Fourier Transform Understanding- Mathquery

Definition Of Fourier Transform:

 

      The Fourier Transform of a function f denoted as f̂ is defined by 

                                ∞

      f̂ (ξ) = 1/√(2π ) ∫ f(x) e ^ -iξx dx    .....(1)

                               -∞

         whenever the integral on the right exists . It is obvious that the integral on the right of (1) exists if
                          ∞
                           ∫ |f(x)| dx    exists .
                         -∞

  If the fourier transform f̂ of a function f is known the function f can be obtained by the following formula , known as the inversion formula :
                              ∞
   f(x) = 1/√(2π )   ∫ f̂(ξ) e^iξx dξ   ..............(2)
                             -∞

   Sometimes it is convenient to use the operator notation F and F⁻¹ for the Fourier transform and its inverse i.e.

         F(f) = f̂     and  F⁻¹(f̂) = f .

   The following important properties of the Fourier transform can be easily verified :

(i) Linearity Of The Transform and its              inverse   

   If f, g are any transformable functions and a and b are real numbers then 

  F(af + bg) = a F(f) + b F(g)

 F⁻¹(af̂ + bĝ) = a F⁻¹(f̂) + b F⁻¹(ĝ) .

(ii) Transform Of nth Derivative 

   If fⁿ⁻¹(x) , fⁿ⁻²(x),.....f(x) all approach zero as x--> ±∞ , then 

      F(f)ⁿ(x) = (iξ)ⁿ f̂(ξ)

and F((-ix)ⁿf(x)) = f̂ ⁿ(ξ) 

(iii) Shift Formula .

      F(f(ax-b)) = 1/|a| e^ -iξb/a f̂(ξ/a)

and F(eⁱᵃˣ f(x)) = f̂(ξ-a) 

(iv) Transform Of The Convolution .

     Defining the Fourier convolution , f*g of f(x) and g(x) as 

                    ∞

   f*g(-x₀) = ∫ f(x₀-x) g(x) dx

                   -∞

  we have   F(f*g) = f̂ * ĝ   

  Here * stands for function composition .

   If f(x) is even or it is only defined over 0<x<∞ in which case we can extend it so as to be even , then we observe that 

    ∞

    ∫ f(x) cos ξx dx = 0

  -∞

 Hence the Fourier transform reduces to 

                         ∞

  f̂(ξ) = i/√(2π ) ∫ f(x) sin ξx dx 

                        -∞

                            ∞

         = -i √(2/π ) ∫ f(x) sin ξx dx

                            0

   Thus if f(x) is given for 0<x<∞ its Fourier sine transform is defined as

                             ∞

     Fₛ(f) = √(2/π ) ∫ f(x) sin ξx dx ,

                             0

and the inversion formula is 

                           ∞

    f(x) = √(2/π ) ∫ sin (ξx) Fₛ(f) dξ

                           0

 Similarly , the cosine transform is defined as  

                             ∞

    Fₒ[f] = √(2/π ) ∫ f(x) cos ξx dx 

                             0

and the inversion formula becomes 

                          ∞

   f(x) = √(2/π)  ∫ cos ξx Fₒ (f) dx

                          0

 Sine and Cosine transforms are useful in treating problems with boundary condition at x= 0 and involving only derivatives of even order as is seen from the properties :

  Fₛ[f''] = f(0) ξ - ξ² Fₒ[f]

  Fₒ[f"] = -f'(0) - ξ²Fₒ[f] , 

 Provided f(x) and f'(x) -->0 as x-->∞

 Thus  sine transform is useful  when f(0) is given and cosine transform is usual when f'(0) is known .

                   Motivational Quote

       " A Thought Is An Idea In Transit " 

                                  By Pythagoras