( C cos at + D sin at)=0

= n

Then (3) becomes,

( , ) sin( ) cos( ) sin( )n x n at n at

y x t B C D

(4)

7 STEP-7


y(x,0) = 0

Then (4) becomes,

STEP-7



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e35

( , ) sin( ) cos0 sin 0n x

y x t B C D

=0

sin( ) 0n x

B C

C = 0

Then (4) becomes,

( , ) sin( ) sin( )n x n at

y x t B D

The most general solution is

1

( , ) sin( )sin( )n

n

n x n aty x t B

(5)

0t

y

t= 0Then (4) becomes,

Differentiating (5) partially w.r.to ‘t’ and put t =0

0

sin( ) sin( ) cos( )t

y n x n at n at n aB C D

t

sin( ) 0n x n a

B D

D = 0

Then (4) becomes,

( , ) sin( ) cos( )n x n at

y x t B C


1

( , ) sin( ) cos( )n

n

n x n aty x t B

(5)

8 STEP-8

Differentiating (5) partially w.r.to ‘t’

1

sin( ) cos( )n

n

y n x n at n aB

t

Using Boundary condition (4),

0t

y

t= f(x)

1

( ) sin( )n

n

n x n af x B

This is the Half Range Fourier Sine Series.

0

2( )sin( )n

n a n xB f x

0

2( )sin( )n

n xB f x dx

n a

STEP-8

Using Boundary condition (4),

y(x,0) = f(x)

1

( ,0) sin( )cos(0)n

n

n xy x B

1

( ) sin( )n

n

n xf x B

This is the Half Range Fourier Sine Series.

0

2( )sin( )n

n xB f x

9 STEP-9

The required solution is

1

( , ) sin( )sin( )n

n

n x n aty x t B

Where

0

2( )sin( )n

n xB f x dx

n a

STEP-9


1

( , ) sin( )sin( )n

n

n x n aty x t B

Where

0

2( )sin( )n

n xB f x dx


Pag

e36

ONE DIMENSIONAL HEAT EQUATION

TWO DIMENSIONAL HEAT FLOW EQUATION

1 The one dimensional heat equation is

22

2

u u

t x

The two Dimensional equation is

2 2

2 20

u u

x y

2 Boundary conditions

1.u(0,t) = 0 for 0t

2.u( ,t) = 0 for 0t

3.u(x,t) = f(x) for 0<x<

Boundary conditions

1.u(0,y) = 0 for 0<y<

2.u( ,y) = 0 for 0<y<

3.u(x, ) = 0 for 0<x<

4.u(x,0) = f(x) for 0<x<

3 The possible solutions are 2 2

2 2

( , ) ( )

( , ) ( cos sin )

( , ) ( )

x x t

t

u x t Ae Be Ce

u x t A x B x Ce

u x t Ax B C


( , ) ( )( cos sin )

( , ) ( cos sin )( )

( , ) ( )( )

x x

y y

u x y Ae Be C y D y

u x y A x B x Ce De

u x y Ax B Cy D

4, The most suitable solution is 2 2

( , ) ( cos sin ) tu x t A x B x Ce (2)

The most suitable solution is

( , ) ( cos sin )( )y yu x y A x B x Ce De (2)

5 Using boundary condition 1

u(0,t) = 0 2 2

(0, ) ( cos0 sin0) tu t A B Ce =0

2 2

( ) tA Ce =0

A = 0

Then (2) becomes 2 2

( , ) ( sin ) tu x t B x Ce (3)

Using boundary condition 1

u(0,y) = 0

(0, ) ( cos0 sin 0)( )y yu y A B Ce De

(0, ) ( )( )y yu y A Ce Be

A = 0

Then (2) becomes

( , ) ( sin )( )y yu x y B x Ce De (3)

6 Using boundary condition 2

u(l,t) = 0 2 2

( , ) ( sin ) tu t B Ce =0

2 2

( sin ) 0tB Ce

n

Then (3) becomes 2 2 2

2

( , ) sin( )

n tn x

u x t B Ce

The most general solution is 2 2 2

2

1

( , ) sin( )

n t

n

n

n xu x t B e

(4)

Using boundary condition 2

u(l,t) = 0

( , ) ( sin )( )y yu y B Ce De

0 ( sin )( )y yB Ce De

n

Then (3) becomes

( , ) ( sin )( )n y n y

n xu x y B Ce De

(4)


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e37

7 Using Boundary condition 3

u(x,0) = f(x)

1

( ,0) sin( )n

n

n xu x B

1

( ) sin( )n

n

n xf x B

This the Half range Fourier sine series

0

2( )sin( )n

n xB f x dx


u(x, ) =0

( , ) ( sin )( )n x

u x B Ce De

0 ( sin )( 0)n x

B C D

C=0

then (3) becomes

( , ) ( sin )( )n y

n xu x y B De


1

( , ) sin( )n y

n

n

n xu x y B e

(5

8 The required solution is 2 2 2

2

1

( , ) sin( )

n t

n

n

n xu x t B e

Where

0

2( )sin( )n

n xB f x dx

Using Boundary condition 4 y(x,0) = f(x)

0

1

( ,0) sin( )n

n

n xu x B e

1

( ) sin( )n

n

n xf x B

This the Half range Fourier sine series

0

2( )sin( )n

n xB f x dx


1

( , ) sin( )n y

n

n

n xu x y B e

Where

0

2( )sin( )n

n xB f x dx

QUESTION WITH ANSWER

1. Classify the Partial Differential Equation i)

2 2

2 2

u u

x y

Answer:


Pag

e38

2 2

2 2

u u

x y here A=1,B=0,&C=-1

B2

- 4AC=0-4(1)(-1)=4>0

The Partial Differential Equation is hyperbolic

2. Classify the Partial Differential Equation

2u u uxy

x y y x

Answer:

2u u uxy

x y y x here A=0,B=1,&C=0

B2

-4AC=1-4(0)(0)=1>0

The Partial Differential Equation is hyperbolic

3. Classify the following second order Partial Differential equation

2 22 2

2 2

u u u u

x y y x

Answer:

2 22 2

2 2

u u u u

x y y x here A=1,B=0,&C=1

B2

-4AC=0-4(1)(1)=-4<0 The Partial Differential Equation is Elliptic


2 2 2

2 24 4 6 8 0

u u u u u

x x y y x y

Answer:

2 2 2

2 24 4 6 8 0

u u u u u

x x y y x y

here A= 4,B =4, & C = 1

B2

-4AC =16 -4(4)(1) = 0

The Partial Differential Equation is Parabolic


i) 2 22 2 3 0xx xy yy xy u xyu x u u u


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e39

ii) 2 2 2 7 0xx yy x yy u u u u

Answer:

i) Parabolic ii) Hyperbolic (If y = 0)

iii)Elliptic (If y may be +ve or –ve)

6. In the wave equation

2 22

2 2

y yc

t x what does c

2

stands for?

Answer:

2 22

2 2

y yc

t x

here 2 T

am

T-Tension and m- Mass

7. In one dimensional heat equation ut

= α2

uxx

what does α2

stands for?

Answer:-

22

2

u u

t x

2

=k

c is called diffusivity of the substance

Where k – Thermal conductivity

- Density

c – Specific heat

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

Answer:

i) Heat flows from higher to lower temp

ii) The rate at which heat flows across any area is proportional to the area

and to the temperature gradient normal to the curve. This constant of

proportionality is known as the conductivity of the material. It is known as

Fourier law of heat conduction


Pag

e40

9. A tightly stretched string of length 2 is fastened at both ends. The midpoint of the

string is displaced to a distance ‘b’ and released from rest in this position. Write the

initial conditions.

Answer:

(i) y(0 , t) = 0

(ii) y(2 ,t) = 0

(iii)

0

0t

y

t

(iv) y(x , 0 ) =

0

(2 ) 2

bx x

bx x

10. What are the possible solutions of one dimensional Wave equation?


Answer:

y(x,t) = (A ex

+ B e- x

) (C eat

+ D e- at

)

y(x,t) = (A cos x + B sin x )( C cos at + D sin at)

y(x,t) = (Ax + B) ( Ct + D)

11. What are the possible solutions of one dimensional head flow equation?

Answer:


2 2

2 2

( , ) ( )

( , ) ( cos sin )

( , ) ( )

x x t

t

u x t Ae Be Ce

u x t A x B x Ce

u x t Ax B C

12. State Fourier law of heat conduction

Answer:

uQ kA

x

(the rate at which heat flows across an area A at a distance from one end of a bar is

proportional to temperature gradient)


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e41

Q=Quantity of heat flowing

k – Thermal conductivity

A=area of cross section ;u

x=Temperature gradient

13. What are the possible solutions of two dimensional head flow equation?

Answer:


( , ) ( )( cos sin )

( , ) ( cos sin )( )

( , ) ( )( )

x x

y y

u x y Ae Be C y D y

u x y A x B x Ce De

u x y Ax B Cy D

14. The steady state temperature distribution is considered in a square plate with sides x

= 0 , y = 0 , x = a and y = a. The edge y = 0 is kept at a constant temperature T and the

three edges are insulated. The same state is continued subsequently. Express the

problem mathematically.

Answer:

U(0,y) = 0 , U(a,y) = 0 ,U(x,a) = 0, U(x,0) = T

15. An insulated rod of length 60cm has its ends A and B maintained 20°C and

80°C respectively. Find the steady state solution of the rod

Answer:

Here a=20°C & b=80°C

In Steady state condition The Temperature ( , )b a x

u x t al

80 2020

60

x

( , ) 20u x t x

16. Write the D’Alembert’s solution of the one dimensional wave equation?

Answer:


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e42

1 1( )

2 2

x at

x at

y x at x at v da

here x f x g x

v x ax f ag

17. What are the boundary conditions of one dimensional Wave equation?

Answer:

Boundary conditions

1. y(0,t) = 0 for 0t

2. y( , t) = 0 for 0t

3. y(x,0) = 0 for 0 < x <

4.

0t

y

t= f(x) for 0 < x <

18. What are the boundary conditions of one dimensional heat equation?

Answer:

Boundary conditions

1.u(0,t) = 0 for 0t

2.u( ,t) = 0 for 0t

3.u(x,t) = f(x) for 0<x<

19. What are the boundary conditions of one dimensional heat equation?

Answer:

Boundary conditions

1.u(0,y) = 0 for 0<y<

2.u( ,y) = 0 for 0<y<

3.u(x, ) = 0 for 0<x<

4.u(x,0) = f(x) for 0<x<


Pag

e43

20.T he ends A and B has 30cm long have their temperatures 30c and 80c until steady

state prevails. If the temperature A is raised to40c and Reduced to 60C, find the

transient state temperature

Answer:


In Steady state condition The Temperature ( , )b a x

u x t al


60 40 240 40

30 3t

xu x

PART-B QUESTION BANK APPLICATIONS OF PDE

1. A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in its

equilibrium position. If it is set vibrating giving each point a velocity 3x (l-x). Find the

displacement.

2. A string is stretched and fastened to two points and apart. Motion is started by displacing

the string into the form y = K(lx-x2

) from which it is released at time t = 0. Find the

displacement at any point of the string.

3. A taut string of length 2l is fastened at both ends. The midpoint of string is taken to a

height b and then released from rest in that position. Find the displacement of the string.

4. A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in its

position given by y(x, 0) =3

0 sinx

yl

. If it is released from rest find the displacement.

5. A string is stretched between two fixed points at a distance 2l apart and points of the

string are given initial velocities where

0 < x < 1

(2 ) 0 < x < 1

cx

lV

cl x

l

Find the

displacement.

6. Derive all possible solution of one dimensional wave equation. Derive all possible solution

of one dimensional heat equation. Derive all possible solution of two dimensional heat

equations.

7. A rod 30 cm long has its end A and B kept at 20o

C and 80o

C, respectively until steady state

condition prevails. The temperature at each end is then reduced to 0o

C and kept so. Find

the resulting temperature u(x, t) taking x = 0.


Pag

e44

8. A bar 10 cm long , with insulated sides has its end A & B kept at 20o

C and 40o

C respectively

until the steady state condition prevails. The temperature at A is suddenly raised to 50o

C

and B is lowered to 10o

C. Find the subsequent temperature function u(x , t).

9. A rectangular plate with insulated surface is 8 cm wide so long compared to its width that

it may be considered as an infinite plate. If the temperature along short edge y = 0 is u (

x ,0) = 100sin8

x 0 < x < 1. While two edges x = 0 and x = 8 as well as the other short

edges are kept at 0o

C. Find the steady state temperature.

10. A rectangular plate with insulated surface is10 cm wide so long compared to its width that

it may be considered as an infinite plate. If the temperature along short edge y = 0 is given

by

20 0 x 5

20(10 ) 5 x 10

xu

x and all other three edges are kept at 0

o

C. Find the steady

state temperature at any point of the plate.


Pag

e45

Unit - 4

FOURIER TRANSFORMS

FORMULAE

1. Fourier Transform of f(x) is isx

-

1[ ( )] f(x)e

2F f x dx

2. The inversion formula -isx

-

1( ) ( )e

2f x F s ds

3. Fourier cosine Transform Fc

[f(x)] = Fc

(s) =

0

2( )cosf x sxdx

4. Inversion formula f(x) =

0

2( )coscF s sxds

5. Fourier sine Transform (FST) Fs

[f(x)] = Fs

(s) =

0

2( )sinf x sxdx

6. Inversion formula f(x) =

0

2( )sinsF s sxds

7. Parseval’s Identity

2

2( ) ( )f x dx F s ds

8. Gamma function 1

0

1, 1 &

2

n xn x e dx n n n

9. 2 2

0

cosax ae bxdx

a b

10 2 2

0

sinax be bxdx

a b

11. 0

sin

2

axdx

x


Pag

e46

12. 2 2

0

&2

x xe dx e dx

13. cos & sin2 2

iax iax iax iaxe e e eax ax

ORKING RULE TO FIND THE FOURIER TRANSFORM

Step1: Write the FT formula.

Step2: Substitute given f(x) with their limits.

Step3: Expand isxe as cos sx + isin sx and use Even & odd property

Step4: Integrate by using Bernoulli’s formula then we get F(s)

WORKING RULE TO FIND THE INVERSE FOURIER TRANSFORM

Step1: Write the Inverse FT formula

Step2: Sub f(x) & F(s) with limit , in the formula

Step3: Expand isxe as cos sx -isin sx and equate real part

Step4: Simplify we get result

WORKING RULE FOR PARSEVAL’S IDENTITY

If F(s) is the Fourier transform of f(x) then

2

2( ) ( )f x dx F s ds is known as Parseval’s identity.

Step1: Sub f(x) & F(s) With their limits in the above formula

Step2: Simplify we get result

WORKING RULE TO FIND FCT

Step1: Write the FCT formula & Sub f(x) with its limit in the formula

Step2: Simplify, we get ( )C

F S

WORKING RULE TO FIND INVERSE FCT


Pag

e47

Step1: Write the inverse FCT formula & Sub ( )C

F S with its limit in the formula

Step2: Simplify, we get f(x)

WORKING RULE TO FIND FST

Step1: Write the FST formula & Sub f(x) with its limit in the formula

Step2: Simplify, we get ( )SF S

WORKING RULE TO FIND INVERSE FCT

Step1: Write the inverse FST formula & Sub ( )sF S with limit in the formula

Step2: Simplify, we get f(x)

WORKING RULE FOR f(x) = axe

Step:1 First we follow the above FCT & FST working rule and then we get this

result

Fc

(e-ax

) = 2 2

2 a

a s F

s

(e-ax

) = 2 2

2 s

a s

By Inversion formula, By Inversion formula,

2 2

0

cos

2

axsxds e

a s a

2 2

0

sin2

axssxds e

a s

TYPE-I : If problems of the form i)2 2

x

x a ii)

2 2

1

x a , then use Inversion formula

TYPE-II: If problems of the form i)

2

22 2

0

xdx

x a ii)

22 2

0

dx

x a, then use Parseval’s Identity

TYPE-III

2 2 2 2

0

dx

x a x b , then use

0 0

( ) ( ) ( ) ( )C Cf x g x dx F f x F g x dx


Pag

e48

UNIT - 4

FOURIER TRANSFORM

1. State Fourier Integral Theorem.

Answer:

If ( )f x is piece wise continuously differentiable and absolutely on , then,

( )1( )

2

i x t sf x f t e dt ds.


Pag

e49

2. StateandproveModulation

theorem.

1cos

2F f x ax F s a F s a

Proof:

( ) ( )

1cos cos

2

1

22

1 1 1 1

2 22 2

1 1

2 2

1cos

2

isx

iax iaxisx

i s a x i s a x

F f x ax f x ax e dx

e ef x e dx

f x e dx f x e dx

F s a F s a

F f x ax F s a F s a

3. State Parseval’s Identity.

Answer:

If F s is a Fourier transform of f x , then

2 2

F s ds f x dx

4. State Convolution theorem.

Answer:

The Fourier transform of Convolution of f x and g x is the product of their Fourier

transforms.

F f g F s G s

5. State and prove Change of scale of property.


Pag

e50

Answer:

If ,F s F f x then

1s

aF f ax Fa

1

2

1; where

2

sa

isx

i t

F f ax f ax e dx

dtf t e t ax

a

1 sF f ax Faa

6. Prove that if F[f(x)] = F(s) then ( ) ( ) ( )

nn n

n

dF x f x i F s

ds

Answer:

1

2

isxF s f x e dx

Diff w.r.t s ‘n’ times

1

2

1( ) ( )

2

nn isx

n

n n isx

dF s f x ix e dx

ds

f x i x e dx

1 1( )

( ) 2

1( ) ( )

2

nn isx

n n

nn n isx

n

nnn

n

dF s x f x e dx

i ds

di F s x f x e dx

ds

dF x f x i F s

ds

7. Solve for f(x) from the integral equation

0

( )cos sf x sxdx e

Answer:


Pag

e51

0

( )cos sf x sxdx e

0

2cos

2

c

s

c

F f x f x sx dx

F f x e

0

0

2

0

2( ) cos

2 2cos

2 2 1cos

1

c

s

s

f x F f x sx ds

e sx ds

e sx dsx

2 2

0

cos

1,

ax ae bx dx

a b

a b x

8. Find the complex Fourier Transform of

1( )

0 0

x af x

x a

Answer:

1

2

isxF f x f x e dx

;x a a x a


Pag

e52

11

2

1(cos sin )

2

a

isx

a

a

a

F f x e dx

sx i sx dx

00

2 2 sin(cos )

2 2

2 sin

aasx

sx dxs

as

s

[Use even and odd property second term become zero]

9. Find the complex Fourier Transform of ( )0 0

x x af x

x a

Answer:

1

2

1

2

1(cos sin )

2

isx

a

isx

a

a

a

F f x f x e dx

xe dx

x sx i sx dx

;x a a x a

2

0 0

2

2 2 cos sin( ( sin ) (1)

2 2

2 cos sin

aai sx sx

x i sx dx xs s

as as asi

s

[Use even and odd property first term become zero]


Pag

e53

10. Write Fourier Transform pair.

Answer:

If ( )f x is defined in , , then its Fourier transform is defined as

1

2

isxF s f x e dx

If F s is an Fourier transform of f x , then at every point of Continuity of f x , we

have

1

2

isxf x F s e ds .

11. Find the Fourier cosine Transform of f(x) = e-x

Answer:

0

0

2

2cos

2cos

2 1

1

c

x x

c

x

c

F f x f x sx dx

F e e sx dx

F es

2 2

0

cosax ae bx dx

a b

12. Find the Fourier Transform of

,( )

0,

imxe a x bf x

otherwise

Answer:


Pag

e54

1

2

1 1

2 2

isx

b bi m s ximx isx

a a

F f x f x e dx

e e dx e dx

1 1 1

2 2

bi m s x

i m s b i m s a

a

ee e

i m s i m s

13. Find the Fourier sine Transform of 1

x.

Answer:

0

0

2sin

2 sin 2

2

1

2

s

s

F f x f x sx dx

sxdx

x

Fx

14. Find the Fourier sine transform of xe

Answer:

0

0

2

2sin

2sin

2

1

s

x x

s

x

s

F f x f x sx dx

F e e sx dx

sF e

s

2 2

0

sinax be bx dx

a b

15. Find the Fourier cosine transform of 2 2x xe e

Answer:


Pag

e55

0

2coscF f x f x sx dx

2 2

0

2

0 0

22 2 cos

2cos 2 cos

x x x x

c

x x

F e e e e sx dx

e sx dx e sx dx

2 2 2 2

2 2 1 2 1 12 2

4 1 4 1s s s s

16. Find the Fourier sine transform of

1 , 0 1( )

0 1

xf x

x

Answer:

0

1

0 1

11

00

2sin

2sin sin

2 2 cos1sin 0

2 1 cos

s

s

F f x f x sx dx

F f x f x sx dx f x sx dx

sxsx dx

s

s

s


Pag

e56

17. Obtain the Fourier sine transform of

, 1

( ) 2 , 1 2

0, 2

x o x

f x x x

x.

Answer:

0

1 2

0 1

1 2

2 2

0 1

2sin

2sin 2 sin

2 cos sin cos sin2

sF f x f x sx dx

x sx dx x sx dx

sx sx sx sxx x

s s s s

2 2 2

2

2 cos sin sin 2 cos sin

2 2sin sin 2

s

s s s s sF f x

s s s s s

s s

s

18. Define self reciprocal and give example.

If the transform of f x is equal to f s , then the function f x is called self

reciprocal.

2

2x

e is self reciprocal under Fourier transform.


Pag

e57

19. Find the Fourier cosine Transform of

0( )

0

x xf x

x

Answer:

0 0

2 2 2

0

2

2 2cos cos

2 sin cos 2 cos 1sin

2 sin cos 1

cF f x f x sx dx x sx dx

sx sx sx s

s s s s s

s s s

s

20. Find the Fourier sine transform of 2 2

x

x a.

Answer:

L et

axf x e

2 2

2ax

s

sF e

s a

Using Inverse formula for Fourier sine transforms

2 2

0

2 2

0

2 2sin

( ) sin , 02

ax

ax

se sx ds

s a

sie sx ds e a

s a


Pag

e58

Change x and s, we get 2 2

0

sin2

asxsx dx e

x a

2 2 2 2

0

2sin

2

2 2

s

as as

x xF sx dx

x a x a

e e

FOURIER TRANSFORM

PART-B

1. (i)Find the Fourier Transform of

21 1( )

0 1

x if xf x

if x and hence

deduce that (i)3

0

cos sin 3cos

2 16

x x x xdx

x (ii)

2

3

0

sin cos

15

x x xdx

x

(ii). Find the Fourier Transform of

2 2

( )0 0

a x x af x

x a . hence

deduce that 3

0

sin cos

4

x x xdx

x

2. Find the Fourier Transform of

1( )

0

if x af x

if x a and hence evaluate

0

sin)

xi dx

x ii)

2

0

sin xdx

x

4. Find Fourier Transform of

1 1( )

0 1

x if xf x

if x and hence evaluate

i)

2

0

sin xdx

x ii)

4

0

sin xdx

x


Pag

e59

5. Evaluate i)

2

22 2

0

xdx

x a ii) 2

2 20

dx

x a

6 i). Evaluate (a) 2 2 2 2

0

dx

x a x b(b)

2

2 2 2 2

0

x dx

x a x b

ii). Evaluate (a) 2 2

0 1 4

dx

x x(b)

2

2 2

0 4 9

t dt

t t

7. (i)Find the Fourier sine transform of

sin ;( )

0 ;

x when o xf x

whenx

(ii) Find the Fourier cosine transform of

cos ;( )

0 ;

x when o x af x

whenx a

8. (i) Show that Fourier transform

2

2

x

e is

2

2

s

e

(ii)Obtain Fourier cosine Transform of

2 2a xe and hence find Fourier sine Transform x

2 2a xe

9. (i) Solve for f(x) from the integral equation

0

( )cosf x xdx e

(ii) Solve for f(x) from the integral equation

0

1 ,0 1

( )sin 2 ,1 2

0 , 2

t

f x tx dx t

t

10. (i) Find Fourier sine Transform of xe , x>0 and hence deduce that

2

0

sin

1

x xdx

x

(ii) Find Fourier cosine and sine Transform of4xe , x>0 and hence deduce

that8 8

2 2

0 0

cos2 sin 2( ) ( )

16 8 16 8

x x xi dx e ii dx e

x x

11.(i)Find &ax ax

S cF xe F xe

(ii) Find &ax ax

S c

e eF F

x x

(iii) Find the Fourier cosine transform of ( ) cosaxf x e ax


Pag

e60

Z - TRANSFORMS

Definition of Z Transform

Let {f(n)} be a sequence defined for n = 0, 1,2 … and f(n) = 0 for n< 0 then its

Z – Transform is defined as

( ) ( ) n

n

Z f n F z f n z (Two sided z transform)

0

( ) ( ) n

n

Z f n F z f n z (One sided z transform)

Unit sample and Unit step sequence

The unit sample sequence is defined as follows

1 0( )

0 0

for nn

for n

The unit step sequence is defined as follows

1 0( )

0 0

for nu n

for n

Properties


Pag

e61

1. Z – Transform is linear

(i) Z {a f(n) + b g(n)} = a Z{f(n)} + b Z{g(n)}

2. First Shifting Theorem

(i) If Z {f(t)} = F(z),

then aT

at

z zeZ e f t F z

(ii) If Z {f(n)} = F(z),

then n z

Z a f n Fa

3. Second Shifting Theorem

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

(i)Z[f(n +1)] = z[ F(z) – f(0)]

(ii)Z[f(n +2)] = 2z [ F(z) – f(0)-f(1)

1z ]

(iii)Z[f(n +k)] = kz [ F(z) – f(0)-f(1)

1z - f(2)2z ………- f(k-1)

( 1)kz ]

(iv) Z[f(n -k)] = kz F(z)

4. Initial Value Theorem

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

F z

5. Final Value Theorem

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

lim ( ) lim( 1) ( )n z

f n z F z

PARTIAL FRACTION METHODS


Pag

e62

Model:I

1 A B

z a z b z a z b

Model:II

2 2

1

( )

A B C

z a z b z bz a z b

Model:III

22

1 A Bz C

z a z bz a z b

Convolution of Two Sequences

Convolution of Two Sequences {f(n)} and {g(n)} is defined as

0

{ ( )* ( )} ( ) ( )n

K

f n g n f K g n K

Convolution Theorem

If Z[f(n)] = F(z) and Z[g(n)] = G(z) then Z{f(n)*g(n)} = F(z).G(z)

WORKING RULE TO FIND INVERSE Z-TRANSFORM USING CONVOLUTION THEOREM

Step: 1 Split given function as two terms

Step: 2 Take 1z both terms

Step: 3 Apply 1z formula

Step: 4 Simplifying we get answer

Note:

12 1

1 .......1

nn a

a a aa


Pag

e63

1

2 11 .......

1 ( )

n

n aa a a

a

Solution of difference equations

Formula

i) Z[y(n)] = F(z)

ii) Z[y(n +1)] = z[ F(z) – y(0)]

iii) Z[y(n +2)] = 2z [ F(z) – y(0)- y(1)

1z ]

iv) Z[y(n +3)] = 3z [ F(z) – y(0)- y(1)

1z + y(2)2z ]

WORKING RULE TO SOLVE DIFFERENCE EQUATION:

Step: 1 Take z transform on both sides

Step: 2 Apply formula and values of y(0) and y(1).

Step: 3 Simplify and we get F(Z)

Step:4 Find y(n) by using inverse method

Z - Transform Table

No.

f(n)

Z[f(n)]

1. 1

1

z

z

2. an

z

z a

3. n

2( 1)

z

z

4. n2

2

3( 1)

z z

z


Pag

e64

6. 1

n log

( 1)

z

z

7. 1

1n log

( 1)

zz

z

8. 1

1n

1log

( 1)

z

z z

9. ean

( )a

z

z e

10. 1

!n

1

ze

11. Cos n

2

( cos )

2 cos 1

z z

z z

12. sin n

2

sin

2 cos 1

z

z z

13.

cos2

n

2

2 1

z

z

sin2

n

2 1

z

z

14. nna

2( )

az

z a

f(t) Z(f(t)

1 t

2( 1)

Tz

z

2. t2

2

3

( 1)

( 1)

T z z

z

3 eat

( )aT

z

z e

4. Sin t

2

sin

2 cos 1

z T

z z T


Pag

e65

5. cos t

2

( cos )

2 cos 1

z z T

z z T

TWO MARKS QUESTIONS WITH ANSWER

1. Define Z transform

Answer:

Let {f(n)} be a sequence defined for n = 0, 1,2 … and f(n) = 0 for n< 0 then

its Z – Transform is defined as

( ) ( ) n

n

Z f n F z f n z (Two sided z transform)

0

( ) ( ) n

n

Z f n F z f n z (One sided z transform)

Find the Z Transform of 1

Answer:

0

n

n

Z f n f n z

0

1 (1) n

n

Z z

1 21 ....z z

111 z

1 11 1

11

z z

z z z

11

zZ

z

2. Find the Z Transform of n

Answer:


Pag

e66

0

0

1 2 3

0

2 22

1 1

2

0 2 3 ...

1 1 11 1

1

1

n

n

n

n

n

n

Z f n f n z

Z n nz

nz z z z

zz z

z z z z

z

z

3. Find the Z Transform of n2

.

Answer:

2 dZ n Z nn z Z n

dz, by the property,

2 2

2 4 3

1 2 1( )

( 1)1 1

z z zd z z zz z

dz zz z

4. State Initial & Final value theorem on Z Transform

Initial Value Theorem

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

F z

Final Value Theorem

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

lim ( ) lim( 1) ( )n z

f n z F z

6. State convolution theorem of Z- Transform.


Pag

e67

Answer:

Z[f(n)] = F(z) and Z[g(n)] = G(z) then Z{f(n)*g(n)} = F(z) · G(z)

7. Find Z –Transform of nna

Answer:

0

0

1 2 3

0

2

2

0 2 3 ...

1

n

n

n n n

n

n

n

Z f n f n z

Z na na z

a a a an

z z z z

a a az

z z z a

8. Find Z – Transform of cos

2

n and

sin2

n

Answer:

We know that

0

n

n

Z f n f n z

2

coscos

2 cos 1

z zZ n

z z

2

22

cos2

cos2 1

2 cos 12

z zz

Z nz

z z


Pag

e68

Similarly 2

sinsin

2 cos 1

zZ n

z z

22

sin2sin

2 12 cos 1

2

zz

Z nz

z z

9. Find Z – Transform of 1

n

Answer:

0

n

n

Z f n f n z

0

1 2 3

1

1

1 1

1....

1 2 3

1 1log 1 log

log1

n

n

n

n

Z zn n

z z zz

n

z

z z

z

z


!n

Answer:


Pag

e69

1

0

0

1 2 3

0

1

1 1

! !

11 ....

! 1! 2! 3!

n

n

n

n

n

n

z z

Z f n f n z

Z zn n

z z zz

n

e e


1n

Answer:

0

0

( 1)

0

2 31

1 1

1 1

1

1

....2 3

1log 1

log1

n

n

n

n

n

n

Z f n f n z

Z zn n

z zn

z zz z

zz

zz

z

12. Find Z – Transform of an


Pag

e70

Answer:

0

0 0

1 2 3

1

1

1 ...

1

n

n

nnn

nn n

Z f n f n z

a aZ a

z z

a a a

z z z

a

z

z a z

z z a

13. State and prove First shifting theorem

Statement: If Z f t F z , then ( )at aTZ e f t F ze

Proof:

0

( ) ( )at anT n

n

Z e f t e f nT z

As f(t) is a function defined for discrete values of t, where t = nT,

then the Z-transform is

0

( ) ( ) ( )n

n

Z f t f nT z F z ).

0

( ) ( ) ( )n

at aT aT

n

Z e f t f nT ze F ze

14. Define unit impulse function and unit step function.

The unit sample sequence is defined as follows:

1 0( )

0 0

for nn

for n

The unit step sequence is defined as follows:


Pag

e71

1 0( )

0 0

for nu n

for n

15. Find Z – Transform of

atZ e

Answer:

0 0

n nat anT n aT n aT

n n

n

aT

Z e e z e z z e

z zz a

z e z a

[Using First shifting theorem]


2tZ te

Answer:

2

2

2

2

2

22

1

1

T

T

t

z ze

z ze

T

T

TzZ te Z t

z

Tze

ze


17. Find Z – Transform of cos2tZ e t

Answer:

2

coscos 2 cos 2

2cos 1T

T

t

z ze

z ze

z zZ e t Z t

z z

2

cos

2cos 1

T T

T T

ze ze T

ze T ze



Pag

e72


2 t TZ e

Answer:

Let f (t) = e2t

, by second sifting theorem

2( )

2

2 2

( ) ( ) (0)

11

1 1

t T

T

T T

Z e Z f t T z F z f

zez z

ze ze

19. Find Z – Transform of sinZ t T

Answer:

Let f (t) = sint , by second sifting theorem

2

2 2

sin( ) ( ) ( ) (0)

sin sin0

2cos 1 2cos 1

Z t T Z f t T z F z f

z t z tz

z t z z t z

20. Find Z – transform of 1 2n n

Answer:

0

n

n

Z f n f n z

2

2 2

2

3 2

1 2 2 2

3 2 3 2 1

3 211 1

Z n n Z n n n

Z n n z n z n z

z z z z

zz z


Pag

e73

QUESTION BANK

Z-TRANSFORMS

1. (i)Find

21 8

(2 1)(4 1)

zZ

z z &

21 8

(2 1)(4 1)

zZ

z zby convolution theorem.

(ii) Find

21

( )( )

zZ

z a z b &

21

( 1)( 3)

zZ

z zby convolution theorem

2. (i) Find

21

2( )

zZ

z a &

21

2( )

zZ

z aby convolution theorem

3. (i ) State and prove Initial & Final value theorem.

(ii) State and prove Second shifting theorem

(i) Find the Z transform of 1

( 1)( 2)n n&

2 3

( 1)( 2)

n

n n

4. (i) Find

21

2( 4)

zZ

z by residues.

(ii) Find the inverse Z transform of

2

21 ( 1)

z z

z zby partial fractions.

5. (i) Find 1

2 2 2

zZ

z z &

21

2 7 10

zZ

z z

6. (i)Find the Z transform of 1

( )!

f nn

Hence find 1

( 1)!Z

n and

1

( 2)!Z

n.

(ii) Find 1

!Z

n and also find the value of sin( 1)n and cos( 1)n .

7. (i)Solve 2 6 1 9 2 0 0& 1 0ny n y n y n with y y

(ii) Solve 2 4 1 4 0y n y n y n y(0) = 1 ,y(1) =0

8. (i )Solve 3 1 4 2 0, 2 (0) 3& (1) 2y n y n y n n given y y

(ii) Solve 3 3 1 2 0, 0 4, 1 0& 2 8y n y n y n y y y ,

9. (i)Find cos & sinZ n Z n and also find cos & sinn nZ a n Z a n

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