Question

Verify Convolution theorem for f(x) = g(x) = e−x2​

Answer 1

Answer:

The proof use the same token as in the case of Fourier transform. Indeed, f and g be 2π-periodic functions. But since the function y↦e−iyxg(y) is 2π−periodic, then using this we have ,

∫2π0e−ix(t−u)g(t−u)dt=y=t−u∫2π−u−ue−ixyg(y)dy=∫2π0e−ixyg(y)dy=2πg^(x)

Inserting this in the following relation we get

f∗gˆ(x)=12π∫2π0e−ixtf∗g(t)dt=12π∫2π0e−ixt∫2π0f(u)g(t−u)dudt=12π∫2π0(e−iuxf(u)∫2π0e−ix(t−u)g(t−u)dt)du=2πg^(x)12π∫2π0e−iuxf(u)du=2πgˆ(x)⋅fˆ(x)

Step-by-step explanation:

Answer 2

Answer:

Convolution theorem is verified using Fourier transforms.

Step-by-step explanation:

Convolution Theorem:

• The convolution operation f*g of two functions f(t) and g(t) having Fourier transforms F(s) and G(s) is defined by the integral

              (f*g)t = ∫f(t-v)g(v) dt      (integral lies between 0 to t)

• The convolution operator along with other Integral transforms, is used to  solve differential equations. The Laplace transforms denoted by Y(s) for a function y(t) is defined by the improper integral.
•   Y(s) = ∫ y(t) (e^(-st) )dt    (integral lies between 0 to infinity)

Here we used Fourier transforms to prove convolution theorem.  

Given f(x) = e⁻ˣ² and g(x) = e⁻ˣ²

    By convolution theorem,

      we need to prove  (f*g)(t) = (g*f)(t)

    Consider (f*g)(t)

   (f*g)(t) = ∫f(t-v)g(v) dt             (integral lies between 0 to t)

             = ∫(e^-(t-v))(e^-t) dt

             = ∫(e^(-t+v-t)) dt

             = ∫(e^(-2t+v) dt

             = [(e^(-2t+v))/-2]                  (v lies from 0 to t)

             = [(e^(-2t+0))/-2] - [(e^(-2t+t))/-2]

             = [(e^(-2t)/-2] - [(e^(-2t+t))/-2]

  (f*g)(t) = (-1/2)[(e^(-2t)) - e^(-t)]             ------------(i)

   Consider (g*f)(t)

     (g*f)(t) = ∫g(t-v)f(v) dt  

               = ∫(e^-(t-v))(e^-t) dt

               = ∫(e^(-t+v-t)) dt

               = ∫(e^(-2t+v) dt

               = [(e^(-2t+v))/-2]              (v lies from 0 to t)

               = [(e^(-2t+0))/-2] - [(e^(-2t+t))/-2]

               = [(e^(-2t)/-2] - [(e^(-2t+t))/-2]

    (g*f)(t) = (-1/2)[(e^(-2t)) - e^(-t)]             ------------(ii)

          from (i) and (ii)

                (f*g)(t) = (g*f)(t)

     Hence, Convolution is proved.

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