Question on logarithmic and linear detectors

Re: Question on logarithmic and linear detectors

On Oct 13, 6:42=A0am, "DrJohnR...@AOL.com"  wrote:
> On Oct 12, 10:20=A0pm, Xiaoxiao  wrote:
>
>
>
>
>
> > Hi,
>
> > I read from thathttp://micro.magnet.fsu.edu/primer/java/digitalimaging/=
processing/bac...
>
> > that "Background subtraction is the method of choice for logarithmic
> > detectors (such as analog video cameras and film), but not for linear
> > detectors (many digital cameras and flatbed scanners), where the
> > proper technique is to divide the image by the background."
>
> > I am not sure whether the detector used for my image capturing is a
> > logarithmic or and linear detector so that I can use the right
> > background subtraction. I have some images taken from a CMOS detector
> > and some from a CCD camera. Can anyone tell me the logarithmic or
> > linear detector types of CMOS and CCD detectors?
>
> > Thanks for the help in advance.
>
> Both CMOS and CCD detectors themselves are inherently linear - the
> output is directly proportional to the number of incident photons. But
> it is common in many devices to apply nonlinear, usually more or less
> logarithmic, functions in the electronics to make the final image look
> more like film (which is inherently logarithmic). Some devices, such
> as Fuji cameras, are linear for the dark part of their range and
> nonlinear for bright light. (So were vidicons). You would have to
> experiment with your detector and a target such as a step-wedge to
> measure your particular camera.- Hide quoted text -
>
> - Show quoted text -

John's right.  Fuji does have two linear pixels at each pixel, each is
linear with a different slope.  And also, many cameras apply a "gamma"
so that they are not linear.  This is done in a post processing step.
You'd be surprised what goes on in a digital camera.  Even a "raw"
image might have some processing done on it - it depends on the
manufacturer.  But even if you were using a linear detector with a
gamma of 1, the linearity still depends on your sample.  Your question
is really more complicated than it looks on the surface and I don't
have time to get into it here but what the web site said was basically
correct.  "Most" of the time you want to do a background division.
Why?  Well if you had 90% as many photons at the corner as you do in
the center, wouldn't you want to divide by 0.9 there?  Of course you
would.  However, where you do background subtraction is in
transmission situations like X-rays.  I'm not going to get into the
math here but if you do the simple math (Beer's Law) you'll see that
you want to do background subtraction there and NOT division if you
reall want to get the mass density of the sample.  That's why you see
"Digital Subtraction Angiography" in the hospitals, not "Digital
Division Angiography" even though they are using a linear detector
under the image intensifier.

John is right when he says that the best way is to have a known
calibration target - this is the gold standard way of doing things.
Regards,
ImageAnalyst

Re: Question on logarithmic and linear detectors

On Oct 13, 8:54=A0am, ImageAnalyst  wrote:
> On Oct 13, 6:42=A0am, "DrJohnR...@AOL.com"  wrote:
>
>
>
>
>
>
>
> > On Oct 12, 10:20=A0pm, Xiaoxiao  wrote:
>
> > > Hi,
>
> > > I read from thathttp://micro.magnet.fsu.edu/primer/java/digitalimagin=
g/processing/bac...
>
> > > that "Background subtraction is the method of choice for logarithmic
> > > detectors (such as analog video cameras and film), but not for linear
> > > detectors (many digital cameras and flatbed scanners), where the
> > > proper technique is to divide the image by the background."
>
> > > I am not sure whether the detector used for my image capturing is a
> > > logarithmic or and linear detector so that I can use the right
> > > background subtraction. I have some images taken from a CMOS detector
> > > and some from a CCD camera. Can anyone tell me the logarithmic or
> > > linear detector types of CMOS and CCD detectors?
>
> > > Thanks for the help in advance.
>
> > Both CMOS and CCD detectors themselves are inherently linear - the
> > output is directly proportional to the number of incident photons. But
> > it is common in many devices to apply nonlinear, usually more or less
> > logarithmic, functions in the electronics to make the final image look
> > more like film (which is inherently logarithmic). Some devices, such
> > as Fuji cameras, are linear for the dark part of their range and
> > nonlinear for bright light. (So were vidicons). You would have to
> > experiment with your detector and a target such as a step-wedge to
> > measure your particular camera.- Hide quoted text -
>
> > - Show quoted text -
>
> John's right. =A0Fuji does have two linear pixels at each pixel, each is
> linear with a different slope. =A0And also, many cameras apply a "gamma"
> so that they are not linear. =A0This is done in a post processing step.
> You'd be surprised what goes on in a digital camera. =A0Even a "raw"
> image might have some processing done on it - it depends on the
> manufacturer. =A0But even if you were using a linear detector with a
> gamma of 1, the linearity still depends on your sample. =A0Your question
> is really more complicated than it looks on the surface and I don't
> have time to get into it here but what the web site said was basically
> correct. =A0"Most" of the time you want to do a background division.
> Why? =A0Well if you had 90% as many photons at the corner as you do in
> the center, wouldn't you want to divide by 0.9 there? =A0Of course you
> would. =A0However, where you do background subtraction is in
> transmission situations like X-rays. =A0I'm not going to get into the
> math here but if you do the simple math (Beer's Law) you'll see that
> you want to do background subtraction there and NOT division if you
> reall want to get the mass density of the sample. =A0That's why you see
> "Digital Subtraction Angiography" in the hospitals, not "Digital
> Division Angiography" even though they are using a linear detector
> under the image intensifier.
>
> John is right when he says that the best way is to have a known
> calibration target - this is the gold standard way of doing things.
> Regards,
> ImageAnalyst

Thanks a lot for John and ImageAnalyst's help. It seems that there are
a lot more than what I thought to be. Thanks for the detailed
explanation again.