|Sep11-12, 02:25 AM||#35|
Fourier Transform question (kind of Urgent)
(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!
|Sep11-12, 04:55 AM||#36|
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 ???
|Sep11-12, 09:57 AM||#37|
(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.
|fourier, frequency domain, system, time, transform|
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