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NOISE, DYNAMIC RANGE, AND TONAL RANGE
Noise. That's where the controversy over Equivalence
begins. People think that the "equivalence argument" is based on the
presumption that Equivalent images have the same noise. Even
different cameras from the same format will have not have the exactly
the same noise,
either in quantity or quality, so, clearly, neither will Equivalent images
from different formats. However, for sensors of the same generation,
Equivalent images will usually be fairly similar in terms of noise,
and certainly more similar than images from different formats with the
same exposure (same f-ratio and shutter speed).
Furthermore, it is of utmost importance to distinguish between noise at
the pixel level, and noise at the image level. Since pixels are the
building blocks of digital images, this section begins with the discussion
of noise at the pixel level, followed by a discussion of how these pixels,
in aggregate, relate to the image as a whole (beginning
here).
When people refer to noise in an image, what they mean is what
this essay calls "apparent noise", which is the density of
the noise in an image (NSR -- noise-to-signal ratio)
and is often represented as a percent. Often, we hear the term "SNR"
(signal-to-noise ratio), which is the reciprocal of the NSR (SNR = 1 /
NSR). However, since a "noisy image" corresponds to a high NSR, and
a clean image corresponds to a low NSR, whereas it is exactly opposite for SNR, it
is less confusing to think in terms of NSR than SNR.
In terms of
photography, there are two
principle sources of noise: photon noise and read noise.
Except for the deep shadows in an image, the
photon noise is the primary source of noise in an image. It is
an inherent characteristic of incoherent light
(the kind of light in almost all situations -- see the diagrams at the
bottom of
this page),
and unavoidable -- one of those "Laws of Physics" things, as opposed to "an
engineering challenge". Light has the properties of both a particle (photon)
and wave,
and the noise is measured in terms of its particle characteristics.
The photons are collected and focused by the lens onto the sensor, where
they are converted into electrons, and the signal is processed and
recorded. The only
role the sensor plays in the photon noise is what proportion of the photons
falling on the sensor are converted into electrons, since the electrons
are the source of the electrical current that is processed by the hardware. The
read noise, discussed in more detail further down, is
how much noise is added when collecting and processing the signal produced
by the photons.
There are, of course, other sources of noise, such as thermal noise, which
plays a central role in long exposures, PRNU (Pixel Response
Non-Uniformity) noise, which plays an important role in the highlights of
the image, as well as other sources of noise. So, noise, is, of
course, even more complicated than this essay makes it appear, and for
some specific forms of photography (such as astrophotography) we may find
that the noise is very different for equivalent images in some situations,
much in the same way that corner sharpness is very different for
equivalent images in some situations.
Let's begin with an analogy to traffic to understand
photon noise. Imagine we drew a line across a busy freeway, and
counted cars crossing the line. The number of cars crossing the line
in any given time interval represents the total light
falling on the sensor during an exposure. If we are talking about short intervals of
time, like seconds, or, at most, minutes, then we can safely assume that
there is a constant average flow of traffic. But we also know
that it is very unlikely that any two equal time intervals will contain
exactly the same number of cars. This variation from the "true
average" is what we call "noise". The larger the time interval, the
larger the noise will be, but the less significant it will be in terms of
the total number of cars counted (apparent noise).
For example, let's say that, on average, 10 cars pass by
our line every second, and let's say we take three one-second samples that
come up with 8, 11, and 13 cars. These three "photographs" would
have a noise of 2, 1, and 3 cars, respectively, which correspond to an
apparent noise of 20%, 10%, and 30%. Now, let's say we extend the
time interval to ten seconds. The expected number of cars would now
be 100 cars (10 cars / second · 10 seconds =
100 cars). Let's say that, once again, we take three "photographs"
and count 93, 98, and 112 cars. The noise is now 7, 2, and 12 cars
-- much more than before. But the apparent noise is 7%, 2%, and 12%
-- much less than before.
So, the question is, then,
how do we know what the "true average" number of cars is? Well, we
don't. But what we do know is that the
arrival of photons for incoherent light is described by a
Poisson Distribution,
that the
standard deviation
for phenomena that is described with a Poisson Distribution
is equal to the square root of the mean (average), and that the standard
deviation is the photon noise (often
called the
Shot
Noise, which is a more general term). Thus, since the magnitude of the noise is equal to the square root of the
number of recorded photons, the noise increases with more light. But
since the NSR is the ratio of the noise and the recorded signal, the
apparent noise decreases with more light. For example,
whenever we double the amount of light, the photon noise increases by 41%,
but the apparent photon noise decreases by 41%.
A good way to visualize the role the lens and sensor play
in photon noise is to think of rain falling
on a flat surface through an opening. The size of the opening
corresponds to the
aperture area, and length of time the rain falls corresponds to
the shutter speed. If a lot of rain is falling (lots of light), then the surface
will quickly be covered in water having a very smooth
appearance (low apparent noise). However, let's imagine a much lighter rain
(low light). At first,
we will see splotches of water here and there in a random and irregular
pattern (noisy image). As more water
falls (the total amount of light increases), either by letting more time
pass (longer shutter speed) and/or by making the opening larger (larger
aperture area), the pattern becomes smoother
(less noisy).
Let's now pack a large number of cups on the surface to
collect the water. The cups are analogous to the pixels on the
sensor. Larger cups will collect more water than smaller cups, but
smaller cups will give us a better idea of the pattern of the rain.
If we compare an array of cups covering the same area, the water in the
larger cups will be more uniformly filled (less noisy) than the smaller
cups. This is why photos from sensors with larger pixels appear less noisy than
sensors with smaller pixels. However, the amount of water in the
smaller cups gives us a much better idea of the pattern of rain that has
fallen (resolve greater detail). If the resulting "image" formed by
the water in the cups is too noisy to our liking with the smaller cups, we
could replace the smaller cups with larger cups and pour the water
collected in the smaller cups into the larger cups, achieving the same
smoothness we had if we had used the larger cups from the beginning
(binning). Alternatively, we could siphon water from one cup
and add it to an adjacent cup to smooth the appearance of the "image".
This method of smoothing (NR -- noise reduction) would retain more
of the detail of the original pattern than if the smaller cups were just
poured into bigger cups (binning).
There are, of course, important considerations. How
closely are the cups packed together? How deep is each cup?
How thick is the glass in each cup? These concerns are all analogous
to the efficiency of the pixel, and has much to do with whether or not a
sensor with more pixels can accurately achieve the lower noise of a sensor
with fewer pixels via binning. For the same technology, it appears
as though this is very much the case. Thus, smaller pixels offer
more options of detail vs noise than do larger pixels. For photos
composed with a large amount of light, smaller pixels basically resolve more detail,
and, while more noisy, still "clean enough" to where the additional detail
likely contributes far more to the IQ of the photo than the greater noise
detracts from the IQ.
It is only for images, or portions thereof, that are created with little
light that sensors with smaller pixels much choose, via post processing,
to have greater detail and more apparent noise, or less detail with less
apparent noise.
Of course, this choice only comes to play if we display
the photo large enough and/or view it closely enough, that we can resolve
the additional detail afforded by more pixels. Otherwise, we do not
need to do anything at all -- the photos would look the same, whether they
came from sensors with large, or small, pixels.
However, it is not merely the total
amount of light that falls on the sensor, but also how efficient
the sensor is. The primary attributes of sensor efficiency are the
QE (Quantum Efficiency -- the proportion of photons falling on the
sensor that are converted into electrons) and the read noise (the
additional noise added by the sensor and supporting hardware). A
Bayer CFA is usually RGGB which means that 25% of the pixels are covered
with red filters, 50% are covered with green filters, and 25% are
covered with blue filters. Of course, the filters actually accept
a range of colors, which overlap (otherwise, yellow photons, for
example, would never make it through the color filters). How much
the filters overlap, and how strong the filters are, also contribute to
the sensor efficiency. For example, if the color filters are weak,
they are subject to more transmission error (when the wrong color photon
passes through), which reduces luminance noise, but increases color
noise. Thus, the QE for the red, blue, and green filters is likely
different (how different, I cannot say, as I have no data on this
matter, but I don't think it's by more than 10% based on transmission
graphs that I've seen).
In addition, the read noise often varies considerably as
a function of the camera's ISO setting, usually getting progressively
less with higher ISOs ("ISOless" sensors have a relatively constant read
noise as a function of the camera's ISO setting). In addition,
some sensors have a base ISO of 100, whereas others have a base ISO of
200. Sensors with a base ISO of 100 can absorb twice the light at
that setting than sensors with a base ISO of 200. This brings up
the saturation capacity of the pixel, which is the number of electrons
that it can release. Both read noise and saturation capacity are
only meaningful, in terms of comparative IQ of the photos between
systems, when taken on a per area basis. For example, consider a
10 MP and 40 MP sensor of the same size and with the same QE. If
the pixels of the 40 MP sensor have 25% the saturation capacity and 50%
the read noise, then they would be "equally efficient", since a four
pixel block (same area as a single pixel of the 10 MP sensor) would
record the same number of photons that landed on it, have the same total
saturation, and the same total read noise.
Lastly, the issue of banding needs to be discussed.
Banding is not noise, since it represents a systematic bias in the
output signal, whereas noise is completely random. In that sense,
banding is usually significantly more a detriment to IQ than is noise,
depending, of course, on how serious the banding is.
For
example, the Canon 5D has a QE of 25%, the Canon 5DII has a QE of 33%, and
the Nikon D3s has a QE of 57% (see
here, but keep in
mind that these figures are for the green channel only, and I am
assuming that there is no significant difference in the other channels,
and that the transmission error for the color filters is "close"). What this means is that the Nikon D3s
will record just over a stop more total light for a given exposure
than the 5D, and about 2/3 than the 5DII. The maximum possible QE is, of
course, 100%, which is just under a stop more than the Nikon D3s. Of
course, that doesn't mean that a 100% QE will record all the light falling
on the sensor -- the Bayer CFA, recording only one color per pixel, will
only pass, at best, 1/4 of the red and blue light, and 1/2 the green light
(since most Bayer CFA filters are RGGB). There is also the matter of
light absorption by the color filter itself, as well as allowing the wrong
color light to pass through (color noise).
The read noise (R), is the sum of the pixel noise (P)
and electronic noise (E). The total
noise (N) is the sum of the photon noise (p) and the read noise (R).
Since noise is a standard deviation, it does not sum linearly -- that is,
R ≠ P + E. For independent random phenomena, like noise, it is
instead the variances (the squares of the standard deviations) that sum
linearly: R˛ = P˛ + E˛. Thus, the read noise is the square
root of the sum of the squares: R = sqrt (P˛ + E˛). We
can represent the total noise (N) as the sum of the photon noise and read
noise, N = sqrt (p˛ + R˛), or as the sum of the three components discussed
(photon noise, pixel noise, and electronic noise): N = sqrt
(p˛ + P˛ + E˛).
For example, the
Nikon D3X pixels have a FWS of 48975 electrons.
For the sake of discussion, let's say that the output voltage of the pixel
is 1V at full saturation. Then one electron would correspond to
1/48975 V ~ 20.4 µV. Thus, a
pixel noise of 61 µV would correspond to 61
µV / 20.4 µV ~ 3.0 electrons. Finally, other signal
processing, such as the conversion of the analog signal into a digital
signal, add in additional electronic noise. Again, for example, if
the electronic noise were 44 µV,
then it would correspond to 44 µV /
20.4 µV ~ 2.2 electrons. The
total read noise can be computed as sqrt (3.0˛ + 2.2˛) ~ 3.7 electrons (as
opposed to the 5.2 electrons we would get for a simple linear sum).
If
250 photons fell on the pixel, and the QE is 40%, then the signal is 250 · 0.4 = 100 electrons. Thus, the
photon noise would be p = sqrt (100) = 10 electrons and would result in a
combined noise of N = sqrt (10˛ + 3.0˛ + 2.2˛) ~ 10.7 electrons for the
pixel, which is considerably less than the 15.2 electrons of a linear
sum, and gives an NSR of 10.7 / 100 ~ 10.7%.
A common myth is that
higher ISOs cause more noise. In fact, what causes the increase in
noise at higher ISOs is the lower exposure. That is, when we
increase the ISO, the camera will either increase the shutter speed, close
down the aperture, or both. The effect of this is that less light
falls on the sensor, and it is the lesser amount of light falling on the
sensor that increases the noise, not the higher ISO, per se.
Unlike film, the
sensitivity of the sensor is fixed. The effect of the ISO is merely
to apply a
gain to the signal that, for some sensors, is a bit more efficient than a
digital push (if the gain is analog as opposed to digital -- see
here
and
here for demonstrations) and results in less
noise (at the possible expense of blowing more highlights). If higher ISOs
were not more efficient than base ISO, then there would be no point in
having higher ISOs, except for the convenience of not having to push to
the desired brightness in post. The camera would only have base ISO, and we
would simply choose the appropriate f-ratio for the DOF and/or sharpness
we desire, and the appropriate shutter speed to account for motion blur
and/or camera shake. The exposure meter would tell us how over/under
exposed we are (less exposure means more noise, more exposure means more
blown highlights), at which point the photographer could adjust the
balance of the f-ratio and shutter speed with the exposure to achieve the
exposure that gave the best balance of DOF / sharpness, motion blur /
camera shake, and noise / blown highlights. We would then
push the image digitally to whatever brightness we desired. This, of course,
represents an ideal situation, but, in fact, is a reality now with some
sensors, such as the new Sony sensor in the
Nikon D7000 and Pentax K5 (see
here for a demonstration).
It's more than worthwhile to note that the
photon noise will dominate the image in the midtones and highlights, so
the read noise plays little role except in the shadows. Many systems
deal with shadow noise simply by clipping them earlier, which, of course,
removes the detail as well. However, much more important than the
shadow noise itself is banding, which is most noticeable and distracting
in the shadows due to the fact that banding is a regular pattern as
opposed to the random nature of noise. For example, a sensor with
low read noise and banding may well produce a significantly worse image in
the shadows than a sensor with more read noise without the banding.
That said, the issue of banding is separate from the issue of noise, and
not discussed in this essay as banding has nothing to do with sensor size.
However, there can be a more common disparity in noise due to the CFA (color
filter array). Color noise arises from the fact that the color filters
are not perfect. For example, the green filter may only admit 60% of
the green light that falls on it, but also admit 10% of other colors that
fall on it. If we use a weaker filter to increase the transmissivity,
we will reduce the luminance noise (more total light will pass through the
filter and onto the sensor), but also increase the transmission error by
concomitantly allowing a greater percentage of other colors to also pass
through. Thus, different manufacturers may strike different balances
between luminance noise vs color noise in their choice of color filters in
the same way they strike a different balance between pixel noise and
electronic noise, as in the D3X and 5DII.
In short, the quantity of noise in a photo is determined by
the following three factors:
• The total amount of light
that falls on the sensor
• The efficiency the sensor captures the
light
• The additional noise added by
the hardware
It is also important to note that the processing has a
huge effect of the apparent noise in a photo, but it must be noted
that NR (noise reduction) also removes detail. However, depending on
the NR algorithm, the resulting photo can range anywhere from a horrid
smear to appearing just as detailed but less noisy. Any NR algorithm
will remove detail. But, depending on what detail is removed, and
what detail is left intact, the results can vary drastically, especially
depending on the scene. In fact, it is possible that some NR (noise
reduction) engines may even introduce false detail to replace the detail
lost in the NR process, as do some upsampling algorithms, to give the
appearance of less noise with no, or little, lost detail. In addition, it needs to be noted that the
amplitude (quantity) of noise depends on the frequency (detail) at which
the photo is viewed, which is discussed in more detail
further down in this section.
For a given
scene, perspective, and framing, the total amount of light is
determined solely by the aperture diameter (where
aperture
diameter
= focal length / f-ratio) and the shutter speed. The role the sensor size plays in total
light is that a larger sensor can absorb more light before oversaturating. By using a larger
aperture, we also force a more
shallow DOF, because the
aperture
plays an integral
role for both total light and DOF. So, if a more shallow DOF is not
desirable (and/or the concomitant effects that can occur, such as softer
corners and more vignetting), then the only way to decrease noise is to
decrease the shutter speed, at the risk of overexposing more of the image
(this technique is known as ETTR -- expose to the right), and works best
for images with a narrow DR (dynamic range). The efficiency by which this light is captured is
a function of many variables, primarily the CFA,
the efficiency of the microlens covering, the percentage of the light
transmitted by the color filters, and the QE (quantum efficiency) of the
sensor. The efficiency of the signal amplification is a function of
not only the sensor, but the supporting hardware, such as the ADUs
(Analog-to-Digital Units).
In terms of the total signal, the amount of light lost by
traveling through the glass of the lens is insignificant, so there is no
improvement to be made there. The microlens covering over the
sensor, which directs the light that falls on the sensor into the pixels
is also near 100% efficiency for modern cameras, so, again, there is no
improvement to be made there, either. However, for Bayer sensors,
each pixel records only one color (usually 25% red, 25% blue, and 50%
green), which results in around another stop of light lost. Modern
DSLRs have a QE of anywhere from 30% - 40%, so there is about another 1.5
stops more gain that can be made there. Thus, modern DSLRs with
Bayer CFA's are around 2.5 stops away from the maximum possible
improvement in photon noise, with one more stop for a scheme that does
not require color filters.
While alternative schemes from Bayer may one day be able
to give us this extra 1-2 stops improvement in photon noise and possibly render
double the detail for a given pixel count as well, so far these
technologies have run into significant problems of their own, such as the
Foveon sensor where each pixel records three colors as opposed to one.
While ostensibly a very good idea, and one that may eventually take hold,
in its current implementation it is actually more noisy than a Bayer CFA.
Another scheme is to use prisms or dichromatic mirrors to direct the
different colors to three sensors ("3CCD"
video cameras, for example), but these alternatives currently have
significant technical problems of their own to overcome.
So far, the discussion of noise has been focused
on noise at the pixel level. However, as we saw above when computing
the tonal range, calculating values at the image level will give us a
different result, and this result is what describes the visual properties
of the final photo. Most believe that because a larger
pixel gathers more light for a given exposure, that larger pixels result
in less apparent noise. However, for a given sensor size, the
smaller the pixel, the more pixels you have. So, while a larger
pixel will have less noise than a smaller pixel since it gathers more
light for the same exposure, the image as a whole will be made from
the same total amount of light regardless of the number of pixels.
This is not to say that the number of pixels has no effect on the apparent
noise in an image. In terms of the IQ of the final image, we are
best served by viewing noise as a vector measure (multiple components)
rather than a scalar measure (single valued). In other words, noise has both an
amplitude and frequency. The amplitude of the noise is the standard
deviation from the mean signal, and the frequency of the noise is the
number of samples taken. The higher the amplitude of the noise, the
lower the IQ. The higher the frequency of the noise, the higher the
IQ. For a given sensor size, smaller pixels
will result in both a higher amplitude and higher
frequency of noise. The problem with noise comparisons on sensors with
different pixel counts is that people often compare noise at different
frequencies by comparing 100% crops rather than equal areas of the image
displayed at the same dimensions.
Consider, for example, a 10 MP sensor and a 40 MP sensor of the same size
and efficiency. If we displayed the photos large enough, or viewed
them closely enough, that we could resolve the individual pixels, then the
40 MP photo would appear more noisy, but also be more detailed (see
here for an example of this point).
In other words, the more detailed image will have more noise at
higher spatial frequencies because it can resolve it. A good
way to think of this is that the noise from the scene already exists, and
the closer you look, the more clearly you see it:
A sensor with more pixels will more clearly resolve the noise that already
exists, whereas a sensor with fewer pixels will simply blur
the existing noise.
On the other hand, if we cannot resolve the individual pixels, then the 40
MP photo would not appear any more noisy than the 10 MP photo. But
neither would it appear more detailed, since we could not resolve the
individual pixels. So, what's the point of 40 MP over 10 MP if 40 MP
will necessarily be more noisy if we can resolve the additional detail?
The answer is rather simple: detail often matters more than noise.
More than that, even if the detail cannot be resolved, it may indirectly
affect other elements of IQ in the photo, such as tonal gradations,
microcontrast, smoothness, etc.
Of course, all this presumes that the 40 MP sensor is just as efficient as
the 10 MP sensor. While the pixel counts have steadily risen, and
the sensor efficiencies have steadily improved, many feel that if the
engineering efforts went entirely into improving efficiency, rather than
divided between increasing pixel count and improving efficiency, we'd have
more efficient sensors still. However, there is no evidence to
support this assertion for CMOS sensors, but may be true for CCD sensors.
Let's work an example with two hypothetical sensors with the same size and
efficiency (and, for the sake of simplicity, assume QE = 100% for each),
but the pixels for Sensor B are half the size (1/4 the area) as the pixels
of Sensor A. Without any loss in generality, let us simplify the
comparison to consider 1 pixel from Sensor A and 4 pixels from Sensor B,
since they will each cover the same area of the sensor (and resulting
image). Let's say the FWC for the pixels in sensor A is 80000
electrons. Then the FWC for the pixels of Sensor B will be 1/4 as
much, since they have 1/4 the area -- 20000 electrons. Likewise, if
the read noise for Sensor A is 8 electrons, then the read noise for Sensor
B will be 2 electrons. The LMS for a pixel on Sensor A will be
[1 + sqrt (1 + 4·8˛) ] / 2 ~ 8.5 electrons,
and the LMS for a pixel on Sensor B will be [1 + sqrt (1 + 4·2˛)
] / 2 ~ 2.6 electrons. Thus, the DR for Sensor A will be log2
(80000 / 8.5) ~ 13.2 stops and the DR for
Sensor B will be log2 (20000 / 2.6) ~ 12.9 stops --
virtually identical.
If we assume, then, that a pixel on Sensor A receives a signal of 64
photons, then, also assuming relatively uniform illumination, each pixel
of Sensor B will receive a signal of 16 photons (1/4 as much). This
will result in a photon noise of sqrt 64 = 8 electrons for Sensor A and
sqrt 16 = 4 electrons for Sensor B. Thus, the total noise for
a pixel of Sensor A will be
sqrt (8˛ + 8˛) ~ 11.3 electrons, and
the total noise for a pixel of Sensor B will be
sqrt (4˛ + 2˛) ~ 4.5 electrons. This
will result in an apparent noise of 11.3 / 64 ~ 18% for each pixel on
Sensor A and 4.5 / 16 ~ 28% for each pixel on Sensor B. Thus, the
price for the increased detail on Sensor B is more apparent noise.
In other words, if we displayed the image from Sensor B with double the
dimensions as the image from Sensor A, and viewed from the same distance,
it would appear to have up to double the linear detail (depending on the
performance of the lens and the effects of diffraction softening), and 56%
more apparent noise.
If, however, we instead downsized the image from Sensor B to the same
output dimensions as Sensor A, then the effect would be the same as if we
binned the pixels together. Thus, the signal for four binned pixels
would the same as one pixel from Sensor A, and the noise for a "binned
pixel" on Sensor A would be sqrt (4.5˛ + 4.5˛
+ 4.5˛ + 4.5˛) ~ 9 electrons. This gives us an apparent noise of 9 /
64 ~ 14% which is actually lower than the apparent noise of a pixel from
Sensor A! However, even more effective would be to instead apply NR
(noise reduction) to the more detailed image, as this process is more
efficient than binning in trading detail for a cleaner image.
Regardless, we need to keep in mind that more pixels do not result in more
noise, but rather show the noise more clearly along with the extra detail.
In more simple terms, it meaningless to
discuss apparent noise without considering the detail of the image.
Just as it makes since to compare the sharpness and detail of images at
the same output size, it only makes sense to compare the noise in images
at the same level of detail. In other words, it makes no sense to
say that one image has less noise than another, when it also has less
detail, since NR (noise reduction) can be applied to the more detailed
image to get a cleaner image at the expense of detail.
Thus, for a meaningful
comparison of apparent noise between images, we would first apply NR to the more
detailed image until it matches the level of detail of the less detailed
image. Interestingly, this means that the sharpness of the lens has
an effect on apparent image noise since a sharp image is able to withstand more NR
than is a soft image. In other words, it is not merely the
efficiency of the sensor, but the
relative sharpness of the lens that has a
major impact on the apparent noise in a photo.
Even more important, in some instances, is
focus accuracy, since even a small focus error can often lead to a
significant loss of sharpness (please see
this article). In any event, it is often more an issue of how the noise is processed,
rather than the quantity of noise, that is the primary issue. For
example, some manufacturers may select a higher black point that gives
cleaner shadows, but also destroys all detail. So, we must take care
to once again consider apparent noise vs detail.
All that said, while we often speak of the
apparent noise of Camera A vs Camera B, what many
overlook is that it is not merely the quantity of apparent noise, but the
quality of
the noise, that is important. It
is not only possible, but likely, that one image may be have more apparent
noise than another, yet have
a much more pleasing look due to the quality of the noise. For
example, color noise is usually much more distracting
than luminescence noise. In addition, higher frequency apparent noise
(finer grain) with the accompanying greater detail is usually
considered much more appealing than lower frequency apparent noise (coarse
grain) with less detail. In other words, a more noisy
image with a finer "grain" may well look better
than a less noisy image with a clumpier "grain", depending on
how close the overall quantities of noise are and the differences in
detail rendered (an excellent demonstration of
this is given here
and here).
To this end, having more
pixels, even at the expense of more apparent noise, can lead to a more appealing
overall image, but this is most certainly subjective. Of course, if
the more detailed image has the same, or even less, apparent noise than the less
detailed image after NR is applied to match the level of detail, then the
system with the more detailed, yet more noisy, image will have a
substantial IQ advantage by being able to better balance noise and detail
in post. Regardless, it is important to consider the types of images where
apparent
noise is even an issue. This, of course, depends greatly on both the
QT (quality threshold) of the viewer which is strongly influenced by print
size and the viewer's "noise floor" -- that is, the point at
which less apparent noise has no noticeable impact on the IQ of the image.
For example, while an ISO 100 image from 35mm FF has less apparent noise than an
ISO 100 image from 4/3, the advantage in apparent noise of 35mm FF may be unnoticeable
to the viewer at ISO 100. Of course, the "noise floor" is
likely a function of the print size and viewing distance as well.
For example, the apparent noise in an image may not be distracting in a 5x7 print,
but become an issue in a 12x18 print. Furthermore, the impact of the
apparent noise is greatly dependent on the scene. The apparent noise
may go overlooked in areas with lots of detail, but stick out in areas
with low detail, such as sky noise.
In addition to the mere quantity of apparent noise, we have to
consider the balance of apparent noise in the different color channels, which is a
function of the CFA (color filter array) that is used on the sensor.
One image may be less noisy than another overall, but exhibit
significantly more apparent noise in one of the color channels which will give it a
less appealing overall look. Furthermore, both photon and read noise
are completely random which makes for a significantly more pleasing
appearance than banding which has a regular pattern. In other words,
while apparent noise is most certainly an important consideration in the IQ of an
image, the quantity of the apparent noise most likely is less important than the
quality of the apparent noise.
Lastly, we have to take into account that different RAW
converters and JPG engines will deal with noise differently, so the
noise/detail present in the final image is not necessarily an accurate
representation of the actual hardware. NR (noise reduction) can be
applied even with a setting of "0", blacks can be clipped early to hide
shadow noise (albeit at the cost of erasing detail), etc. In fact,
even the same RAW converter might use different settings for different
cameras since the programmers decided that such-and-such a look was
"better". So, for sure, it is the final image that matters.
But the final image is not necessarily representative of the capability of
the hardware. In fact, many compare images on the basis of in-camera
jpgs, which, of course, is a very poor way in which to compare the
hardware performance. But it is the best way to compare if you shoot
jpg!
Thus, the advantage in apparent noise
of larger sensor systems is limited to situations when they can use a lower shutter
speed than the smaller sensor systems, such as good light, tripod use where
motion blur is not a factor, flash photography when the balance of the
light from the flash and the ambient light is not an issue, or when a more
shallow DOF is used by trading f-ratio for ISO. And, once again, all
these factors only matter if we are talking about sensors that have the
same, or nearly the same, efficiency.
Regardless, it is likely that it is the quality of the apparent noise, more so
than quantity of the apparent noise, that is the primary factor in distinguishing
between the IQ of Equivalent images in terms of apparent noise.
Just
as with any element of IQ, apparent noise is very subjective, and different people will
reach different conclusions about which image is more pleasing, even if the
numbers clearly point to one image or the other as having more overall
apparent noise.
Dynamic Range
The
Dynamic Range (DR) tells us the maximum range of
light levels where we can record detail, and is the number of stops from
the noise floor to the saturation point: DR = log2 (Saturation
Point / Noise Floor). The saturation point is the maximum number
of electrons that can be released, and the noise floor is the smallest
number of electrons that can be released for detail to be recorded.
For digital cameras, the DR is a pixel level measure where the noise floor
is the read noise of the pixel, and the saturation
point is the saturation limit for a pixel: DR = log2 (Pixel
Saturation / Read Noise). However, a problem with
this definition is that for a perfect sensor with zero read noise, the
DR would be infinite. Thus, a more meaningful noise floor, in
terms of the visual properties of the captured photo, is to use the 100%
NSR, where the 100% NSR is the number of electrons that results in equal parts noise and signal
for a read noise, R: 100% NSR = [1 + sqrt (1 + 4·R˛)
] / 2. Thus, DR100 = log2 (Pixel Saturation / 100% NSR).
For
example, the read noise at base ISO (ISO 100) the
Nikon D7000 is 3.1 electrons, and the 100% NSR is [1 + sqrt (1 + 4·3.1˛)
] / 2 = 3.64 electrons. The pixel saturation is 49058 electrons
so the DR at base ISO for
the D7000 using the read noise for the LMS is log2 (49058
/ 3.1) ~ 13.9 stops, and the DR using the 100% NSR for the LMS is log2
(49058 / 3.64) ~ 13.7 stops -- no significant difference.
It is also important to note that two systems with the same DR do not
necessarily look the same. One complicating factor is, since DR is
a pixel-level measure, is if two photos are made from a different number
of pixels (discussed below). The other factor is that, even for
the same number of pixels, one sensor with pixels that have a noise
floor of 10 electrons and a saturation limit of 40960 electrons (12
stops) does not have the same properties as another sensor with a noise
floor of 2 electrons and a saturation limit of 8192 electrons (12
stops).
In addition, many confuse bit depth with DR. The DR is the number of bits of
information about the scene that the sensor can collect per pixel.
The bit depth is the number of bits of information the supporting
hardware can process and record per pixel. If the DR is greater
than the bit depth, the supporting hardware is not taking full advantage
of the sensor output, that is, not all levels of the captured DR will be
distinguishable. On the other hand, if the bit depth is greater
than the DR, more memory is being used than necessary to store the
photo. The
tone curve applied to the photo maps the linear capture of the RAW
file into a processed image file that may, or may not, reflect
the DR of the initial capture.
This brings us to how DR fares for jpg vs RAW. Since jpgs are 8
bits, they can display at most 8 levels of the DR. The tone curve
decides which levels of the DR are represented in the jpg file.
The advantage of RAW over in-camera jpgs is that RAW allows the
photographer to choose the "appropriate" tone curve after the fact,
but if the in-camera jpg has the desirable tone curve, then the RAW file
has no advantage in this regard since most display media (prints and
computer monitors) can only display 8-9 levels.
In a sense, RAW vs jpg is analogous to Auto vs M mode. In Auto mode, the
camera chooses the f-ratio and shutter speed for the capture; in M mode,
the photographer chooses. Just as some photographers may feel that
the camera makes the right choices in Auto mode, some photographers may
feel that the tone curve applied in-camera is the right choice. If
the camera has multiple options for tone curves, this is analogous to
shooting in an AE (auto exposure) mode such as Av, where the
photographer chooses the f-ratio, but the camera chooses the shutter
speed, or Tv mode, where the camera chooses the f-ratio, but the
photographer chooses the shutter speed. The analogy breaks down,
however, in that that photographer cannot change the f-ratio and/or
shutter speed after the fact, but the photographer can choose the tone
curve after the fact if they shoot RAW.
An excellent example of
applying the "appropriate" tone curve after the fact, are the
following two photos here:
http://forums.dpreview.com/forums/read.asp?forum=1034&message=36903045
Note how the bottom photo has compressed the DR of the scene by revealing
details in the highlights that were blown in the top photo, since the
higher ISO of the bottom photo shifted the DR of the capture four stops
lower. So,
even though the medium (jpg displayed on a computer monitor) cannot display
all 14 stops of the DR of the initial RAW capture,
the DR of the capture can be mapped into the fewer levels of the display
medium by an "appropriate" tone curve to produce a
better photo. This is not unlike how we see. Our eyes take
several exposures of scenes, and our brain merges these exposures into
an HDR-like memory of what we "saw".
However, as discussed above, the DR is not an ideal measure for comparing
the IQ of photos from different systems,
since:
• DR is a pixel
level measure as opposed to image level measure, which is an issue
if the systems have significantly different pixel counts
• A perfect sensor with zero read
noise would have an infinite DR, which is unrealistic in terms of a
measure of the visual properties of the photo.
Enter the DR100 /
µphoto, where a µphoto is one-millionth of a photo: DR100 / µphoto =
log2 [(Saturation / µphoto) / (100% NSR / µphoto)] .
Since one µphoto represents the same portion
of the photo regardless of how many pixels it is made of, the DR100
/
µphoto is an
image level measurement. It is worthwhile to note that a photo
displayed at 1200 x 900 pixels on a computer monitor is 1.08 MP.
Hence one µphoto represents one pixel on a computer monitor for a photo
displayed at that size.
So, let's compute the DR100 /
µphoto for the
D7000 at ISO 100 (16.37 MP sensor, read noise = 3.1 electrons,
pixel saturation
= 49058 electrons). One µphoto is 16.37 pixels, so the read
noise / µphoto is 3.1 electrons · sqrt 16.37 ~ 12.5 electrons. The 100%
NSR / µphoto is [1 + sqrt (1 + 4·12.5˛)
] / 2 ~ 13.1 electrons. The saturation capacity / µphoto is
16.37 · 49058
electrons = 803079 electrons. Hence, the DR100 /
µphoto = log2 (803079 / 13.1) ~ 15.9 stops -- two stops higher than the
DR pixel-level measurement.
The significance of the difference between the DR and the DR100 /
µphoto can been seen
in an example comparing the
60D to the
40D at ISO 100. The 40D (10.3 MP,
read noise = 19.6 electrons, pixel saturation = 40311 electrons) has a DR
of 11.0 stops at base ISO -- slightly higher than the 60D (read noise =
13.2 electrons, pixel saturation = 24322 electrons). However,
the DR100 /
µphoto for the 40D works out to be 12.7 stops -- slightly
lower than the 60D. Thus, we can see how the DR100 /
µphoto reflects the IQ
differential from the slightly more efficient sensor of the 60D, whereas
the DR does the exact opposite.
Increasing sensor efficiency by lowering the read noise / area and/or
increasing the saturation capacity / area will increase both the DR and
the DR100 /
µphoto. However, increasing the number of pixels will have
no effect on the DR, but will increase the DR100 /
µphoto of
a system (and hence the IQ of a photo), so long as the sensor efficiency does not suffer. The maximum
possible DR100 /
µphoto for a system is log2 (saturation capacity / µphoto)
which can only be achieved on an ideal sensor with zero read noise.
However, this is not to say that two systems with the same DR100 /
µphoto will produce
equally pleasing photos. A good example is to compare the DR of
the
Canon 5D at ISO 100 and ISO 400, which is virtually the same, and,
since we are comparing on the same sensor, the DR100 /
µphoto would also be the
nearly same. Clearly, we would not claim that the ISO 400 photo
has the same IQ of the ISO 100 photo. While the range of lighting
in which the 5D could capture detail would be essentially the same, the
noise would not, since the ISO 400 photo would be capturing 1/4 the
light as the ISO 100 photo.
In other words, DR100 / µphoto is simply another measure to measure the IQ of a photo,
but must be taken in context. Just as two cameras with the same
pixel count will not necessarily record the same amount of detail, two
systems with the same DR100 /
µphoto will not necessarily produce photos with the
same "smoothness".
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