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<math> X(w) = \int_{-\infty}^{1}e^{(t-1)}e^{-jwt}dt+\int_{1}^{\infty}e^{-(t-1)}e^{-jwt}dt</math><br><br>
 
<math> X(w) = \int_{-\infty}^{1}e^{(t-1)}e^{-jwt}dt+\int_{1}^{\infty}e^{-(t-1)}e^{-jwt}dt</math><br><br>
 
<math> X(w) = \int_{-\infty}^{1}e^{-1}e^{(1-jw)t}dt+\int_{1}^{\infty}e^{1}e^{(1+jw)t}dt</math><br><br>
 
<math> X(w) = \int_{-\infty}^{1}e^{-1}e^{(1-jw)t}dt+\int_{1}^{\infty}e^{1}e^{(1+jw)t}dt</math><br><br>
<math> X(w) = {\left.\frac{e^{-1}e^{(1-jw)t}}{1-jw}\right]_{\infty}^{0}</math>
+
<math> X(w) = {\left.\frac{e^{-1}e^{(1-jw)t}}{1-jw}\right]_{\infty}^{0}}</math>

Revision as of 17:49, 7 October 2008

$ x(t) = e^{-|t-1|} \, $

$ X(w) = \int_{-\infty}^{\infty}e^{-|t-1|}e^{-jwt}dt $

$ X(w) = \int_{-\infty}^{1}e^{(t-1)}e^{-jwt}dt+\int_{1}^{\infty}e^{-(t-1)}e^{-jwt}dt $

$ X(w) = \int_{-\infty}^{1}e^{-1}e^{(1-jw)t}dt+\int_{1}^{\infty}e^{1}e^{(1+jw)t}dt $

$ X(w) = {\left.\frac{e^{-1}e^{(1-jw)t}}{1-jw}\right]_{\infty}^{0}} $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010