Difference between revisions of "E infinty & P infinty - Ryne Rayburn (rrayburn)" - Rhea

Revision as of 07:25, 17 June 2009

Work By Ryne Rayburn (rrayburn)

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

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

   $ E\infty=\int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt $ (due to sqrt limiting to positive Real numbers)
   $ E\infty=.5*t^2| $(from $ 0 $ to $ \infty $)
   $ E\infty=.5(\infty^2-0^2)=\infty $

$ P\infty=lim\bullet T\rightarrow\infty*1/(2T)\int_{-T}^{T} |x(t)|^2\,dt $

   $ P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^T |\sqrt{t}|^2\,dt=\int_0^T t\,dt $
   $ P\infty=lim_{T \to \infty} \ 1/(2T)*.5t^2| $(from $ 0 $ to $ T $)
   $ P\infty=lim_{T \to \infty} \ 1/(2T)*(.5T^2) $
   $ P\infty=lim_{T \to \infty} \ (.25T)=\infty $

$ x(t)=\cos(t)+\jmath\sin(t) $

$ |x(t)|=|\cos(t)+\jmath\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=1 $

$ E\infty=\int|x(t)|^2dt $ (from $ -\infty $ to $ \infty $)

   $ E\infty=\int|1|^2dt=t| $(from $ -\infty $ to $ \infty $)
   $ E\infty=\infty $

$ P\infty=lim\bullet T\rightarrow\infty~~1/(2T)\int|x(t)|^2dt $ (from $ -T $ to $ T $)

   $ P\infty=lim\bullet T\rightarrow\infty~~1/(2T)\int|1|^2dt $
   $ P\infty=lim\bullet T\rightarrow\infty~~1/(2T)*t| $(from $ -T $ to $ T $)
   $ P\infty=lim\bullet T\rightarrow\infty~~1/(2T)*(T-(-T)) $
   $ P\infty=lim\bullet T\rightarrow\infty~~1 $
   $ P\infty=1 $

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