(New page: <math>\,\mathcal{X}(\omega)= \frac{\pi}{j} \delta (w - 2\pi) + \frac{2\pi}{j} \delta (w + 2\pi)</math>)
 
Line 1: Line 1:
 +
Compute the Inverse Fourier Transform of:
 +
 
<math>\,\mathcal{X}(\omega)= \frac{\pi}{j} \delta (w - 2\pi) + \frac{2\pi}{j} \delta (w + 2\pi)</math>
 
<math>\,\mathcal{X}(\omega)= \frac{\pi}{j} \delta (w - 2\pi) + \frac{2\pi}{j} \delta (w + 2\pi)</math>
 +
 +
<math>x(t) = \frac{1}{2\pi}\int^{\infty}_{-\infty} \mathcal{X} (\omega) e^{jwt}dw</math>
 +
 +
<math>x(t) = \frac{1}{2\pi} \frac{\pi}{j}\int^{\infty}_{-\infty} \delta(w-2\pi)e^{jwt} dw + \frac{1}{2\pi}\frac{2\pi}{j} \int_{-\infty}^{\infty} \delta(w+2\pi)e^{jwt} dw]</math>
 +
 +
<math>x(t) = \frac{1}{2j}e^{j2\pi t} + \frac{1}{j}e^{-j2\pi t}

Revision as of 16:18, 8 October 2008

Compute the Inverse Fourier Transform of:

$ \,\mathcal{X}(\omega)= \frac{\pi}{j} \delta (w - 2\pi) + \frac{2\pi}{j} \delta (w + 2\pi) $

$ x(t) = \frac{1}{2\pi}\int^{\infty}_{-\infty} \mathcal{X} (\omega) e^{jwt}dw $

$ x(t) = \frac{1}{2\pi} \frac{\pi}{j}\int^{\infty}_{-\infty} \delta(w-2\pi)e^{jwt} dw + \frac{1}{2\pi}\frac{2\pi}{j} \int_{-\infty}^{\infty} \delta(w+2\pi)e^{jwt} dw] $

$ x(t) = \frac{1}{2j}e^{j2\pi t} + \frac{1}{j}e^{-j2\pi t} $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett