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# Chapter 4 (maths 3)

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### Chapter 4 (maths 3)

1. 1. CHAPTER 4 FOURIER TRANSFORMSINTEGRAL TRANSFORM b The integral transform of a function f (x ) is defined by ∫ f ( x).k (s , x)dx where a k(s , x) is a known function of s and x and it is called the kernel of the transform. When k(s , x) is a sine or cosine function, we get transforms called Fourier sine orcosine transforms.FOURIER INTEGRAL THEOREM If f (x) is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then ∞ ∞ 1 π ∫ −∫ f ( x) = f (t ) cos λ (t − x) dt dλ 0 ∞At a point of discontinuity the value of the integral on the left of above equation is1 { f ( x + 0) − f ( x − 0)}.2EXAMPLES 1 for x ≤ 1 1. Express the function f ( x) =  as a Fourier Integral. Hence evaluate 0 for x > 1 ∞ ∞ sin λ cos λx sin λ∫ λ0 dλ and find the value of ∫ 0 λ dλ .Solution: We know that the Fourier Integral formula for f (x) is ∞ ∞ 1 f ( x) = π ∫ ∫ f (t ) cos λ (t − x) dt dλ 0 −∞ ……………….(1) Here f (t ) = 1 for t ≤ 1 i.e., f(t) = 1 in -1 < t < 1 f (t ) = 0 for t > 1 f (t ) = 0 in − ∞ < t < −1 and 1 < t < ∞ ∞ 1 1∴ Equation (1) ⇒ f ( x) = π ∫ ∫ cos λ (t − x) dt dλ 0 −1 ∞ 1 1  sin λ (t − x)  π ∫ =   dλ 0 λ  −1 ∞ 1 sin λ (1 − x) − sin λ (−1 − x) π∫ = dλ 0 λ ∞ 1 sin λ (1 − x) + sin λ (1 + x ) = ∫ dλ π 0 λ 1
2. 2. ∞ 2 sin λ cos λx π∫ ∴ f ( x) = dλ .………………(2) 0 λ [Using sin (A+B) + sin (A-B) = 2 sin A cos B]This is Fourier Integral of the given function. From (2) we get ∞ sin λ cos λx π ∫ 0 λ dλ = 2 f ( x) ……………….(3  1 for x ≤ 1 But f ( x) =  ………………..(4) 0 for x > 1 Substituting (4) in (3) we get π ∞ sin λ cos λx  for x ≤ 1 ∫ λ dλ =  2 0 0 for x > 1  ∞ sin λ πPutting x = 0 we get ∫0 λ dλ = 22. Find the Fourier Integral of the function 0 x<0 1  f ( x) =  x=0 2 e − x x > 0 Verify the representation directly at the point x = 0.Solution: The Fourier integral of f (x) is ∞ ∞ 1 f ( x) = π ∫ ∫ f (t ) cos λ (t − x) dt dλ 0 −∞ ……………….(1) 1   ∞ 0 ∞ = ∫ −∫∞ π 0 f (t ) cos λ (t − x )dt + ∫ f (t ) cos λ (t − x)dt d λ 0  1   ∞ 0 ∞ = ∫  ∫ 0. cos λ (t − x)dt + ∫ e −t cos λ (t − x)dt dλ π 0 − ∞ 0  ∞ ∞ 1  e −t = ∫ 2 [ − cos( λt − λx ) + λ sin(λt − λx)]  dλ  π 0 λ +1 0 ∞ 1 cos λx + λ sin λx π∫ f (x) = dλ ……….………(2) 0 λ2 + 1 2
3. 3. Putting x = 0 in (2), we get ∞ 1 f (0) = ∫ 2 1 π 0 λ +1 1 d λ = tan −1 ( λ ) 0 π ∞ [ ] 1 [ = tan −1 ( ∞ ) − tan −1 (0) π ] 1 π  1 =  = π 2 2 1The value of the given function at x = 0 is . Hence verified. 2FOURIER SINE AND COSINE INTEGRALS The integral of the form 2∞ ∞ f ( x) = ∫ sin λx ∫ f (t ) sin λt dt dλ π0 0is known as Fourier sine integral. The integral of the form ∞ ∞ 2 f ( x) = ∫ cos λx ∫ f (t ) cos λt dt dλ π 0 0is known as Fourier cosine integral.PROBLEMS1. Using Fourier integral formula, prove that 2(b 2 − a 2 ) ∞ u sin xu ∫ (u 2 + a 2 )(u 2 + b 2 ) du (a, b > 0) − ax − bx e −e = π 0Solution: The presence of sin xu in the integral suggests that the Fourier sine integral formulahas been used. Fourier sine integral representation is given by ∞ ∞ 2 f ( x) = ∫ sin ux ∫ f (t ) sin ut dt du π 0 0 ∞ ∞  sin ux du  ∫ ( e − at − e −bt ) sin ut dt  2 e − ax − e −bx = ∫ π 0 0  3
4. 4. ∞ ∞ 2  e − at − bt  = ∫ sin ux du  2 { − a sin ut − u cos ut} − 2e 2 { − b sin ut − u cos ut}  π 0 a + u 2 b +u 0 ∞ 2  u u  = ∫ sin ux du  a 2 + u 2 − b 2 + u 2  π 0   ∞ 2(b 2 − a 2 ) u sin ux = π ∫ (u 2 + a 2 )(u 2 + b 2 ) du 02. Using Fourier integral formula, prove that 2 ∞ (λ 2 + 2 ) cos xλ ∫ dλ −x e cos x = π 0 λ2 + 4Solution: The presence of cos xλ in the integral suggests that the Fourier cosine integralformula for e − x cos x has been used. Fourier cosine integral representation is given by ∞ ∞ 2 π∫ f ( x) = cos λx ∫ f (t ) cos λt dt dλ 0 0 2∞ ∞ − t  ∴e −x cos x = ∫ cos xλ dλ  ∫ e cos t cos λt dt  π 0 0  2 ∞ 1 ∞  = ∫ cos xλ dλ  ∫ e −t { cos(λ + 1)t + cos(λ − 1)t } dt  π 0 2 0  ∞  = 2 π 0  1 −t [ ∫ cos xλ dλ  (λ + 1) 2 + 1 e { − cos(λ + 1)t + (λ + 1) sin(λ + 1)t} ] ∞ 0  + 1 (λ − 1) + 1 2 [ e −t { − cos(λ − 1)t + (λ − 1) sin(λ − 1)t } 0 ∞ ] ) ∞ 1  1 1  = ∫ +  cos xλ dλ π 0  (λ + 1) + 1 (λ − 1) + 1 2 2 ∞ 2 (λ2 + 2) cos xλ = ∫ dλ. π 0 λ2 + 4COMPLEX FORM OF FOURIER INTEGRALS 4
5. 5. The integral of the form ∞ ∞ 1 ∫ e − iλx ∫ f (t ) e iλt f ( x) = dt d λ 2π −∞ −∞is known as Complex form of Fourier Integral.FOURIER TRANSFORMSCOMPLEX FOURIER TRANSFORMS ∞ 1 The function F [ f ( x)] = ∫∞ f (t ).e dt is called the Complex Fourier transform ist 2π −of f (x ) .INVERSION FORMULA FOR THE COMPLEX FOURIER TRANSFORM ∞ 1 ∫∞F [ f ( x)].e ds is called the inversion formula for the −isx The function f ( x) = 2π −Complex Fourier transform of F [ f ( x)] and it is denoted by F −1 [ F ( f ( x))].FOURIER SINE TRANSFORMS ∞ 2 The function FS [ f ( x )] = ∫ f (t ).sin st dt is called the Fourier Sine Transform of π 0the function f (x ) . ∞ 2 The function f ( x) = π ∫ F [ f ( x)]. sin sx ds is called the inversion formula for the 0 SFourier sine transform and it is denoted by FS −1 [ FS ( f ( x))].FOURIER COSINE TRANSFORMS ∞ 2 The function FC [ f ( x)] = π ∫ f (t ). cos st dt is called the Fourier Cosine 0Transform of f (x) . ∞ 2 The function f ( x) = π ∫ F [ f ( x)]. cos sx ds 0 C is called the inversion formula for theFourier Cosine Transform and it is denoted by FC −1 [ FC ( f ( x))].PROBLEMS1. Find the Fourier Transform of 1 − x 2 in x ≤ 1  f ( x) =  0  in x > 1 5
6. 6. ∞ sin s − s cos s s 3π Hence prove that ∫ 0 s 3 cos ds = 2 16 .Solution: We know that the Fourier transform of f (x) is given by ∞ 1 F [ f ( x )] = ∫ f ( x).e isx dx 2π −∞ −1 1 ∞ 1 1 1 = ∫ f ( x).e dx + ∫ f ( x).e dx + ∫ f ( x).e isx isx isx dx 2π −∞ 2π −1 2π 1 −1 1 ∞ 1 1 1 = ∫ 0.e dx + ∫ (1 − x ).e dx + ∫ 0.e isx 2 isx isx dx 2π −∞ 2π −1 2π 1 1 1 = ∫ (1 − x 2 ).e isx dx 2π −1 1 1  e isx e isx e isx  =  (1 − x 2 ) − ( −2 x ) 2 2 − 2 3 3  2π  is i s i s  −1 1  − 2 is 2 is − 2 −is 2 e −is  =  2 e + 3e + 2 e −  2π  s is s i s3  1  − 2 is 2  = s (e + e −is ) + 3 (e is − e −is ) 2π  2 is  1 − 4 4  1 4  =  s 2 cos s + s 3 sin s  =  s 3 (sin s − s cos s ) 2π   2π  By using inverse Fourier Transform we get ∞ 1 1 4 f ( x) = 2π . 2π ∫ −∞ s 3 (sin s − s cos s ).e −isx ds ∞ 1 4 = 2π ∫s −∞ 3 (sin s − s cos s ).(cos sx − i sin sx ) ds ∞ 1 4 = 2π ∫ −∞ s3 (sin s − s cos s ) cos sx ds ∞ 1 4 − 2π ∫ −∞ s3 (sin s − s cos s ) i sin sx ds 6
7. 7. The second integral is odd and hence its values is zero. ∞ 2 sin s − s cos s π −∫ ∴ f ( x) = cos sx ds ∞ s3 ∞ 4 sin s − s cos s = ∫ cos sx ds π 0 s3 ∞ sin s − s cos s π i.e., ∫ 0 s 3 cos sx ds = f ( x) 4 1Putting x = , we get 2 ∞ sin s − s cos s s π  1  π  1  3π ∫0 s 3 cos ds = 2 f   = 1 −  = 4  2  4  4  16 . ∞ sin s − s cos s s 3π ∫ 0 s 3 cos ds = 2 16 .2. Find the Fourier sine transform of e − x , x ≥ 0 (or) e − x , x > 0. Hence evaluate ∞ x sin mx ∫ 1 + x 2 dx. 0Solution: The Fourier sine transform of f(x) is given by ∞ 2 FS [ f ( x )] = ∫ f ( x).sin sx dx π 0 −x −xHere e =e for x > 0 ∞ [ ] FS e − x = 2 π ∫e −x . sin sx dx 0 2 s ∞ − ax b  =  ∫ e sin bx dx = 2  π s2 +1 0 a + b2 Using inverse Fourier sine transform we get ∞ ∫ F [e ]. sin sx ds 2 −x f ( x) = s π 0 ∞ 2 2 s = π ∫ 0 . 2 π s +1 . sin sx ds ∞ 2 s = ∫ s 2 + 1 sin sx ds π 0 7
8. 8. ∞ π s i.e., f ( x) = ∫ 2 . sin sx ds 2 0 s +1 ∞ s. sin sx π i.e., ∫0 s +1 2 ds = e − x 2Replacing x by m we get ∞ s. sin ms π i.e., ∫ 0 s +1 2 ds = e − m 2 ∞ x. sin mx π i.e., ∫ 0 x +1 2 dx = e −m 2 [since s is dummy variable, we can replace it by x] e − ax3. Find the Fourier cosine transform of . xSolution: ∞ 2 We know that FC [ f ( x )] = ∫ f ( x). cos sx dx π 0 − ax eHere f ( x) = . x ∞ 2 e − ax ∴ FC [ f ( x )] = ∫ x . cos sx dx π 0Let FC [ f ( x)] = F ( s ) ∞ 2 e − axThen F (s) = π ∫ x . cos sx dx 0 ………………(1)Differentiating on both sides w.r.t. ‘s’ we get, ∞ dF ( s ) d 2 e − ax ds = ds π ∫ x . cos sx dx 0 ∞ 2 ∂  e −ax  = ∫ ∂s  x . cos sxdx π0   ∞ ∞ 2 e −ax 2 − ax = ∫ x (− sin sx).x dx = − π ∫ e sin sx dx π0 0 dF ( s ) 2 s  ∞ b  ∫e − ax =− . 2  sin bx dx =  ds π a + s2  0 a + b2  2Integrating w.r.t. ‘s’ we get 8
9. 9. 2 s F (s) = − .∫ 2 ds π s + a2 2 1 1 =− . . log ( s 2 + a 2 ) = − . log ( s 2 + a 2 ) π 2 2π e − ax − e −bx4. Find the Fourier cosine transform of . xSolution: We know that the Fourier cosine transform of f(x) is ∞ 2 FC [ f ( x )] = ∫ f ( x). cos sx dx π 0 − ax e − e −bxHere f ( x) = x ∞  e − ax − e −bx  2 e − ax − e − bx ∴ FC   x =  π ∫ x . cos sx dx 0 ∞ ∞ 2 e − ax 2 e −bx = π ∫ x . cos sx dx − π ∫ x cos sx dx 0 0  e −ax   e −bx  = Fc   − Fc    x   x  1 1 =− log ( s 2 + a 2 ) + log ( s 2 + b 2 ) 2π 2π 1  s2 + b2  = log  2  s + a2  2π   e − as5. Find f (x) , if its sine transform is . Hence deduce that the inverse sine s 1transform of . sSolution: We know that the inverse Fourier sine transform of FS [ f (x )] is given by ∞ 2 f ( x) = π ∫ F [ f ( x)]. sin sx ds 0 S e − asHere FS [ f ( x )] = s 9
10. 10. ∞ 2 e − as ∴ f ( x) = π ∫ s . sin sx ds 0Differentiating w.r.t. ‘x’ on both sides, we get, d [ f ( x )] ∞ 2 e − as ∂ dx = π ∫ s . ∂x (sin sx) ds 0 ∞ ∞ 2 e − as 2 − as = π ∫ s . cos sx s ds = π ∫ e . cos sx ds 0 0 2 a  ∞ −ax a  =  ∫ e cos bx dx = 2  π a + x2 2  0 a + b2  d [ f ( x )] 2 a = dx π x + a2 2 2 1 2 1 −1  x  ∴ f ( x) = a π ∫ x 2 + a 2 dx = a π a tan  a    2  x ∴ f ( x) = tan −1   π a 1To find the inverse Fourier sine transform of : s Put a = 0, in (1), we get 2 2 π π f ( x) = tan −1 (∞) = . = π π 2 2PROPERTIES 1. Linearity Property If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then F [ a f ( x ) + b g ( x)] = a F ( s) + b G ( s )Proof: ∞ 1 F [ a f ( x) + b g ( x)] = ∫∞[ a f ( x) + b g ( x)] e dx isx 2π − ∞ ∞ 1 1 = ∫ a f ( x).e dx + ∫ b g ( x).e isx isx dx 2π −∞ 2π −∞ ∞ ∞ a b = ∫ f ( x).e isx dx + ∫ g ( x).e isx dx 2π −∞ 2π −∞ = a F ( s ) + b G ( s) 10
11. 11. 2. Change of Scale Property 1 s If F(s) is the Fourier transform of f (x ) then F [ f (ax)] = F  ,a>0 a aProof: ∞ 1 F [ f (ax)] = ∫ f (ax).e isx dx 2π −∞Put ax = y dy a dx = dy i.e., dx = aWhen x = −∞, y = −∞ and x = ∞, y=∞ ∞ y ∞ s 1 dy 1 1 i  y F [ f (ax)] = is 2π ∫ f ( y).e −∞ a . = a a 2π ∫ f ( y).e −∞ a .dy 1 = F ( s a) a3. Shifting Property ( Shifting in x ) If F(s) is the Fourier transform of f (x ) then F [ f ( x − a )] = e ias F ( s )Proof: ∞ 1 F [ f ( x − a)] = ∫ f ( x − a).e isx dx 2π −∞Put x-a = y dx = dyWhen x = −∞, y = −∞ and x = ∞, y=∞ ∞ ∞ 1 e ias F [ f ( x − a)] = ∫ f ( y ).e is ( y + a ) . dy = ∫ f ( y).e isy .dy 2π −∞ 2π −∞ ∞ e ias = ∫ f ( x).e .dx = e isa F ( s ) isx 2π −∞4. Shifting in respect of s If F(s) is the Fourier transform of f (x ) then F e iax f ( x) = F ( s + a ) [ ]Proof: ∞ Fe[ iax ] f ( x) = 1 ∫e iax f ( x) e isx dx 2π −∞ 11
12. 12. ∞ 1 ∫ f ( x).e i( s+a) x = dx = F ( s + a) 2π −∞5. Modulation Theorem 1 If F(s) is the Fourier transform of f (x ) then F [ f ( x) cos ax ] = [ F ( s + a ) + F ( s − a )] 2Proof: ∞ 1 F [ f ( x) cos ax ] = ∫ f ( x). cos ax.e isx dx 2π −∞ ∞ 1  e iax + e −iax  = ∫ 2π −∞ f ( x).e isx    2 dx   ∞ ∞ 1 1 1 1 ∫∞ f ( x).e dx + 2 . 2π ∫ f ( x).e i( s+a ) x i ( s −a ) x = . dx 2 2π − −∞ 1 1 1 = f ( s + a ) + f ( s − a ) = [ f ( s + a ) + f ( s − a )] 2 2 2 1 F [ f ( x) cos ax ] = [ F ( s + a ) + F ( s − a)] 2COROLLARIES 1(i ) FC [ f ( x) cos ax ] = [ FC ( s + a) + FC ( s − a)] 2 1(ii ) FC [ f ( x) sin ax ] = [ FS (a + s) + FS (a − s )] 2 1(iii ) FS [ f ( x) cos ax ] = [ FS ( s + a ) + FS ( s − a )] 2 1(iv ) FS [ f ( x) sin ax ] = [ FC ( s − a ) − FC ( s + a)] 26. Conjugate Symmetry Property If F(s) is the Fourier transform of f (x ) then F f ( − x) = F ( s) [ ]Proof: ∞ 1 We know that F ( s ) = ∫ f ( x). e isx dx 2π −∞Taking complex conjugate on both sides we get ∞ 1 ∫∞ f ( x). e dx −isx F (s) = 2π − 12
13. 13. Put x = -y dx = -dyWhen x = −∞, y = ∞ and x = ∞, y = −∞ −∞ 1 ∴ F (s) = ∫ f (− y) .e (− dy ) isy 2π ∞ −∞ 1 =− ∫ f (− y). e isy dy 2π ∞ [ ] ∞ 1 = ∫ f (− x). e dx = F f (− x) isx 2π −∞7. Transform of Derivatives If F(s) is the Fourier transform of f (x ) and if f (x) is continuous, f ′(x) is piecewisecontinuously differentiable, f (x ) and f ′(x) are absolutely integrable in (−∞ , ∞) and lim [ f ( x)] = 0 , thenx → ±∞ F ( f ′( x ) ) = −is F ( s )Proof: By the first three conditions given, F { f (x)} and F { f ′(x)} exist. ∞ 1 F { f ′( x)} = ∫ f ′( x) e isx dx 2π −∞ ∞ = 1 [e isx ] ∞ f ( x) −∞ − is ∫e isx f ( x) dx, on int egrating by parts. 2π 2π −∞ = 0 − isF { f ( x)} , by the given condition. = −is F ( s).The theorem can be extended as follows. If f , f ′, f ′′, , f ( n −1) are continuous, f (n ) is piecewise continuous, f , f ′, f ′′, , f ( n )are absolutely integrable in (−∞ , ∞) and f , f ′, f ′′,  , f ( n −1) → 0 as x → ±∞ , then F ( f ( n ) ( x) ) = (−is ) n F ( s ) 13
14. 14. 8. Derivatives of the Transform dF ( s ) If F(s) is the Fourier transform of f (x ) then F [ x. f ( x )] = (−i ) dsProof: ∞ 1 F (s) = ∫ f ( x )e isx dx 2π −∞ ∞ ∫ ds [ f ( x)e ]dx dF ( s ) 1 d ∴ = isx ds 2π −∞ ∞ i = ∫ [ x. f ( x)]e dx = iF [ xf ( x)] isx 2π −∞ dF ( s) ∴ ( −i ) = F [ x. f ( x)] ds [ ]Extending, we get, F x n . f ( x) = (−i ) n d n F (s) ds nDEFINITION ∞ 1 ∫ f ( x − u ) g (u )du is called the convolution product or simply the convolution 2π −∞of the functions f (x) and g (x) and is denoted by f ( x) * g ( x ) .9. Convolution Theorem If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then theFourier transform of the convolution of f(x) and g(x) is the product of their Fouriertransforms. i.e., F [ f ( x ) * g ( x)] = F ( s ).G ( s)Proof ∞ 1 F [ f ( x ) * g ( x )] = ∫∞ f ( x) * g ( x)e dx isx 2π − 1  1 ∞ ∞  isx = ∫∞  2π −∫∞ f ( x − u) g (u )du  e dx 2π −   1 ∞  1 ∞  = ∫∞  2π −∫∞ f ( x − u)e dxdu, isx g (u )  2π −  on changing the order of int egration. 14
15. 15. ∞ ∫ g (u )[e ] 1 = ius F ( s) du , by the shifting property. 2π −∞ ∞ 1 = F ( s). ∫ g (u ).e ius du 2π −∞ = F ( s).G ( s )Inverting, we get F −1 [ F ( s ).G ( s)] = f ( x) * g ( x ) = F −1 { F ( s )} * F −1 { G ( s )}10. Parseval’s Identity (or) Energy Theorem If f (x) is a given function defined in (−∞ , ∞) then it satisfy the identity, ∞ ∞ ∫ ∫ F ( s) 2 2 f ( x) dx = ds −∞ −∞where F(s) is the Fourier transform of f (x ) .Proof: We know that F −1 [ F ( s ).G ( s)] = f ( x) * g ( x ) ∞ ∞ 1 1 ∫ F (s).G(s)e ∫ f (t ) g ( x − t )dt −isx ds = 2π −∞ 2π −∞Putting x = 0, we get ∞ ∞ ∫ F (s).G(s)ds = ∫ f (t ) g (−t )dt −∞ −∞ ………………..(1) Let g ( − t ) = f (t ) .……………….(2) i.e., g (t ) = f (−t ) ………………..(3) ∴ G ( s ) = F [ f (− x)] = F ( s ) by property (9) i.e., ∴ G ( s) = F ( s) ………………..(4) Substituting (2) and (4) in (1) we get ∞ ∞ ∫ F (s).F (s) ds = −∞ ∫ f (t ). f (t ) dt −∞ [ F (s).F (s) = F (s) ] ∞ ∞ ∫ ∫ 2 2 2 F ( s) ds = f ( x) dx −∞ −∞11. If f (x) and g (x) are given functions of x and FC [ f ( x)] and FC [ g ( x)] are theirFourier cosine transforms and FS [ f ( x)] and FS [ g ( x)] are their Fourier sine transforms then 15
16. 16. ∞ ∞ ∞(i) ∫ 0 f ( x ) g ( x)dx = ∫ FC [ f ( x)].FC [ g ( x)]ds = ∫ FS [ f ( x )].FS [ g ( x)]ds 0 0 ∞ ∞ ∞ ∫ f ( x) dx = ∫ FC [ f ( x)] ds = ∫ FS [ f ( x )] ds , 2 2 2(ii) 0 0 0which is Parseval’s identity for Fourier cosine and sine transforms.Proof: ∞ ∞  2∞ (i) ∫ FC [ f ( x )].FC [ g ( x )]ds = ∫ FC [ f ( x)]  ∫ g ( x) cos sx dx  ds 0 0  π 0  ∞  2∞  = ∫ g ( x)  ∫ FC [ f ( x)] cos sx ds  dx, 0  π 0  Changing the order of integration ∞ = ∫ f ( x) g ( x)dx 0Similarly we can prove the other part of the result.(ii) Replacing g ( x) = f * ( x) in (i) and noting that FC [ f ( x)] = FC [ f ( x)] andFS [ f ( x)] = FS [ f ( x)] , we get ∞ ∞ ∞ ∫0 f ( x ). f ( x).dx = ∫ FC [ f ( x)].FC [ f ( x)] ds = ∫ FS [ f ( x )].FS [ f ( x )] ds 0 0 ∞ ∞ ∞ ∫ f ( x) .dx = ∫ FC [ f ( x)] ds = ∫ FS [ f ( x)] ds 2 2 2i.e., 0 0 012. If FC [ f ( x)] = FC ( s ) and FS [ f ( x)] = FS ( s ) , then d(i) { FC ( s)} = − FS { xf ( x)} and ds d(ii) { FS ( s)} = − FC { xf ( x)}. dsProof: ∞ 2 π∫ FC ( s) = f ( x) cos sx dx 0 ∞ d { FC ( s)} = ∫ f ( x)(− x sin sx)dx ds 0 ∞ = − ∫ {xf ( x)}sin sx dx 0 = − FS {xf ( x)}Similarly the result (ii) follows. 16
17. 17. PROBLEMS a 2 − x 2  x <a1. Show that the Fourier transform of f ( x ) =  is 0  x >a>0 ∞ 2  sin as − as cos as  sin t − t cos t π2   . Hence deduce that ∫ dt = . Using Parseval’s π s 3  0 t 3 4 ∞ 2  sin t − t cos t  πidentity show that ∫  3  dt = . 0  t 15Solution: We know that ∞ 1 F [ f ( x )] = ∫ f ( x).e isx dx 2π −∞ 1   −a a ∞ =  ∫ f ( x)e dx + ∫ f ( x)e dx + ∫ f ( x )e dx  isx isx isx 2π  −∞ −a a  1   a a 1 =  0 + ∫ ( a 2 − x 2 )e isx dx + 0 = ∫ (a 2 − x 2 ).e isx dx 2π  − a  2π −a a 1  2  e isx   e isx   e isx  =  ( a − x 2 )  is  − (−2 x) 2 2  − 2 3 3   i s  i s  2π        − a = 1  − 2a isa  2 e +e [−isa 2i ] [  + 3 e −isa − e isa  ] 2π  s s  1  − 2a 4  =  2 [2 cos as ] + 3 sin as  2π  s s  4  sin as − as cos as  = 2π   s3   2  sin as − as cos as  F (s) = 2 π  s3   2  sin s − s cos s  When a = 1, F ( s ) = 2 ………………..(A) π   s3  Using inverse Fourier Transform, we get 17
18. 18. ∞ 1 2 1 f ( x) = 2π .2. π ∫s −∞ 3 [sin as − as cos as].e −isx ds ∞ 1 2 1 { sin as − as cos as}{ cos sx − i sin sx} ds π −∫ s 3 = .2. 2π ∞ ∞ 2 sin as − as cos as π −∫ f ( x) = cos sx ds ∞ s3[The second integral is odd and hence its value is zero] ∞ 4 sin as − as cos as = ∫ cos sx ds π 0 s3[since the integrand is an even function of s]Putting a = 1, we get ∞ 4 sin s − s cos s f ( x) = ∫ cos sx ds π 0 s3 ∞ 4 sin t − t cos t f ( x) = ∫ cos tx dt π 0 t3Putting x = 0, in the given function we get ∞ 4 sin t − t cos t π∫ dt = f (0) = 1 0 t3 ∞ sin t − t cos t π ∴ ∫ 0 t 3 dt = 4 ∞ ∞ ∫ ∫ F (s) 2 2Using Parseval’s identity, f ( x) dx = ds [Using (A)] −∞ −∞ 2  2  ∞  2. (sin s − s cos s)  1 π ∫  − ∞ s3  ds = ∫ (1 − x 2 ) 2 dx  −1     ∞ 2 1 8  sin s − s cos s  ∫∞  ds = 2 ∫ (1 − x ) dx 2 2 π− s 3  0 ∞ 2 16  sin s − s cos s  8 ∫ π 0 s 3  ds = 2.  15 18
19. 19. ∞ 2  sin s − s cos s  π i.e., ∫ 0 s 3  ds =  15 ∞ 2  sin t − t cos t  π i.e., ∫ t3 0  dt =  152. Find the Fourier Transform of f (x) if 1 − x , x < 1  f ( x) =  0,  x >1 ∞ 4  sin t  πHence deduce that ∫   dt = 0 t  3Solution:We know that ∞ 1 F [ f ( x )] = ∫∞ f ( x).e dx isx 2π − 1 ∫ (1 − x ).e 1 = isx dx 2π −1Since x > 1, f ( x) = 0, i.e., in − ∞ < x < −1, and 1 < x < ∞, f ( x ) = 0. 1 ∫ [1 − x ] [ cos sx + i sin sx] dx 1 = 2π −1 1 ∫ [1 − x ] cos sx dx 1 = 2π −1The second integral becomes zero since it is an odd function. 1 2 = 2π ∫ (1 − x) cos sx dx 0 [ [1 − x ] cos x is an even function] 1 2   sin sx   − cos sx  = (1 − x ) s  − (−1) s 2  π      0 2  − cos s 1  =  + 2 π  s2 s  2 1 i.e., F ( s) = . (1 − cos s ) π s2 19
20. 20. Using Parseval’s identity ∞ ∞ ∫ ∫ F ( s) 2 2 f ( x) dx = ds −∞ −∞ 1 ∞ [1 − x ] dx = 2 ∫ (1 − cos s) ds 2 ∫ 2 −1 π −∞ s4 1 ∞ 2 (1 − cos s) 2 2 ∫ [1 − x ] dx = π −∫ 2 ds 0 ∞ s4 ∞ 2 2 (1 − cos s ) 2 3 π −∫ = ds ∞ s4 put s = 2t when s = ∞, t = ∞ ds = 2dt when s = −∞, t = −∞ π ∞ (1 − cos 2t ) 2 3 −∫ = .2 dt ∞ 16t 4 π ∞ (1 − cos 2t ) 2 3 −∫ = dt ∞ 8t 4 ∞ π (1 − cos 2t ) 2 = 2∫ dt 3 0 8t 4 π ∞ sin 4 t 3 ∫ t4 = dt 0 ∞ 4  sin t  π i.e., ∫  t  dt = 3 0  ∞ dx3. Evaluate ∫ (x 0 2 + a )( x 2 + b 2 ) 2 using transforms.Solution: 2 a We know that the Fourier cosine transform of f ( x) = e − ax is . 2 . π s + a2 2 bSimilarly the Fourier cosine transform of f ( x) = e − ax is . 2 . π s + b2 20
21. 21. ∞ ∞We know that ∫ FC [ f ( x)].FC [ g ( x)] ds = ∫ f ( x).g ( x) dx 0 0 ∞ ∞ 2 a 2 b i.e., ∫ 0 . 2 . . 2 π s +a π s +b 2 2 . ds = ∫ e − ax .e −bx dx 0 ∞ ∞ 2 ab ∫ (s 2 + a 2 )(s 2 + b 2 ) ds = ∫ e −( a +b ) x i.e., ds π 0 0 ∞  e −( a + b ) x  1 1 =  = 0− =  − ( a + b)  0 − ( a + b) a + b ∞ dx π i.e., ∫ (x 0 2 2 2 2 = + a )( x + b ) 2ab( a + b)4. Find the Fourier transform of e − a x and hence deduce that ∞ cos xt π −a x(i) ∫ 2 2 dt = e 0 a +t 2a [(ii) F xe −a x ]=i 2 2as π (s + a 2 ) 2 2Solution: ∞ 1 F [ f ( x )] = ∫ f ( x).e isx dx 2π −∞ 1   0 ∞ = ∫ f ( x)e isx dx + ∫ f ( x)e isx dx  2π  −∞ 0  e − ax  if 0 ≤ x < ∞ Here f ( x ) =  ax e  if − ∞ < x < 0 1  ax isx  0 ∞ = ∫ e. e dx + ∫ e −ax .e isx dx  2π  −∞ 0  1  ( a +is ) x  0 ∞ =  ∫ e. dx + ∫ e −( a −is ) x dx  2π  −∞ 0  21
22. 22. 1  e ( a +is ) x   e −( a −is ) x   0 ∞ =   +   2π  (a + is)  −∞  − (a − is )  0    1  1 1  =  a + is + a − is  2π   Fe[ −a x ]= 2 a π s + a2 2Using inversion formula, we get ∞ 1 2 a f ( x) = 2π ∫ −∞ . 2 π s +a 2 e −isx ds ∞ a cos sx − i sin sx π −∫ = ds ∞ s2 + a2 ∞ a cos x = π−∫∞ s 2 + a 2 ds ∞ cos sx π π −a x ∫s 0 2 +a 2 dx = 2a f ( x) = 2a .e (or ) ∞ cos tx π −a x ∫s 0 2 +a 2 dt = 2a .ePutting a = 1, we get, Fe[ ]=−x 2 1 . π s2 +1 ∞ ∞ cos sx π −x cos tx π −x and ∫ s 2 + 1 ds = 2 e 0 (or ) ∫t 0 2 +1 dt = e 2FINITE FOURIER TRANSFORMS If f (x) is a function defined in the interval (0 , l) then the finite Fourier sinetransform of f (x) in 0 <x < l is defined as 22
23. 23. nπx l FS [ f ( x)] = ∫ f ( x). sin dx where ‘n’ is an integer 0 l The inverse finite Fourier sine transform of FS [ f (x )] is f (x) and is given by 2 ∞ nπx f ( x) = ∑ FS [ f ( x )] sin l n =1 l The finite Fourier cosine transform of f (x ) in 0 < x < l is defined as nπx l FC [ f ( x)] = ∫ f ( x). cos dx where ‘n’ is an integer 0 l The inverse finite Fourier cosine transform of FC [ f (x)] is f (x) and is given by 1 2 ∞ nπx f ( x) = FC (0) + ∑ FC [ f ( x )] cos l l n =1 lPROBLEMS1. Find the finite Fourier sine and cosine transforms of f ( x) = x 2 in 0 < x < l.Solution: The finite Fourier sine transform is nπx l FS [ f ( x)] = ∫ f ( x). sin dx 0 lHere f ( x) = x 2 [ ] nπx l FS x 2 = ∫ x 2 . sin dx 0 l l   nπx   nπx   nπx    − cos   − sin   cos  = x 2 l  − 2 x l  + 2 l    nπ   n 2π 2   n 3π 3           l   l2   l 3  0  − l3 2l 3 2l 3 = cos nπ + 3 3 cos nπ − 3 3 nπ nπ nπ = l3 nπ 2l 3 [ (−1) n +1 + 3 3 (−1) n − 1 nπ ]The finite Fourier cosine transform is 23
24. 24. nπx l FC [ f ( x)] = ∫ f ( x). cos dx 0 lHere f ( x) = x 2 [ ] nπx l FC x 2 = ∫ x 2 . cos dx 0 l l   nπx   nπx   nπx    sin   − cos   − sin  = x 2  l  − 2 x l  + 2 l    nπ   n 2π 2   n 3π 3    l         l2   l3  0 2l 3 = cos nπ n 2π 2 2l 3 = (−1) n nπ 2 22. Find the finite Fourier sine and cosine transforms of f ( x ) = x in (0 , π ) .Solution: The finite Fourier sine transform of f ( x) = x in (0 , π ) is π FS [ f ( x)] = ∫ f ( x ). sin nx dx 0Here f ( x) = x in (0 , π ) π π   − cos nx   − sin nx  FS [ x ] = ∫ x. sin nx dx =  x  − 1 2  0   n   n  0 π π =− cos nπ = (−1) n +1 . n nThe finite Fourier cosine transform of f ( x) = x in (0 , π ) is π FC [ f ( x)] = ∫ f ( x). cos nx dx 0Here f ( x) = x in (0 , π ) π π   sin nx   − cos nx  FC [ x ] = ∫ x. cos nx dx =  x  − 1 2  0   n   n  0 = 1 n 2 1 1 [ cos nπ − 2 = 2 (−1) n − 1 n n ] 24
25. 25. 2π (−1) p−13. Find f (x) if its finite sine transform is given by , where p is positive p3integer and 0 < x < π .Solution: We know that the inverse Fourier sine transform is given by ∞ 2 f ( x) = ∑ FS [ f ( x )] sin px ………………..(1) π p =1 2π (−1) p−1Here FS [ f (x )] = ………………..(2) p3Substituting (2) in (1), we get 2 ∞ 2π (−1) p −1 f ( x) = ∑ sin px π p =1 p3 ∞ (−1) p −1 = 4∑ sin px p =1 p3  2 pπ  cos   3  find FC [ f ( p )] if 0 < x <1. −14. If f ( p) = (2 p + 1) 2Solution: ∞ nπx We know that FC −1 [ f ( p)] = 1 FC (0) + 2 ∑ FC [ f ( x)] cos l l n =1 l  2 pπ  cos Here f ( p) =  3  (2 p + 1) 2Let FC [ f ( x)] = f ( p) ∞ nπx ∴ FC −1 [ f ( p)] = 1 f C ( 0) + 2 ∑ f ( p ) cos [ l = 1] l l n =1 l  2 pπ  ∞ cos  = 1 + 2∑  3  . cos nπx n =1 (2 p + 1) 2 25
26. 26. UNIT-4 PART A1. State the Fourier integral theorem.Ans: If f (x) is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then ∞ ∞ 1 f ( x) = π ∫ ∫ f (t ) cos λ (t − x) dt dλ 0 −∞2. State the convolution theorem of the Fourier transform.Ans: If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then theFourier transform of the convolution of f(x) and g(x) is the product of their Fouriertransforms. i.e., F [ f ( x ) * g ( x)] = F ( s ).G ( s)3. Write the Fourier transform pair.Ans:F [ f (x)] and F −1 [ F ( S )] are Fourier transform pairs.4. Find the Fourier sine transform of f ( x) = e − ax (a > 0).Ans: ∞ 2 FS [ f ( x)] = π ∫ f ( x ). sin sx dx 0 2 ∞ 2 s   ∞ −ax b  ∫ e . sin sx dx =  ∫ e sin bx dx = 2 − ax =  π 0 π  s2 + a2     0 a + b2  2 s  = π  s2 + a2   5. If the Fourier transform of f (x ) is F(s) then prove that . F [ f ( x − a )] = e isa F ( s)Ans: ∞ 1 F [ f ( x − a)] = ∫∞ f ( x − a).e dx isx 2π −Put x-a = y dx = dyWhen x = −∞, y = −∞ and x = ∞, y = ∞ ∞ ∞ 1 e ias F [ f ( x − a)] = ∫ f ( y).e is ( y + a ) . dy = ∫ f ( y).e isy .dy 2π −∞ 2π −∞ ∞ e ias = ∫ f ( x).e .dx = e isa F ( s ) isx 2π −∞6. State the Fourier transforms of the derivatives of a function. 26