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At first glance, finding its Fourier transform seems impossible. The Fourier transform of a function ( f(t) ) is:
confirming the result. | Function | Fourier Transform | |----------|------------------| | ( u(t) ) (unit step) | ( \pi\delta(\omega) + \frac1i\omega ) | | ( \textsgn(t) ) (sign) | ( \frac2i\omega ) | | Constant ( 1 ) | ( 2\pi\delta(\omega) ) | fourier transform step function
The Fourier transform of ( \textsgn(t) ) is ( 2/(i\omega) ) (without a delta, since its average is zero). Thus: At first glance, finding its Fourier transform seems
[ \boxed\mathcalFu(t) = \pi \delta(\omega) + \frac1i\omega ] \quad \alpha >
[ \int_0^\infty e^-\alpha t e^-i\omega t dt = \int_0^\infty e^-(\alpha + i\omega) t dt = \frac1\alpha + i\omega ]
(its value at ( t=0 ) is often set to ( 1/2 ) for Fourier work), it represents an idealized switch that turns “on” at time zero and stays on forever.
[ u(t) = \lim_\alpha \to 0^+ e^-\alpha t u(t), \quad \alpha > 0 ]