By definition, H(x) is 0 if x < 0, 1 if x > 0, so setting x -> -x gives H(-x) is 0 if x > 0, 1 if x < 1. By inspection, you should be able to relate H(-x) to H(x), by H(-x) = 1 - H(x), which will help you with your Fourier transform.
#3
squenshl
468
4
Sweet.
My Fourier transform is f(w) = 1/(1-iw)
#4
squenshl
468
4
So is the Fourier transform of H(x)exp(-2x) + H(-x)exp(x)
f(w) = 1/(2+iw) + 1/(1-iw)?