Objectives: To understand;
1
Contrast, signal, and noise and the physical factors affecting them
2
Calculation of exposure requirements using the Rose model
3
Optimization of signal to noise ratio by proper choice of X-ray
energy
4
Detective Quantum Efficiency - dependence on fractional signal
amplitude
5
Minimum detectable contrast
6
Digitization noise, digitization requirements
Imaging systems are characterized by:
1.
The ability to detect small attenuation changes, i.e., low contrast,
faint objects. This increases as signal increases and as noise
decreases.
2.
The ability to see small objects, fine detail. Increases as the point
spread function gets narrower.
3.
The ability to stop motion and to sample motion. Increases as the
exposure time decreases and the sampling rate increases.
This section addresses the factors which contribute to contrast
resolution. These generally relate to overall system signal to noise
ratio and the various factors affecting this quantity. Consider the
differential attenuation due to some object as shown in Figure 1
No
N1
N2
Subject contrast is defined as
C = (N1 - N2) / N1
(1)
assuming no scatter is detected. This may also be defined in terms
of intensity. For monoenergetic beams the two definitions are
equivalent. For polyenergetic beams the contrast would be
somewhat different for the two definitions. In this discussion we
will work with the definition in terms of fluence since we will be
discussing numbers of x-rays.
Christensen defines
C = N2 / N1.
Another definition which is sometimes used is
C = (N1 - N2) / ( N1 + N2)
You can use any of these but you must be consistent in the derivation
of results.
C =( N1 - N2)/ N
approaches 0 as N1 approaches N2, and approaches 1 as N2
approaches 0.
However it is defined, it refers to the difference between an object and
its surroundings.
When these fluences (or intensities) are detected by a detector a
signal is generated, for example, video signal or optical density.
Refering to Figure 2,
No
Object area A
N1
N2
Detector resolution
element of area a
the input signal S observed by each detector element a is given by
S = Na
The size of the detector element a is determined by whatever defines
the resolution of the system, for example a picture element ( pixel ) in
a digital image, or the point spread function in a film image.
Perception of an object in an image is dependent upon the differential
signal S representing the difference between the detected signal,
integrated over the object size A, and a similar area in the region
surrounding the object defined by
S = N1A - N2A = (N1 -N2)A
{2}
Whether this differential signal is perceptible depends on the degree
of fluctuation on the signal in the area in the immediate vicinity of
the object. These fluctuations, which may be due to a variety of
causes are called noise.
Noise = undesirable signal = n
Noise masks small signals and, along with the differential signal,
determines the contrast resolution of the system. The noise relevant
to the task of object detection is the average noise over an area equal
to the object size.
Image quality is often described in terms of signal to noise ratio.
S / n = SNR ( signal to noise ratio)
The differential signal divided by the noise is also referred to as
signal to noise ratio in some contexts. We will refer to this as
differential signal to noise ratio.
S / n = SNR ( differential SNR )
This is also sometimes referred to as contrast to noise ratio. We will
not use this term since it is somewhat confusing in view of our
definition of contrast as S/S.
What determines differential signal and noise?
Signal
subject Contrast
kVp
filtration
object
scatter
detector
display
Noise
quantum noise
system noise
anatomical noise
observer noise
artifacts
Consider exposure requirements for contrast and spatial resolution in
an IDEAL SYSTEM - i.e. one in which all noise due only to quantum
noise, ( statistical fluctuations in the number of transmitted x-rays). Xray noise is described by a Poisson distribution as shown in Figure 3.
n
Probability
of #x-rays
in area A
N1
n = (N1A)1/2
= N
Fluence N
The differential signal is given by equation 2 as
S = ( N1 - N2 ) A
For very small contrasts
and
(3)
Therefore we can write
S =NCA
(4)
The noise is given by
(5)
so
(6)
According to equation 6, the detected fluence is related to the s/n by
(7)
The detected fluence is related to the incident fluence N0 by the
transmission factor T as
N = N0 T
(8)
(For non-ideal systems the detection efficiency of the detector would
also multiply the transmission factor in the above equation)
Therefore the incident fluence is related to the imageSNR by
(9)
The ROSE MODEL is a model which assumes that the statistical
fluctuations calculated above are the only determinant of perception.
The effects of edge gradients, which are known to be important, are
not considered. The Rose model states that small contrasts will be
visualized when
SNR = 3 - 5
We can use the Rose model to obtain approximate estimates of
required exposure. We can illustrate the Rose model using a raindrop
analogy. Suppose you had a marble and a basketball on your driveway
as shown in Figure 4.
Figure 4
After a small amount of rain, if you removed these objects, you could
detect the presence of the basketball, but not that of the marble. This is
illustrated in Figure 5.
Figure 5
As more rain falls, and the average spacing of the raindrops becomes
smaller than the marble, you can begin to see the dry spot under the
marble when it is removed. This is illustrated in Figure 6 where we
have only included the region around the marble.
Figure 6
A lot more rain is required to define the marble than the
basketball. A higher fluence (exposure) is needed to
visualize small objects.
So for an ideal system, No is proportional to 1/ A, 1 / C2,
1/ T, and ( s / n )2.
DEPENDENCE OF IMAGE QUALITY
ON X-RAY EXPOSURE
0.025 mR
Kruger and Reiderer
Basic Concepts of Digital Subtraction Angiography p 80-81
0.400 mR
For some time after the initial introduction of digital subtraction
angiography techniques radiologists investigated the possibility of
obtaining angiograms using intra-venous injections of contrast
material, usually into a vein in the arm, but sometimes using a
catheter placed in the superior vena cava or right atrium. Although
considered less invasive than intra-arterial catheter placement, this
technique required high system contrast sensitivity because of the fact
that the contrast was diluted by a factor of twenty or so, depending on
cardiac output, before reaching the arteries of interest.
Suppose we have low iodine contrast in a vessel following intravenous
injection and we want to detect 1 mm long narrowing of the vessel as
shown in Figure 7.
1 mm
1 mm
N2
N1
Suppose:
I = 5 mgm/cm3
iodine density at the artery after iv injection
t = 1 gm/cm3
tissue density
t = 20 cm
µtissue = µt = 0.25 cm2 / gm
µiodine = µI =15 cm2/gm
Keff = 35 keV
Then
C = (N1 - N2) / N1
=
(e-µt
*
20
-
µ
t
µt*20∆
I
I)
e-
/ (e-µt* 20)
= 1 - e - µI ∆tI ~ 1 - ( 1 - µI ∆tI .... )
Therefore, in general, for small contrasts
(10)
Plugging in, we get
C = µI ∆tI = µI pl ∆xI
= 15 cm2/gm* .005 gm/cm3* .1 cm = .0075
The transmission factor is
Plugging into equation 9 with s/n = 3 we get
Remembering that
1R ≈ 2x1010 photons/cm2 at 35 keV
we get a required incident exposure of
E0 ~ .12 R
This is the exposure required to image the 1mm length of
a contrast filled vessel. This should also be approximately
the exposure required to see a total stenosis with a length
approximately equal to the vessel diameter.
This shows that exposure is not only related to contrast
resolution, but also can limit spatial resolution, i.e., in
order to see something very small (of low contrast), a
large exposure is required.
For example, what if ∆tI decreases to 0.1 mm, down by a
factor of ten?
Then contrast decreases by a factor of ten to 0.0075, area
goes down by ten, assuming we are still looking at a 1mm
length, and T stays the same since it is dominated by the
tissue transmission.
Then
N0 = 2.39 x 109 x 102 x 10 = 2.39 x 1012
E0 = 119 R !
The effect of increasing the number of quanta in an image
are illustrated in Figure 8 for the case of light photons.
This is from Rose’s book called Vision.
3 • 10
3
ROSE W0MAN
9.3 • 10
3.6 • 10
4
5
1.2 • 10
4
7.6 • 10
5
7
2.8 • 10
An illustration of the same principle using x-rays is illustrated in
Figure 9. Figure 9A shows an attempt to image the arteries in the
head using an intravenous injection. For this image there was a
malfunction of the exposure regulation circuitry and the exposure
was approximately a factor of 100 less than anticipated. Figure 9B
shows the examination repeated with the correct exposure. The
increase in image quality is evident.
A
Figure 9
B
We can use the Rose model to calculate optimal beam energies for
various situations. Again, we assume quantum statistical noise
dominates image noise. We will try to determine the optimum x-ray
energy for a particular imaging problem.
CASE I: Small tissue thickness t on a larger tissue thickness t
(gm/cm2) as shown in Figure 10.
N0
t
∆t
N1
N2
From equation 9
No ≈ (s /n )2 / C2 AT
so for fixed s /n and A
where µt is the effective attenuation coefficient over distance t,
ignoring the details of beam hardening for now.
but
Therefore,
Now we can find the beam energy which minimizes exposure which
is  No using the usual calculus max / min procedure of setting the
first derivative to zero.
No  (e µtt) / µt 2
(dNo) / (dµt) = (teµt t) / (µt 2) + (-2eµt t) / (µt 3) = 0
(11)
(12)
which gives
µt = 2 / t
That this is a minimum and not a maximum can be checked by
looking at the second derivative.
(13)
Chest imaging: t ≈ 20 cm
The attenuation coefficient associated with the
optimal energy is given by the above equation as
µT = 2/20 = 0.1 cm2/gm
This corresponds to an effective beam energy of
about 100 keV and a kVp of something like 200
depending on the filtration.
Mammography t ≈ 5 cm
The same analysis results in
µt = 2/5 cm2/gm
Keff = 25 keV for tissue and 23 keV for fat
Bear in mind that these are ideal system results. For actual
mammography, where object contrast must be kept
somewhat higher than the ideal to compete with addition
system noise such as film grain, the optimal energies
actually used are somewhat lower.
Daffodil Imaged at 5 and 20kV
5
Gilardoni et al. Radiology-Electromedicine
20
From the examples, we see that the best energy is very
thickness dependent. There is a tradeoff between
transmission and contrast.
For large tissue thicknesses, transmission is the most
important factor, i.e., it is desirable to use the highest kVp
possible until the contrast is so small that it becomes
comparable to the system noise. In computed
tomography, with high device SNR, studies show 140
kVp to be superior to 80-120 kVp for tissue tumor
detection.
Refer to Figure 10 below, with a small iodine thicknesstI substituted
for the small tissue thickness t. In this case, for fixed A and tI, the
required fluence is given by
No ≈ (s/n)2 / (C2  1 / (C2 e-µt t)
N0
t
∆t
N1
N2
(14)
but since
C ≈ µItI
No (e+µt t) / (µI2∆tI 2)
(e µt t) / µI 2
(15)
Now we have an expression involving two attenuation coefficients,
both of which are dependent upon energy. Therefore, the
differentiation involved in the minimization procedure must be with
respect to energy. In order to accomplish this we will have to
parameterize the attenuation coefficients in terms of the energy k and
fill in the appropriate terms in the expression.
dN / dk = (N / µI) (dµI/dk) + N /µT)(dµT/dk) = 0
(16)
For tissue the Compton and photoelectric contributions are both ≈
0.25 cm2/gm at 25 keV. Modeling the Compton contribution as a
constant with energy and inserting the known energy dependence of
the photoelectric component we get
µT ≈ 0.25 [1 + (25/k)3]
(17)
For iodine we can make the approximation that the only significant
contribution is photoelectric because of the high atomic number of
iodine. Normalizing to the value of 36cm2/gm at 33 keV we get,
µI ≈ 36 (33/k)3
valid for keV > 33
(18)
From these expressions for the attenuation coefficients we get
(dµT / dk) = (0.25)(25)3 (-3)/(k4)
(dµI)/(dk) = (36)(33)3 (-3) / k4
Inserting into equation 16, we get,
dN / dk =[ e µt t (-2) / (µ3I) ](36)(33)3 (-3)/(k4)
+ [te µt t / µ2I] (0.25)(25)3 (-3) / (k4) = 0
Simplifying,
[2(36)(33)3] / (µI) = t (0.25)(25)3
Substituting for µI gives,
[2(36)(33)3] / [(36)(33)3/k3] = t (0.25)(25)3
solving for k we get,
k3 = t [(0.25) / 2] (25)3
k = 25 (0.25/2)1/3 t1/3
k = 12.5 t1/3
for k > 33 keV
(19)
For k < 33 keV the model does not apply. At 33 keV t = 18
cm.
For t < 18 cm 33 kev will minimize the exposure. For t >
18cm the above formula predicts that the optimal energy
increases away from the k-edge very slowly with thickness.
For example at t = 27 cm,
k =12.5(27)1/3 = 12.5(3) = 37.5 keV
This dependence is quite different from that of tissue where the
optimal energy for detecting small tissue contrasts was considerably
higher at the same tissue thickness, e.g.
CASE I (tissue detection): t = 20
k ≈ 100 keV
CASE II( iodine
detection):
k ≈ 34 keV
In our discussion of ideal systems we only considered quantum noise.
In real systems additional noise can be contributed by the imaging or
display components such as television cameras or film. Let us assume
that the imaging system transforms an input fluence to an output
signal as illustrated in Figure 11.
Input x-ray signal N
+ quantum noise
Imaging system
Output signal S
+ quantum noise
+ system noise
It is useful to define a quantity called Detective Quantum
Efficiency , DQE, which is a measure of how well the system
utilizes the incident x-rays to produce an image.
This concept includes the actual detection efficiency of the detector
which we will designate as .
The DQE is defined in terms of the input and output signal (S) to
noise (n) ratios S/n as
DQE = (S/n)2out / (S/n)2in
(20)
Note that in the above equation S represents the entire signal and
not the differential signal associated with any particular object.
The input signal is given by , the input fluence multiplied by the
characteristic area a of the system image element e.g. pixel (picture
element) size.
Sin =  = Na
(21)
The DQE we calculate will depend on the size of a. Choosing larger
a will increase the averaging of system noise and increase DQE.
In some analyses, DQE is actually plotted as a function of spatial
frequency, which will be discussed later.
For the purposes of this discussion we will just assume a fixed
fundamental image element size a.
The input quantum noise is given by the square root of the number of
x-rays in a.
nin =  1/2
22)
giving
(S/n)in = 1/2
(23)
Therefore,
DQE = (S/n)2out /  = NEQ / 

where NEQ is called the Noise Equivalent Quanta and represents the
number of photons an ideal system ( with DQE =1) would have to
detect to produce the same output S/n as the real system. Because the
real system detects only a fraction  of the incident x-rays and contains
system noise, NEQ is always less than the actual number of incident xrays.
In order to illustrate the role of detection efficiency and system
noise on DQE and the detection of objects in an image, we will now
consider the details of a video camera detector such as those used
for fluoroscopy or digital radiography and digital angiography.
In these applications, the transmitted image is detected by an image
intensifier, the output of which is viewed by a television camera
tube.
The output video signal is shown in Figure 12.
noise
Vm
blanking
Sync pulse
Superimposed throughout the video signal is noise (illustrated only
at the top of diagram).
The total noise is due to a combination of the camera noise Vc and
the electrical representation of the quantum noise.
We will look in detail at how these are combined a little later.
The dynamic range DR of the camera tube is defined as the ratio of
the peak signal Vm and the camera noise.
DR = Vm / Vc
(25)
Modern cameras typically have dynamic range values of 1000 or
more.
Cameras typically have a linear response over a certain range of
exposures.
In order to reduce the role of the video camera noise, the light
incident upon the camera tube is adjusted by means of an aperture
placed between the image intensifier output and the TV camera
input so that Vm is near the top of the linear range .
aperture
patient
Vm
video
signal
image
intensifier
tv
camera
large
aperture (low dose/image)
small
aperture (high dose/image)
detected X-rays/pixel
The relationship between camera signal and, the number of
detected photons per pixel (in this case determined by the
bandwidth of the electronics) is shown below.
The maximum signal Vm corresponds to the maximum
number of photons per pixel m in the linear range as shown in
Figure 13.
Saturation region
Vm
Camera
signal
noise
Detected x-rays per pixel m
Figure 13
The output voltage (not including noise) is related to the number of
detected x-rays by
V = CC 0 
(26)
where  is the detection efficiency.
The error in V due to quantum noise is given by
q=C 1/2
(27)
The x-ray to voltage conversion factor does not get included in the
square root because it is just a scaling factor.
Intuitively, it might be expected that q should be times the noise
in the incident x-rays rather than the square root of the number of
detected photons. Professor VanLysel will discuss the reasons for
equation 27 in the CT/DSA course
The error in V due to camera noise is given by
C = Vc = Vm/DR
(28)
The overall noise variance is the quadrature sum of these.
n2 = 2total = 2q + 2C = C2  + Vm2 / DR2
From equation 26 we have
C = Vm /  m
so,
(29)
(30)
The output signal to noise ratio is then given by equations 26,29 and
30 as,
(31)
Now we can calculate DQE. The input signal is given by

Sin = /
The input quantum noise is given by
(q)in = (/)1/2
giving
(33)
(S/n)2in =/
From the definition of DQE (eq.24) we obtain,
(34)
The m / DR2 term is a constant for a given aperture.
In other words for a given TV camera and aperture,m is
determined. Therefore the aperture is a means of adjusting the
exposure and therefore the amount of noise in the image.
For large apertures, the maximum signal is achieved with a small
x-ray exposure and quantum noise will dominate over camera
noise. (The percent noise is largest at low exposure).
For small apertures, a large exposure is required to achieve
maximal video signal and the fractional quantum noise fluctuations
will be small. In this situation, camera noise becomes a larger
factor.
Lets consider the minimal detectable contrast and DQE for large,
intermediate, and small video signals.
V= Vm, ( = m)
Assume DR = 1000, A = 1mm2, and that Edetected = 1 mR produces
Vm.
Then m = 2 x 107 x-rays/cm2 x 0.01 cm2 = 2 x 105 x-rays per pixel.
Plugging into equation 31 we get
Therefore
n/S = .0025, or n = .0025 S
The Rose criterion for signal detection says that the differential
signal S must be 3-5 times larger than n.
Taking S = 4n, the minimal detectable contrast is given by
Cmin = S/S = 0.01
i.e. , a 1% contrast can be seen at a 1mR detected exposure for a
1mm2 object in the brightest portion of the image.
From equation 34, assuming that the x-ray detection efficiency
 = 0.5, we get
V= Vm /10
In this case
giving
n/S = 0.012
and, according to the Rose criterion,S=4n ~ .05S
and
Cmin = S/S = 0.05
Therefore 5% contrast is required for detection at V = Vm / 10. The
DQE in this case is reduced to 17%.
V = Vm / 100
Using the same equations, it is found that the minimum required
contrast is increased to 40% and the DQE is reduced to 2.4%.
This points out the importance of removing bright spots by means
of bolus materials such as water bags or other devices.
Figure 14 illustrates the effect of bright spot removal.
Without bright spot removal, the video aperture or the exposure
must be adjusted so that the central bright spot must be at the top of
the unsaturated video range. This relegates the signal corresponding
to the artery, which is superimposed on a thick region, to the small
video signal regime where SNR and DQE are poor.
When the bolus material is added the transmitted image becomes
more uniform and can be brought up to the maximum video level
either by means of an exposure increase or enlargement of the video
aperture.
In either case, the larger arterial signal will compete more favorably
with the camera noise.
Without brightspot removal
With brightspot removal
“artery”
“patient”
video
“patient” + bolus
Amplified video
The bolusing operation in Figure 14 alters the large area contrast of
the image by removing the contrast of the area creating the brightspot.
However, in applications such as blood vessel imaging (angiography),
especially when subtraction techniques are used to remove nonvascular anatomy, the net effect is improved vessel signal.
As we will discuss later, the scatter fraction measured at the vessel
will also be reduced due to the reduction in scattered radiation from
the highly transmissive region.
Figure 15 provides another illustration of the role of bright spots in
setting up the peak video signal via aperture or mAs changes.
In Figure 15A there were lines of high intensity coming through
cracks in the phantom which consisted of a barium strip placed,over
a pressed board step phantom. When the peak video signal was set
using this large signal, the rest of the image information was placed
in the low signal amplitude portion of the image and noise became
dominant as shown in Figure 15A.
In Figure 15B the bright spots were simply ignored and the peak
video was set to match the rest of the image. This resulted in
improved signal to noise ratio for the barium strip.
Effect of Aperture
Bright Spots
at Peak Video
Bright Spots
Ignored
A further illustration of the effects of system noise is shown in
Figure 16 where in 16A a video camera has been operated with the
brightest portion of the video at peak video signal and in 16B the
brightest portion of the scene has been placed at 15% of peak
amplitude. This simulates the difference in image quality that
would be perceived using cameras differing in dynamic range by a
factor of 6.6.
100%
15%
Figure 16
A
B
As x-ray exposure is increased, fractional quantum noise is reduced
relative to camera noise and image quality is improved. It is
important to appreciate that as exposure is increased the quantum
noise, which is the square root of the detected exposure, increases in
absolute value. However, when the optical aperture is reduced to
keep Vm fixed, the percentage of the video signal occupied by
quantum noise is reduced. Since the camera noise is not affected by
the aperture, it becomes more dominant at increased exposure and
further increases in exposure produce smaller gains.
Consider the behavior of the DQE at maximum video signal.
Consider the case of
V = Vm.
From equation 31 for the signal to noise ratio we get
Figure 17 shows the SNR at Vm for two different
camera dynamic ranges.
The 1000:1 camera has low noise and the 250:1 camera
has high noise.
In the case of DR = 250 where system noise becomes
dominant very quickly as exposure is increased, it makes
little sense to increase exposure beyond 0.5 mR per
video image.
For DR =1000, image quality continues to improve with
exposure, leveling off at much higher exposures.
Figure 17
DR1000
1000
DR
DR = 250
For modern image intensifier-television based digital
angiography systems such as used for digital subtraction
angiography (DSA), detected exposures on the order of
0.5 mR are common.
For cardiac cine’ angiography where 30-60 frames per
second are typically recorded on film or real time digital
recording devices, the exposure per frame is typically .05
mR per frame and the images are dominated by quantum
noise.
The exposure dependence of the DQE is illustrated in
Figure 18 for V =Vm. In this case DQE is given by
Figure 18
DR = 1000
DR = 250
Notice that for the DR =250 camera, DQE falls off
quickly with exposure signifying decreased gains in
image quality as exposure is increased.
Because of the large camera noise component additional
x-rays are less inefficiently utilized than in the DR = 1000
camera.
For the previous decision we have looked at the
maximum video signal.
The lower signal portions of the image will already have
lower DQE at any given exposure and will improve SNR
less rapidly than the maximum signal region as exposure
is increased.
However, it is usual to try to set the exposure based on
the large signal region of the video signal and try to use
bolus materials to make the video signal distribution as
uniform as possible.
The quantum sink of an imaging system is defined as the point in the
system where the signal corresponding to quantum fluctuations reach
their largest fraction of the total signal.
Ideally, the quantum sink should be at the front end of the system
where the x-rays are detected. In a well-designed system, the relative
importance of quantum fluctuations due to x-rays or, for example,
optical photons produced by the x-rays, should not become more
dominant in a fractional sense than they are at the point of x-ray
detection. Let’s look at two examples to illustrate this concept.
Image-intensified fluoroscopy
Figure 19
The number of detected x-rays for an area A is given by ( = NA)
(35)
Following detection, the x-rays produce large numbers of electrons.
The number of electrons per x-ray is sufficiently large that the
quantum fluctuations are completely dominated by the x-ray
fluctuations.
When these electrons impinge on the output screen of the image
intensifier, light photons are produced.
The number of these associated with each incident x-ray is still
sufficiently large that the fluctuations in the number of light photons
subsequently presented to the video camera are dominated by the
initial fluctuations in the x-rays detected at the input screen.
The combination of electron production and subsequent photon
production leads to an overall gain factor G, giving a number of
photons associated with input screen area A of
(36)
The signal to noise in and the signal to noise out are both equal to
Because of the large numbers of electrons and photons produced,
the quantum sink for the case of image intensified fluoroscopy
remains at the input of the system as it should.
Non-Intensified Fluoroscopy
Before the introduction of image intensified fluoroscopy as shown in
Figure 20, radiologists looked directly at the output of a fluorescent
screen detector.
This had several disadvantages. Because of the low light levels, dark
adaption was required and rod vision, with its lower visual acuity, had
to be employed.
Furthermore, as we will see, the quantum sink moved from the input
screen to the eye of the observer, indicating that the information
perceived was less than that available at the fluorescent screen.
Figure 20
0
[ 0 ± √ (0)]• g
R
screen
Eye
The factor g corresponds to the number of light photons
per detected x-ray.
Since this is large, the quantum fluctuations after the
screen are dominated by the x-ray quantum statistics as in
the case of the image intensifier.
However, following the conversion to light, which is
spatially isotropic, the eye, which we will assume has
area Ae, intercepts only a fraction of the total light
photons.
This fraction is equal to the ratio of Ae to the area of the
sphere of radius R.
Therefore, the number of photonse detected at the eye is given by
(37)
This number can be less than one per x-ray depending upon the
brightness of the screen.The signal to noise ratio is (if # photons/xray is still large)
at best (38)
which is the SNR after the detector multiplied by the second square
root factor .
Therefore the quantum sink moves from the screen to the eye
whenever
(39)
(in which case S/n is worse than eq. 38)
So far we have discussed two general sources of noise, quantum
noise and system noise. In many systems the analog signal
produced by the system is digitized. This is the case in DSA,
phosphor plate imaging, computed tomography (CT) and many
other applications.
Since the digitization process also introduces a sort of noise, it
will be useful to break the system noise into its analog component
a, including quantum noise and analog system noise, and the
digital component d.
Digitization is accomplished by an analog to digital (A/D)
converter, which transforms the analog signal into 2N levels where
N is the number of digital bits in the output of the A/D converter.
For example, for an eight bit converter the analog signal is
converted to 256 discrete levels from 0 to 255 as shown in Figure
21 for the case of a single horizontal line of video which has been
chosen to have a monotonically increasing (ramp) signal for the
purposes of illustration.
The analog video signal Va is on the left. The digitized signal Vd is
on the right.
Figure 21
Analog
Digital
The digital spacing, which we will designate as  , must be chosen
on the basis of the amount of analog noise in the system as we will
discuss below.
The assignment of a given analog voltage to a digital level is made
on the basis of voltage thresholds.
As illustrated in Figure 22, if the analog voltage is equal is greater
than or equal to /2 above the previous level, it is assigned to the
next digital level.
Because of this the digital representation of the image is discrete
and in some circumstances can have a blotchy appearance in which
large areas of the scene get caught below a threshold and then, at
some adjacent point in space, abruptly make the transition to the
next level.
Figure 22
i+1
u

Analog signal
i
i-1
Digital levels
Thresholds
We can calculate the RMS error associated with the
digitization process by considering the error associated
with each digital level, indicated by u.
If we assume that the analog signal has equal probability
of falling between two digital levels, then we can
calculate the RMS digital error dig by integrating over u
as follows.
The total noise variance including analog and digital noise is
given by the usual quadrature sum
(41)
The ratio of the total noise to the analog noise as a function of  /a is
shown in Figure 23.
For small digital level spacing, the digitization process adds little
noise to the analog noise. As the level spacing increases, the
digital noise can greatly increase the total noise. Usually the
criterion that
(42)
is chosen to ensure that the digitization process increases overall
system noise by less than 20%.
The number of bits needed to satisfy the above criterion depends on
the system analog noise and the range of signals to be digitized.
Referring to our video camera example, suppose that we wish to
digitize the video signal between the levels Vm and Vm/10. It is
common to choose such a lower level below which one might not
hope to see much anyway because of the dominance of noise in the
lower signal regions. For the case discussed earlier with a detected
exposure of 1 mR , a 1mm2 pixel area and Vm/10, we had
a = .012 ( Vm / 10 )
If we use the criterion that
 = 2a= .0024 Vm
Then the number of digital levels #d required is given by
Since 28 = 256 and 29 = 512, this requires a 9 bit A/D converter.
Notice that the level spacing criterion must be applied
where a is the smallest, in this case at Vm / 10.
If the criterion is satisfied in the small signal portions of
the image, the criterion will be satisfied in the large signal
regions where the absolute size of the noise a at Vm is
larger because of the larger absolute size of the quantum
noise contribution.
These concepts are illustrated in Figure 24, which shows
a series of cardiac images digitized with increasing
numbers of bits. For the lower numbers of bits, the signal
in the dark portion of the image appears blotchy due to
the fact that the level spacing is too large.
As the number of bits is increased, this blotchiness
disappears. Notice that in the bright regions of the image
there is no blotchiness because the absolute size of the
analog noise in these regions is large enough to satisfy
the level selection criterion.
Basically the noise helps signals jump between levels so
that large areas do not get stuck in a single level over a
large spatial region.
Figure 24
ADDED NOISE
NONE
0.8 %
3.1 %
12.5 %
50 %
BITS
8
7
5
3
1
Figure 25 illustrates the effects of having an insufficient number of
digital levels for the case of a digital chest film.