| 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 10:23, 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 $
