[ u(t) = \begincases 0, & t < 0 \ 1, & t > 0 \endcases ]
This gives ( 1/(i\omega) ), but this is not the whole story. Something is missing: the step function has a nonzero average value (1/2 over all time, if we consider symmetric limits), which implies a DC component. It turns out that the Fourier transform of the unit step function is: fourier transform step function
At first glance, finding its Fourier transform seems impossible. The Fourier transform of a function ( f(t) ) is: [ u(t) = \begincases 0, & t <
[ \boxed\mathcalFu(t) = \pi \delta(\omega) + \frac1i\omega ] The Fourier transform of a function ( f(t)
[ \int_0^\infty e^-\alpha t e^-i\omega t dt = \int_0^\infty e^-(\alpha + i\omega) t dt = \frac1\alpha + i\omega ]
The Fourier transform of ( \textsgn(t) ) is ( 2/(i\omega) ) (without a delta, since its average is zero). Thus: