## 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).

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

## Alumni Liaison

Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics