Velocity Reviews > FF sensors: is 80MP needed?

# FF sensors: is 80MP needed?

Ilya Zakharevich
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Posts: n/a

 11-07-2006
[A complimentary Cc of this posting was sent to
acl
<(E-Mail Removed)>], who wrote in article <(E-Mail Removed) .com>:
> > > But this is not surprising. No matter how smoothly you cut off the
> > > higher frequencies, you'll have ringing.

> > Nope (assuming that "ringing" means Gibbs). E.g., in one-dimensional
> > case, the Cesaro-cutoff function (max(0,1-|f|)), has no Gibbs
> > phenomenon.

> What is f in this expression? Could you be a little more precise?

f is for "frequency". Filtering is done by multiplying by the above
MTF. Since this MTF is, in fact, a convolution of two brick-wall
functions, its FT is square of sinc(), thus non-negative.

Positivity guaranties absense of overshoot; actually, it guaranties
much more: image of a step function is monotonic, so there is no
(1-dim analogue of) ringing at all: neither undershooting ringing, nor
overshooting ones. Note also that positivity is not necessary for
absense of overshooting: what is necessary [for even functions] is
INT[A to INFTY] FT(x) dx >= 0 for A>0; and this is very delicate property...

Yours,
Ilya

Ilya Zakharevich
Guest
Posts: n/a

 11-07-2006
[A complimentary Cc of this posting was sent to
acl
<(E-Mail Removed)>], who wrote in article <(E-Mail Removed)>:
> OK. If by impulse you mean the filter we need to apply to the fourier
> transform of the signal (*), then I agree: if it's positive everywhere,
> no ringing.

not one.[*]

> But the point is that this filter must be of infinite extent
> (or extend? Hmmm...) in frequency space to avoid overshooting: it cannot
> vanish above some frequency.

This was already shown to be false: Cesaro filtering gives no
overshooting.

> I mean, for example: Take a delta-fn in
> position space, FT it; you get a constant. Then, multiply this constant
> by a "filter" exp(-k^2/D^2), and FT back; you'll get something
> proportional to exp(-D^2 x^2), ie, no ringing, like you said.

So far, correct.

> But the moment you truncate the filter (ie make it vanish above some
> frequency), you'll get ringing (in real space), no matter how high
> the maximum frequency included. Try it.

I do not think this is true - but cannot be ABSOLUTELY sure right
now. Anyway, we are not discussing "Gauss multiplied by a cut-off
step function", but a more general case of a function with finite support.

> All of the above applies also to edges (as opposed to points)

[*] Here comes the crucial distinction when one consider
"point-sources" of light (e.g., subpixel speckles), and edges
(e.g., disk-like sources of light). Positive PSF will not get
ringing near edges, or any overshoots. However, if the PSF has
"ringed" structure (consider one of diffraction), there will be

In pictorial photography, point-like sources are so rare that
the ringing about them does not matter (much).

Yours,
Ilya

Ilya Zakharevich
Guest
Posts: n/a

 11-09-2006
[A complimentary Cc of this posting was sent to
acl
<(E-Mail Removed)>], who wrote in article <(E-Mail Removed) .com>:
> > I expect it would be very education to listen to the opinion of

^^al

> > somebody who "understands probability" (I presume you meant yourselves
> > there) on what is deviation(A+B) when B = -A... Please go ahead.

>
> No, I misunderstood. I did not realise you meant that there is

....

> Who are "ourselves", though? I'm just one.

Consider them just a figure of speech.

Ilya