Download I am assembling a - RIT Center for Imaging Science

Q1…Focus: imaging systems
Question: Two variants of the HD format are the 720p (1280H x 720V
progressive scan) and 1080i (1920H x 1080V interlaced scan).
An observer is seated 2m from a 65" LCDTV (16:9 aspect ratio, screen width =
144cm), viewing a 'slide show' displaying very high quality still images.
Incorporating the properties of the human visual system, the viewing conditions,
and the HD formats, discuss the tradeoffs between using the 720p and 1080i
formats in this situation. Recommend one for this purpose.
[[Answer: 65" 16:9 LCD TV has a width of 144cm. at a viewing distance of
200cm, that subtends a half angle of atan(72/200) = 20 degrees, or a full angle of
40 degrees. 1080i = 1920 pixels wide = 960 cycles wide = 960/40 cycles/degree =
24 cycles/deg
Compare to 720p, which is 1280 x 720: = 640/40 = 16 cycles/degree, well within
the visible range. Since there is a static image (therefore no concern about
temporal aliasing with the interlace) there is an advantage to 1080i.
The discussion should include the dis/advantages of progressive and interlaced
scanning (e.g., spatial vs. temporal artifacts), the net difference in bandwidth
between the two standards (1080i is slightly higher), and the fact that the
sensitivitiy of the human visual system is limited to spatial frequencies below 60
cpd. The answer should recognize that the issue is cycles/angle (vs.
cycles/distance), so the viewing distance has to be taken into account. A
thoughtful answer might also talk about the effects of aliasing the 'very high
quality" images.]]
Q2…Focus: imaging systems
Question: In a classical experiment, the line-spread function of the human eye
was measured using a double-pass optical technique. The Figure below illustrates
the effective on-axis LSF of the combined cornea and eyelens system at three
pupil apertures; 1.5mm, 2.4mm, and 6.6 mm. The LSF is given in terms of visual
angle subtended at the eye (1 arc min = 1/60th degree).
a. Explain the likely mechanisms responsible for the variation in LSF as a
function of aperture.
b. Sketch the spatial-frequency response of the eyes at these three pupil sizes.
Label and scale the axes, clearly mark your curves, and justify your response.
[[[Answers:
a. We are treating the eye as we would any optical system, so its performance is
limited by diffraction and aberrations. The optical performance has nothing to do
with rods & cones or sampling issues - diffraction limits the response at the
smallest aperture (1.5mm), aberrations increase the LSF as pupil size increases.
The answer should note that the optical quality increases between 1.5 and 2.4mm
diameters, then decreases for larger pupils as the aberrations overwhelm the
diffraction.
b. The answer should recognize that the MTF is the FT of the LSF. The 2.4mm
MTF curve should extend to the highest spatial frequencies (~30-60
cycles/degree), with both larger and smaller pupils dropping below that. All
should be unity at low spatial frequencies (the drop in low-frequency CSF is due
to neural computations, so it does not affect the MTF).]]]
Q3…Focus: imaging systems
Question: High-definition digital television displays come in two common
varieties, 1080i and 720p. A 1080i display nominally has 1920x1080 pixels in an
interlaced format, while a 720p display has 1280x720 pixels ina progressive scan
format. Quantitatively describe how you would compare the expected image
quality ofthe two displays. State all your assumptions. Draw a conclusion on
which display type is better.
Answer: First one can compute the nominal number of pixels. That’s 2.0736
million for 1080i and 0.9216 million for 720p. It appears that 1080i has a huge
advantage (2.25 X). However, the 1080i display, being interlaced, is only
refreshed at half the rate of the 720p display (assume equal frame rate).
Thus the pixel count of the 1080i display must be cut in half for a fair comparison
(some say a reduction to 75% is more appropriate, but anything in the range of
50-75% is defensible and 50% is the true information content). Thus the pixel
count becomes 1.0368 million for 1080i, now a factor of 1.125 times 720p and
much more comparable. This would still suggest that 1080i would have better
image quality since there are 12.5% more pixels. However, that assumes that the
interlaced display doesn’t introduce objectionable artifacts. If interlace is
objectionable, as it often is for static graphics or fast-moving subjects), then it is
quite likely that the 720p display will be perceived to be of higher overall quality.
At a large enough viewing distance, the interlace artifacts might become more
difficult to perceive, but at such a distance, the difference in pixel count would
also be impossible to perceive. Given the small increase in pixel count that would
be more than offset be the deleterious effects of interlace, it is clear to me that
720p provides higher image quality. (It is reasonable to give the opposite answer
as long as the potential problems of interlace have been addressed and the
conclusion is that they are not important ... personally, I think they are.)
Q4…Focus: imaging systems
Question: The recent Mars Rovers had a pair of cameras mounted 30cm apart
with 1-degree of toe-in (both pointing in 1-degree from parallel). Each camera has
a field of view of 16.8 x 16.8 degrees and sensors with1024x1024 pixels. One
claimed use for the dual-camera configuration is to provide stereoscopic
depthinformation. Given the above configuration, over what distances from the
cameras can useful stereoscopic images be obtained? If the stereoscopic
capabilities of this system are limited, what was the truevalue of having two
cameras in such a configuration for this mission?
Answer:There are two aspects to this, overlap in the field of view and angular
differences in corresponding pixels. The FOV overlap is a matter of simple
geometry and with some criterion for overlap (say 50% image area), one can
easily compute the minimum and maximum distances given the camera FOV
and geometry. It is also important to mention that it would be assumed the
cameras can focus over that range. Angular differences for each pixel will also
require some criterion. A reasonable value might be 5 pixels of angular difference
at the furthest useful distance for stereopsis. The FOV of a single pixel can be
computed and then the distance at which, the angular separation of the cameras is
5 times (or whatever the selected criterion) that value would be the far limit. The
system is surprisingly limited for stereo and could be much better with a single
camera taking two views after moving the rover slightly. The real reason for the
two cameras is probably a back up and to have some more spectral channels
(although that could have been done with one). It also makes the rover more
human-like, probably good for NASA PR.
Q5…Focus: imaging systems
Question: Anti-aliasing is an important process in digital cameras and computer
graphics. Describe what an anti-aliasing filter in a digital camera accomplishes
and how it functions. Speculate on how a similar process would be performed in
computer graphics rendering of images and why it might be important.
Answer: Aliasing in image capture is caused by the sampling interval being at a
lower spatial frequency than supported by the sampling function (sensor active
area and optical point-spread function). This is particularly a problem with fixed
area sensors such as CCDs that collect light over an area smaller than the interpixel area and is further exacerbated by the use of color filter arrays in single-chip
cameras. An anti-aliasing filter is placed in contact with, or close proximity to, the
sensor array and essentially blurs the incoming image with the objective being to
make the sensor response function span an area greater than the raw sensor site on
the chip (preferably overlapping with adjacent pixels to truly eliminate aliasing).
This can be accomplished with simple blurring filters or sometimes with
microlens arrays that serve to focus incoming energy from a larger area on each
sensor pixel site (also increasing sensitivity for the package).
In computer graphics, a rendered pixel area might be made up of fractional areas
of several objects. The ideal pixel value would be a weighted average of the
values within the pixel area (and perhaps slightly beyond). However, a rendering
algorithm normally samples only a single point in the center of the pixel. Such a
rendering will result in jaggy edges (as sometimes observed with text on a display
that is not antialiased). A process similar to the anti-aliasing filter in a digital
camera could solve this problem. Instead of blurring the rendered image, it is
instead rendered at a higher resolution (perhaps as many as 8x8 pixels rendered
for each pixel displayed) and then an appropriate averaging function (a Sinc
function for those that are very particular) is used to compute the final displayed
image.
Q6…Focus: imaging systems
Question: Blackbody (or Planckian) radiators that emit in the visible wavelength
region can have their color (spectral power distribution) described with just one
number, their absolute temperature. How is this possible? D aylight follows a
similar trend in that the color of various phases of daylight can be described by
one number (correlated color temperature, analogous to the color temperature of
blackbodies). However, the spectral power distributions of daylight are much
more complex. Principal components analysis (PCA) of relative daylight spectra
suggests that accurate reconstructions can be made with a mean vector and two
additional vectors. Outline a system that would allow one to specify the correlated
color temperature of daylight and then reconstruct a relative spectral power
distribution. How can you go from the one-dimensional data of correlated color
temperature to the two-dimensions required for determining the weights of the
PCA vectors?
Answer: The spectral power distribution of a blackbody radiator can be computed
from its absolute temperature (Kelvin) through enumeration of Planck’s equation.
For daylight, if the color can be described by a similar temperature value, then
there must be a daylight locus of chromaticities similar to the blackbody locus. A
path traced through 2-D chromaticity space with absolute intensity as a free
variable. Since it is stated that two-dimensions are required to reconstruct a
spectral power distribution, that path must be curved in linear chromaticity space
(in fact both the Planckian and daylight loci are curves. Thus, the temperature
could be used as an index into the curve and then the 2D chromaticity coordinates
could be used to transform into the necessary weights on each of the PCA vectors
to recreate the desired spectrum.
Q7…Focus: Basic Principles of Img Sci I
Question: Consider the use of a sharp edge imaged onto an analog detector. Derive an
equation relating the edge spread function to the line spread function. Verify this
equation using (a) convolution theory, (b) probability theory.
Solution:
Mathematically, the line spread function
integrated with respect to y.
One can get the edge spread function
in the x direction is simply the point spread function
at
by adding all the line spread functions at
This equivalent to integrating a single line spread function from -
.
to
(a) Assuming a linear, shift invariant system, the output
is a convolution of the input knife-edge
exposure
with the impulse response function, the line-spread function
. In 1D we have
Using the commutative property of convolution
The figure on the next page illustrates the convolution.
To find the value of
we stop the convolution at the point
(b) An alternative method of understanding Eq. (14.2) is to use probability theory. Because the area
of
is normalized to unity, we can view it as a probability density function. This function is derived
from the probability distribution function
by differentiation.
The distribution function represents the cumulative probability as a function of with a value of
zero at
and a value of one at . It represents the cumulative probability, or area under the
probability density function. For a given value of
we can find the value of
We can recognize the similarity with Eq. (14.2) in the notes and then identify
distribution function.
by
as a probability
Q8…Focus: Basic Principles of Img Sci I
Question: (a) Calculate the number of effective pixels in a 35-mm frame (24x36 mm) of
an ISO 50 monochrome film whose 50% MTF is at 65 cyc/mm and assuming square
pixels.
(b) Assuming the same pixel size, calculate the number of pixels in a 2/3” (8.8x6.6
mm) optical format CCD.
(c) Discuss the differences in pixel count in terms of system performance.
(d) What size pixel would be needed in the CCD to match the pixel count in film?
(e) Comment on the system performance in this case, continuing with the
assumption that both imaging systems are monochrome.
Now consider a monochrome CCD array is illuminated by a light source with
emission at 450, 550, and 650 nm, each with an irradiance of 1 W/cm2. Each pixel is
9
-square, and the array has a 100% fill factor. The CCD is designed to be read out
10 times/s. The collection efficiency of the CCD is 0.2 at 450 nm, 0.35 at 550 nm,
and 0.3 at 650 nm. (f) What is the minimum full-well capacity required to accurately
detect the irradiance from the light source?
Solution:
(a) The pixel size is calculated using Eq. (14.38) giving
The frame area is 24x36 mm = 864 mm2 = 8.64x108
. The number of pixels is then
(b) A 2/3” format has a sensor size of 8.8x6.6 mm. The number of pixels is then
(c) At equal print magnification the CCD will have poorer resolution, although it may be sufficient
at low magnification (4x6 inch) to give acceptable images.
(d) 107 pixels are now distributed over a 58 mm2 area (8.8x6.6 mm). The pixel size is then
Assuming square pixels this means the pixel edge is 2.4
.
(e) The CCD resolution is now better, although the CCD image will have to be magnified about 15x
[(9.3)2/(2.4)2] more than the film image to produce approximately the same print pixel size. This could
lead to pixelation artifacts in the CCD image, depending on the final print size. However, the sensitivity
is now reduced by (2.4)2/(9.3)2 or about 15x. In addition, the dynamic range will be lowered because
the full-well capacity is reduced by (2.4)2/(9.3)2, or also about 15x.
(f) For this problem we use
. We need the photon energies which are 2.76,
2.25, and 1.91 eV for the 450, 550, and 650 nm wavelengths, respectively. An irradiance of 1
converts to 6.24x1012 eV/cm2 s. Since the array is sampled every 0.1 s it sees only
6.24x1011 eV/cm2 per sample period. Multiplying by the pixel area gives 5.05x10 5 eV/pixel for each
wavelength. To get the number of photons we need to divide this energy by the photon energy to get
conversion to photons
photons
450
5.05x105 eV/ 2.76 eV/photon
1.83x105
550
5.05x105 eV/ 2.25 eV/photon
2.25x105
650
5.05x105 eV/ 1.91 eV/photon
2.65x105
, nm
Q9…Focus: Basic Principles of Img Sci I
Question: From the Kubelka-Munk equation for reflectance, derive a much simpler
equation assuming
,
,
.
Solution:
For a Beer-Lambert system placed against a Lambertian reflecting surface, and provided that the
reflectance is collected with a hemispherical detector, we have the conditions:
We cannot immediately reduce the above equation, since infinities result when
can recognize that as
we get
And, since
, then
Now we replace
with
Expanding and replacing
. However, we
in the given equation to get
with
in the hyperbolic trig function gives
Since
, all of the terms are very large, except for the 1 in the numerator and
denominator. These two terms can therefore be dropped in the limit of
. Thus
Now we represent the hyperbolic trig functions in their exponential format
in the
Q10…Focus: Basic Principles of Img Sci I
Question: Derive an equation relating the granularity variance to imaging particle area
and transmission density. An equation giving the transmission density is
Solution:
For the granualrity variance we are concerned with fluctuations in density so we can recast the above
equation into
since the only variation from point-to-point as the aperture scans the image will be the number of image
grains and their image density. We square both sides of Eq. (15.1) and take
definition of standard deviation and then divide by
summation to each side of the equation, we have
where the
. If we do likewise with
as the numerator in the
, and then apply a
s represent the variances of the indicated quantities. If the positioning of the imaging particles
is random, then the distribution of the
is Poisson, in which case the variance of the distribution is equal
to its mean, that is
. This brings us to the following equation as an expression for the rms
granularity in terms of the average number of imaging particles and their size.
The noise level will also be a function of the signal level. To see this we recast the Nutting equation so that
appears on the left-hand side
This expression is then substituted into Eq. (15.3) so that
which can be rearranged to get
where we have introduced a new granularity constant
coefficient
. Specifically,
that is analogous to the Selwyn granularity
.
From Eq. (15.6) we see that the granularity increases both with increasing imaging particle area and with
increasing density.
Q11…Focus: Basic Principles of Img Sci I
Question: Calculate the number of photons/s incident on a detector 75square at a
wavelength of 4
from a human 1 km away. Assume the human is 180 cm tall by 40
cm wide.
Solution:
This problem is solved by assuming the human is a perfect blackbody radiator radiating at 98.6 F or
310.16 K. The photon energy is 1240/4000 nm = 0.31 eV.
Eq. (5.6) we have
for this temperature is 0.0275 eV. Using
where we have dropped the wavelength unit since we are considering monochromatic light. The
irradiance at the detector is given by Eq. (5.18). We first get the radiant intensity by first deriving the
flux from the radiant exitance by multiplying by the area of the source (the dimensions of the human)
and then dividing by the total solid angle 4 . Then we divide by the 1 km distance to get the irradiance.
Note we are using
, not
, because
in the first equation assumes isotropic source. Now we
multiply by the area of the detector and convert to photons/s
The ability to detect this signal depends upon how it compares with the background. In midday in
July the background will be approaching the temperature of the human and detection will be difficult.
At 1 am in January on a cloudless night the background will be considerably colder and the possibility
of detecting the human will be higher.
Q12…Focus: Basic Principles of Img Sci II
Question: Below is the histogram data for a 4 bit image of a black-and-white soccer
ball on a gray background. This soccer ball is 230 mm in diameter. What is the pixel
spacing? What fraction of the total image area is occupied by the soccer ball? Propose
a processing scheme that will detect the circular edge of the soccer ball. In other
words, ideally the final image should be black every where except that the edge pixels
should be white. If you plan to chain together several different processing steps, then
give details for each one (i.e. if you use a convolution kernel, then write it down and
justify it). Discuss possible problems and non-ideal nature of the final image.
Pixel count[0,520,920,490,30,40,5910,24040,6050,80,20,80,440,960,420,0]
black
gray
white
Answer: Area of the ball = 41,526 mm2
No. of pixels in the ball = 3860  area of a pixel = 10.76 mm2
Pixel spacing = 3.28 mm
Total no. of image pixels = 40,000
Fractional area = 3860/40,000 = .0965
Possible processing steps:
(i)
(ii)
(iii)
Histogram transformation: LUT level 0 to 4 go to 15, 5 to 5 to 10 go to
0, and level 11 to 15 go to 0. Image should have all the ball pixels white
and the background black.
Now use a Laplacian kernel or a combination of x and y first derivative
kernels. Kernel coefficient values must add up to zero. Take the absolute
value at each pixel location. Image should have zero (almost) pixel
values every where except at the edge locations.
Re-map pixel values from 0 to 15, examine the histogram and pick a
threshold to generate a binary output image.
Q13…Focus: Noise & Random Processes
Question: Show that the relationship
is true for a linear, shift
invariant system with a WSS input. is the power density spectrum. Assume the
mean output is known to be a constant.
Solution:
Next, we write the equation for the output autocorrelation function of our linear, shift-invariant
system.
Now express
in terms of the input and the impulse response function
Using the commutative property of convolution again, we can rewrite this as
Note that if we had followed the procedure used in getting the first integral to get the second
integral, we would have had
write this more simply as
Because we have assumed
. But because we have assumed a shift-invariant system we can
. Swapping the
and
is WSS we rewrite this as
where we used
equation says
operators again gives
to get the argument of
is WSS if
is WSS because
only depends on
constant from Eq. (9.2). Now we compute the output power spectral density
theorem gives
. Note that this
and
is a
. Using the EWK
Making a change of variable using
we get
where
is the system transfer function as defined earlier,
is its complex conjugate, and
is a deterministic gain ( in film systems). If is a random variable such as in the photon amplifier,
then we shall use . Equation (9.9) says in a WSS system the output power spectrum is just the input
power spectrum times the squared magnitude of the transfer function. The same relationship is found
for discrete systems. In terms of the NPS and when the system adds noise we have
where
is the additive noise term.
Q14…Focus: Noise & Random Processes
Question: Given the following
for a digital radiographic scintillator-detector,
shown in the accompanying figure, in which the Poisson excess was assumed to be
negligible, derive an expression for the
of a high-resolution scintillator and
discuss the implications. The NPS after going thru the detector is also given below.
Solution:
A high-resolution scintillator implies that the system MTF in the direction is limited by the
detector-array aperture function rather than by the scintillator MTF. There is no correlated noise on
the detector in this case because all the scattered photons from a single x-ray photon in the screen are
collected in the same pixel. This corresponds to system designs incorporating high-resolution
scintillator materials, as well as systems designed for “direct-detection” flat-panel detectors (X-rays
are converted directly into electrons in the detector). In this case
over the frequencies passed by
and the
simplifies to
is approximately constant
of
It has been shown elsewhere (reference 2) that the sum of
is always equal to a constant given by
where
is the detector fill factor in the
and its aliases at harmonics
direction. We then get
for a one-dimensional detector. For two dimensions we get
The one-dimensional
detecto
along the
of this two-dimensional
r, evaluated
axis of the two-dimensional detector, is therefore given by
For this special case of a photon-noise limited detector with a high-resolution scintillator screen,
the
is proportional to the screen quantum efficiency
and the detector fill factor
, and
always has a shape given by
. Fig. 10.1 illustrates the calculated
for a twodimensional detector in the -direction with a high-resolution scintillator for various fill factor values
, assuming
. As the fill factor decreases for a constant
, higher frequencies are
passed by the detector-element aperture function and noise aliasing increases. This is directly
responsible for the decreasing
.
Q15…Focus: Noise & Random Processes
Question: A processed film has a true density
and a density
measured
with a square aperture having area
. Find an expression for the scale value of
the NPS and discuss its implications.
Solution:
Aperture spatially integrates true density to give measured density.
Then
and
can be converted into autocorr func. Then, after applying EWK
The conversion factor in (12) is the squared magnitude of the system transfer func. The
removes the complex exponentials, leaving a real func.
Via cent ord thm, the
If both
and
s have unit amplitude at (0,0) (
are large wrt any correlation distance in
be a constant for frequencies where
. Variance is
has unit vol). So
,
will
where double integral evaluates to b/c via cent ord thm area in freq domain is height in spatial
domain,
Thus
and area of
= area of
. (13) was used in last step.
(15)
which is Eq. (15.16) of BP I with
the granularity constant.
So, when aperture (sampling function) large wrt any correlation distance, zero-freq value of NPS is
measured variance scaled by aperture area. Sampling func is such that we can’t detect change in
correlation. If sampling aperture made small enough to detect change in correlation, then
Q16…Focus: Noise & Random Processes
Question: Consider the case of a negative contact printed to make a positive as shown
in the figure below. Find an expression for the
of the system, including spatial
frequency aspects, and discuss its implications.
Solution:
To include the spatial frequency aspect, a schematic of the system using symbols appropriate to the
first part of the problem would be as follows
The
for the negative stage would be, using the standard formula for film
The NPS transfer equation would be
Finally, the system
would be
In those cases where second term in the denominator is much less than 1, the overall
will be
that of the first stage with negligible reduction due to the second print stage. Now suppose at all
frequencies
and
clearly negligible. But, if at some frequency
. Then, if
the second term in the denominator is
, the second term in the denominator will not
be negligible and
. In other words, we can no longer neglect the pos stage
because its MTF may be such that stage does impact the system
.
Q17…Focus: Noise & Random Processes
Question: A single-stage photon amplifier is used to enhance the low-light sensitivity
of a CCD array, as shown in the figure below. Find an expression for the
at the
output of this amplifier, that is, before the photons enter the fiber optic, and discuss its
implications. Spatial frequency aspects can be ignored. Recall that
for a photon amplifier.
Solution:
The degree of photon amplification is limited by the sensitivity of the input stage to arriving quanta
(the fraction of light absorbed and the efficiency of their conversion to secondary emission) and by
noise that is introduced in each stage of amplification. The noise that is added in the first element of
each stage is customarily called background noise and will be symbolized by . The output of the first
element of the stage is
These
the stage
emissions from the first element are then multiplied in the second element. The output of
is then
To calculate the
of the image intensifier we need to know the variances for each element of
each stage.
Throughout the analysis we assume that the signal-propagated events and background events are
statistically independent. Then the variance can be formed simply as the sum of the individual variances
(Recall Peebles Example 5.1-3, the variance of the sum is the sum of the variances when the RVs are
uncorrelated, and if they are statistically independent they are certainly uncorrelated.) to get for the
first element of the stage
To go further we need to know the statistics of both these contributions. For simplicity we assume
that the background events are Poisson distributed. Then
. The signal undergoes a binomial
selection process (i.e., a Bernoulli trial) in the first element of each stage, and then undergoes a Poisson
amplification process. To calculate the variances we can use the equation given in the problem
statement.
First element. Assuming no correlation between
output of this stage is
and
, we can write the variance of
, no background noise, and knowing the
as
where we used the fact that the variance of a binomial process is the mean times one minus the
mean. The last equation in Eqs. (8.10) represents a preserved Poisson process with a new mean. Adding
the background noise, we finally have
Note that the variance equals the mean for this first element, as expected, since both the signal and
noise processes are Poisson, and the “gain” is binomial.
Second element. After the second element the output is
we get
where we have used the fact that both
and
are Poisson and used Eq. (8.7) for
background noise for this stage were zero, the variance would be
. If the
which is analogous to Eq. (8.52) except the mean input is now
instead of . Note that the
second term of this equation indicates the variance is greater than that of a Poisson process with mean
. This increased noise variance was seen in Eq. (8.52) and is due to the randomness of the
amplification process.
With this information we are now ready to calculate the
element the
is given by
where the subscript
is given by
for the two elements. For the first
indicates the first element of the first stage. The
of the second element
Suppose
. Then Eq. (8.15) becomes
Figure 8.16 shows that the maximum
increases above this lower limit
background noise on
is
, but only in the case when
decreases for low exposures (
. This effect of
exposure is increased. Of course
on
. As
) showing the effect of
diminishes, and eventually vanishes, as the
is just proportional to (background noise)/(photon noise)
since we are assuming Poisson statistics for both noise sources. As shown in Fig. 8.17, when
the system is background-noise limited, and when
the system is photon-noise limited. Referring
back to Eq. (8.15), if we instead assume no background noise then we have
which is the same as Eq. (8.56) with
replacing
.
Q18…Focus: Noise & Random Processes
Question: Consider an ideal array of photon detectors. Each element of the array acts as
an individual photon detector and the imaging system has the property that each photon
detected can be read out individually — thus an ideal photon detector Assume the array
consists of individual photon detectors, each having identical properties. Each detector is
assumed to record incident photons in a manner independent of its neighbors and has a
unique and distinguishable image state dependent only on the number of incident photons
on that detector.
Also, assume that the exposure over the array is uniform and that the number of photons
incident on a detector obeys Poisson statistics. If the mean number of photons/detector is
, then the actual number of photons incident on a detector will have probability
(1)
Calculate the variance of the noise for this idealized detector array.
Solution: Given the Poisson probability formula in Eq. (1), it is easy to calculate the fraction of detectors
that have reached the saturation level . This is given by
This summation will provide an overall measure of the efficiency of the imaging process.
First, we consider the relationship between the mean count level, , of the array and the mean exposure,
Using the definition of the mean we have
.
where the first term takes into account those detectors that have not reached saturation and the second term
takes into account those that have reached saturation. Figure 3 shows a plot of the mean count level in
relation to the saturation level, as a function of the mean exposure level. As might be expected, there is a
linear relationship between input quanta and number of image counts when
. This linearity breaks
down as
. At exposures high enough for all receptors to have received at least photons, each
detector records exactly counts.
Equation (3) may be rewritten as
There are summations on the right-hand side of this equation and, since these are all summed to , each
term will be equal to unity less the missing initial terms. Hence, we can rewrite Eq. (4) as minus an
expression for the missing terms. Then we get
where
There will be spatial fluctuations in the image which arise in the present ideal case from a statistical
distri ution of the incident photons, that is, photon noise. We can calculate the variance of this noise if we
have the first and second moments of the random variable . We already have the first moment in Eq. (5).
The second moment of the distribution about the origin
moment (Eq. (3)). Thus
is written in a form analogous to the first
(7)
This expression can be rearranged to
where
Now the variance of
can be written as
noise merely mirrors the input noise, as we would expect for an ideal detector.