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| + | Emily Thompson erthomps@purdue.edu | ||
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<math>x(t)=\sqrt(5t)</math> | <math>x(t)=\sqrt(5t)</math> | ||
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<math>E\infty=\int_{-\infty}^\infty |x(t)|^2dt</math> | <math>E\infty=\int_{-\infty}^\infty |x(t)|^2dt</math> | ||
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<math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |x(t)|^2dt</math> | <math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |x(t)|^2dt</math> | ||
Revision as of 18:58, 21 June 2009
Emily Thompson erthomps@purdue.edu
$ x(t)=\sqrt(5t) $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2dt $
$ E\infty=\int_{-\infty}^\infty |\sqrt(5t)|^2dt $
$ E\infty=\int_{-\infty}^\infty |5t|dt $
$ E\infty=\int_{0}^\infty 5tdt $
$ E\infty=\frac{5}{2}(t^2|_{0}^{\infty}) $
$ E\infty= \infty-0 $
$ E\infty=\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(5t)|^2dt $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|5t|dt $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T5tdt $
$ P\infty=\lim_{T \to \infty}\frac{|\frac{5}{2}t^2|_{0}^{T}}{2*T} $
$ P\infty=\lim_{T \to \infty}\frac{5}{4}\frac{T^2-0}{T} $
$ P\infty=\lim_{T \to \infty}\frac{5}{4}T $
$ P\infty=\infty $
