# graph of lowpass filter

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 • Please, use proper english – MaximGi Mar 9 '16 at 14:25

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 $(0,0)$.