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| − | <math> \mathcal{X}(\omega) =-(3+ \frac{1}{(4+ j\omega)^2} )e^{-(12+ | + | <math> \mathcal{X}(\omega) =-(3+ \frac{1}{(4+ j\omega)^2} )e^{-(12+ j3\omega)} </math> |
Revision as of 13:42, 7 October 2008
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).
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)} $
