(New page: {| | align="left" style="padding-left: 0em;" | Parseval's property |} <math> \mathcal{X}(\omega)=\mathcal{X}(2\pi f) \ </math> <div align="left" style="padding-left: 0em;"> <math> \begin{...) |
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| − | + | =How to obtain the Parseval's property in terms of f in hertz (from the formula in terms of <math>\omega</math>) = | |
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| − | + | To obtain X(f), use the substitution | |
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| + | <math>\omega= 2 \pi f </math>. | ||
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| + | More specifically | ||
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<math> \mathcal{X}(\omega)=\mathcal{X}(2\pi f) \ </math> | <math> \mathcal{X}(\omega)=\mathcal{X}(2\pi f) \ </math> | ||
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<div align="left" style="padding-left: 0em;"> | <div align="left" style="padding-left: 0em;"> | ||
<math> | <math> | ||
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</math> | </math> | ||
</div> | </div> | ||
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<math>Since\ X(\alpha)=\mathcal{X}(2\pi \alpha) </math> | <math>Since\ X(\alpha)=\mathcal{X}(2\pi \alpha) </math> | ||
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| + | ---- | ||
| + | [[ECE438_HW1_Solution|Back to Table]] | ||
Latest revision as of 11:18, 15 September 2010
How to obtain the Parseval's property in terms of f in hertz (from the formula in terms of $ \omega $)
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \mathcal{X}(\omega)=\mathcal{X}(2\pi f) \ $
$ \begin{align} \int_{-\infty}^{\infty} |x(t)|^2 dt &= \frac{1}{2\pi} \int_{-\infty}^{\infty} |\mathcal{X}(2\pi f)|^2 d2\pi f \\ &= \int_{-\infty}^{\infty} |\mathcal{X}(2\pi f)|^2 df \\ &= \int_{-\infty}^{\infty} |X(f)|^2 df \end{align} $
$ Since\ X(\alpha)=\mathcal{X}(2\pi \alpha) $
