Difference between revisions of "Calculating E infinity and P infinity - Alex Gaul (argaul)" - Rhea

Latest revision as of 03:52, 22 June 2009

Solve for $ P\infty $ and $ E\infty $ of $ x(t) $

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

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

   $  P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T |\sqrt{t}|^2\,dt=\int_0^T t\,dt $
   $  P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}t^2)|_0^T $
   $  P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}T^2) $
   $  P\infty=lim_{T \to \infty} \ \frac{T}{4}=\infty $ 

$ E_\infty=lim_{T \to \infty} \int_{-T}^T |x(t)|^2\,dt $

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

   $  E\infty= \int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt $ 
   $  E\infty=(\frac{1}{2})*t^2|_{-\infty}^\infty $
   $  E\infty=(\frac{1}{2})\infty^2-0^2)=\infty $ 

Note that $ P\infty $ is allowed to be equal to $ E\infty $ which is equal to $ \infty $.