1. ## Fourier Transform

Hi there,

Just wondering if anyone can give me some clues as to how to go about getting the fourier transform using the formula integral from - infinity to + infinity f(t) e^-iwt dt

f(t) = { 1 - t/2 : 0 =< t <= 2

1 + t/2 : -2 =< t <= 0

0 : |t| > 2.

i have got as far as splitting it into four integrals and started integration by parts on 1/2 * integral between -2 and 0 of t.e^-iwt dt. it gets messy from there.. Just wondering if integration by parts is the way to go. I think i should end with an answer containing a few sinc functions.

Many thanks
Jack

2. Originally Posted by JackLondon
Hi there,

Just wondering if anyone can give me some clues as to how to go about getting the fourier transform using the formula integral from - infinity to + infinity f(t) e^-iwt dt

f(t) = { 1 - t/2 : 0 =< t <= 2

1 + t/2 : -2 =< t <= 0

0 : |t| > 2.

i have got as far as splitting it into four integrals and started integration by parts on 1/2 * integral between -2 and 0 of t.e^-iwt dt. it gets messy from there.. Just wondering if integration by parts is the way to go. I think i should end with an answer containing a few sinc functions.

Many thanks
Jack
I can't imagine why that would "get messy". To integrate $\displaystyle \int_{-2}^0 t e^{i\omega t}dt$ let u= t, $\displaystyle dv= e^{i\omega t}$ so that du= dt and $\displaystyle v= \frac{-i}{\omega}e^{i\omega t}$ so the integral becomes $\displaystyle \frac{-2i}{\omega}e^{-2i\omega}+ \int_{-2}^0 e^{i\omega t}dt$$\displaystyle = \frac{-2i}{\omega}e^{-2i\omega}- \frac{i}{\omega}(1- e^{-2i\omega}) 3. Originally Posted by HallsofIvy I can't imagine why that would "get messy". To integrate \displaystyle \int_{-2}^0 t e^{i\omega t}dt let u= t, \displaystyle dv= e^{i\omega t} so that du= dt and \displaystyle v= \frac{-i}{\omega}e^{i\omega t} so the integral becomes \displaystyle \frac{-2i}{\omega}e^{-2i\omega}+ \int_{-2}^0 e^{i\omega t}dt$$\displaystyle = \frac{-2i}{\omega}e^{-2i\omega}- \frac{i}{\omega}(1- e^{-2i\omega})$

Thanks HallsofIvy

It was 1/2 * integral between -2 and 0 of t.e^-iwt dt. You missed the minus sign.

Thanks anyway. dv = e^-iwt dt I was getting v = e^-iwt / -iw

Correct me if I am wrong.

So I assume integration by parts is the best way to get the Fourier Transform. The Maths Lecturer mentioned Partial Fractions.

Thanks again
Jack

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# was ist integral e^-iwt

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