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<math>= \frac{1}{2\pi}\int_{-\infty}^{\infty}(\frac{\frac{1}{2j}}{(2 - j4 + jw)^{2}} - \frac{\frac{1}{2j}}{(2 + j4 - jw)^{2}})e^{j\omega t}\,d\omega</math>
 
<math>= \frac{1}{2\pi}\int_{-\infty}^{\infty}(\frac{\frac{1}{2j}}{(2 - j4 + jw)^{2}} - \frac{\frac{1}{2j}}{(2 + j4 - jw)^{2}})e^{j\omega t}\,d\omega</math>
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by looking at the table on p. 329 of the book some observations can be made:
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the form <math>\frac{1}{(2 - j4 + jw)^{2}}</math> means that there is a term of the form <math>\ te^{-2t}</math>

Revision as of 12:59, 8 October 2008

$ \mathcal{X}(\omega) = \frac{\frac{1}{2j}}{(2 - j4 + jw)^{2}} - \frac{\frac{1}{2j}}{(2 + j4 - jw)^{2}} $

$ \ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega $

$ = \frac{1}{2\pi}\int_{-\infty}^{\infty}(\frac{\frac{1}{2j}}{(2 - j4 + jw)^{2}} - \frac{\frac{1}{2j}}{(2 + j4 - jw)^{2}})e^{j\omega t}\,d\omega $

by looking at the table on p. 329 of the book some observations can be made:

the form $ \frac{1}{(2 - j4 + jw)^{2}} $ means that there is a term of the form $ \ te^{-2t} $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010