(New page: <math>x(t)=\sqrt{t}</math> ---- <math> \int_{-\infty}^\infty t,dt </math> <math> E_\infty=lim_{T \to \infty} \int_{-T}^T |x(t)|^2\,dt</math> <math> E_\in...) |
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<math>x(t)=\sqrt{t}</math> | <math>x(t)=\sqrt{t}</math> | ||
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<math> E_\infty=lim_{T \to \infty} \int_{-T}^T |x(t)|^2\,dt</math> | <math> E_\infty=lim_{T \to \infty} \int_{-T}^T |x(t)|^2\,dt</math> | ||
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<math> E\infty= \int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt</math> | <math> E\infty= \int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt</math> | ||
<math> E\infty=(\frac{1}{2})*t^2|_{-\infty}^\infty</math> | <math> E\infty=(\frac{1}{2})*t^2|_{-\infty}^\infty</math> | ||
− | <math> E\infty=(\frac{1}{2})\infty^2-0^2)=\infty</math> | + | <math> E\infty=(\frac{1}{2})\infty^2-0^2)=\infty</math> in Joules |
<math> P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt</math> | <math> P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt</math> | ||
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<math> P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}t^2)|_0^T</math> | <math> P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}t^2)|_0^T</math> | ||
<math> P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}T^2)</math> | <math> P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}T^2)</math> | ||
− | <math> P\infty=lim_{T \to \infty} \ \frac{T}{4}=\infty</math> | + | <math> P\infty=lim_{T \to \infty} \ \frac{T}{4}=\infty</math> in Watts |
− | + | --[[User:Freya|Freya]] 16:44, 21 June 2009 (UTC) | |
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− | --[[User:Freya|Freya]] 16: | + |
Latest revision as of 12:44, 21 June 2009
$ x(t)=\sqrt{t} $
$ 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 $ in Joules
$ 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 $ in Watts
--Freya 16:44, 21 June 2009 (UTC)