Line 15: Line 15:
 
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j)]e^{j\omega t}d\omega</math>
 
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j)]e^{j\omega t}d\omega</math>
  
<math>=\frac{2-3j}{2}\int_{-\infty}^{\infty}delta(\omega - 4\pi)e^{j\omega t}d\omega + \frac{2+3j}{2}\int_{-\infty}^{\infty}delta(\omega + 4\pi)e^{j\omega t}d\omega </math>
+
<math>=\frac{2-3j}{2}\int_{-\infty}^{\infty}\delta(\omega - 4\pi)e^{j\omega t}d\omega + \frac{2+3j}{2}\int_{-\infty}^{\infty}\delta(\omega + 4\pi)e^{j\omega t}d\omega </math>

Revision as of 09:23, 3 October 2008

Inverse Fourier Transform

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



$ X(\omega) = \pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j) $



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

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

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett