Thread: Convolution Theorem - Fourier Transform

1. Convolution Theorem - Fourier Transform

Hi Guys,

I have the following question to solve (of which I have attempted over and over again to no avail):

Find the function:

g(x,y) = inversefourier{i/k(1+abs(k)x)exp(-abs(k)x}

and hence use the convolution theorem to find a solution for s(x,y) where;

S(x,k)=iF(k)/k(1+abs(k)x)exp(-abs(k)x)

If anyone has any idea please give me a hand. I appreciate that it may be hard to read/understand in the above format so if you have difficulty understanding the way it is written let me know and I'll email you the question.

Patrick.

2. Re: Convolution Theorem - Fourier Transform

Hey bluesboy91.

What is the frequency variable for your transform? Is it k?

3. Re: Convolution Theorem - Fourier Transform

Hey Chiro,

The original function was s(x,y) and was being operated on by a biharmonic operator;

We were asked to transform in terms of y, so - S(x,k) =F(s(x,y)). I solved the system subject to a set of initial conditions and got the following:

S(x,k)=iF(k)/k(1+abs(k)x)exp(-abs(k)x)

The question then asks us to consider the function:

g(x,y) = inversefourier{i/k(1+abs(k)x)exp(-abs(k)x)}

and then using the knowledge gained from that exercise solve S(x,k) via the convolution theorem.

Does that help?

Patrick.

4. Re: Convolution Theorem - Fourier Transform

Ohh ok I see where this is going.

Well if you have an transform that is in the form of G(p)F(p) then the inverse will be the convolution corresponding to the integral of g(t-x)*f(x) or the integral of f(t-x)*g(x)with respect to dx for 0 to t. wher F(p) is the fourier transform of the function f, and G(p) is the fourier transform of the function g.

Now if you know the fourier transform of F and G separately with respect to the corresponding functions f and g, you can just use this integral expression to evaluate the inverse by plugging in those function definitions.

So since you have g(x,y) has all the terms without the F(k), it means that F(k) is the fourier transform for some function f.

So if you know how to get the inverse of F(k) giving a corresponding f then the inverse of the entire thing is going to be Integral (0 to t) f(t-x)g(x)dx or Integral (0 to t) g(t-x)f(x)dx and if you have an explicit value of f or F(k) and get f, then you can calculate the integral.

5. Re: Convolution Theorem - Fourier Transform

Hey mate.

Thanks for that.

I'm struggling to work out what the g(x,y) is though. I can't find anything in my notes that helps me in that department.

Could you help me take the inverse of g(x,k) when g(x,k)=(i/k)(1+abs(k)x)exp(-abs(k)x))?

6. Re: Convolution Theorem - Fourier Transform

g(x,y) = InverseFourier(G(k)) = InverseFourier(i/k(1+abs(k)x)exp(-abs(k)x)).

f = InverseFourier(F(k)) and you are trying to find

S = InverseFourier(F(k)*G(k)) which is why you use the convolution theorem since you are given g in terms of G and if can get f then you can use the fact that InverseFourier(F(k)*G(k)) = Integral (0 to t) f(t-x)g(x)dx since Fourier(Integral (0 to t)f(t-x)g(x)dx) = F(k)*G(k).

7. Re: Convolution Theorem - Fourier Transform

Awesome,

F(k) is just defined by the boundary conditions in the problem as S_y=f(y) and so when I solved the system in the beginning and took FT's of the BC's that's where F(k) came from.

So when putting into the integral of the convolution, F(k)=f(y) it is more or less undefined until the 3rd part of the question which I need this result to do.

So the solution, will then be the following?

InverseFourier(F(k)*G(k)) = Integral (0 to t) f(t-x)g(x)dx since Fourier(Integral (0 to t)f(t-x)g(x)dx) = F(k)*G(k).

8. Re: Convolution Theorem - Fourier Transform

Yeah that's the crux of it and the main reason is your very last sentence: without that you couldn't do it.

9. Re: Convolution Theorem - Fourier Transform

Having said and understood all of that the only thing I can't do and need help desperately is inverting g(x,k). Can you help me with that?

10. Re: Convolution Theorem - Fourier Transform

For the g function you know s(x,y) it's fourier transform and I assume you know the inverse transform of F(k) as well.

So you know FourierTransform(s(x,y)) = S(x,y) = F(k)*G(k).

You also know that s(x,y) = InverseFourierTransform(F(k)*G(k)) which is Integral (0 to t) [f(t-x)g(x)]dx. So you know the value of the integral, you know what f(y) is for some y which means you can look at an Integro-Differential equation for g(x) where you have s(x,y) = Integral (0 to t) f(t-x)g(x)dx. Now s(x,y) is an explicitly known function, as is f(t-x) so you can apply the chain rule to the integral to get g(x) in terms s'(x,y) and f(x) for various values inputted into the functions f and s.

This wiki does a way better job that I can at the moment:

Differentiation under the integral sign - Wikipedia, the free encyclopedia

11. Re: Convolution Theorem - Fourier Transform

Hi Chiro.

Thanks for all your help. I'll work it from here.

All the best,

Patrick.