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Mathematics 14 Online
OpenStudy (curry):

Fourier Transform

OpenStudy (curry):

OpenStudy (curry):

@ganeshie8 @dan815

OpenStudy (ben00):

NEED HELP?

OpenStudy (curry):

Yes please!

OpenStudy (ben00):

Well I can't.

OpenStudy (kainui):

What have you tried?

OpenStudy (curry):

Well my initial thought was going to be to recover the function then apply the fourier on that. But this seems like just a linear piece wise function, so I'm lost on where to start.

OpenStudy (curry):

Because the problems during lecture were usually an input signal, a transfer function, and using that, we applied the fourier transform. So let's say the function we have is the input function. Should I convert that to a linear piece wise?

OpenStudy (kainui):

Wait, is this the entire function, is it zero for all other values less than 0 and greater than 2? Yeah, you should break it into piecewise to make your integral possible to evaluate, since you can break it up into separate integrals over your region

OpenStudy (curry):

Ye that is the entire region. Ok, so once I evaluate, can you help me from there? The piece wise would just be +2x for every period, given the period is 1, and starts from -1. Ofc, I'm stating that naively, but once I have that, how do I proceed?

OpenStudy (curry):

Since it's not in the exponential form, is it just Re(e^(it)), or something of that sort?

OpenStudy (kainui):

Wait I think there's some miscommunication, this is not a periodic function because it is just a single wiggle. Write out your best guess at the Fourier transform. I'm not sure what you are saying about "Not in the exponential form".

OpenStudy (curry):

well it's not a periodic function, because if it was, then it can be written using euler's formula.

OpenStudy (kainui):

yep I agree

OpenStudy (curry):

Ok, so let the unit function unit (t, a, b) have the value 1 on the interval a≤t<b and the value 0 otherwise. f(t) = (t)unit(t, 0, 0.5) + (-t)unit(t, 0.5, 1.5) + (t)unit(t, 1.5, 2).

ganeshie8 (ganeshie8):

.

OpenStudy (fifciol):

Tip: use the property \[F\{f'\} = j\omega F\{j\omega\}\] Simply differentiate each continuous piece of function to get square signal, then apply the equation: \[F_{\square}\{j\omega\} = Asinc(\omega a)e^{-j\omega b}\] |dw:1456912692007:dw| A is the area of a pulse

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