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     <math>= lim_{T \to \infty} \ 1/(2T) \int_{-T}^T |\sqrt(t)|^2\,dt</math>
 
     <math>= lim_{T \to \infty} \ 1/(2T) \int_{-T}^T |\sqrt(t)|^2\,dt</math>
     <math>lim_{T \to \infty} \ 1/(2T) \int_{-T}^T .5*t^2|_{-T}^T</math>
+
     <math>= lim_{T \to \infty} \ 1/(2T) \int_{-T}^T .5*t^2|_{-T}^T</math>

Revision as of 11:01, 21 June 2009

$ x(t) = \sqrt(t) $

$ x(t) = \cos(t) + \jmath\sin(t) $


$ E_\infty = \int_{-\infty}^\infty |x(t)|^2\,dt $

    $ =\int_{-\infty}^\infty |\sqrt(t)|^2\,dt $
    $ =\int_0^\infty t\,dt $
    $ =.5*t^2|_0^\infty $
    $ =.5(\infty^2 - 0^2) $

$ E_\infty = \infty $


$ P_\infty = lim_{T \to \infty} \ 1/(2T) \int_{-T}^T |x(t)|^2\,dt $

    $ = lim_{T \to \infty} \ 1/(2T) \int_{-T}^T |\sqrt(t)|^2\,dt $
    $ = lim_{T \to \infty} \ 1/(2T) \int_{-T}^T .5*t^2|_{-T}^T $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010