Line 12: Line 12:
 
     <math>=.5(\infty^2 - 0^2)</math>
 
     <math>=.5(\infty^2 - 0^2)</math>
 
<math>E_\infty = \infty</math>
 
<math>E_\infty = \infty</math>
 +
 +
<math>P_\infty = lim_{T \to \infty}</math>

Revision as of 10:46, 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} $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010