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| − | + | =How to obtain the convolution 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{U}(\omega)=\mathcal{X}(\omega)\mathcal{Y}(\omega) \ </math> | |
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| − | + | <math> U(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> | <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 11:13, 15 September 2010
How to obtain the convolution property in terms of f in hertz (from the formula in terms of $ \omega $)
Denoting
$ \mathcal{U}(\omega)=\mathcal{X}(\omega)\mathcal{Y}(\omega) \ $
$ U(f)=X(f)Y(f) \ $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} U(f) &= \mathcal{U}(2\pi f) \\ &=\mathcal{X}(2\pi f)\mathcal{Y}(2\pi f) \\ &= X(f)Y(f) \end{align} $
$ Since\ X(\alpha)=\mathcal{X}(2\pi \alpha),Y(\alpha)=\mathcal{Y}(2\pi \alpha) $
