# Homework Help: A Wave with complex values

1. Dec 1, 2013

### sapz

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

I have a basic question about how to mathematically treat waves.

I've seen many times a guess of a wave of the form $Ae^{iwt}$
My question is, how do I actually know what is the amplitude in t=t0? Do I take the real value only? The imaginary value only?

For example, here:
$ψ(t)=A*(Heaviside(t+T)-Heaviside(t-T))*e^{-iwt}$

How do I know if the wave is an even or an odd function? And how do I know it's amplitude at t=3?
Thank you (:

2. Dec 1, 2013

### rude man

Graph the function ψ. What is the definition of an odd function? Of an even function? So which is ψ?
The convention is to read exp(-iwt) as Re{exp(-iwt)} = cos(-wt) = cos(wt).

3. Dec 1, 2013

### sapz

So in our case (Heaviside(t+T)−Heaviside(t−T)) is an even function, and the Re{exp(-iwt)} = cos(-wt) = cos(wt) is an even function, so the function of their multiplication is also even?

4. Dec 2, 2013

### rude man

Right. U(t+tau) - U(t-tau) is 0 except = 1 from -tau to +tau. So that function is even, and multiplying it by cos(wt) keeps it even: ψ(-t) = ψ(t).

This is a strange question, actually. We are not distinguishing between exp(iwt) and exp(-iwt) here, yet the two are different functions of t. On the other hand, I know of no such thing as even and odd complex functions. So the question seems a bit academic.

5. Dec 2, 2013

### sapz

I'm trying to figure out a solution of a question I saw... So this is where the question comes from.
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).
(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.

But it seems its an even function? How would you approach this?

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

### rude man

Have you had analytic functions and the Hilbert transform?

7. Dec 2, 2013

Nope :(

8. Dec 2, 2013

### rude man

OK, I'll try to outline an approach for you:

The first thing that's apparent is that your B(w) is not the Fourier transform of a real function of time. If it were, then the inverse transform of B(w), which you've already done, and done correctly I believe, would be real, but it's not.

Yet you're looking for a real function of time.

My take, and here I must warn you that I'm not an expert in this area & that I'm guessing a bit:

Assume B(w) is a one-sided transform of a new function, call it z(t). z(t) is called analytic because Z(w) = B(w) is one-sided.
By definition, Z(w) = 0 for w < 0.

And that z(t) = your b(t) + jb^(t)
where b^(t) is the Hilbert transform of b(t).

In other words, you have already solved for F-1(Z(w)} = z(t)
so your b(t) is just the real part of z(t). No need to even compute the Hilbert transform.

Try that & see what answer you get. If you don't get additional responses in this forum you might want to re-post on the advanced math forum.

I notice the answer given you is not exactly the same as what you'd get with the above approach. But remember cos(wt) = sin(wt + π/2) and that a spectrum function has no phase information.

BTW don't write B(t) as the inverse FT of B(w). B(t) = B(w) with w replaced by t. Write b(t) instead.
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I also looked at F-1(w) = (1/2π)∫ from 0 to ∞ of F(w)exp(iwt) dw since F(w) is assumed one-sided, with w > 0 only, but ran into problems performing that integral.

9. Dec 2, 2013

### sapz

Thank you rude man

10. Dec 4, 2013