1. ## Fourier transformation

Find the Fourier transformation of f(x)=(1-|x|)(H(x+1)-H(x-1)) where H(x) is the funcion of the Heaviside.
I can't even guess where to start.

2. Originally Posted by totalnewbie
Find the Fourier transformation of f(x)=(1-|x|)(H(x+1)-H(x-1)) where H(x) is the funcion of the Heaviside.
I can't even guess where to start.
H(x+1) = 0 for x<-1 otherwise it is 1.
H(x-1) = 0 for x< 1 otherwise it is 1.

This tells us that,
H(x+1) - H(x-1) = 0 - 0 = 0 for x<1
H(x+1) - H(x-1) = 1 - 0 = 1 for -1<x<1
H(x+1) - H(x-1) = 1 - 1 = 0 for 1<x

Thus, the function H(x+1)-H(x-1) is non-zero on the interval -1<x<1

So when you integrate on the interval (-oo,+oo) you can just do it on (-1,1)
Because it is zero everywhere else and it does not matter.

But furthermore, on the interval (-1,1) the Heaviside functions add up to 1.
Which makes no difference when you multiply a number by them.

Thus,
1/sqrt(2*pi) * INT(-1,1) (1-|x|)*exp(-ist) dx
Is the required Fourier Transform.

3. Originally Posted by ThePerfectHacker
H(x+1) = 0 for x<-1 otherwise it is 1.
H(x-1) = 0 for x< 1 otherwise it is 1.

This tells us that,
H(x+1) - H(x-1) = 0 - 0 = 0 for x<1
H(x+1) - H(x-1) = 1 - 0 = 1 for -1<x<1
H(x+1) - H(x-1) = 1 - 1 = 0 for 1<x

Thus, the function H(x+1)-H(x-1) is non-zero on the interval -1<x<1

So when you integrate on the interval (-oo,+oo) you can just do it on (-1,1)
Because it is zero everywhere else and it does not matter.

But furthermore, on the interval (-1,1) the Heaviside functions add up to 1.
Which makes no difference when you multiply a number by them.

Thus,
1/sqrt(2*pi) * INT(-1,1) (1-|x|)*exp(-ist) dx
Is the required Fourier Transform.
What do u mean by exp(-ist) ?

4. Originally Posted by totalnewbie
What do u mean by exp(-ist) ?
It is a type for exp(-i x t) which is the kernel of the Fourier Transform (up to
a randomly positioned constant, and arbitrary convention on what is the
forward and what the backward transform) where the variable in the
transformed domain is t.

RonL