[ \boxed\mathcalFu(t) = \pi \delta(\omega) + \frac1i\omega ]
[ u(t) = \begincases 0, & t < 0 \ 1, & t > 0 \endcases ]
At first glance, finding its Fourier transform seems impossible. The Fourier transform of a function ( f(t) ) is: fourier transform step function
[ u(t) = \lim_\alpha \to 0^+ e^-\alpha t u(t), \quad \alpha > 0 ]
[ \int_0^\infty e^-\alpha t e^-i\omega t dt = \int_0^\infty e^-(\alpha + i\omega) t dt = \frac1\alpha + i\omega ] [ \boxed\mathcalFu(t) = \pi \delta(\omega) + \frac1i\omega ]
The unit step function, often denoted ( u(t) ), is one of the most fundamental, yet mathematically troublesome, signals in engineering and physics. Defined as:
Here, ( e^-\alpha t ) ensures convergence for ( \alpha > 0 ). Then: Then: [ F(\omega) = \int_-\infty^\infty f(t) e^-i\omega t
[ F(\omega) = \int_-\infty^\infty f(t) e^-i\omega t dt ]
