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| <math> \mathcal{U}(\omega)=\mathcal{X}(\omega)\mathcal{Y}(\omega) \ </math>

Revision as of 19:32, 9 September 2010

convolution property
$ \mathcal{U}(\omega)=\mathcal{X}(\omega)\mathcal{Y}(\omega) \ $
$ U(f)=X(f)Y(f) \ $

$ \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) $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010