Understanding Roel Snieder's Fourier Transform Conventions

[billiards]

Homework Statement

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. Homework 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)

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.

 

[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}.

 

[billiards]

Hi hikaru, thanks for quick reply.

Just to be clear, by

...the response to the input e^{-i\omega _0t} is G(\omega _0)e^{-i\omega _0t}

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

 

[hikaru1221]

Yup.

 

[hikaru1221]

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

 

[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?

 

[hikaru1221]

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

 

[billiards]

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

Really, I just checked them in the textbook 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...?

 

[hikaru1221]

You can see the first two here: http://mathworld.wolfram.com/FourierTransform.html (equation (7)-(10))
There are reasons why they should be that way but not the way you wrote them.

 

[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.

 

[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

 

[billiards]

HaHa

Thanks

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

 

[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.