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<math> \mathcal{X}(\omega) = 4 \pi \delta(\omega - 3) + 4 \pi \delta(\omega + 3) - 8 \pi \delta(\omega - 7) </math> | <math> \mathcal{X}(\omega) = 4 \pi \delta(\omega - 3) + 4 \pi \delta(\omega + 3) - 8 \pi \delta(\omega - 7) </math> | ||
| + | |||
| + | By the integral formula: | ||
| + | |||
| + | <math> x(t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \mathcal{X}(\omega) e^{-j\omega t}\,d \omega</math> | ||
| + | |||
| + | Therefore: | ||
| + | |||
| + | <math> 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</math> | ||
Revision as of 16:11, 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 $
