Difference between revisions of "HW5.2 Jonathan Morales ECE301Fall2008mboutin" - Rhea

Latest revision as of 11:34, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform

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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