FOURIER INTEGRALS Que1. Solve the integral equation

1   , 0    1 f ( x ) cos  x d x   0  1  0, 

Solution: We have given

0

Hence evaluate

sin 2 t 0 t 2 d t

f ( x) cos  x d x  Fc ( )

1   , 0    1 Fc ( )    1  0,

Therefore

(1)

and by inversion formula we have,

2



f ( x) 

0

Fc ( ) cos  x d  

2

1

(1   ) cos  x d   0

1  1 2 sin x sin x ( 1   )   ( 1) d     x 0 0  

2 cosx 2(1  cos x )   x x 0  x2 1

 Now 

0

0

Fc ( )   f ( x ) cos x d x  

2(1  cos x ) cos  x d x  x2

( 2)

From (1) and (2), we have

2



0

Now for

2



1   , 0    1 (1  cos x ) cos x d x   2  1 x  0,

 0

0



 (1  cos x ) (1  cos x )  d x1  dx 2 2 0 x x 2

0

sin 2 x  dx 2 x 2 Dr. Jogendra Kumar

Page 1

FOURIER INTEGRALS Que2. Using Fourier integrals, show that

0

1  sin x, 0  x   sin  sin  x d 2 2  1  x   0,

Solution. Clearly the given problem is Fourier sine integral of some function f (x)

 sin x, 0  x   x   0,

Let f (x)  

Then Fourier sine integral representation of f (x) is given by 

f (x)   B( )sin  x d  0

Where B( ) 

2



0

f (x)sin  x dx 

 2  sin x sin  x dx   0. sin  x dx       0

1  1  2 sin x sin  x dx  [cos (1   ) x  cos (1   ) x]dx  0  0

1  sin(1   ) x sin(1   ) x       1  1   

0

1  2

 (1   )sin(1   ) x  (1   )sin (1   ) x    1 2

1  (1   ) sin x cos  x  cos x sin  x  (1   ) sin x cos  x  cos x sin  x      1 2  1 2sin   2sin      1   2  (1   2 )

0

 f (x) 

1



0

B( )sin  x d  

2



0

0

sin   sin  x d 1  2

Dr. Jogendra Kumar

Page 2

FOURIER INTEGRALS 

0

sin   sin  x    sin x, 0  x   d   f (x)   x  1  2 2 2  0,   sin x , 0 x   2  x   0,

1 for x  1 Que3. Express the function f (x)   as Fourier integral. Hence Evaluate  0 for x  1

0

sin  cos  x

d .

Solution: Clearly the Fourier integral representation of

 1 for x  1 i.e.for  1  x  1 f (x)   is given by  0 for x  1 i.e.for x  1& x  1 

f (x)   [ A( ) cos  x  B( )sin  x] d  where 0

A( ) 

 A( ) 

1





1



f (x) cos  x dx , B( ) 



f (x) cos  x dx 

1



1

1

1





1

1



1



1



1



f (x)sin  x dx

1



1

1

 1

1

1

1.cos  x dx 

1 sin  x



f (x) cos  x dx 

0.cos  x dx 

cos  x dx 

1

1

f (x) cos  x dx 



1

1



1

f (x) cos  x dx

0.cos  x dx

2sin 



Dr. Jogendra Kumar

Page 3

FOURIER INTEGRALS B ( ) 

1

f (x) sin  x dx



1



1

1

1

1



1



1

f (x) sin  x dx  0.sin  x dx 

sin  x dx 

1

1



1

1

1

1.sin  x dx 

1

1 cos  x

f (x) sin  x dx 

1



1

1

1

f (x) sin  x dx

0.sin  x dx

1

0 1

 f (x)   [ A( ) cos  x  B( )sin  x] d  0

  2sin     cos  x  0.sin  x  d  0     2 sin    cos  x d 



0

sin  cos  x

0

d 

 2

   1 for x  1  for x  1 f (x)    2 2  0 for x  1   0 for x  1

Now since f (x) is discontinuous at x  1

 f (x) 

Thus

1 1    f (1  0)  f(1  0)    0   2 2 2  4   2 for x  1   f (x)   for x  1  4  0 for x  1   

Dr. Jogendra Kumar

Page 4

FOURIER INTEGRALS Que4. Using the Fourier integral representation, show that

0

 sin  x  d   e  x (x  0) . 2 1  2

Solution. Clearly the given problem is Fourier sine integral representation of some function f (x) . Let

f (x)  e  x . Therefore the Fourier sine integral representation of 

f (x)   B ( ) sin  x d  ,where B( )  0 We can rewrite the formula as

2

0

f (x) will be

f (x)sin  x dx .

  f (x)   B( )sin  x d  and B( )  2  f (x)sin  x dx 0  0

Now,

B( )  

2

0

f (x)sin  x dx 

0

0

0

e x e sin  x dx    sin  x   cos  x   1 2 0 2

x

2  (1   2 )

 f (x)   B( )sin  x d 



2

2   sin  x d  0 1   2

 sin  x   x d   f (x)  e 1 2 2 2

Note. It means  can be replaced by  in the formula and we will get required solution. Que5. Using the Fourier integral representation, show that

0

cos  x  d   e  x (x  0) . 2 1  2

x

Solution. Let f (x)  e . Therefore the Fourier cosine integral representation of f (x) will be 

f (x)   A( ) cos  x d  , where A( )  0

2



0

f (x) cos  x dx .

Dr. Jogendra Kumar

Page 5

FOURIER INTEGRALS 

2

0

f (x)   A( ) cos  x d and A( ) 

We can rewrite the formula as

0

f (x) cos  x dx

Now,

A( )  

2

0

f (x) cos  x dx 

2

0

e x e cos  x dx    cos  x   sin  x   1 2 0 2

x

2  (1   2 ) 

 f (x)   A( ) cos  x d  0



2



0

cos  x d 1 2

cos  x   d  f (x)  e x 2 1  2 2

0

Que6. Using the Fourier integral representation, show that

0

sin  cos  x  d   , (0  x  1)  2

Solution. Clearly the given problem is Fourier sine integral representation of some function f (x) . Let f (x)  1 for (0  x  1) . Therefore, the Fourier cosine integral representation of f (x) will be given by f (x) 

0

A( ) cos  x d  , where A( ) 

We can rewrite the formula as

2



0

f (x) cos  x dx .

2

0

f (x)   A( ) cos  x d and A( ) 

0

f (x) cos  x dx

Now,

2

2

1

f (x) cos  x dx  0 1 2 1 2 sin  x 2sin    cos  x dx   0    0 

A( ) 

0

f (x) cos  x dx 

f (x) isdefined for 0  x  1only 

Dr. Jogendra Kumar

Page 6

FOURIER INTEGRALS 

2 sin 

0

 f (x)  

sin  x d

0

sin  cos  x

d 

  f (x)  2 2

Que7. Using the Fourier integral representation, show that

0

1  sin  , 0     sin   sin   d  2 2  1    0,

Solution. This is the same question as Que2. In solution of Que2 just replace

 by  and x by  .

Hints: (I) In using Fourier integral and show Type question always suppose f (x) independent from constant and in same conditions which is given for integral as we have taken in above problems. (ii) Observe the term in integral which is defined in term of the variable w.r.t. which function is being integrated and if it is sine term then consider Fourier sine integral and if it is cosine term then consider Fourier cosine integral. (iii) Learn all formulae and definitions.

Dr. Jogendra Kumar

Page 7

## fourier integrals

1 sin(1 )x sin(1 )x. 1 (1 )sin(1 )x (1 )sin(1 )x. 1. 1. 2. 1. (1 ) sin cos cos sin. (1 ) sin cos cos sin. 1. 1. 1 2sin. 2sin. 1. (1. ) x x x x x x x x Ï Ï Ï Î» Î» Î» Î» Î» Î» Ï.

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