Line 20: Line 20:
 
     <math>= lim_{T \to \infty} \ 1/(2T) * .5(T^2 - 0^2)</math>
 
     <math>= lim_{T \to \infty} \ 1/(2T) * .5(T^2 - 0^2)</math>
 
     <math>= lim_{T \to \infty} \ 1/(2T) * .5T^2</math>
 
     <math>= lim_{T \to \infty} \ 1/(2T) * .5T^2</math>
 +
    <math>= lim_{T \to \infty} \ 1/(4T)*T^2</math>

Revision as of 11:09, 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) .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 $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett