# Question about Waves

1. Dec 2, 2013

### sapz

Hello,
I'm trying to figure out a solution of a question I saw, I'll ask it here in this forum because perhaps this is where it belongs.

Let's say I have a Pulse wave that I am given it's frequency spectrum: $B(w) = V_0\frac{Sin[(w-w_0)T]}{(w-w_0)T}$ Where V0, T, w0 are constants.

So I do an inverse fourier transform to get the actual wave, and I get:
$B(t) = F^{-1}( B(w) ) = \frac{V_0}{T}\sqrt{\frac{\Pi}{2}}e^{-iw_0t}[Heaviside(t+T)-Heaviside(t-T)]$

And now I'm asked, what is the intensity of the pulse for w=w0?

So the solution I've seen, claims that when w=w0, you get:
$B(t)_{w=w_0} = B(w_0)Sin(w_0t)$ (Where B(w0) is when you put w0 in B(w))
It states this without an explanation of why this is correct.

(And after you have that, you can find the intensity pretty easily, but its here that I get stuck)

To me, this statement ($B_{w=w_0}(t) = B(w_0)Sin(w_0t)$) makes sense only if the entire wave B(t) was an odd function, because then it would comprise only of Sin waves.
And then, out of all the continuous values of w I could take only the one I'm interested in, which is w0, and putting it in B(w) would give me the amplitude of that specific wave.

How is this statement justified? Is B(t) an odd function?

2. Dec 2, 2013

### Simon Bridge

Sketch it out - is it?

The difference between cosine and sine is just phase though.
Technically it only needs to be odd at t=0 since it is saying that B(t) is oscillating.
Can you tell if the pulse is composed entirely of sine waves or not by examining B(w)?

3. Dec 2, 2013

### sapz

I don't really know how to sketch it since it has an imaginary part... Do I take only the real part?
I which case it is not odd, but even.

As to B(w), it is not even and not odd, I think. So can it tell me something about the form of B(t) just by examining it?

Last edited: Dec 2, 2013
4. Dec 2, 2013

### Simon Bridge

When $\omega=\omega_0$, $B(\omega)=?$
The frequency spectum is a sinc function right? It's symmetric about...?

Don't you also expect $B_\omega$ to be different for different values of $\omega$?
Do you have a general expression for this?

Last edited: Dec 2, 2013
5. Dec 3, 2013

### sapz

I'm really not sure I'm getting where you're going with this.

When w=w0, B(w)=V0. So its symmetric around w0. I do expect Bw to be different for different values of w, but that is true whether b(t) is even or odd or neither.
Does it tell me anything about the symmetric properties of b(t)?

6. Dec 3, 2013

### Simon Bridge

The hard part about this is trying to nudge you into the answers without actually telling you the answer ...

You are concerned that:
$B_\omega(t)=\mathcal{B}(\omega)\sin\omega t$

Should be
$B_\omega(t)=\mathcal{B}(\omega)\cos\omega t$

... or... something... I was hoping you'd tell me your understanding ;)
i.e. symmetric or anti-symmetric in time.

These two differ only by phase - i.e. by the choice of t=0.
So you are looking for a way to tell what the appropriate phase is - of if that even matters.

7. Dec 3, 2013

### sapz

My understanding is that the phase is extremely important, since the next step would be to find the intensity by integration of z|b'(t)|^2 (which is the instantaneous intensity) over [0,T] (because the wave exists in [-T,T] and there is no negative time], and T is not neccesarily the period, its a given constant. So phase does matter - I can't shift the time axis.

So it is important to decide whether we get sin or cos, and the way I could think of to do that, would be to determine if the pulse is odd or even.

Last edited: Dec 3, 2013
8. Dec 3, 2013

### Simon Bridge

... hmmm, first time you've mentioned instantanious intensity.

I take it, then, you are trying to calculate the contribution to the instantanious intensity vector $I(\vec{r},t)$ for 1D pulse propagating along the z axis - and you suspect the book made a mistake in the phase of the components of the time-domain pulse?

Perhaps the best thing for your understanding would be to procede as if you have it right and see what difference it makes.

9. Dec 4, 2013

### Simon Bridge

OK - I've had a chance to review.
Your question seems to be related to your other thread: