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In article <jecome01.715798061@starbase.spd.louisville.edu>, jecome01@rigel.spd.louisville.edu (Eric Comer) writes:
    	if the lens caps is place on a ccd camera and an image is capture,
    	why aren't all of the pixel values 0 (black)?

Because of the "dark current": charge accumulates on a CCD due
to BOTH light falling on it (zero in your case), and due to thermal
effects. The dark current builds up over time, so you normally
want to keep exposure times short (if you can). If you cannot
(like in astronomy) you can instead reduce the dark current by
cooling the CCD array, charge build-up falls off exponentially
(roughly) with temperature.

Anyway, dark current is quite predictable at constant temperature,
so it can be predicted and subtracted out.

	when the image is
    	normalized it seems like there are faint light diagonal lines
    	throughtout the whole image.  the image acquired was a gray scale
    	image the min value was 0 and the max value was 6.  any ideas?
    
The CCD array is not perfect - meaning that the individual cells
have different response characteristic and different dark current.

When you have subtracted the dark current you also should
calibrate the image, accounting for each pixel's individual
response characteristic. If you use the CCD array as an ordinary
camera the calibration may not be necessary of course.

torbjorn.

In article <jecome01.715798061@starbase.spd.louisville.edu> jecome01@rigel.spd.louisville.edu (Eric Comer) writes:
>	if the lens caps is place on a ccd camera and an image is capture,
>	why aren't all of the pixel values 0 (black)?  when the image is
>	normalized it seems like there are faint light diagonal lines
>	throughtout the whole image.  the image acquired was a gray scale
>	image the min value was 0 and the max value was 6.  any ideas?

Noise, mostly.  There may also be imperfections in the CCD -- a
perfect chip, in fact, is fairly rare (and expensive), particularly at
higher resolutions.  Finally, there is a charge accumulation (dark
current) that occurs even when the shutter is closed or the lense cap
is on.

Excerpting from the Photometrics Series 200 manual, "[all] electronic
circuitry generates noise.  Photon shot noise, dark current, and read
noise are primary CCD camera noise sources.

"Photon shot noise is a funamental property of the quantum nature of
light...

"Dark current, or thermally generated charge, cannot be distinguished
from photon induced charge.  Dark current can be measured and
subtracted, but its noise component cannot be ignored...

"The third primary source of noise, read noise, sets the detection
limits of a CCD camera.  Read noise is generated by the on chip output
ampilfier...

"Three types of integration measure the charge that has integrated on
the CCD:  exposure, bias, and dark integrations. ..."

There's a whole lot more to this, obviously, but that should at least
give you some ideas on what's going on.

The bottom line to all of this is that you have to be certain to
remove the effects of the bias, dark current, and, possibly, light
source non-linearities and artifacts when you grab an image.

							spl

-- 
Steve Lamont, SciViGuy -- (619) 534-7968 -- spl@dim.ucsd.edu
UCSD Microscopy and Imaging Resource/UCSD Med School/La Jolla, CA 92093-0608
"There was a time when tyrants feared/The new world of the free.
Now freedom is afraid and shrieks/At tyranny." - "Brave New World," A. Macleish

In <jecome01.715798061@starbase.spd.louisville.edu> jecome01@rigel.spd.louisville.edu (Eric Comer) writes:
>	if the lens caps is place on a ccd camera and an image is capture,
>	why aren't all of the pixel values 0 (black)?  when the image is
>	normalized it seems like there are faint light diagonal lines
>	throughtout the whole image.  the image acquired was a gray scale
>	image the min value was 0 and the max value was 6.  any ideas?

Well, two things. The CCD doesn't need to give you a value of 0 without
light. This depends on the offset and gain of the video amplifier. Second,
your system obviously catches some periodic stray signal. If you know
your pixel rate then you can translate the number of pixels into the
offending frequency in the time domain. Most likely this will be a 60Hz
noise signal.

Regards,
-- 
Michael van Elst
UUCP:     universe!local-cluster!milky-way!sol!earth!uunet!unido!mpirbn!p554mve
Internet: p554mve@mpirbn.mpifr-bonn.mpg.de
                                "A potential Snark may lurk in every tree."

Would anyone like to comment on the problem of calibrating the
radiometry of individual pixels in a CCD array?  

Essentially the problem is that for applications which require accurate 
radiometry (ie accurate grey levels are needed) the response of each pixel 
becomes important.

This is the classic issue of "flat fielding".  That is, if you supply a
perfectly uniform input, the response of every pixel can be calibrated.
Do this at two light levels and an offset and gain for each pixel can be
computed.

How do you create such a flat field?  Perfectly uniform light sources
don't grow on trees!

If I get any individual replies I will post a summary.

Mark.


____________________________________________________________________
Dr. Mark R. Shortis,                 Mark_Shortis@mac.unimelb.edu.au
Senior Lecturer,
Dept. of Surveying and Land Information,
University of Melbourne,                     Telephone +613 344 6401
Parkville 3052, AUSTRALIA.                   Facsimile +613 347 2916

"Any sufficiently advanced technology will be indistinguishable from
 magic" : Robert A. Heinlein
____________________________________________________________________

In article <3093@ariel.its.unimelb.EDU.AU> mark@sunrise.UUCP (Mark Shortis) writes:
>Would anyone like to comment on the problem of calibrating the
>radiometry of individual pixels in a CCD array?  
>
>Essentially the problem is that for applications which require accurate 
>radiometry (ie accurate grey levels are needed) the response of each pixel 
>becomes important.
>
>This is the classic issue of "flat fielding".  That is, if you supply a
>perfectly uniform input, the response of every pixel can be calibrated.
>Do this at two light levels and an offset and gain for each pixel can be
>computed.
>
>How do you create such a flat field?  Perfectly uniform light sources
>don't grow on trees!
>

The way I do it is to put a light diffuser (a piece of white transluscent 
glass or plastic) *between the lens and the ccd*.  Then, point the lens at
a fairly uniformly-lit wall, and put it way out of focus.  If you can't put
the diffuser after the lens, put it right before the lens, and make sure to
focus far from the diffuser.  (The problem with doing it this way is that 
the lens will introduce some non-uniformity, though this should be at very
low spatial frequency, which may be acceptable to you.)

Another way to go is to remove the lens completely, but you can only do this
if the lens and shutter are separable.

-Stephen Blackstock
 Leaf Systems, Inc.

In article <3093@ariel.its.unimelb.EDU.AU>, mark@sunrise.sli.unimelb.edu.au
(Mark Shortis) writes:
> Would anyone like to comment on the problem of calibrating the
> radiometry of individual pixels in a CCD array?  
> 
> Essentially the problem is that for applications which require accurate 
> radiometry (ie accurate grey levels are needed) the response of each pixel 
> becomes important.
> 
> This is the classic issue of "flat fielding".  That is, if you supply a
> perfectly uniform input, the response of every pixel can be calibrated.
> Do this at two light levels and an offset and gain for each pixel can be
> computed.

An alternative approach is to produce a calibration image for each
exposure which will not only remove the effects of CCD response
but nonuniform illumination.   

In my case, I have to deal with nonuniformities in the light box
when digitizing x-ray radiographs.  Thus, for each acquisition
session, I digitize the lightbox with no film and a neutral density
filter on the lens.  A calibrated image is then produced by dividing 
the image of interest by the calibration image (after bias and
dark current subtraction from both images).

In any case, if your illumination is constant over your measurements
and calibrations, there is no need for it to be perfectly uniform.
The illumination nonuniformities will be removed along with the CCD
nonuniformities.
 
--
+--------------------------------------------------------------------+
|Dick Wallingford      | Internet: dickw@iastate.edu                 |
|Center for NDE        | Packet  : wd0anb@n0an.ia.usa.na             |
|Iowa State University | "This river don't go to Aintree" - S.H.     | 
+--------------------------------------------------------------------+

In article <18e1vpINN9m2@network.ucsd.edu> spl@szechuan.uucp (Steve Lamont) writes:
>In article <jecome01.715798061@starbase.spd.louisville.edu> jecome01@rigel.spd.louisville.edu (Eric Comer) writes:
>>	if the lens caps is place on a ccd camera and an image is capture,
>>	why aren't all of the pixel values 0 (black)?  when the image is
>>	normalized it seems like there are faint light diagonal lines
>>	throughtout the whole image.  the image acquired was a gray scale
>>	image the min value was 0 and the max value was 6.  any ideas?
>
>Noise, mostly.  There may also be imperfections in the CCD -- a
>perfect chip, in fact, is fairly rare (and expensive), particularly at
>higher resolutions.  Finally, there is a charge accumulation (dark
>current) that occurs even when the shutter is closed or the lense cap
>is on.
>
>Excerpting from the Photometrics Series 200 manual, "[all] electronic
>circuitry generates noise.  Photon shot noise, dark current, and read
>noise are primary CCD camera noise sources.
>
>"Photon shot noise is a funamental property of the quantum nature of
>light...
>
>"Dark current, or thermally generated charge, cannot be distinguished
>from photon induced charge.  Dark current can be measured and
>subtracted, but its noise component cannot be ignored...
>
>"The third primary source of noise, read noise, sets the detection
>limits of a CCD camera.  Read noise is generated by the on chip output
>ampilfier...
>
>"Three types of integration measure the charge that has integrated on
>the CCD:  exposure, bias, and dark integrations. ..."
>
>There's a whole lot more to this, obviously, but that should at least
>give you some ideas on what's going on.
>
>The bottom line to all of this is that you have to be certain to
>remove the effects of the bias, dark current, and, possibly, light
>source non-linearities and artifacts when you grab an image.
>

There are ways of "beating the noise" to some degree:

Combining any noisy signal with any other noisy signal always results in
increased noise.  So you will increase the noise of the dark-subtracted
and bias-subtracted images unless you take precautions.  You improve the
precision of the dark frame by combining many darks together and then
scaling the pixel values by the ratio of image exposure time to total
summed-dark exposure time.  This scaled frame must then be subtracted
from the image of interest.  Since you've added many darks together and
the dark count/unit time should not change as long as the detector
temperature remains constant, the increased precision in the composite dark
dark frame will result in less noise per unit of dark signal being added
to the image when you subtract the dark count (the composite dark frame
has a higher signal to noise ratio, S/N).  The same principle applies to
bias as well.  In addition to the older, brute force approach of taking
bias *frames*, newer chips will be over- or under-scanned (= reading
fictituous pixels to measure the electronic bias and noise in the absense
of signal) to determine the bias to be applied to each particular row of
the image.  By combining many over/underscan pixels per row, you can beat
down the uncertainty in the bias level for each row just as you would by
combining many bias *frames* together to beat down the uncertainty of
the bias in each *pixel*.  Subtracting this higher S/N bias frame or over/
underscan vector will do less damage to the resulting bias-subtracted image.
Note that the bias must be subtracted from *every* frame before you use it
for anything, and this includes the individual dark frames I was suggesting
be combined (discussed above).  Finally, note that if your dark count
appears constant and featureless in a relatively long exposure, you are
better off subtracting a *constant* number (which represents the dark count
expected during the exposure time of the image) from the entire image as this
is an almost noiseless quantity and hence adds virtually no noise to the data.

I've written a paper which addresses these noise issues and discusses
maximizing the signal to noise ratio for data obtained with CCD's.
Although this is aimed at astronomers, it is very general in nature; just
replace references to the word "sky" with the word "background".  See
"Publications of the Astronomical Society of the Pacific", vol.103, p.122,
January 1991.

Mike Newberry
Steward Observatory
newberry@astro.as.arizona.edu

<1992Sep7.150506.6704@mpifr-bonn.mpg.de>

In <1992Sep7.150506.6704@mpifr-bonn.mpg.de> mlelstv@specklec.mpifr-bonn.mpg.de
(Michael van Elst) writes:

>In <jecome01.715798061@starbase.spd.louisville.edu>
jecome01@rigel.spd.louisville.edu (Eric Comer) writes:
>>      if the lens caps is place on a ccd camera and an image is capture,
>>      why aren't all of the pixel values 0 (black)?  when the image is
>>      normalized it seems like there are faint light diagonal lines
>>      throughtout the whole image.  the image acquired was a gray scale
>>      image the min value was 0 and the max value was 6.  any ideas?

>Well, two things. The CCD doesn't need to give you a value of 0 without
>light. This depends on the offset and gain of the video amplifier. Second,
>your system obviously catches some periodic stray signal. If you know
>your pixel rate then you can translate the number of pixels into the
>offending frequency in the time domain. Most likely this will be a 60Hz
>noise signal.

I tend to agree with Michael, although I'm from the optical side.  So 
count me in as one more vote for the electrical noise, rather than 
optical.

Well, what I meant by optical noise is really the dark current non-
uniformity everybody else mentioned.  What I meant by electrical noise
is the stray signal (such as 60Hz) as quoted above.

The key here is that the levels of signal is rather uniform here, and
form diagonal lines.  For a non-uniformity, usually the first thing you
see is the blemishes, which are like holes in a satuation picture, and
not much in a dark picture.  The second thing you might see, again for
non-uniformity, is the chip granding turns.  These again probably shows
up in lighted pictures better.

Most CCD cameras are not auto-gain-controlled, I believe.  Dark current
noise level (non-uniformity noise) should have been set at a rather low
level, probably 0-3 levels, I would guess.

Am I venturing too far on my guesses?


-- 
Duanfeng He (Jackson)           Duanfeng.He@AtlantaGa.NCR.com