Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
$ x(t) = e^{-2t} , t>2 \, $
$ x(t)= e^{-4t} , 0 \le t \le 2 $
$ x(t)= 0 , t < 0 \, $
We can summarize it as
$ x(t)= e^{-2t} u(t-2) + e^{-4t}( u(t-2)-u(t))\, $
$ X(\omega) = \int^\infty_\infty e^{-2t}e^{-j\omega t} dt + \int^2_0 e^{-4t}e^{-j\omega t} dt\, $
$ X(\omega) = \int^\infty_\infty e^{-(2+j\omega)t} dt + \int^2_0 e^{-(4+j\omega) t} dt\, $
$ X(\omega) = {\left. \frac{e^{-(j\omega + 2)t}}{-(j\omega +2)} \right]^{\infty}_0 } + {\left. \frac{e^{-(j\omega + 4)t}}{-(j\omega +4)} \right]^2_0 }\, $
$ X(\omega) = \frac{e^{-(2j\omega + 4)t}}{j\omega +2} - \frac{e^{-(2j\omega + 8)t}}{-j\omega +4} + \frac{1}{4+j\omega} \, $
$ X(\omega) = \frac{e^{-(2j\omega + 4)t}}{j\omega +2} + \frac{1 - e^{-(2j\omega + 8)t}}{-j\omega +4} \, $
