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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
| <math> \mathcal{U}(\omega)=\mathcal{X}(\omega)\mathcal{Y}(\omega) \ </math>
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<math> \mathcal{U}(\omega)=\mathcal{X}(\omega)\mathcal{Y}(\omega) \ </math>
| <math> U(f)=X(f)Y(f) \ </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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<div align="left" style="padding-left: 0em;">
 
<math>
 
<math>
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<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 12:13, 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) $


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