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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Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva