(New page: Solve for <math>P\infty</math> and <math>E\infty</math> of <math>x(t)</math> <math>x(t)=\sqrt{t}</math> ---- <math> P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-...) |
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− | Note that <math> P\infty </math> is allowed to be equal to <math>E | + | Note that <math> P\infty </math> is allowed to be equal to <math> E\infty </math> which is equal to <math> \infty </math>. |
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 $.