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<math>E_\infty = \int_{-\infty}^\infty |x_1(t)|^2\,dt</math>
 
<math>E_\infty = \int_{-\infty}^\infty |x_1(t)|^2\,dt</math>
  
     <math>= \int_{-\infty}^infty 1 \,dt</math>
+
     <math>= \int_{-\infty}^\infty 1 \,dt</math>
 +
    <math>= t|_{-\infty}^\infty</math>

Revision as of 11:23, 21 June 2009

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

$ x_1(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) .5*t^2|_0^T $
    $ = lim_{T \to \infty} \ 1/(2T) * .5(T^2 - 0^2) $
    $ = lim_{T \to \infty} \ 1/(2T) * .5T^2 $
    $ = lim_{T \to \infty} \ 1/(4T)*T^2 $
    $ = lim_{T \to \infty} T/4 $

$ P_\infty = \infty $


$ |x_1(t)| = \sqrt{\cos^2(t)+\sin^2(t)}=1 $



$ E_\infty = \int_{-\infty}^\infty |x_1(t)|^2\,dt $

    $ = \int_{-\infty}^\infty 1 \,dt $
    $ = t|_{-\infty}^\infty $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood