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I am unclear on what is being "figure out". But, the signal processing
theory for the type
of sampling that CCD elements do is well established in old textbooks on
communication theory. The old one on my
shelf is by Taub and Schilling entitled Principles of Communication Systems,
McGraw-Hill 1971 (but I took the course in
1978 - !!! ;o)). Page 167 gives a nice simple diagram. This type of
sampling is called "natural sampling" according to the
communication theorists of that time. At least according to ideal theory,
the signal can be recovered by theoretical techniques
in the book so long as Nyquist is satisfied and there's no noise (which of
course is not the case). But, at least we know that, in
the ideal case, the information is preserved. I don't know if that's what
you are trying to figure, or maybe I'm just making it
more confusing ;o)
-----Original Message-----
From: Confocal Microscopy List [mailto:[log in to unmask]] On
Behalf Of Guy Cox
Sent: Monday, April 16, 2012 8:16 AM
To: [log in to unmask]
Subject: Re: A pixel is not a little square
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Christophe,
As I said in my original message, if you want to produce an
image of your sensor, with the squares coloured in, that is indeed the case
- no argument. But that is not an image of your sample, it's an image of
your detector. I was (and am) writing about the digital image in general,
as I thought I made clear, regardless of the detection technology. Your CCD
image does NOT tell you that intensity X exists across the whole of one
detector element, and then suddenly intensity Y exists across the whole of
the next detector element. As you said yourself "There is no information
about where did the photons arrive on the square detector portion
corresponding to a pixel". So we are still dealing with samples and have to
handle the problem of relating these samples to the real world.
Guy
-----Original Message-----
From: Confocal Microscopy List [mailto:[log in to unmask]] On
Behalf Of Christophe Leterrier
Sent: Monday, 16 April 2012 10:52 PM
To: [log in to unmask]
Subject: Re: A pixel is not a little square
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Guy,
But in the CCD case the sampling is integrated over the whole pixel area of
the sensor (assuming pixel coverage of 100% which is the case for EMCCDs for
exemple). There is no information about where did the photons arrive on the
square detector portion corresponding to a pixel. The value X you get as
intensity for this pixel is the integrated density over the whole detector
portion, so why would you take that value as a point-measurement
corresponding to the center of this area ? In that CCD case, isn't the mage
indeed an histogram with a bin that has (quite literally) a width equal to
the pixel size, not a sample at mathematical points corresponding to the
centers of the pixels ?
Christophe
On Mon, Apr 16, 2012 at 14:36, Guy Cox <[log in to unmask]> wrote:
> *****
> To join, leave or search the confocal microscopy listserv, go to:
> http://lists.umn.edu/cgi-bin/wa?A0=confocalmicroscopy
> *****
>
> Mark,
>
> You are continuing to confuse the samples with the
> representation of those samples.
>
> Let's imagine we have a series of data points: 100 90 80 70 60
> 50 40
>
> We are mapping these on an image where each is separated by a defined
> distance. So we need to fill in this distance.
>
> You are saying that the 'correct' representation is:
> 100 100 100 100 100 90 90 90 90 90 80 80 80 80 80
> 70 70 70 70 60 60 60 60 60 50 50 50 50 50
40
>
> I am saying that this is a wildly implausible and totally unjustified
> interpretation, and the best representation we can derive from the data
is:
> 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72
> 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42
> 40
>
> EITHER way we are interpolating the sampled data - we have no option - so
> let's just get over this. Your proposed representation includes detail
> that we could not possibly detect, mine does not. Remember, these are
> SAMPLES. Neither representation changes our recorded data. End of
> story, IMHO.
>
> How did we get into this mess? Why does everyone then 'do it wrong'?
> Well, actually, everyone doesn't. Scanning probe microscopes always
> remap
> - because by the time they appeared the computing power to do it was
> available. When confocal microscopes first became widely available,
> in 1987, the data they produced completely overwhelmed available computing
> power (believe me, I was there and writing software). So we got used to
> the 'quick and dirty' approach. Consumer digital cameras do a sort of
> remap because the Bayer mosaic requires it, but modern sensors so far
> exceed the resolution of the camera optics that we never get to see
> any spurious frequencies anyway. Computer games consoles always
> remap. So do X-ray and EM tomography systems.
>
> Guy (arrogant bastard)
>
>
>
> ----Original Message-----
> From: Confocal Microscopy List
> [mailto:[log in to unmask]]
> On Behalf Of Mark Cannell
> Sent: Monday, 16 April 2012 7:37 PM
> To: [log in to unmask]
> Subject: Re: A pixel is not a little square
>
> *****
> To join, leave or search the confocal microscopy listserv, go to:
> http://lists.umn.edu/cgi-bin/wa?A0=confocalmicroscopy
> *****
>
> Sorry Guy, I still think you don't see the point I'm trying to make.
> The camera actually says "The mean signal from x to x+dx is ..."
> (where dx is the sensor pixel size). It does NOT say the signal at x
> is 'K' and that is where I think the confusion lies. The camera output
> is a 2D 'histogram' and showing little boxes with the same intensity
> is (I say again) a perfectly accurate representation of the data (
> i.e. F(x) for x -> x + dx = K). With respect, it is not, as you say
> inaccurate -even if it is unaesthetic. If you fit a sinusoid you have
> just carried out a fitting exercise... That is not a "more accurate"
> presentation of the data despite what your Smith says (even if it may
> be a more accurate representation of the object which has been
> discretized). One should not loose sight of the fact that you have made
some (possibly large) assumptions in the fitting process.
>
> Put mathematically, if you smooth out the displayed pixel edges you
> extend the actual sampling frequency (note how you are putting new
> unrecorded samples between recorded data values -which is what
> drawing a line between points actually does) -you are adding
> information to the data that was NOT present in the RAW data. It may
> be that your additional information is correct and adds value (e.g.
> the band limit of the microscope is...) but one should not loose sight
> of distinction between the addition of data/information by the
> experimenter (which may or may not be wrong) and that reported by the
> instrument (the closest to truth the experimenter can get).
>
> At the risk of boring some readers on this list, let me emphasize my
> point
> : The camera actually says "The mean signal from x to x+dx is ..."
> (where dx is the sensor pixel size). It does NOT say the signal at x
> is 'K' . This can be portrayed as a square with constant color and I
> can think of no other truer portrayal of the measured data. Hopefully
> dx is less than the resolution of the viewer at final display
> resolution but if it is not, then the only choice (IMHO) is between
> aesthetics (or some other goal) and truthfully displaying the recorded
data -there is no middle ground.
>
> Cheers Mark
>
> PS My CD player can't output square waves because the detector etc.
> has a rather finite bandwidth... Even if it could, my ears are too
> many dB down at 44 kHz to sample it correctly and hear the artifacts
> introduced by digital sampling ... :-)
>
> On 16/04/2012, at 9:18 AM, Guy Cox wrote:
>
> > *****
> > To join, leave or search the confocal microscopy listserv, go to:
> > http://lists.umn.edu/cgi-bin/wa?A0=confocalmicroscopy
> > *****
> >
> > " There are _no_ 'higher harmonics' present in the data, only in
> > ones
> 'artistic' interpretation for display purposes." That is exactly what
> I said!
> >
> > I also never said that the data is a continuous function, I said it
> > is a
> series of discrete samples of a continuous function. So when you
> choose to display it you have to do something. Drawing little boxes
> is NOT 'doing nothing' and neither is it 'displaying the raw data'.
> On the contrary - it is corrupting the data with frequencies which
> shouldn't be there AND confusing the human eye (for which, presumably, we
are doing the drawing).
> The raw numbers are useful - indeed essential - for the computer but
> fundamentally cannot just 'be displayed' to the human eye as an image.
> Our sampling rationale is based on sine-wave frequencies and
> therefore, as Alvy Ray Smith said, sinusoidal mapping is the truest
> (not the most aesthetic, though this is also true) way of displaying
> the data. It doesn't add any spurious higher harmonics, it presents
> the data as accurately as our sampling permits. Drawing little boxes
> may be easier, but it is just as much mapping the measured samples to
> a displayed image - the difference is that this method is both inaccurate
and un-aesthetic.
> >
> > If your CD player spat out square waves to the speakers, you'd take
> > it
> back to the shop pretty promptly!
> >
> >
> Guy
> >
> > -----Original Message-----
> > From: Confocal Microscopy List
> > [mailto:[log in to unmask]]
> On Behalf Of Mark Cannell
> > Sent: Monday, 16 April 2012 5:27 PM
> > To: [log in to unmask]
> > Subject: Re: A pixel is not a little square
> >
> > *****
> > To join, leave or search the confocal microscopy listserv, go to:
> > http://lists.umn.edu/cgi-bin/wa?A0=confocalmicroscopy
> > *****
> >
> > I think I see the problem, the spurious frequencies arise from your
> thinking the _data_ is a continuous function and treating it as such
> (by "drawing a line ..."), but it is not, it is discrete and can be
> faithfully represented by a _discrete_ Fourier transform (which folds
> at Fs/2). The hiighest frequency in the DFT is Fs, but we know we
> shouldn't look at that right? There are _no_ 'higher harmonics'
> present in the data, only in ones 'artistic' interpretation for display
purposes.
> >
> > If it looks jagged, that is because in reality sampled data really is!
> The problem really arises because you do not know how to fill in the
> space between data samples. You can interpolate (or not). If you
> interpolate you are making a statement about the model underlying the
> data and have just carried out a fitting exercise. Fitting is NOT raw
> data presentation. If you just plot data values you make no assumption
> about what should join the data, no model has been fit to the data.
> Every scientist should know the difference between a histogram and a
> continuous distribution and not be fooled by the vertical lines at the
> histogram boundaries (which is what you show in a pixel image).
> >
> > The choice is yours, in one case you faithfully show unadulterated
> sampled data (the histogram looks less 'pretty' than a curve) or you
> fit a model and interpolate. The trouble with the latter is that the
> model is probably wrong and you hide the defects in the data (e.g.
> camera pixel
> size) from the keen eyed reviewer... Of course if the data points are
> really close together, the myopic reviewer can't see defects in you
> data
> :-) ! From Guy's reasoning, it would be impossible to represent any
> digitally sampled data because you are always pixelating a continuous
> function (all pictures get mad up of little squares -the printer dumps
> blobs of ink etc). So, where does the pixelation become acceptable?
> This is now aesthetic and has nothing to do with science or
> mathematics (those with perfect vision will always see discretization
'artifacts' more easily) .
> >
> > Cheers Mark
> >
> > On 16/04/2012, at 3:31 AM, Guy Cox wrote:
> >
> >> *****
> >> To join, leave or search the confocal microscopy listserv, go to:
> >> http://lists.umn.edu/cgi-bin/wa?A0=confocalmicroscopy
> >> *****
> >>
> >> OK, having slept on it, I now feel that just maybe I can explain
> >> what
> this is all about. If only the list would let us include pictures it
> would be much easier!
> >>
> >> Let's assume we have a digital image, from any source, consisting
> >> of
> pixels with a spacing s. The smallest spacing we can resolve in this
> image is 2s, and this will correspond, in frequency space, with a
> frequency f. f represents the bandpass limit of this system, no
> higher frequencies can be passed. Now imagine we have a row of pixels
> containing the following
> values:
> >>
> >> 255 0 255 0 255 0 255 0 255
> >>
> >> If we represent these pixels by little squares, we'll have
> >> something
> like a chessboard. Taking a line along this chessboard will give us a
> square wave. Now this square wave cannot be represented within the
> bandpass limit of the system, defined by the frequency f. To
> represent a square wave we need an infinite series of sine waves f + 3f +
5f +7f .....
> To get even a crude approximation to a square wave we need f + 3f -
> that is a frequency three times higher than the image can contain.
> >>
> >> In other words, we've introduced a whole series of spurious
> >> frequencies
> into our image that not only were not there to start with, they could not
> possibly have been there. Does this matter? After all, we know they
> can't be real. It does matter, because we are talking about a visual
> representation of our data - that's why we drew the little boxes in
> the first place. Our eyes are very sensitive to edges* and the edges
> will take over if we let these frequencies come within the bandwidth of
our eyes.
> We will find it very hard to actually see the finest detail in our
> picture (defined by 2s, remember) because if we enlarge it enough to
> see this easily we'll also get the edges created by these spurious
> frequencies. In everyday terms, the pixellation takes over from the
picture.
> >>
> >> Note that in all this discussion I have not mentioned microscopes,
> cameras or anything - we are just talking about a digital image from
> any source. It applies to confocal, widefield, and electron
> microscopes, telescopes, X-ray images and your holiday snaps. Coming
> back to the microscopic world, if we oversample to the point where r,
> our minimum resolved distance, is substantially greater than 2s, we
> may not need to enlarge to the point where we see the spurious
> frequencies. This is probably why some contributors to this
> discussion have advocated considerable levels of oversampling (though
> they probably didn't realise this, they just knew they got good
> pictures that way). But oversampling in fluorescence can be very hard on
our specimens.
> >>
> >> "But I'm using a CCD detector so my image is made up of little
> squares". Yes, you can produce a 'coloured in' picture of your
> detector that way. I'm assuming the image is actually what you want
> to see, though, not the detector.
> >>
> >> *Amusingly, the human eye does the same thing to emphasize edges as
> computer image processing does - it makes the dark side of the edge
> darker than it is and the light side lighter.
> >>
> >>
> Guy
> >>
> >> PS. This has doubtless confirmed my reputation among some people
> >> as an
> arrogant bastard. They are probably right, but at least I'm an
> arrogant bastard who tries to help. It's taken me two hours to write
this.
>
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