Difference between revisions of "HW5.2 Scott Hamilton ECE301Fall2008mboutin" - Rhea

Latest revision as of 12:30, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform

Question

Specify a signal x(t) and compute its Fourier transform using the integral formula. (Make sure your signal is not trivial to transform; it should be hard enough to be on a test).

Answer

Defining x(t):

$ x(t) = te^{-4t}u(t-3) $

By the integral formula:

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

Therefore:

$ \mathcal{X}(\omega) = \int_{-\infty}^{\infty}te^{-4t}u(t-3)e^{-j\omega t}\,dt\, $

$ \mathcal{X}(\omega) = \int_{3}^{\infty}te^{-4t}e^{-j\omega t}\,dt\, $ (maybe remove)

$ \mathcal{X}(\omega) = \int_{3}^{\infty}te^{-t(4+ j\omega)}\,dt\, $

Integrating by parts

$ \int UdV=UV - \int VdU $

Where $ U=t; dU=1; V= \frac{-1}{4+j \omega} e^{-t(4+j \omega)}; dV= e^{-t(4+j \omega)} $

Therefore:

$ \mathcal{X}(\omega) = \left. te^{-t(4+ j\omega)} + \frac{1}{(4+ j\omega)^2}e^{-t(4+ j\omega)} \right]_{3}^{+\infty} $

$ \mathcal{X}(\omega) = \left. (t+ \frac{1}{(4+ j\omega)^2} )e^{-t(4+ j\omega)} \right]_{3}^{+\infty} $

$ \mathcal{X}(\omega) = [0] - [(3+ \frac{1}{(4+ j\omega)^2} )e^{-3(4+ j\omega)}] $

Therefore: $ \mathcal{X}(\omega) =-(3+ \frac{1}{(4+ j\omega)^2} )e^{-(12+ j3\omega)} $

Back to Practice Problems on CT Fourier transform