Fourier transform convolution Question! Please

[cyw1984]

Given F[f(t)*g(t)] = F(ω)G(ω) , where * = convolution
F(f(t)) = F(ω) ; F(g(t)) =G(ω)

How to prove F[f(t)g(t)] = 1/2pi [ F(w)*G(w)]

I have to do -
F^-1(F(ω)) = 1/2pi ∫F(ω)e^iωt dω = f(t)
F^-1(G(ω)) = 1/2pi ∫G(ω)e^iωt dω = g(t)


F[f(t)g(t)] = F[F^-1(F(ω))F^-1(G(ω))]
= ∫[1/2pi ∫F(ω)e^iωt dω 1/2pi ∫G(ω)e^iωt dω ] e^-iωt dt
......
.....
.....

 

[claudiocamera]

First you have to consider two variables for frequency domain, thus

f(t)=1/2pi ∫F(ζ)e^iζt dζ
g(t)=1/2pi ∫G(η)e^iηt dη

so F¯1 {f(t).g(t)} = 1/(2pi)² ∫∫F(ζ)G(η)ej(η+ζ)tdζdη

Now change the variable η=ω-ζ to obtain
1/2pi ∫ [1/2pi∫F(ζ)G(ω-ζ)dζ]e^jωtdω

regonizing the expression into the brackets as convolution you have the answer you want

 

[Dmitrij]

The idea, i'm going to introduce, is may be not a strict proof, but the physical sence of this very important equality.

This expression is known to be the Parseval equation, which is widely spread in applications of Signal Processing. It is especially useful and helpful when you need to compute the energy of the process either in time or frequency domains:

Imagine f1(t) = f2(t) = f(t).

Then Parseval equation converts to Plansherel equation (the private case):

int (s(t)^2) = 1/2pi (int(S(w)^2)), where S(w) - amplitude spectrum of the signal. Both left and right parts are the energy of the process. Thus this very important characteristic may be estimated in 2 domains.

Besides, the mathematical sense of this equation is that Fourier operator doesn't change the scalar multiplication of 2 functions, which can be computed in 2 domains.

With repect,

Dmitrij

 

[cyw1984]

First you have to consider two variables for frequency domain, thus

f(t)=1/2pi ∫F(ζ)e^iζt dζ
g(t)=1/2pi ∫G(η)e^iηt dη

so F¯1 {f(t).g(t)} = 1/(2pi)² ∫∫F(ζ)G(η)ej(η+ζ)tdζdη

Now change the variable η=ω-ζ to obtain
1/2pi ∫ [1/2pi∫F(ζ)G(ω-ζ)dζ]e^jωtdω

regonizing the expression into the brackets as convolution you have the answer you want

O....I have some idea now

But I want to start at F {f(t)g(t)}
= ∫[1/2pi ∫F(ω)e^iωt dω 1/2pi ∫G(ω)e^iωt dω ] e^-iωt dt

How to express to 1/2pi ∫ [1/2pi∫F(ζ)G(ω-ζ)dζ]e^jωtdω
>__<