Difference between revisions of "HW5.3 Scott Hamilton ECE301Fall2008mboutin" - Rhea

Revision as of 18:58, 7 October 2008

Specify a Fourier transform X(w) and compute its inverse 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).

Define X(w):

$ \mathcal{X}(\omega) = 4 \pi \delta(\omega - 3) + 4 \pi \delta(\omega + 3) - 8 \pi \delta(\omega - 7) $

By the integral formula:

$ x(t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \mathcal{X}(\omega) e^{-j\omega t}\,d \omega $

Therefore:

$ x(t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} (4 \pi \delta(\omega - 3) + 4 \pi \delta(\omega + 3) - 8 \pi \delta(\omega - 7)) e^{-j\omega t}\,d \omega $

$ x(t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} 4 \pi \delta(\omega - 3) e^{-j\omega t}\,d \omega + \frac{1}{2 \pi} \int_{-\infty}^{\infty} 4 \pi \delta(\omega + 3) e^{-j\omega t}\,d \omega - \frac{1}{2 \pi} \int_{-\infty}^{\infty} 8 \pi \delta(\omega - 7) e^{-j\omega t}\,d \omega $

Pull out the constants:

$ x(t)= \frac{4 \pi}{2 \pi} \int_{-\infty}^{\infty} \delta(\omega - 3) e^{-j\omega t}\,d \omega + \frac{4 \pi}{2 \pi} \int_{-\infty}^{\infty} \delta(\omega + 3) e^{-j\omega t}\,d \omega - \frac{8 \pi}{2 \pi} \int_{-\infty}^{\infty} \delta(\omega - 7) e^{-j\omega t}\,d \omega $

Simplifying:

$ x(t)= 2 \int_{-\infty}^{\infty} \delta(\omega - 3) e^{-j\omega t}\,d \omega + 2 \int_{-\infty}^{\infty} \delta(\omega + 3) e^{-j\omega t}\,d \omega - 4 \int_{-\infty}^{\infty} \delta(\omega - 7) e^{-j\omega t}\,d \omega $

Remember that:

$ \int_{-\infty}^{\infty} \delta(\omega - T_0) e^{-j\omega t}\,d \omega = e^{-j T_0 t} $

Therefore:

$ \ x(t)= 2e^{-j3t} + 2e^{j3t} -4e^{-j7t} $

Recall that:

$ Cos(\omega t) = \frac{e^{j \omega t} + e^{-j \omega t}}{2} $

Therefore:

$ \ x(t) = 4cos(3t) - 4e^{-j7t} $