## Fourier Transform

$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$

$x(t)=(t-1)e^{-6t+6}u(t-1) \,\$

$X(\omega)=\int_{-\infty}^{\infty}x(t)=(t-6)e^{-6t+6}u(t-6) e^{-j\omega t}dt \;$

$x(t) \,\$looks like $te^{-6t}u(t) \,\$ so we evaluate that

the F.T of $te^{-6t}u(t) \,\$ is

$\int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \;$

$=\int_{0}^{\infty}te^{-6t-j\omega t}dt \;$

$=\int_{0}^{\infty}te^{-t(6+j\omega t)}dt \;$

Do integration by parts

$={\left. \frac{-te^{-t(6+j\omega )}}{6+j\omega }\right]_{0}^{\infty}} - \int_{0}^{\infty}\frac{-te^{-t(6+j\omega )}}{6+j\omega }dt \;$

$={\left. \frac{-e^{-t(6+j\omega )}}{(6+j\omega)^2 }\right]_{0}^{\infty}}$

$= \frac{1}{(6+j\omega)^2}$

And now we use the time shift property and get

$X(\omega)=\frac{e^{-j\omega}}{(6+j\omega)^2}$

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