# Question on Convolution theorem

1. Nov 26, 2011

### dimension10

I have a question on the Convolution theorem for Fourier Transforms. The convolution theorem states that

$$\mathscr{F}\{f(t) g(t)\}=\mathcal{F}\{f(t)g(t)\}=\mathcal{F}\{f(t)\}\ast \mathcal{F}\{g(t)\}-\mathscr{F}\{f(t)\} \ast \mathscr{F}\{g(t)\}$$

$$\mathscr{F}\{f(t) \ast g(t)\}=\mathcal{F}\{f(t)\ast g(t)\}=\mathcal{F}\{f(t)\}\mathcal{F}\{g(t)\}= \mathscr{F}\{f(t)\}\mathscr{F}\{g(t)\}$$

and the same for the Laplace Transform.

However, my sources use the definition of the Fourier Transform as

$$\mathscr{F}\{f(t)\}=\mathcal{F}\{f(t)\}=\int_{-\infty}^{\infty}{e}^{-2i\pi\xi t}f(t) \mbox{d}t$$

But my question is that if I were to use the definition,

$$\mathscr{F}\{f(t)\} = \mathcal{F}\{f(t)\}=\frac{1}{ \sqrt{2\pi} }\int_{-\infty}^{\infty}{e}^{-i\omega t}f(t) \mbox{d}t$$

then will the convolution theorem for Fourier transforms still hold? There is one article that states that it does hold. However, when I checked Wikipedia, they say that the convolution theorem for Fourier transforms does not hold for the definition which uses angular frequency instead of ordinary frequency.

So, does the Convolution theorem for Fourier Transforms still hold for the definition involving omega instead of xi.

2. Nov 26, 2011

### I like Serena

Hi dimension10!

Yes, the convolution theorem holds for all variants of the Fourier transform and the Laplace transform.

It is an arbitrary choice whether you include 2pi in the exponent or not.
It is also an arbitrary choice where you put the normalization constant, which can be in the Fourier transform, the Inverse Fourier transform, or in both.
As long as you're consistent of course.

3. Nov 26, 2011

### dimension10

Ok thanks a lot.

4. Nov 26, 2011

### I like Serena

I was curious where you may have found that the convolution theory would not hold, so I tried to look it up in wikipedia.

What I found is this:
http://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships
In the table, you can see that the convolution theorem holds, but in the case of angular frequency with a normalization constant of $1 \over\sqrt{2\pi}$ in the Fourier transform, you also need to multiply by $\sqrt{2\pi}$ for the convolution theorem to hold.

5. Nov 26, 2011

Thanks.