# Fourier Transform

• Feb 13th 2009, 04:10 PM
JackLondon
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
• Feb 13th 2009, 06:21 PM
HallsofIvy
Quote:

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$$\displaystyle = \frac{-2i}{\omega}e^{-2i\omega}- \frac{i}{\omega}(1- e^{-2i\omega}) • Feb 14th 2009, 02:44 AM JackLondon Quote: Originally Posted by HallsofIvy I can't imagine why that would "get messy". To integrate \displaystyle \int_{-2}^0 t e^{i\omega t}dt let u= t, \displaystyle dv= e^{i\omega t} so that du= dt and \displaystyle v= \frac{-i}{\omega}e^{i\omega t} so the integral becomes \displaystyle \frac{-2i}{\omega}e^{-2i\omega}+ \int_{-2}^0 e^{i\omega t}dt$$\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