Difference between revisions of "E infinity and P infinity - Evan Witkoske" - Rhea

Revision as of 04:29, 22 June 2009

[1]

$ f(t)=j*5*sin(t) $

$ E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt) $

$ E_\infty = \int_{-\infty}^\infty 25sin(t)^2u(t)\,dt) $

$ E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt) $

$ E_\infty =\frac{25t}{2} + \frac{25sin(t)}{4}\bigg]_{-\infty}^\infty) $

$ E_\infty =\infty-0 = \infty $

$ P_\infty calculation $

$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt $

$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt $

$ = lim_{T \to \infty} \frac{1}{2T}*(\frac{25t}{2} + \frac{25sin(t)}{4}|_{-T}^T) $

$ = lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{4}-\frac{25sin(-T)}{4} $

$ = lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{2}) $

$ P\infty= \frac{25}{2} + 0 $