Revision as of 07:58, 22 June 2009 by Thompso7 (Talk | contribs)

$ x(t)=\sqrt(2t)u(t) $


Compute $ E\infty $

$ E\infty=\int_{-\infty}^\infty |x(t)|^2dt $

$ E\infty=\int_{-\infty}^\infty |\sqrt(2t)u(t)|^2dt $

$ E\infty=\int_{-\infty}^\infty |2tu(t)|dt $

$ E\infty=\int_{0}^\infty 2tdt $

$ E\infty=t^2|_{0}^{\infty} $

$ E\infty= \infty-0 $

$ E\infty=\infty $


Compute $ P\infty $

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

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

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

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T2tdt $

$ P\infty=\lim_{T \to \infty}\frac{|t^2|_{0}^{T}}{2*T} $

$ P\infty=\lim_{T \to \infty}\frac{T^2-0}{2*T} $

$ P\infty=\lim_{T \to \infty}\frac{T}{2} $

$ P\infty=\infty $


-Tylor Thompson

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang