1. ## Fourier Transforms

1. Obtain from first principles, the Fourier transform of the function:

f(t) = 1, |t|<1 and 0, |t|>1

- I'm unsure of where to begin with this question, is it by using the formula (picture attached) and I don't know how to use the fact that |t|<1 or |t|>1

2. Find the fourier transform of

f(t) = u(t)e^(-at), a>0

- I know this can be read easily off tables but I tried using first principles and I found this statement (picture in red). A little off-topic but why is this statement true?

Any help would be greatly appreciated! Thanks in advance

2. Originally Posted by blueturkey
1. Obtain from first principles, the Fourier transform of the function:

f(t) = 1, |t|<1 and 0, |t|>1

- I'm unsure of where to begin with this question, is it by using the formula (picture attached) and I don't know how to use the fact that |t|<1 or |t|>1
Yes, the formula $F(\omega) = \int_{-\infty}^\infty\!\!f(t)e^{-i\omega t}dt$ is the place to start. The function f is 1 in the interval $-1\leqslant t\leqslant1$, and zero outside that interval. So the integral becomes $F(\omega) = \int_{-1}^1\!\!e^{-i\omega t}dt$, which you should be able to do.

Originally Posted by blueturkey
2. Find the fourier transform of

f(t) = u(t)e^(-at), a>0

- I know this can be read easily off tables but I tried using first principles and I found this statement (picture in red). A little off-topic but why is this statement true?
Not sure about this, but I suspect that u(t) is meant to be the Heaviside function, defined by u(t) = 1 if t>0, and u(t) = 0 if t<0. If so then this function f(t) is zero when t<0, so the integral over negative values of t is going to be zero.

3. thanks!