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| − | + | =How to obtain the multiplication property in terms of f in hertz (from the formula in terms of <math>\omega</math>) = | |
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| − | + | Denoting | |
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| − | + | <math> \mathcal{Z}(\omega)=\mathcal{X}(\omega)*\mathcal{Y}(\omega) \ </math> | |
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| − | + | <math> Z(f)=X(f)*Y(f) \ </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> Z(f)=\mathcal{Z}(2\pi f)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(\theta)\mathcal{Y}(2\pi f-\theta)d\theta \ </math> | ||
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| + | <math>Let\ \varphi =\frac{\theta}{2\pi},\ then\ \theta=2\pi \varphi \ </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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</div> | </div> | ||
<math>Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) </math> | <math>Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) </math> | ||
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| + | ---- | ||
| + | [[ECE438_HW1_Solution|Back to Table]] | ||
Latest revision as of 12:10, 15 September 2010
How to obtain the multiplication property in terms of f in hertz (from the formula in terms of $ \omega $)
Denoting
$ \mathcal{Z}(\omega)=\mathcal{X}(\omega)*\mathcal{Y}(\omega) \ $
$ Z(f)=X(f)*Y(f) \ $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ Z(f)=\mathcal{Z}(2\pi f)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(\theta)\mathcal{Y}(2\pi f-\theta)d\theta \ $
$ Let\ \varphi =\frac{\theta}{2\pi},\ then\ \theta=2\pi \varphi \ $
$ \begin{align} Z(f) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{X}(2\pi \varphi)\mathcal{Y}(2\pi f-2\pi \varphi)d2\pi \varphi \\ &= \int_{-\infty}^{\infty} \mathcal{X}(2\pi \varphi)\mathcal{Y}(2\pi (f-\varphi))d\varphi \\ &= \int_{-\infty}^{\infty} X(\varphi)Y(f-\varphi)d\varphi \end{align} $
$ Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) $
