(→Inverse Fourier Transform) |
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:<math>x(t) = \frac{1}{2\pi} \int_{-\infty}^{ \infty} X(jw)e^{jwt}dw \,</math> | :<math>x(t) = \frac{1}{2\pi} \int_{-\infty}^{ \infty} X(jw)e^{jwt}dw \,</math> | ||
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| + | Suppose we have <math>2 \pi \delta(w - 2\pi)</math> (From the 'not so easy' question in class) | ||
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| + | Substituting that into the formula: | ||
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| + | :<math>x(t) = \frac{1}{2\pi} \int_{-\infty}^{ \infty} 2 \pi \delta(w - 2\pi) e^{jwt}dw \,</math> | ||
| + | |||
| + | :<math>x(t) = \int_{-\infty}^{ \infty} \delta(w - 2\pi) e^{jwt}dw \,</math> | ||
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| + | :<math>e^{j2 \pi t}\,</math> | ||
Revision as of 17:22, 7 October 2008
The formula of the inverse transform is:
- $ x(t) = \frac{1}{2\pi} \int_{-\infty}^{ \infty} X(jw)e^{jwt}dw \, $
Suppose we have $ 2 \pi \delta(w - 2\pi) $ (From the 'not so easy' question in class)
Substituting that into the formula:
- $ x(t) = \frac{1}{2\pi} \int_{-\infty}^{ \infty} 2 \pi \delta(w - 2\pi) e^{jwt}dw \, $
- $ x(t) = \int_{-\infty}^{ \infty} \delta(w - 2\pi) e^{jwt}dw \, $
- $ e^{j2 \pi t}\, $
