(New page: Suppose we have a signal: :<math>e^{-2(t-1)}u(t-1)\,</math> The formula of Fourier Transform is: :<math>X(w) = \int_{-\infty}^{ \infty} x(t)e^{-jwt}dt\,</math> Substituting: :<math>X(...) |
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Integrating yields: | Integrating yields: | ||
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| + | :<math>X(w) = {\left. -\frac{e^{2-(2+jw)t}}{2+jw} \right]_{1}^{\infty}}\,</math> | ||
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| + | :<math>X(w) = 0 - -\frac{e^{2-(2+jw)}}{2+jw} \,</math> | ||
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| + | :<math>X(w) = \frac{e^{2-2-jw}}{2+jw} \,</math> | ||
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| + | :<math>X(w) = \frac{e^{-jw}}{2+jw} \,</math> | ||
Revision as of 15:57, 7 October 2008
Suppose we have a signal:
- $ e^{-2(t-1)}u(t-1)\, $
The formula of Fourier Transform is:
- $ X(w) = \int_{-\infty}^{ \infty} x(t)e^{-jwt}dt\, $
Substituting:
- $ X(w) = \int_{-\infty}^{ \infty} e^{-2(t-1)}u(t-1)e^{-jwt}dt\, $
From the step function, the range becomes 1 to $ \infty $, so the equation becomes:
- $ X(w) = \int_{1}^{ \infty} e^{-2(t-1)}e^{-jwt}dt\, $
- $ X(w) = \int_{1}^{ \infty} e^{2-(2+jw)t}dt\, $
Integrating yields:
- $ X(w) = {\left. -\frac{e^{2-(2+jw)t}}{2+jw} \right]_{1}^{\infty}}\, $
- $ X(w) = 0 - -\frac{e^{2-(2+jw)}}{2+jw} \, $
- $ X(w) = \frac{e^{2-2-jw}}{2+jw} \, $
- $ X(w) = \frac{e^{-jw}}{2+jw} \, $
