Fourier Transform question (kind of Urgent)

toneboy1

Just experimenting, I noticed if w = 3, width = 2 then
(2/w)*sin(2*w/2) is ALMOST the same as (2.pi/w)*sin(2*w/2.pi),
so you seem right about it being a notational thing, but what I can't understand is that the one with the pi in it seems more accurate because in one instance the number was finite and the other kept going. Can you see why this would be?


H'mm, I'm still not clear, I'll elaborate on where I got stuck:

x(t) = (1/2pi)* ∫ (exp(jwt)-exp(-jwt) /2j)*exp(jwt) dw -∞ to ∞

∴ (1/2pi)* ∫(exp(2jwt)-exp(0) / 2j) dw -∞ to ∞

∴ (1/j4pi)*[exp(2jwt)/2jw - w ] -∞ to ∞

And I'm not sure where to go from here. (especially with your time delay)

Alternatively using duality and linearity I attempted to make

sin(ω) = 2pi*sin(-w)/2pi

inverse transformed into: j.pi/2pi * [δ(1-t)-δ(t-1)] which is wrong according to wolfram.

Cheers for any input!

rude man

Not sure what you're getting at here.

In the below, where on Earth did you get that equation? You're apparently trying to do an inverse transform on sin(wt)?? On a function containing t ???

I think I need to go to bed!
Cheers!

toneboy1

[I was saying
(2/w)*sin(2*w/2) ≈ (2.pi/w)*sin(2*w/2.pi)
for reasons I cannot reckon]

Woops, good call, it should have been 't'less, I.e:
x(t) = (1/2pi)* ∫ (exp(jw)-exp(-jw) /2j)*exp(jwt) dw -∞ to ∞


Heh, heh, fairenough, Night.

rude man

Why are you trying to determine F^-1{(sin(w)} anyway? Its inverse is not at all a common time function ....

If I had to do it I would study the way sin(wt) was transformed, then try to apply that technique to the inversion integral.