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Check the equation 4.8-1 $H(u,v)=1$ if $D(u,v)\le D_0$

my question is the high frequency is at center.

$D(u,v)$ is smaller if it is close to center.

why $H(u,v)=1$ in that case. Thanks

lowpass

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  • $\begingroup$ Please, use proper english $\endgroup$ – MaximGi Mar 9 '16 at 14:25
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I suspect a confusion between the ordinal domain (where the frequencies indexed by $(u,v)$ live), and the cardinal domain (the amplitude of the filter for each $(u,v)$).

An ideal low-pass filter in 2D has amplitude $1$ (white disk) in some frequency domain $\mathcal{D}$ around the zero frequency, and $0$ outside (black), as illustrated in the central figure. The theory somehow requests that the domain $\mathcal{D}$ possesses central symmetry around point $(0,0)$: if $(u,v)\in \mathcal{D}$, then $(-u,-v)\in \mathcal{D}$. One of the most natural domain has circular symmetry: it is defined as a disk centered at $(u_0,v_0) =(0,0)$, and radius $D_0$.

Hence, the amplitude is "high" ($1$), this is not a high frequency, in a domain of low frequencies around $(0,0)$.

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  • $\begingroup$ Perfect, axes of frequencies and amplitude require a little attention $\endgroup$ – Laurent Duval Mar 10 '16 at 15:51
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There is no high frequency in the center. In the center of your 2D spatial frequency distribution is the 0-frequency (constant DC component of the Fourier-transformation). The text excerpt is in itself consistent. Therefore I suppose you misunderstood the problem or the initial assumptions of its solution.

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