Compute the Fourier transform of the following CT signal using the integral formula:
$ \,x(t)=e^{-5(t+3)}u(t-1) + e^{-j\pi t}\delta(t-\frac{\pi}{2})\, $
Answer
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}e^{-5(t+3)}u(t-1)e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}e^{-j\pi t}\delta(t-\frac{\pi}{2})e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=\int_{1}^{+\infty}e^{-5t}e^{-15}e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}\delta(t-\frac{\pi}{2})e^{-j(\omega +\pi)t}\,dt\, $
$ \,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-(j\omega +5)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\left. \frac{e^{-15}}{-(j\omega +5)}e^{-(j\omega +5)t}\right]_{1}^{+\infty} + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\frac{e^{-15}}{-(j\omega +5)}(e^{-j\omega(\infty)}e^{-5(\infty)}-e^{-(j\omega +5)}) + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\frac{e^{-15}}{(j\omega +5)}e^{-(j\omega +5)} + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\frac{1}{(j\omega +5)}e^{-(j\omega +20)} + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
