Line 44: Line 44:
 
     <math>= lim_{T \to \infty} \ 1/(2T) * (T- (-T))</math>
 
     <math>= lim_{T \to \infty} \ 1/(2T) * (T- (-T))</math>
 
     <math>= lim_{T \to \infty} \ 1/(2T) * (2T)</math>
 
     <math>= lim_{T \to \infty} \ 1/(2T) * (2T)</math>
     <math>= lim_{T \to \infty} \ 1</math
+
     <math>= lim_{T \to \infty} \ 1</math>
 
<math>P_\infty = 1</math>
 
<math>P_\infty = 1</math>

Revision as of 11:38, 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|^2 \,dt $
    $ = t|_{-\infty}^\infty $

$ E_\infty = \infty $


$ P_\infty = lim_{T \to \infty} \ 1/(2T) \int_{-T}^T |x_1(t)|^2\,dt $

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

$ P_\infty = 1 $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva