# Fourier transform help

1. Jun 22, 2011

### billiards

1. The problem statement, all variables and given/known data

Show that for a fixed value of $\omega$ that $G(\omega)e^{-i\omega t}$ is the response of the system to the input signal $e^{-i\omega t}$.

(From Roel Snieder's book 'A Guided Tour of Mathematical Methods for the Physical Sciences', pg 233 (Section 15.7, Problem e))

2. Relevant equations (I think)

The Fourier transform pair:

$f(t)=\int^{\infty}_{-\infty}F(\omega)e^{-i\omega t}d\omega$

$F(\omega)=\frac{1}{2\pi}\int^{\infty}_{-\infty}f(t)e^{i\omega t}dt$

Convolution relations:

$o(t)=\int^{\infty}_{-\infty}g(t-\tau)i(\tau)d\tau$

$O(\omega)=2\pi G(\omega)I(\omega)$

3. The attempt at a solution

I'm not really sure I understand the question. I am assuming that the 'response of the system' is $g(t)$ and that 'input signal' is $i(t)$. I could well be wrong on that.

So I am reading the question as saying: show that for the case $i(t)=e^{-i\omega t}$ ...

$g(t)=G(\omega)e^{-i\omega t}$.

I have tried playing around with the symbols but I have not managed to get thing to click, which makes me suspect that I am misunderstanding the question. I have been sitting on this problem for weeks and have managed to progress through the book no problem, but not getting this question is bugging me.

Any help would be greatly appreciated.

2. Jun 22, 2011

### hikaru1221

g(t) is the impulse response of the system, as you pointed out in the convolution relations
It would be clearer I think if we rewrite the question like this: show that for a system of impulse response g(t), the response to the input $e^{-i\omega _0t}$ is $G(\omega _0)e^{-i\omega _0t}$.

3. Jun 22, 2011

### billiards

Hi hikaru, thanks for quick reply.

Just to be clear, by

Do you mean the output, $o(t)$ is $G(\omega _0)e^{-i\omega _0t}$?

4. Jun 22, 2011

### hikaru1221

Yup.

5. Jun 22, 2011

### hikaru1221

Oh I forgot this: the system should be linear time-invariant. Otherwise, it would be absurd to mention G(w) without any clarification.

6. Jun 22, 2011

### billiards

OK so let me see ...

Rewrite the convolution theorem such that:

$o(t)=\int^{\infty}_{-\infty}g(\tau)i(t-\tau)d\tau$

Let $i(t-\tau)=e^{-i\omega (t-\tau)}$, (in compliance with the original question,) such that:

$o(t)=\int^{\infty}_{-\infty}g(\tau)e^{-i\omega (t-\tau)}d\tau$

Consider the Fourier transform of $g(\tau)$:

$G(\omega)=\frac{1}{2\pi}\int^{\infty}_{-\infty}g(\tau)e^{i\omega \tau}d\tau$

And plug this into above to get:

$o(t)=2\pi G(\omega)e^{-i\omega t}$.

I'm off by a factor of $2\pi$, where did I go wrong?

7. Jun 22, 2011

### hikaru1221

You should check the relations you wrote in your post 1 section 2 They're wrong.

8. Jun 22, 2011

### billiards

Really, I just checked them in the text book and they're not wrong (unless the book's wrong).

I was thinking that the $2\pi$ dropped out from my choice of scale factor in my Fourier transform, and so it is arbitrary...?

9. Jun 22, 2011

### hikaru1221

10. Jun 22, 2011

### billiards

OK so if I changed that scale factors around for my Fourier transforms it would work, and I'm kind of happy with that because it means I understand it (up to a fair level). But something is still out of place, making me feel uneasy and uncertain about whether I really do understand it.

Why would the book be wrong? The book sticks to the convention and it works in all the other examples perfectly... so what is it about this particular problem that makes this issue stick out?

And what are the reasons for not having the scale factors this was around? I was under the impression that it didn't matter, as long as the product of the scale factors is equal to $1/2\pi$. If I can get these issues straight in my head I will be happy.

11. Jun 22, 2011

### hikaru1221

The book is not so wrong; it just uses a different set of conventions (but not very practically applicable to physics/ science though, in my opinion). Under those conventions, the factor of 2*pi should be there. However, the question follows the popular convention, and so, the factor is not there in the question. Perhaps the author messes up himself just kidding

12. Jun 22, 2011

### billiards

HaHa

Thanks

Maybe I will email the author and see what he has to say

13. Jun 22, 2011

### hikaru1221

I think you should also ask him about the motivation that he decided on using his rather "own" set of conventions for the F-transform pair. To me, his way is quite counter-intuitive.