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<math>x(t) = \int_{-\infty}^{\infty}X(\omega )e^{j\omega t}d\omega\,</math>
 
<math>x(t) = \int_{-\infty}^{\infty}X(\omega )e^{j\omega t}d\omega\,</math>
  
<math> =  \int_{-\infty}^{\infty}\delta(\omega )e^{j\omega t}d\omega + \int_{-\infty}^{\infty}\delta(\omega - 5)e^{j\omega t}d\omega + \int_{-\infty}^{\infty}\delta(\omega + 5)e^{j\omega t}d\omega\,</math>
+
<math> =  \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega )e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega - 5)e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega + 5)e^{j\omega t}d\omega\,</math>

Revision as of 13:22, 7 October 2008

$ X(\omega ) = \delta(\omega ) + \delta(\omega - 5) + \delta(\omega - 5)\, $

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

$ = \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega )e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega - 5)e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega + 5)e^{j\omega t}d\omega\, $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn