# ma2211

May 22, 2018 | Author: Saravanan MJ | Category: Fourier Series, Sine, Fourier Transform, Trigonometric Functions, Integer

#### Short Description

all unit formula...

#### Description

Engineering Mathematics Material

2013

SUBJECT NAME

: Transforms and Partial Differential Equations

SUBJECT CODE

: MA2211

MATERIAL NAME

: FormulaMaterial FormulaMaterial

MATERIAL CODE

: JM08AM3005

Unit – I (Fourier Series) 1)

Dirichlet’s Conditions: Any function f ( x ) can be expanded as a Fourier series

a 0

bn  sin n x where   a0 , an , b n  are constants provided the

a n cos n x

2

n 1

n  1

following conditions are true.  f ( x ) is periodic, single – single – valued  valued and finite.

2)

f ( x ) has a finite number of discontinuities in any one period.

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

The Fourier Series in the interval (0,2π): f (x)

a 0

2

Where a0

a n cos nx nx n 1

1

bn  sin n x   n  1

2

f ( x )dx ,  an

1

2

f ( x ) cos n x dx ,  bn  0

0

3)

1

2

f ( x ) sin n x dx   0

The Fourier Series in the interval (-π,π): (-π,π): f (x)

a 0

2

Where a0

a n cos nx nx n 1

2

bn  sin n x   n  1

f ( x )dx ,  an  0

2

f ( x ) cos n x dx ,  bn

2

f ( x ) sin n x dx   0

0

In this interval, we have to verify the function is either odd function or even function. If it is even function then then find only a0  and a n  ( b n  0 ). If it is odd function then find only b n  ( a0

a n

0 ).

Engineering Mathematics Material

2013

If the function is neither odd nor even then you should find 1

a0 , an and b n  by using the following formulas a0

1

an

4)

f ( x ) cos n x dx ,  bn

1

f ( x )dx ,

f ( x ) sin n x dx .

The half range Fourier Series in the interval (0,π):  The half range Cosine Series in the interval (0,π): a 0

f (x)

a n  cos n x

2

n  1

2

Where a0

f ( x )dx ,  an

2

0

f ( x ) cos n x dx   0

The half range Sine Series in the interval (0,π): f (x)

bn  sin n x   n  1

2

Where bn

f ( x ) sin n x dx   0

5)

The Parseval’s Identity in the interval (0,2π):

1

2

2

2

[ f ( x )] dx 0

6)

4

2

bn  ]

n  1

2

a 0

2

[ f ( x )] dx

[an2

4

0

2

bn  ]

n  1

The Parseval’s Identity for half range cosine series in the i nterval (0,π): 2

2

a 0

2

[ f ( x )] dx

2

a n

4

0

8)

[an2

The Parseval’s Identity in the interval ((-π,π):

2 7)

a 0

n  1

The Parseval’s Identity for half range sine series in the interval (0,π): 2

2

2

[ f ( x )] dx

bn    n  1

0

Change of interval: 9)

The Fourier Series in the interval (0,2ℓ): f (x)

a 0

2

Where a0

a n cos

n x

n 1

1

n  1

2

f ( x )dx ,  an  0

n x bn  sin

1

2

f ( x ) cos 0

n x

dx ,  bn

1

2

f ( x ) s in 0

n x

dx

Engineering Mathematics Material 10)

2013

The Fourier Series in the interval (-ℓ (-ℓ, ℓ): f (x)

a 0

n x

a n cos

2

n x bn  sin

n 1

2

Where a0

n  1

f ( x )dx ,  an

2

0

f ( x ) cos

n x

2

dx ,  bn

f ( x ) sin

n x

dx

0

0

In this interval, you have to verify the function is either odd function or even function. If it is even function then find only a0  and a n  ( b n  0 ). If it is odd odd function then find only b n  ( a0

11)

a n

0 ).

The half range Fourier Series in the interval (0, ℓ): 

The half range Cosine Series in the interval (0, ℓ): f (x)

a 0

a n  cos

2

n x

n  1

2

Where a0

f ( x )dx ,  an

2

0

f ( x ) cos

n x

dx

0

The half range Sine Series in the interval (0, ℓ): f (x)

b n  sin

n x

n  1

2

Where bn

f ( x ) sin

n x

dx

0

12)

The Parseval’s Identity in the interval (0,2ℓ): (0,2ℓ): 1

2

2

2

[ f ( x )] dx 0

13)

4

2

bn  ]

n  1

2

2

[ f ( x )] dx 0

a 0

2

[an

4

2

bn  ]

n  1

The Parseval’s Identity for half range cosine series in the i nterval (0,ℓ): (0,ℓ): 2

2

2

[ f ( x )] dx 0

15)

2

[ an

The Parseval’s Identity in the interval (-ℓ (-ℓ,ℓ): 2

14)

a 0

a 0

2

an

4

n  1

The Parseval’s Identity for half range sine series in the interval (0,ℓ): (0,ℓ):

2

[ f ( x )]2 dx 0

2

bn    n  1

Engineering Mathematics Material 16)

2013

Harmonic Analysis: The method of calculation of Fourier constants by means of numerical calculation is called as Harmonic analysis.

f (x)

a 0

where 2 a0 n

b1

a n cos nx nx

2

y ,

2

n  1

a1

y sin x ,

n

bn  sin n x

n 1

2

y cos x ,

n

b2

2

2

a2

n

y sin 2 x ,

n

y cos 2x , a 3

2

b3

When the values of x is given as numbers the

2 n

. .. y cos 3x , ..

, ...  y sin 3x ,.

is calculated by

2 x

. T  Where T is period, n is the number of values given. If the first and last y values are same we can omit one of them. Complex form of Fourier Series: 17)

The Complex form of Fourier Series in the interval (0,2π): in x

f (x)

cn e

where cn

18)

in x

cn e

where cn

2

in x

dx

0

1

f ( x )e

2

in x

dx

The Complex form of Fourier Series in the i nterval (0,2ℓ (0,2ℓ): in x

f (x)

cn e

where cn

20)

f ( x )e

The Complex form of Fourier Series in the interval ( -π,π): f (x)

19)

2

1

1

2

2

in x

f ( x )e

dx

0

The Complex form of Fourier Series in the interval ( -ℓ,ℓ): in x

f (x)

cn e

where cn

1 2

in

f ( x )e

dx

Unit – II (Fourier Transforms) 1)

Fourier Integral theorem The Fourier integral theorem of f ( x ) in the interval f (x)

1

f ( x ) cos ( t 0

)dx d

,

is

Engineering Mathematics Material 2)

2013

Convolution Theorem If F [ s ]  and G [ s ]  are the Fourier transform of the functions f ( x ) and g ( x ) respectively, then F [ f ( x ) * g( x )]

3)

F s .G s

The Fourier Transform of a function f ( x ) is given by F [ f ( x )] is denoted by F [ s ] .

1

isx

4)

Fourier Transform F [ s]

5)

Inverse Fourier Transform f ( x )

6)

The Fourier transforms and Inverse Fourier transforms are called Fourier transforms pairs.

7)

Fourier Sine Transform F s [ s]

F [ f ( x )]

f ( x )e dx

2 1

F [ s]e

2

F s [ f ( x )]

2

isx

ds

f ( x ) sin sx dx   0

8)

Fourier Cosine Transform F c [ s]

F c [ f ( x )]

2

f ( x ) cos sx dx   0

9)

If f ( x )

ax

then the Fourier Cosine and Sine transforms as follows

a) F c [ f ( x )] b) F s [ f ( x )] 10)

F s [ x f ( x )]

b) F c [ x f ( x )]

a

2

2

2

s  s

a

2

2

d  ds  d  ds

F c [ f ( x )]

F s [ f ( x )]

Parseval’s Identity 2

a)

F ( s) ds ds

b)

2

f ( x ) dx   2

f ( x ) g( x )dx    (Or)

F c ( s)Gc ( s)ds 0

12)

Property a)

11)

2

0

)] Condition for Self reciprocal F [ f ( x )]

0

f ( s)

2

F c ( s) ds ds

f ( x ) dx   0

Engineering Mathematics Material

2013

Unit – III (Partial Differential Equation) 1) Lagrange’s Linear equation The equation of the form Pp

Qq

dx

dy

dz

P

Q

then the subsidiary equation is

2) Homogeneous Linear Partial Differential Equation of higher order with constant coefficients: 2

The equation of the form a

x

2

z

b

2

2

z

c

x y

f ( x , y)

2

y

The above equation can be written as aD

2

bD D

2

x

2

f ( x , y )  …………………………….. (1)

z

2

2

where D

cD

2

, D

and D

2

y

, D

2

The solution of above equation is z = C.F + P.I  Complementary Complementary Function (C.F) :

1

m1

To find C.F consider the auxiliary equation by replacing  D  by m and  D by 2 bm c   0 , solving this equation 1.The equation (1) implies that am we get two values of m. m. The following table gives C.F C.F of the above equation. Nature of m Complementary Complementary Function m 2 C.F = f 1 ( y m 1 x ) f 2 ( y m 2 x )

2

m1

m 2

3

m1

m2

4

m1

5

m1

Sl.No.

C.F = f 1 ( y

mx )

m 3

C.F = f 1 ( y

m1 x )

m2

m 3

C.F = f 1 ( y

mx )

m2 ,

m 3 is different different

xf 2 ( y f 2( y xf 2 ( y

C.F = f 1 ( y m x )

xf 2 ( y

2

cD   .

Particular Integral (P.I) : To find P.I consider ( D , D ) 0 , then P.I

Type: 1If 1 If

f ( x , y )

Type: 2

If f ( x , y ) P .I

ax by

e

1 ( D , D  )

ax by

0.

bD D

2

m x )  m2 x ) mx )

mx mx )

f 3( y 2

x f 3(y

f 3( y

m 3 x )  m x )

m 3 x )

Engineering Mathematics Material

2013

Replace D   by a and D  by b. If If

( D , D  )

0 , then it is P.I.

0 , then diff. denominator w.r.t D   and multiply x in

( D , D  )

numerator. Again replace D   by a and D   by b. If again denominator equal to zero then continue the same procedure. Type: 3

If f ( x , y )

si n ( a x

1

P .I

( D , D  )

b y ) ( o r ) c os ( a x

by )

by ) (or ) cos(ax

by )

sin(ax

2 2 2 2 Here replace D  by a  , D   by b   and DD   by ab  . Do not replace for D and D  . If the denominator denominator equal to zero, then apply the same producer as in Type: 2.

Type: 4

m

If f ( x , y ) 1

P .I

m

( D , D  ) 1

1 1

x y

x y m

g ( D , D  )

x y  1

g( D , D )

m

x y

Here we can use Binomial formula as follows: 1

i) 1

x

ii) 1

x

iii) 1

x

iv) 1

x

v) (1

x)

vi) (1

x)

1

2 2

3 3

x

x

1

x

x

3

...

4 x 3   ...

1 2x

3x 2

4 x 3   . ..

1 3x

6x

1 3x

6x2

2

10 x

3

...

10 x 3  ...

V , where

by) (or) cos(ax

m

by) (or ) x y

a x by

( D , D  )

V

ax First operate e

a x by

Type: 4.

x

. ..

3x 2

1

2

3

1 2x

e

V=sin(ax

P .I

x

ax by

Type: 5 If f ( x , y )

P .I

2

1

(D

by

by replacing D   by D a   and D  by D a  . 1 V , Now this will either Type: 3 or a, D b )

Engineering Mathematics Material Type: 6 If f ( x , y )

y sin ax (or) y cos ax

1

P .I

2013

y sin ax

( D , D  )

1 D

m1 D

D

1

D

m2D

y

c

m2 x

m 2 x si   sin ax dx (Apply Bernouili’s method)

c

m 1 D

y sin ax

3) Solution of Partial Differential Equations: Standard Type: 1

Equation of the form Assume that

z

ax

equation.put p

Standard Type: 2

f ( p, q )

0

c   be the solution the above

by

a  and q

b in

equation (1), we get

f (a , b )

0 . Now, solve this, we get b

z

(a ) y

ax ax

c which   is called

Equation of the form

z

The Complete solution is

Complete solution.

px

qy

z

ax

f ( p, q )  by

Singular integral diff. partially w.r.t and eliminate Standard Type: 3

(Clairaut’s form)

f (a , b)  . To find a  & b  , equate to zero

a   and b  .

Equation of the form f 1 ( x , p) The solution is z

Standard Type: 4

( a ) .

f 2 ( y , q )

qdy  .

pdx

Equation of the form f ( z, p, q )  0 In this type put u

x

ay a y ,  then p

dz du

,q

Unit – IV (Application of Partial Differential Equation) 1) The One dimensional Wave equation: e quation: 2

t

y 2

2 2

y  2

The three solutions of the above equation are

px px

i)

y( x , t )

ii)

y( x , t )

A cos px

iii)

y( x , t )

Ax

Ae

Be

B

px

Ce

pat

B sin px

Ct

pat

De

C cos pat

D sin pat

dz   du

2013

But the correct solution is ii),

y( x , t )

A cos px

B sin px

C cos pat

D sin pat   .

2) The One dimensional Heat flow equation: 2

u

2

u  2

t

2

Thermal Conductivity Density

where

Specific Heat

The three solutions of the above equation are px

i)

u( x , t )

Ae

ii)

u( x , t )

A cos px

iii)

u( x , t )

Ax

px

Be

2

Ce

2

p t

B sin px Ce

2

2

p t

B C

But the correct solution is ii), u ( x , t )

A cos px

B sin px Ce

2

2

p t

3) The Two dimensional Heat flow equation: 2

u

2

2

x

u  2

0

The three solutions of the above equation are i)

u( x , y)

Ae

px

Be

px

C cos py

D sin py

(Applicable when given value is parallel to y-axies) ii)

u( x , y)

A cos px

B sin px

Ce

py

py

De

(Applicable when given value is parallel to x-axies) iii)

u( x , y )

Ax

B

Cy

(Not applicable)

Unit – V (Z - Transform) 1) Definition of Z-transform: Let Z

f (n ) be the sequence defined for all the positive integers n such that n

f (n )

f (n )z   n  0

Engineering Mathematics Material

2013

2)

Sl.No

Z f ( n )

1.

Z  1

F [ z ]

z  z  1

2.

Z  ( 1)

3.

Z a

4.

Z n

z  1 z

z

a  z  2

z  1 2

Z n  1

5.

7.

Z  sin

log

Z  cos

2

z  1

1

6.

8.

z  1

z  2

2

1

2

z  2

2

1

3) Statement of Initial value theorem: If Z

f (n )

F [ z]  , then L t F [ z] z

L t f ( n )

0

4) Statement of Final value theorem: If Z 5)

6)

f (n )

F [ z]  , then L t f ( n ) n

Z a f (n ) Z n f (n )

Z f (n )  z

d  dz

L t ( z 1)F ( z)

1

z  a

Z f (n )

7) Inverse Z-transform 1

Sl.No

Z

1.

1

2.

1

3.

1

F ( z ) z  z  1 z  z  1

z  z

f ( n )

1 n

( 1) n

Engineering Mathematics Material

4.

5.

6.

7.

1

2013

z

a  z

1

2

z  1 z

1

z

n  1

n a

2

a  z

1

z

n

2

8.

2

cos

1

2

9.

10.

z

2

a  cos

2

1

11.

1

2

sin

1

2

z

2 n

2

2

1

n  1

2

1

n  1

2

sin

2

8) Inverse form of Convolution Theorem 1

Z [F ( z).G ( z)]

Z

1

1 [F (z )] Z [G ( z ) ] n

and by the defn. of Convolution of two functions f ( n )

g( n )

f ( r ) g( n r  0

9) a) Z [ y( n )]

F ( z)

b) Z [ y( n

1)]

zF ( z)

c) Z [ y( n

2)]

z F ( z)

d) Z [ y( n

3)]

z F ( z)

zy(0   )

2

z y(0)

3

z y(0)

----

2

zy(1  )

3

z y(1)

2

zy(2  )

----

r )