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What's that noise? Part one: Shedding some light on the sources of noise

Published Apr 27, 2015 | Richard Butler

How would you react if I told you that the aperture and shutter speed you choose make more difference to image noise than the ISO setting? You might be surprised to discover that a lot of the noise in your images doesn't come from your camera at all: it comes from the light you're capturing.

To really understand what's going on in your camera and, hence, how to get the best results out of it, it helps to understand a little bit about where noise comes from. Noise is widely misunderstood and yet it lies at the heart of most technical assessments of image quality and camera capability.

In simple terms, noise is any variation from the true 'signal' that you're trying to capture. Visually it's immediately recognizable as pixels of unexpected brightness or color, most easily seen in areas where you might otherwise expect a smooth result. The first thing to realize is that there are several mechanisms that contribute towards noise and that all of them can play a different role, even within a single image.

In this multi-part article, we're going to look at two major sources of noise: shot noise and electronic noise, and what they mean for the way you shoot. And because this first source of noise is so significant, we're not going to talk about electronic noise at all in this first part. By the end of both parts you should understand where noise comes from, where the strengths of your camera lie, and hence how to get the best out of it (and when it's likely to let you down).

Shot noise

Probably the most significant and certainly least recognized source of noise is what we call 'shot noise' or 'photon shot noise'. In the simplest terms, this is you being able to see the impact of light's inherent randomness.

You might remember talk of photons from science classes, and the key thing to remember is that, although we perceive light as being pretty uniform, it actually travels as a series of packets. That is: light is quantized. These packets (photons) arrive at your eye or your camera sensor at random intervals. Because of the way the eye and the brain work, you don't notice this, but when you look at a scene, you're being bombarded by little packets of randomly occurring light from every part of that scene.

'So what?' you might say. Well, let's try a little thought experiment. Rain is also a series of discrete packets, that fall randomly over time, so it makes a fairly robust metaphor for the way light behaves. So let's see what happens if we imagine trying to measure rain, using a series of test tubes.

Stand outside in the rain, with your group of test tubes covered with a plastic card. Remove the card for a fraction of a second, then quickly put it back over the tubes. You probably wouldn't be at all surprised to find no raindrops in one tube, two in another and one in the other two tubes because, just like light, raindrops occur randomly. Try the experiment again and leave the tubes exposed for a longer period of time: now you'll find all your test tubes are nearly full. If you tried measuring in detail, you might find that you had 420 drops in one tube, 380 in another and 400 in the other two tubes, but generally, you'd probably look at them and conclude they're all pretty much the same.*1


Short exposure	Long exposure

This randomness, even though it's raining similarly hard over all the test tubes, is noise: it's variation from the underlying signal.

An important thing to know is that, although the absolute differences in raindrops collected are larger for the longer exposure (it's a difference of 20 raindrops!), it only accounts for a small proportion of the total number (+/- 5%). With the very short exposure to the rain, the differences were only +/- 1 raindrop between test tubes, but proportionately it made a huge difference (+/- 100%). Or, to use the correct terminology, the signal you captured in the short exposure was small, relative to the amount of noise you experienced: you had much lower signal-to-noise ratio.

This is exactly how shot noise contributes to your photographs. A darker exposure gives you less chance to catch photons, so you're more likely to be able to see the random nature of them hitting your pixels. And this doesn't just apply to bright or dark exposures, it also applies to bright and dark areas within the same image. Bright parts of the image are made up from more photons hitting them during the exposure, so it's harder to perceive any variance between neighboring pixels whereas, in the darker parts of the image, fewer photons hit your sensor, so you're more likely to be able to see the underlying randomness.

The important thing to realize is that this type of noise is present whenever you try to capture light. Whether you use film or digital, medium format or a smartphone, all of the light you're capturing has shot noise built into it. And the solution is always the same: the more light you are able to capture, the less you'll be able to see that noise.

The effect of exposure on noise

Understanding the sources and effects of noise in your final image requires a good understanding of the process of capturing light and the steps it passes through on the way to your final image. The diagrams below show an image of a scene you might want to photograph, with the stripe just below the image showing the scene brightness. The light blue dots within this stripe represent the shot noise that you might experience: there are fewer of them at the dark end of the wedge, but they're more visible.

A key thing to note is that: once captured, the signal-to-noise ratio of any tone can't be improved upon.*2 It can get worse as electronic noise is added, but if you try boosting or pushing the signal, you end up boosting the noise by the same amount and the ratio stays the same. This is why your initial exposure is so important.

The diagram then shows how the exposure you choose maps the brightness of the scene to tones captured by your sensor. Because the image represents the scene, its brightness doesn't change even the exposure is changed:

This information captured by the sensor then passes through an amplifier before being converted to digital values and written into the Raw file (in this case we're showing base ISO with minimal amplification, so the brightest tone from the sensor is being mapped to roughly the highest value in the Raw file):

Then, finally, a tone curve is applied, to map this Raw data into a final image. This can be a standard tone curve applied by your camera's JPEG engine or the result of manual Raw conversion and the adjustments made in processing software.

Changing exposure

We'll add the contribution from electronic noise in the next article. For now, though, let's look at how noise is affected by exposure:


Initial Exposure, showing how the highlights, midtones and shadows in the scene are mapped all the way through to the JPEG.	-1EV Exposure. The camera now captures its highlight, midtone and shadows from brighter parts of the scene.

First, let's look at what happens when you reduce the exposure (using a shorter shutter speed or narrower aperture), without making any other changes. The darker exposure captures extra highlight information and you collect each output tone from a brighter point in the scene. Shot noise remains the same if you look at any given brightness level in the final image...

...however, since you captured less light, any given area in the scene (the original highlights, midtones and shadows) will appear darker than your original exposure. And, because each part of the scene is made up from less signal, it will have a worse signal-to-noise ratio and so will appear noisier.

This becomes apparent if you try increasing the ISO to match the original image brightness using this reduced exposure...

Changing ISO

Combine this reduced exposure with increased ISO sensitivity (so that we get the same final image brightness) and the following happens:


Initial Exposure	1EV ISO increase

Just as in the example above, the reduced exposure means any point in the scene appears darker on the sensor (you've captured less signal at every point in the scene). However, the camera is then applying additional amplification to make up for this, pulling the brightness back up again. And, as mentioned earlier, these amplified tones maintain their signal-to-noise ratios, which we've already seen are lower, compared to our original exposure. This result should be immediately familiar: turn the ISO up on your camera and you get the same image brightness from less exposure but at the cost of noisier images. Electronic noise can play a role in this, but the key lesson from this article is that a lot of the noise you see isn't coming from your sensor or your camera: it's inherent in the randomness of the light you captured and it's primarily dependent on the exposure you chose.

Take a look at what happens with an even higher ISO setting and it explains a lot of why your high-ISO images are noisy: most of your image is being made up from the shadow regions of your Raw file (that was captured with poor signal-to-noise ratios), amplified to be brighter.

Another interesting thing to note is that, at least initially, the camera captured the same extra highlight information that we saw when we just changed exposure. However, increasing the amplification meant they no longer fit in the Raw file, so they 'clipped' to white. This is also why higher ISO images have increasingly little dynamic range: you're amplifying-away a large chunk of the information your sensor captured. This will become relevant when we discuss some camera's dynamic range modes and when we discuss our ISO Invariance test, in the next article.

The effect of sensor size

There are three factors that affect how much light is available for your sensor to capture: your shutter speed, f-number and the size of your sensor.

As we showed in our equivalence article, a full frame camera shot at 85mm F5.6 and a Four Thirds camera at 42.5mm F2.8 will have the same angle of view and the same aperture size (15.2mm diameter) and hence will be exposed to the same amount of light if exposed for the same amount of time. Or, put another way, at the same f-number (both cameras set to F2.8), the full frame camera will see four times as much light as a camera with a Four Thirds sensor, since it is exposed to the same light-per-unit-area but has a sensor with four times the area.

As a result, when you shoot two different sized sensors with the same shutter speed, f-number and ISO, the camera with the smaller sensor has to produce the same final image brightness (which the ISO standard demands) from less total light. And, since we've established that capturing more light improves your signal-to-noise ratio, this means every output tone from the larger sensor will have a better signal-to-noise ratio, so will look cleaner. Click here if you want to see an example.



Larger sensor	Smaller sensor

On a bright day, you're unlikely to notice any significant difference in the highlights and midtones since, even on very small sensors, you're usually capturing enough light to give a good signal-to-noise ratio such that the noise isn't particularly visible. However, in shadow regions, which are made up of less and less signal and, therefore, lower and lower signal-to-noise ratio, the differences will become apparent. And in lower light, an increasing part of the image will be made from tones captured with low signal-to-noise ratios and the noise will start to become obvious. This is the reason smaller sensor cameras tend to produce noisier images.

In turn, this is why we talk about different sensor sizes representing an image quality/size/price balance: because, so long as the sensor's electronic performance is similar, the effect of shot noise means that sensor size is the major determinant of image quality. Yes, pixel count can make some difference, but shot noise tends to play a much bigger role, if you compare images at the same output size.

How does this affect my photography? Exposing to the right

Whatever your sensor size, the most effective way of achieving the best results is to 'expose to the right.' It's a phrase you may have seen used, either in full or as the abbreviation ETTR. In the most simple terms, it means using the brightest exposure you can possibly get away with, without information clipping to white (at which point it's impossible to recover with any color accuracy). The name comes from the appearance of the camera's histogram (the graph showing the distribution of image brightness in the image), where the right-hand side represents bright tones: you're trying to expose so your image data is as far to the right as possible without too much clipping.

This technique only really works with Raw files, since you're setting your exposure based on the brightest point in the scene, not on whether anything in the image is the 'correct' brightness. This means that you'll have to process the Raw file afterwards to adjust the image brightness the way you want it.

The other downside is that, because your camera's exposure meter and exposure guides are all based around putting scene midtones at the correct JPEG brightness level, they're not as helpful as they could be for assessing when you your Raw file has been correctly exposed to the right.

However, the benefit of this approach is that you get your sensor to capture as much light as possible without important tones clipping. This ensures that every tone you want to capture in the scene is made up of as much signal, and the least amount of shot noise contribution, as possible, and hence has the best possible signal-to-noise-ratio. This will get you the best possible noise performance and the most flexible Raw file.*3

Of course there will be situations in which there isn't enough light to expose all the way to the right (or your subject is moving so fast that you need a really fast shutter speed). In this situation you have two choices: to raise the amplification or to under-expose a lower ISO and push the Raw file later. The option you choose should depend on the specific behavior of your camera's sensor, and that's something we'll look at in more detail in part 2.

*1: The figures we've used in this example are deliberate: shot noise contributes a variation of +/- the square root of the signal. [Jump back to text]

*2: Technically speaking, you can improve the signal-to-noise ratio with noise reduction, which attempts to remove noise while preserving the signal. However, this typically comes at the cost of detail retention. [Jump back to text]

*3: There's a common misconception that you ETTR so as to make use of 'more available levels' at the higher end of the Raw file, since the greater number of levels there should record finer tonal transitions. The reality, though, is that shot noise dithers these finer transition, so all those 'extra bits' at the upper end of the Raw file go towards just more finely representing noise. In fact, Nikon's NEF compression takes advantage of the principle of shot noise to more efficiently encode a wide dynamic range into a lower bit-depth Raw file in a visually lossless manner.

Acknowledgements: Thanks to Rishi Sanyal for the extensive input into this article and the many discussions it builds upon, and to Simon Joinson for prettifying the tone flow diagrams.

Tags: fundamentals, noise, technology

Comments

Total comments: 705

Thank you so much. This article was just awesome.

Comment edited 27 seconds after posting

0 upvotes

Thank you Richard, very informative article!

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ETTR is quite misguided technique as one may think that by raising ISO and moving the histogram to the right he is improving the image. While the only way to improve the noise is to increase the exposure, which has nothing to do with the ISO. The histogram doesn't show the exposure, but brightness of some jpeg that is fed to the software that draws the histogram.

They should have better advertised ETM (expose to the max), i.e. choose the maximum exposure under the existing constraints (F-stop, shutter speed, sensor saturation). Such linear constraint tasks always have the solution on the boundaries. e.g. either you choose the max F-stop the lens allows, and/or the max shutter speed that still stops the motion or prevents the camera shake, and/or the min ISO camera is capable of.

Comment edited 2 minutes after posting

2 upvotes

Completely agree.

But could we call it ETTL? As in 'expose to the limit!'? Not to be confused with 'expose to the left'.

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So upon re-reading my previous post and calling it kazunga, I was reminded of what my Photography 101 instructor said in 1974. He said expose for the BRIGHTEST AND MOST IMPORTANT part of the image. I think that is the same as saying ETTR and don't worry about blowing out the spectral highlights. That was true for film and I guess it is true for digital too.

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Film as opposed to a digital sensor has a "shoulder": it becomes less sensitive for new photons the more saturated it is (non linear response)
That's why film is inherently better at preserving highlights and hence more tolerant to overzealous ETTR than digital
More than five years ago I started a thread about shot noise in relation to Nikon's reason for choosing lossy raw compression and everybody thought I was nuts for stating that low light levels are inherently noisy by nature
Glad that this has now been generally accepted here

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I think the ETTR "rule" should be changed slightly... my change in uppercase

"ETTR. In the most simple terms, it means using the brightest exposure you can possibly get away with, without information YOU CARE ABOUT clipping to white."

For example, I took a closeup photo of someone's face, with a bright background. I actually wanted the background to clip and become white.

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First my apologies for not wading through 700+ comments to date. My challenges to your thesis may have been made and defended before this.

1. Rain analogy: You discuss only the number of drops, not the size/volume of drops, the fact that rain falls in waves (a factor of light as well not addressed), and that the diameter of the tube will limit the volume of large drops collected.
2. Light packets are modified by the atmosphere, magnetic fields, and interference.
3. The light captured is from an object not the source. Some of it is reflected. some is absorbed (thus absent) and some is re-radiated.
4. Light may hit sensors as packets, but it also comes as wavelengths (hence the color of light). Most light packets are absorbed by a subject and only some wavelengths reflected. That is why leaves are green to our eye.
All of these other effects further randomize or as I prefer, smooth the randomness of light.
While your hypothesis may be correct, I suggest the net effect is negligible.

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I'm afraid you're confusing things that don't apply here.
Your interest in science and different aspects of light though laudable still need some more polishing, the article is correct, and we should all take advantage from it.
Good (re)read.

3 upvotes

I really don't understand comments like chewdoggydog!
Knowing the physics behind doesn't make photography less artistic at all.
You may think that Turner didn't know anything about pigments and how the physics of color works? That Da Vinci didn't know anything about perspective and geometry?
Should I continue the list? Each great artist masters the physics of its domain after years of trying and working and even collaborating with scientists or engineers. This is so obvious that some people forget about it.

1 upvote

You folks are some damn nerds......lol. who the hell cares about all that crap. Shoot as low iso as you need to. Take more photos and less time pretending to make rockets.

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Given that photons or their packets travel at 186,000 miles per sec, at 1/1000 sec exposure you would have photons coming in for the last 186 miles. The random distribution for photons seems to be listed as poisson, so photons arrive randomly clumped together and with short exposures the effect is more noticeable. The longer the exposure the more averaging occurs so the effect of the shot noise is reduced. My question is simple, maybe too simple but at what shutter speeds given a moderate illumination level does the effect of shot noise start to become negligible? I am into bird photography and particularly birds in flight (BIF), with pictures taken generally in broad daylight, and take photos at high shutter speed less than 1/1000 sec to limit motion blur,

0 upvotes

Whatever the speed of light, it is the number of photons that hit your sensor that you have to take into account.
Short exposure in a bright scene or long exposure in a dark one is exactly the same if you are well exposed at ISO 100.

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The rise of shot noise being visible in an image is the amount of photons that hit a pixel in the sensor with respect to the other pixels in the same sensor. Imagine photographing a purely grey or purely luminous homogeneous scene. The appearance of the shot noise is visible when other pixels in the image receive more or less photons, given rise to brighter or darker luminosity. Over a long period of time these variations will average out so the shot noise effect will cancel out according to the law of large numbers in statistics. My question is; given a moderately bright day, with a moderate f-stop at a moderate ISO, at what shutter speed will the effect of shot noise becomes noticeable. I like shooting birds in flight and have noticed some noise in the images and am not sure of the contributions of photon shot noise to sensor noise or to downstream capture noise. I try to keep the shutter speed between 1/1000 and 1/4000 sec. I use either my Sony A99 or A77ii at ISO 400-1200 LT 1/1000

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Shot noise does not depend on the shutter speed but on the number of photons that hit the sensor.
In other words, longer duration or wider aperture wil be the same as soon as you are well exposed at a given ISO speed because the number of photons will be the same.
Therefore if you want to reduce shot noise the only way is to lower the ISO speed. Because a low ISO speed will impose a large number of photons to be well exposed.
Then, for your specific case, I would recommand you to lower the ISO as much as possible.
:D

Comment edited 1 minute after posting

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The point I am trying to make is how to choose a shutter speed in a generally well lit scene that will reduce the effect of the shot noise natural characteristic of light (not noise from the sensor or the camera's electronics). As I understand it light itself has shot noise that is distributed in a statistical manor ( I believe a poisson distribution). If we collect this light over a longer period of time the effect of shot noise will begin average out to a lower level because we are collecting more photons. The fewer photons reaching the sensor by virtue of shorter shutter speeds, the more apparent their random distribution will be, thus increasing the resulting effect of visible "grain".

Comment edited 6 minutes after posting

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Light is naturally wobbly from nuclear explosions on the sun. The energy of photons at the photo site varies with time, and the wavelength of the light is affected by changes in reflection of the subject if it moves during exposure. The camera tries to adjust to these variations, and does a better job if the exposure is longer. You may not want to reduce noise, because to do this the camera has to reduce detail. It is a fact that the more the noise, the more detail in the raw image. In-camera NR (or PP) can reduce details which are needed to make subjects believable.

High ISO exposures are noisier in the shadows (fewer photons available). Additional noise comes from multiplying the fewer available electrons(amplification) to make the image bright. All high ISO images have loss in detail and smaller DR, so use high ISO technique when these attributes are unimportant.
To reduce image noise, make the exposure longer or use a big aperture to capture more photons.

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Thank you !

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Ok. I understand what you are saying. However, I have a technical question. Since photons travel in random patterns and according to your explanation; the more volume of photons that hit the sensor the better.

What happens when one adds time into the equation? Do I benefit more from having a longer exposure (slower shutter speed) than I do with a larger aperture (for any given correct exposure, whether it is TTR or not)?

My thinking is this; There is more change of having more photons hitting your sensor with a longer exposure than there is with a larger aperture. Is this correct?

Comment edited 2 times, last edit 5 minutes after posting

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Both result in more photons hitting your sensor - this is the reason you get the same brightness increase by opening up your aperture by 1 stop as you do for doubling the shutter time.

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Thanks everyone for great comments...

So just let me please rephrase and revisit the question then:

I think - considering that there are relatively large number of photons per pixel (maybe hundreds and/or at least hundred+ rather than five, ten) even at 50% capture rate, I'd guess still "Shot noise" might not be barely noticeable by human eyes, if ever.

So my mew (rephtased) question is this:

Assuming something like 50% capture by sensor, at what point (number of photons per pixel) would we start noticing uneven distribution - and, hence "effects of photon bursts clumping together?

If we can figure thatone out, then maybe we can compare that to any practical level of light human eyes can sense.

ALSO, another thought - - - although each individual photons may" burst", WOULDN'T THESE "BURSTS" STILL BE RELATIVELY EVENLY DISTRIBUTED?

Thnks everyone!

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I'm confused on the extra highlight part where it says it captured the same extra highlight but ultimately had to clip it to fit into the raw file, i can not seem to find where this happens. please help.

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It happens every time you increase the ISO setting on your camera.

Set your camera in P, A or S mode and shoot an image at low ISO. Now increase the ISO setting by 1 stop. The exposure will be reduced, yet your Raw file and final image still clip at the same point as your original, low ISO exposure.

Now switch the ISO back down and match the exposure values that the camera chose for the second shot. You should notice an extra top of highlight information (and everything else 1EV underexposed). Why weren't those extra highlights in your second image?

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so basically the simple way of putting it is that the since you are raising the iso and amplifying the noise the highlight is "brighter and makes it seem like there is more information?

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Thank you for the fascinating information.

I was wondering about the fact that I usually use flash and similar exposure, f8, 1/160 shutter speed, ISO 400, for portraits. Most of the time, the photos are great. But sometimes, even though the lights and midtones are well exposed, the shadows and blacks are very dark, noisy when brightened by post processing and sometimes very grainy.

I was wondering if this is because the shots are taken in different indoor venues, sometimes with the ceiling being low or high, dark beams, etc. and sometimes I set my light further or nearer to the subject. Do light placement and different environments cause the photons to travel in different random patterns, consequently causing the photons in dark areas of the scene to not hit the sensor?

Comment edited 48 seconds after posting

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Very interesting article ! Thanks very much DPR.

As many people here, I was wondering if the number of photons can be so low that shot noise would be significant.

Based on the so-called "sunny 16 rule", I've found that for a proper exposition on a 24 MPix FF sensor, around 50,000 photons should hit each pixel. This lead to a shot noise with an amplitude of 1.4 %.

This kind of noise amplitude should to small to be noticed. However, in darker conditions, you may have to increase the ISO speed let's say for instance 6400. It means that you have fewer photons. The corresponding shot noise amplitude will be around 3.5 %. We reach up to 14 % for ISO 100,000!

And that's the bad news: 1,000,000 ISO speed will never be as good as 100 ISO whatever the quality of the sensor because of the shot noise.

So bad...

(for those who are interseted in the computation, please mail me, you may point out some mistakes ;-) )

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Is that figure of 50,000 photons the figure for a highlight tone or a midtone?

Dark regions will receive progressively fewer photons (hence being darker), so the SNR will drop down in darker regions of the image.

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I'd say for midtones. Indeed, shot noise will be greater in proportion for darker zones of a picture.

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I did the same calculation and it was for direct sunlight!

I think the number of photons involved in catching a normal white area lit by the sun is MUCH lower than that.

And then calculate 8 stops less light for the low lights. Then SN really becomes an issue - certainly if you increase the shutter speed for example at sporting events at 1/8000 for example and with the current Megapixel hype that has fortunately stalled at the moment.

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For the initial photon flux I took the solar constant: 1.4 kW/m² and then I kept 1% of this value before calculating the power passing through the pupil. 1% is a quite low albedo value.

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I can't confirm/deny your calculations but I *WILL* note that it assumes 100% efficiency of capturing incoming photons. That's way too high.

I'm not the expert to describe the percentage that are captured, but in many sensors, a good fraction of photons fall on support electronics that are insensitive, and lost. More importantly, the fraction that hits the sensitive parts may fail to be recorded for a variety of reasons.

Using a SWAG (scientific, wild-assed guess) of 5% detection efficiency, your 1.4% noise rises to 6.3% (the square root of 1/.05). This doesn't seem unlikely for a well-exposed scene.

Note that RAW conversion techniques essentially ALWAYS employ noise-reduction. For example, a green pixel might take the average of the 4 nearest green pixels; this would smush fine details, but reduce the noise by half (again, the square root). In areas with lower signal, the RAW conversion might spread the average further, hoping to create a more acceptable image.

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WaltFrench - I believe 50-70% is a better assumption.

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Indeed, WaltFrench, after passing through the lens, the light intensity decreases. We can take into account: the AA/low pass filter (T > 85 %), Bayer filter (T ~ 25 %), quantum efficiency (~ 30 % FSI or ~ 70 % for BSI), fill factor (T > 40 %, CMOS with lens array). These values can vary a lot from one sensor to another. Thanks for your comment ;-).
Therefore, taking this into account and after having a second look at my calculation (mainly I've changed the albedo value to a much more realistic value of 5 %), it decreases the number of photons at 100 ISO to 15,000. Thus leading to a shot noise of 1/sqrt(15000) = 0.82 % (my mistake for 50,000 photons the shot noise is around 0.45 %).
It also means that for ISO 409,600 the average value is 3.6 photons oO. Sony and Nikon must have improved the global transmittance!
This may be the reason why more and more manufacturers suppress the AA filter. Next step suppress the Bayer filter :D

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Thanks for the article, but I'm not so sure if the theory - PHOTONS CAUSING VISIBLE SHOT NOISE? - is infact relevant...or maybe I'm missing something here:

Comment edited 8 minutes after posting

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Someone asked about speed of light, and yeah sure it all depends how many "photons" there are to capture. HOWEVER, WHAT I DON'T GET IS THIS: even when it's quite DARK AND there's relatively smallER number of "photons" - wouldn't it be that there'd be STILL RELETIVELY HIGH NUMBER OF PHOTONS?

Unlike the "raindrop" analogy, the size of each photon is much much less coarse and I'd think there'll be ALMOST NEVER a case (at least in practical photography) where you'd end up with 1-0-2 kind of variations across tubes( or in this case each pixel).

Simply put, I think (even when it's very dark from human eyes' standard PLUS evn though each pixel on various sensors maybe miniscule by human standards) MY GUESS IS THAT STILL THERE'LL BE SUFFICIENTLY PLENTY OF PHOTONS PER PIXEL - enough to normalize and render any effects of these "burst" effects of photons irrelevant. (PLUS EACH "PHOTON BURST" MUST NOT BE SO COARSE and so large ENOUGH TO CAUSE NOISE)

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Additionally, as someone mentioned (but kinda lost focus) THE SPEED OF LIGHT MIGHT HAVE BEARING TO THIS FEASIBILITY OF "PHOTON BURSTS" CAUSING NOISE AS WELL - because unless we are talking about 1 or 2 photons per pixel (which I think highly unlikely) these photons are moving at literally speed of light - meaning that a photon travels and it's as if simultaneously exists across the distnce of over 100 metres at 1/3000 exposure. So, unless theres literally only a single photon - AND I BELIEVE WE NEED A WHOLE LOT MORE for either to see or record on visible light sensors - It'll be as if countless number of them would be SIMULTANEOUSLY traveling through sensor's pixel plain.

BECAUSE IT'S EACH EXTREMELY TINY PHOTON BEHAVING AS BURST, NOT A WHOLE LOT OF BUNCH OF PHOTONS BUNDLE TOGETHER AND ACT AS A BURST.

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Someone asked about speed of light, and yeah sure it all depends how many "photons"... HOWEVER, WHAT I DON'T GET IS THIS:

Even when it's quite DARK AND there's relatively smallER number of "photons" - wouldn't it be that there'd exist STILL RELETIVELY HIGH NUMBER OF PHOTONS?

Unlike the "raindrop" analogy, I think there'll be ALMOST NEVER a case (at least in practical photography) where you'd end up with mere a few drops kind of variations across tubes( or each pixel): each photon is too small and we need far more than a few photons to see or photograph. So, the alleged noise due to photon's burst effect must be tooo small to measure practucally?

Simply put, (even when it's very dark in human eyes PLUS even though each pixel may be miniscule by human standards) I THINK THAT STILL THERE'LL BE SUFFICIENTLY PLENTY OF PHOTONS PER PIXEL - enough to normalize and render any effects of these "burst" effects of photons irrelevant.
(EACH "PHOTON BURST" MUST NOT BE SO COARSE TO CAUSE NOISE)

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BECAUSE IT'S EACH EXTREMELY TINY PHOTON BEHAVING AS BURST, NOT A WHOLE LOT OF BUNCH OF PHOTONS BUNDLE TOGETHER AND ACT AS A BURST.

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You'd be surprised. Pixels are very very tiny. And shutter speeds of 1/1000 of a second are really very very short exposures. And dark areas really do not emit lots and lots of photons. And pixels do need lots and lots of photons to generate an electric signal.

Regarding your question about the speed of light: You cannot have x photons per second coming off your subject at your camera and then if you could change their speed, suddenly more or less than that are coming through your lens. Where would the rest be or come from?

If their speed is high, there would be a lots of distance between them. If their speed would be low, they would move closer together. But the same amount would cross a certain line.

Suppose you let light shine through a tube with water. Light moves less fast through water. But at the end of the tube, photons would move fast again through air. Does this mean the light would flicker? No.

So the speed of light has no impact here only how many photons are coming.

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I'm sure that if you'd do some calculations, you'd probably be able to approximately calculate how many photons are actually arriving in a pixel at 1/125 and f/8. It would be interesting to see somebody do this calculation.

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Yes, how you shoot is very important.

I have already covered this (in more technical detail) on my blog:

https://stephenstuff.wordpress.com/2014/08/01/digital-photo-random-noise-and-solutions/

I usually shoot at "normal" exposure and again 1/3 or 2/3 stops over and select the best image in raw photo processing. There are differences in the shadows!

SLT.

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Good article, thanks!
I'll test / try these at home.. :)

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Thank you for this great and very interesting article!

When will the other parts of this article be published?
I am looking forward to reading them.

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I'm just trying to put the finishing touches to part 2 but some of the things that have been raised in the comments have complicated this, so I'm trying to make sure it addresses as many points as possible.

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Ok, thank you Richard!

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Of course - I've seen properly-exposed ISO 6400 and even 12,800 images from a 7D that were perfectly usable and ISO 400 and less photos that were full of noise from the same model, because they were underexposed and pushed too far to compensate. That being said, I would love to see truly usable extreme high ISO capability- birders/wildlife photographers and those of us interested in astrophotography would benefit tremendously, but even if sensor noise were completely licked, there will be photon noise due to the chaos nature of light.

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Please, post like this deserve add articles to favorites. Thanks!

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Here's an idea for the next article to show the real-world application of these principles. Using various exposure methods shoot four typical scenarios: A beautiful landscape a few minutes after dawn, a window-light portrait, a busy Seattle farmer's market at high noon, and rock show in a dark, smoky bar. Then show us the results.

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Hey Richard, I know this is going to be a tough one: But in your next article, could you try to do the following:

Shot noise is really impossible to make visible, because it needs to be captured by the sensor first, which adds all kinds of different noise factors.

Would it be possible for you to calculate the number of photons caught for a certain exposure and some tonal values and then calculate the S/N ratio for those tonal values and subsequently SIMULATE that S/N for a 1:1 simulated crop of that tonal value?

So basically take the tonal value in photoshop (without noise) and add the noise based on the calculated shot noise?

That would be so great to see in your next article.

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This might answer, at least partially, your question:

http://www.dpreview.com/forums/post/55781353

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Thanks for an interesting article that certainly states what I think is (is becoming?) the kernal of questions about noise. I.e., sensors being relatively equal the larger will give less noise problems, and exposing to the right (anathema to those of us who shot transparency film for decades) is the better way to expose digital sensors.

However, I often encounter areas of endeavor where a sizable segment of the image I want to capture is at the low end of the spectrum, and yet I don't want to clip what highlights there are. Think architectural photos of large spaces where lights would be fairly useless. How does the use of HDR affect the overall noise in those areas with less light? Yes, one is merging several images, one or two of which expose the darker areas of the scene at a brighter, less noisy, exposure value, but how does merging it with the darker frames affect the noise ratio in the darker portions of the image?

Comment edited 59 seconds after posting

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Following the article, I think it might make sense to over expose the entire range to make sure the important highlights of your darkest frame are just not clipping.

After that it becomes hocus pocus, because I have no clue what algorithms HDR merge programs use to compile the end result.

Assuming the pictures have a dynamic range of 8 stops, the middle exposure has 3 stops brighter and 3 stops darker than the mid tonal value 5. If the HDR software is aware of the theory of this article, it should use information from the exposures "to the right" - so use tonal values 4, 5, 6 and 7 from the mid exposure for example (1 left, middle and 2 right).

Assuming that HDR software do not "use to the right" to secure the best parts of the pictures, perhaps it would make sense to darken all exposures before merging to force the HDR algorithm to "use to the right". The end result should be brightened again.

Would be interesting to hear Richard's input here...

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I don't believe any of the current HDR software is capable of "seeing to the right," which would certainly be a great idea. There are, however, ways to use two photos to combine one shot to prevent highlight clipping and a second shot to dig out the data on the dark end. An example is Photo Merge in Photoshop. However, these aren't always so great in merging, so I have used stacked images with masks to combine the photos.

It would be great if there were a merging software/plugin that would retain both the lowest and highest values between two images, then combine the two middles by blending, without the necessity for creating masks. It would also be great to have some control over that blending algorithm. Even using Smart Object snapshots (Photoshop) requires a mask to combine the two images.

Without such a program, HDR is all that's left. Hence the question. I'll be interested in the reply as well.

Thanks for writing.

Comment edited 51 seconds after posting

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Hello Richard. Thanks for being so active in the questions. I learnt a lot by reading the discussions.

In your 2nd part, could you also dive into the aspect of energy spilling from a pixel to surrounding pixels?

I notices in a few tests with over exposure, that the edge contrast between very light and very dark parts in overexposed photos becomes a bit worse. It looks like light spills over to surrounding pixels?

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Great article. Thanks.

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Very good explained and helpfool, looking forward to the 2nd article.
I have learned thinks I have not be aware before.

thank you !

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This is a very interesting article, but I feel is incomplete as it does not address how the sensor / system uses the light available across the Dynamic Range. Let me try and explain:

12 Bit with 4,096 Levels from Black (Left) to White (Right) - think Histogram

The first stop on the white side uses 2,048 bits of information.
Leaving 2,048
The next stop uses 1,024
Leaving 1,024
The next stop uses 512
Leaving 512
The next stop 261
Next stop has 130 left
Next stop has 65 left
Next stop has 32.5

As you can see as we move from brighter to darker there are less and less available bits of data to make up the shadows. Thus creating more noise in the shadow areas of an image. That is why ETTR is so important and effective.

I would like to see this addressed more accurately than how I have explained it. It is the same for all sensors though larger sensors and higher pixel counts provide more data to use therefore less noise.

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I'll need to check on the precise reasoning, but I'm assured that this isn't why ETTRing is effective.

I think it's because, as with the test tube example in this article, a larger signal will have larger noise variation in absolute terms, and so all the additional bits you're devoting to highlights just end up describing this noise in more detail (which you tend not to be able to see, anyway).

While it's true that linear Raw devotes half the Raw file to the first stop of light, this is something of a red herring when it comes to noise, despite the long-standing articles drawing attention to it.

4 upvotes

I would say the reason for having less tonal values the darker the tones become is exactly what Richard shows in his article. It does not make sense to try to make.a difference of 2048 tonal values if the S/N ratio is so bad in the darker tones... I'd say this technique of recording less and less tonal values the darker the tonal values become even reduces noise at the sacrifice of details.

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I certainly have something to learn and look forward to Part 2.

Hopefully there will be some significant solution (compromise) to the conundrum of "Noise" we all face. Particularly if "ETTR" is not the end all be all that we had been led to believe.

Comment edited 1 minute after posting

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Got this link from www_zeeshan_de: http://www.zeeshan.de/fotografie_apsc_vs_vollformat/0_slta99.htm

Interesting. What this guy is saying though is that his camera under exposes by default by 1.3EV. And indeed you see the effect mentioned by this article that if you over-expose by 1.3EV and then bring the light intensity back in post-processing, the resulting image has less noise - even at ISO100.

BUT. The article above suggests this guy should over-expose by at least 2 stops, because the first stop is to compensate for his under exposure. And already in the 3rd, 4th and 5th pictures that have been over-exposed by 1 stop you can see that highlight clipping took place (look at the histogram). If he would expose 2 stops or 3 as suggested by this article, he would lose information in white areas of his picture completely.

2 upvotes

So, in the end the choice you have is to choose between risk of losing information in your highlights, that CANNOT be remedied by post-processing, or losing information in your lowlights, for which post-processing has some tools.

Like I said: For me "expose to the right" only makes sense in low contrast, low key images. You can reduce the noise substantially.

In normal contrast scenes, I'd just expose normally.

And in high contrast scenes, I'd even expose to the left or make HDR with multiple images!

2 upvotes

I guess this becomes more of an issue if you're into astro and night photography or if you often crop out parts of your image.

Also I'm not sure how much of the noise shown is due to shot noise and how much due to read-noise. To assess that, you cannot look at RAW pictures, but you should calculate how much photons were used for the pixels. For me the distinction does not matter. The message is clear: longer exposures produce less noise - no matter where the noise comes from.

If this means you should over-expose by default to gain more information in your low lights and thereby introducing the risk of COMPLETELY loosing information in your highlights, that is a choice everybody needs to make for themselves I guess.

Comment edited 1 minute after posting

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DuncanDovovan I shoot with Fujifilm X Pro 1 and have shot with the XT-1. This system seems to have a conservative bias of at least 1 1/2 EV's to protect the highlights and Exposing To The Right works even with a sensor that is well known for low noise. This combined with my earlier post I think adds to the discussion and helps explain why noise is more prevalent in the shadows - the entire process / system has less and less data for mid tones to shadow detail available thus more noise.

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Dear DPR, I appreciated this article very much. In the future, would you consider an article on lens distortion/software correction and the impact on IQ? Maybe also a little about why T-stop tails off before f/stop on many ultra-fast mirrorless lenses? (or something about vignetting and dynamic range, etc.) Basically the relationship between optics and digital sensors. Thanks!

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Celsus, if you have questions about those topics or want to discuss them you can post in the Photographic Science & Technology forum.

http://www.dpreview.com/forums/1061?utm_campaign=internal-link&utm_source=mainmenu&utm_medium=text&ref=mainmenu

Comment edited 2 times, last edit 7 minutes after posting

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Mr. Butler,
Pour fa vour, if you would run a test shot with Nikon 1 with their adapter for the FX lens, and same FL/f-stop/SS shot with the CX lens, FWIW, you can make me shut up :-)
Regards,

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I only suggest Nikon 1 because I know there is an adapter that let you mount FX lens. Any other setup will do - if you can mount a Pentax MF lens on a Q camera, even a better test set up.

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Have a look at this.

Nikon D810 with 85mm lens at F5.6

Nikon D7000 with 50mm lens at F5.6 (very nearly equivalent focal lengths, so that they're shot from essentially the same distance).

Nikon D810 in APS-C crop mode at F5.6.

Please read carefully before responding. If you prefer to think of the re-sampling in terms of magnification, that's absolutely fine.

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Thanks for you response - I get the shot noise and why FF camera is less noisy than the crop camera. However, if you would indulge me is that the notion of larger sensor being less noisy (or smaller sensor being noisier) illogical because one could have a perfect large sensor and a perfect small sensor (neither of them is noisy) but the image from a smaller sensor is still noisier.
To me, smaller sensor is not any noisier than a larger sensor but the input to the smaller sensor was noisier. And conversely, it's not that a larger sensor is less noisier, the input to the larger sensor is less noisier due to the FF lens projecting a larger image. So if the noise level of smaller sensor is due to lower SNR input, what happen if it's illuminated with higher SNR input?
If you are saying to yourself, why did I bother to respond, I feel your pain :-(

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On the basis the shot noise isn't coming from the sensor - it's a reflection of how much light was captured - then, strictly speaking, it's not the sensors themselves that make images from small sensors noisier.

And yes, in situations where you can provide a small sensor with more light (without overexposing), then you can get the same noise performance. However, this tends to be difficult, because, if a small sensor can cope with this extra exposure, then you can probably do the same for the large sensor, too...

1 upvote

I went on thinking down that route as well mostlyboringphotog. But then I thought this can't be true.

The lens has to make sure that f/2.8 is f/2.8 - no matter if it is attached to a FF or APS-C body.

So at f/8 the same amount of light for a specific view angle must arrive on the projection plane - no matter how big or small that sensor is.

Think of photographing a huge 18% grey wall.

If at f/8 suddenly more light would come to the sensor, then that would not be f/8 anymore.

So the amount of "rain" remains the same, but in APS-C case you have a smaller area with smaller buckets and in the FF case you have a larger area with larger buckets.

Conclusion: the larger buckets will gather more rain and therefore will have a better S/N ratio.

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Duncan,
Yea, I'm having the "it cannot be true" moment.
1. Saying larger sensor is cleaner than a crop sensor is illogical since both sensors convert the photons that fall on it the same way.
2. A larger number of photons (less bright spread over larger area or brighter concentrated on a smaller area) are considered less noisy due to the higher SNR, then isn't each photo detector less likely to be in error, regardless how many detectors there are?
3. Larger number of test tubes does not change the probability of error for each test tube.
4. I suppose it may not work that way :-)

While saying#3, I think I can finally answer my own question:
5. If there are more number of test tubes, each collection can be averaged out improving the accuracy.
6. So is the cleaner output from a larger sensor looks cleaner because we are averaging out the noise? If we look at a crop section, we are averaging less, so noisier? I think more math oriented folks call this "integrating"?

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After giving this more thought I'd like to ask another question based on our comparison to rain falling into tubes. Let's assume the same exposure at iso3200 - one image is taken at f2.0 and 1/1000th sec while the other is taken at f16 and 1/15th sec - both giving the same exposure. At F2.0 we have 1/1000th sec of large raindrops but they are 'let in' for such a short time the photon "packets" are more random. If we got the same final exposure at f16 and 1/15th sec would the fact that the shutter was open for longer mean that we would achieve a closer average between the 'tubes' compared to the faster shutter speed? As with rain if we put something in the rain briefly the drops are spread out but if it is left for longer eventually the whole surface is covered since there has been enough time for it to become evenly spread thereby eliminating the random aspect.

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When using the rain metaphor to understand the photons passing through the lens, you have to see the tubes as doing exactly the same as the lens does. Therefore, in my mind, the F2.0 lens would equate to using larger tubes versus the F16 lens opening.

So by using larger tubes for a shorter time should let in the same amount of rain as smaller tubes for longer. This would mean that the same random element is present and therefore the same SNR for both "experiments".

Think of this in another way: the rain drops (photon packets) are spread out randomly in the air at any given time. When doing either experiment, both tube sizes are exposed to the same "volume", because of the different times.

On average however, the same volume have the same expected amount of rain drops E(X) as well as the same variance VAR(X). Because you are capturing the same "volume" (or exposure) with both experiments, the E(X) and the VAR(X) are the same, so the SNR should be the same.

2 upvotes

I don't think the duration of the exposure has much bearing to the shot noise within the everyday photography context. Astrophotography has all kind of other issues to deal with.
BTW, the rate at which the whole area gets wet is the same whether one looks at a large area or smaller area. So, me thinks, it's the number of the rain drops dropping and less with the number of test tubes.
To take the test tube analogy a bit further, even if only few test tubes are sparsely placed and a lot of rain drop is wasted, more rain drops mean more likely the test tubes have the accurate raindrop count.

Comment edited 26 seconds after posting

1 upvote

Size of the tubes DOES have an impact, because the more drops you gather in the same amount of time, the higher the S/N ratio will be (= less noise). See Poisson distribution.

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Yes, size will have some impact but if the rain is light misty, then not much difference, I would think. And the disadvantage of larger tube is one need more rain to make some difference in the measurement. So it could go both ways.

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I was thinking further about this and changing the aperture would be like having a smaller hole for the same size drops to fall through so it will smply take longer to happen. It would be an interesting test to do for those with time to play :)

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I think the tubes/rain drops analogy is perhaps misleading.

The sensor is one tube.

Photons are photons – unlike rain drops, they are all the same size and they travel at the same rate. The quality of single photons does not change.

The lens aperture itself has no effect on the final qty of photons due to adjustment of exposure time used to achieve same exposure.

Bigger sensor = more photons per exposed picture of 1s.
Smaller = less photons per exposed picture of 1s.

Less photons = more noise due to more variance.

Less exposure = more noise due to digital amplification*

Higher ISO = more noise due to electronic signal amplification of fewer photon light information

*One thing this article does not seem to discuss is the idea that the qty of bits used to encode light information at the low end of the luminance graph is lower than that used at the top end (It's linear - so 1, 2, 4, 8 bits etc.) Lower exposure uses less available bits to encode the whole image. Next article?

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Desmond: You would have less rain if the aperture hole becomes smaller, but therefore you have to expose longer. So the end result is probably about the same.

1 upvote

GreB:

The Forums have their weirdnesses.

But the answer is to see the quoting convolutions a limited public (not really my term) will go thru to preserve the models of reality with which they're comfortable.

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I presume you are referring to everyone ;-) After all, all we have are the models of reality.

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Double split up post, owing to dpr's webware.

But right we all get locked into models, and methods of description.

Great B was insisting on the near perfectness of some quantum mechanics and general relativity models that I was poking holes in.

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@HowaboutRAW:

Re: "Great B was insisting on the near perfectness of some quantum mechanics and general relativity models that I was poking holes in."

Demonstrating a misunderstanding of the most successful scientific theory of all time (QM -- Quantum Mechanics) is not an example of "poking holes in it". Again, I await your thread in the Photographic Science and Technology Forum:

http://www.dpreview.com/forums/1061

where you explain to those DPR members that are familiar with QM and how it applies to digital cameras that QM is wrong and thus the relationship between the total amount of light making up a photo and the noise in the photo is wrong. Feel free to throw in how GR (General Relativity) is also wrong. It should be very "entertaining". 'Course, you could start it in Off Topic, where you'll get equal input from knowledgeable people and nutters -- even more "entertaining"!

I simply cannot imagine you being shy to start such a thread -- why might that be? : )

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GB:

I'm not starting a thread.

And I wasn't claiming to know much about the claims quantum mechanics types make about digital photography.

There remain big holes in QM, one of which I already gave you.

The GR thing I already gave you too, and if you'd read, and apprehended what I posted, you'd note that I didn't say there are wholes in GR, just that it follows models from mid19th century math.

(There maybe holes in GR--I can posit one. But for now the problem is confusing the math with an explanation for the source of the curve of space/time.)

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@HowaboutRAW

Re: "I'm not starting a thread."

I know. : )

Re: "And I wasn't claiming to know much about the claims quantum mechanics types make about digital photography."

You mean you didn't say, "Anyhow I'm not exactly a big believer in everything quantum--maybe every time."

Re: "There remain big holes in QM, one of which I already gave you."

You mean when you claimed, "As regards the Sun and nuclear fusion--not sure that accounts for the entirety of the Sun's energy output."

Basically, all you've said is a bunch of nutter stuff, so it's no surprise, whatsoever, that you wouldn't want this discussed in a forum where people knowledgeable about these things can chime in. You want a one-on-one so that it will appear that your "knowledge" appears to be on equal footing with the one you are debating with. This is a common strategy, but easily seen through.

3 upvotes

GB:

Reread that quotation, you're missing a big thing in there.

Yep. I gave you one of those holes. But no it wasn't about the sun.

You've dug yourself in pretty deep.

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Start the thread. Seriously. Why are you hiding here?

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GR:

Because I'm lazy.

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These infos are available if you read books about light....
Its cool to have someone at dpreview giving us something to read.
It can be very technical, some people like me would need the very basic of science to dig the infos from books, but reading this makes it a little easy to comprehend.
Thx

1 upvote

Maybe I'm just misreading it but much of noise discussion becomes if a larger sensor is less noisy. As the shot noise is an attribute of the photons and as the SNR is a function of sqrt of number of photons (regardless of whether the photons are captured or not), the shot noise SNR is then a property of the lens and its size of image circle and not the size of a sensor.
For example, of one uses same DX lens on FF and APS-C, the photon shot noise SNR should be same; however, it's counter intuitive to think that conversely, FX lens on FF and APS-C should also have the same SNR.
So I'm very curios if the example in the article used same lens for larger sensor and for smaller sensor?

Comment edited 3 minutes after posting

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The format the lens was designed for makes no difference at all (so long as it at least covers the format you're working on). A 50mm F2.8 is still a 50mm F2.8.

However, the question is whether the sensor is able to see all of the light (or the sensor-shaped area of it).

The diagram at the top of this page should make it clear:

An APS-C/DX sensor doesn't 'see' all the light projected by a full frame/FX lens, so the smaller sensor can't tell whether it's behind an FX or a DX lens.

And, if this is going where I think it is:

The APS-C image would have the same SNR as an APS-C crop from the full frame sensor.

However, if you used all of the full frame sensor (which would mean changing lenses, if you wanted to shoot the same scene), then the full frame image would have higher SNR and so would look cleaner when viewed at the same size.

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Could you please prove your theory with crops from real pictures? Your rule to expose to the right would only allow 1 or at max 2 stops of over exposure before highlights will clip.

Could you show us a visible reduction of noise between a photo at 1/500 and one at 1/250 at same ISO and aperture where the 1/250 photo was made darker with a tone curve to match the result of the 1/500 photo?

I don't believe the difference in S/N ratio will be visible?

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It's neither my theory nor have I stated any rules.

I can't run out and shoot the images right now, but I hope to include some examples in the next part of the article.

How big a difference this makes will depend on what tone you're looking at: if your signal is high enough, then you might well be able to halve it (reduce it by 1EV) and still have a high-enough SNR that the difference is negligible.

However, with a dark tone made up from very little signal, halving it will do your SNR no good at all (or, conversely, doubling the signal could be really beneficial).

3 upvotes

"A 50mm F2.8 is still a 50mm F2.8." ; however, FX 50mm f/2.8 will project a larger image (more photons) thus higher SNR and DX 50mm f/2.8 will project smaller image (less photons) thus lower SNR.
So the larger image is cleaner every part of the image and if a crop sensor was used to take a part of cleaner image, it should still be cleaner image.
The smaller image is noisier because its lower SNR and even if one would use a larger sensor to capture it, it's still noisier, I'm thinking.

Comment edited 46 seconds after posting

0 upvotes

This is a long-standing and well proven theory. A search for ETTR will give you many examples of said shutter speeds, as well as more reasons and more math to prove it. Here are two such examples.

http://schewephoto.com/ETTR/
http://digital-photography-school.com/exposing-to-the-right/

2 upvotes

Remember that a 50mm 2.8 on DX isn't a 50mm 2.8 on FF. It's a ~30mm cropped image. To get a the same angle of view, you need to use a narrower lens ~80mm. Using DX lenses on FF basically makes your sensor a DX sensor, since you are only using a DX-sized portion of the sensor. You can use heat as an analogy. If you have two areas, one large and one small, that are both the same temperature, the larger area will have more total heat than the smaller area, even though at any given point they are the same temperature. Since heat (light) is what you want, you get a cleaner image. For sound it's the same way. If you record low and boost, you get a hiss. This happens because the noise is amplified to the same degree as your signal.

Comment edited 2 minutes after posting

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I've seen the effect of exposing to the right and it's not a theory. If I was less lazy i could provide plenty of examples from my own experience where images exposed to the right and "dimmed back down" have more detail in the shadows.

But this is also why we do HDR exposure blending; you probably have your own examples at home. You can take the mid range exposure of bracket and try jacking up the shadows but you'll find lots of noise when you do. By blending in a longer exposure you get more detail in your shadows.

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I would say the guideline of this article "expose to the right" only makes sense if your photo is all about low key.

If you want to draw the attention of the viewers of your photo to mid tones or in high key photography, I'd say forget about this rule, because the risks of highlight clipping probably are probably larger than the gain in S/N in "unimportant" parts of your photo.

Looking forward to some examples Richard!

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@ DuncanDovovan:

Have a look at

http://www.zeeshan.de/fotografie_apsc_vs_vollformat/0_slta99.htm

and read it wisely.
After that you will understand whats the difference between EV+0 and EV+1 (or EV+1,3 on SONY SLT) about NOISE WITHOUT DENOISING BY SOFTWARE.
Dont forget, with applying denoising by software you also LOSE DETAILS as well.

Taking picures with EV+1,3 on SONY SLT-A99 you get NOISE FREE pictures WITHOUT DENOISING, it means all DETAILS ARE REMAINING.

Its absolutely important to take pics with EV+1,0 (or EV+1,3 on SONY SLTs) especially when you use a APS-C CAM with 24MP (ALWAYS set ISO100 MOST UMPORTANT).

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Thanks for the link. Interesting. What this guy is saying though is that his camera under exposes by default by 1.3EV. And indeed you see the effect mentioned by this article that if you over-expose by 1.3EV and then bring the light intensity back in post-processing, the resulting image has less noise - even at ISO100.

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Like I said: For me "expose to the right" only makes sense in low contrast, low key images. You can reduce the noise substantially.

In normal contrast scenes, I'd just expose normally.

And in high contrast scenes, I'd even expose to the left or make HDR with multiple images!

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@ Duncan:

"
...over-expose by 1.3EV and then bring the light intensity back in post-processing...
"

there is NO POST-PROCESSING applied at all. with EV+1,3 what i have is a CORRECTLY EXPOSURED SENSOR OF MY SLT-A99, there is NO ADDITIONAL brighten up required anymore.

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@ Duncan:
"
In normal contrast scenes, I'd just expose normally.
And in high contrast scenes, I'd even expose to the left or make HDR with multiple images!
"
There are NO normal Contrast scenes. All you have to do is to EXPOSURE the Sensor of Your Cam CORRECTLY. IF the default EV0 results in underexposured (you can determine it with
dcraw -v -h -w -T -W
) then you have to set the correct EV+x value for your cam.
In General, compact cams ARE NOT underexposured with EV+0,0.
DSLR are underexposure by EV-1,0 (when you take pictures with EV+0,0) and brightened up afterwards.
Sony SLTs are underexposure by EV-1,3 (when you take pictures with EV+0,0) and brightened up afterwards.

For Exposing to the left, its URGENT to have NOISE FREE RAW HEADROOM. You only have this on Sensors with BIG PIXELSIZE (Sony A7S, Sony A99 just for Example). The Areas you underexposed extremely are brightened up LOCALLY and DENOISED LOCALLY as well, which leads to LOSING DETAIL LOCALLY as well. Not my style.

0 upvotes

Given the examples that the author uses, I assume he is addressing normal photography and not say astrophotography or night exposures.

Does the author really believe that while keeping the ISO, and aperture the same, exposing a photo 1 stop longer (assuming 2 stops will introduce clipping of highlights) and correcting this down in post-processing will actually lead to a *noticeable* (not theoretical) improvement in noise for normal portrait/landscape photography?

I'd like to see proof of that instead of theoretical talk.

Is in theory it is easy to understand he is right, but will the difference be noticeable to the human eye?

0 upvotes

I'd like to see a comparison between a photo shot at:

ISO100 f/8 1/250 sec with a tone curve to darken
ISO100 f/8 1/500 sec

...for the dark tones, mid tones and bright tones.

I doubt there will be a visible difference.

0 upvotes

I also think his reasoning for the sensor size is faulty. His explanation is only true if you accept that the image the lens produces is the same for both sensor sizes and therefore the smaller sensor crops out a smaller area with the same amount of pixels. Obviously the smaller pixels of the smaller sensor will then receive less pixels.

But in the real world, I will choose another lens that condenses the light to make sure the smaller sensor will "see" the same scene as the larger sensor would. This will make the amount of light available for the smaller pixels higher.

I think the actual noise increment for smaller sensors comes from the fact that the space allocated to capturing light in relation to space allocated to the structure of the chip is worse for smaller sensors. And of course if you accept the crop factor. So same ISO, speed, aperture and lens will give you more noise.

0 upvotes

@DuncanDonovan:

Re: "I think the actual noise increment for smaller sensors comes from the fact that the space allocated to capturing light in relation to space allocated to the structure of the chip is worse for smaller sensors. And of course if you accept the crop factor. So same ISO, speed, aperture and lens will give you more noise."

Perform the following experiment: take a "properly exposed" photo at, say, f/5.6, 1/100, ISO 200. Now put the camera in manual mode and shoot the same scene at f/5.6, 1/1600, ISO 200. Display both photos at the same brightness. What do you notice? Ideally, you will notice that the photo taken at 1/1600 is made with 1/16 as much light and thus 4x as noisy.

1 upvote

Yes you are right. But the difference between 1/100 and 1/1600 is hardly "exposing to the right side"! Those are 4 stops! All of your highlights will be blown out if you do that.

The method described can only mean you overexpose 1 or perhaps 2 stops. And I would like to see some samples of the noise between 1/100 and 1/200 at the same ISO and aperture, where the 1/100 picture is brought back to the same brightness using a tone curve.

To me this is highly theoretical and of no use in our daily lives as photographers...

I also believe that electronic noise is by far a bigger problem, masquerading the mentioned effect completely.

How many photons per pixel are arriving in modern day cameras for normal shots at IS0100 and 1/500 shutter speed? I would assume not 400 but rather thousands or tens of thousands? Doubling that amount will hardly result in a visible improvement of the S/N ratio?

1 upvote

Also, does a (in my view) theoretical and non-visible gain in S/N ratio outweigh the substantial risk of losing details in your highlights?

0 upvotes

The more light that makes up the photo, the lower the noise and higher the DR of the photo, but the greater the risk of blown highlights.

To get more light on the sensor, you either have to use a longer exposure time (which increases the risk of motion blur and/or camera shake) or a wider aperture (which results in a more shallow DOF and/or softer photo).

To maximize the IQ of the photo, all these factors must be balanced according to your aesthetics.

0 upvotes

The section on sensor sizes specifies the lenses and the apertures used and makes clear that they offer the same field of view (and hence the same scene).

A 42.5mm F2.8 lens will condense the same total amount of light onto a sensor 1/4 as large as an 85mm F5.6 lens would onto a full frame sensor. (Same field of view, same aperture diameter).

By contrast, an 85mm F2.8 lens will project the same intensity of light onto the large sensor, so 4 times the total amount of light.

Pixel size makes very little difference (and those differences will be discussed in part 2). The end result will be much more closely tied to sensor size and the total light available than to pixel size.

2 upvotes

But isn't the lens a reversed projector? And we know that if we spread the projected image over a larger area, the light intensity for a certain area is being reduced? How can you say a larger image sensor receives more light when all of that light is determined by exposure and aperture? In my view both sensors get the same amount of light?

0 upvotes

With the same aperture diameter (not f-number), you're spreading the same light over a larger area. (As per your projector example).

With the same F-number, you have four times the light with the 85mm lens, and are spreading it over a larger area (such that each unit area gets the same amount of light).

1 upvote

Thank you for this excellent article! It is giving me the clue to understand the rationale behind what I was observing already so many times in the past.
I am looking forward to the next part!

2 upvotes

I'm confused, using the 'rain' analogy if we have a deluge, low f number, and we hold our test tubes under this deluge the test tubes will collect a certain amount of water. If it's a 'light' rain shower, high f number, and now hold our test tubes under this 'light shower' for a correspondingly longer time we can collect the same number of rain drops - can't we?

Sorry if i'm being a bear of little brain about this :)

1 upvote

Absolutely. The intensity of the rain in this metaphor is the brightness of the scene. There's nothing representing aperture (so it's fair to assume that the light/rain is unrestricted: the aperture is wide-open).

But yes, in low light/light rain you can keep your aperture open and keep the test tubes exposed for longer. In bright weather/heavy rain, you would need to shorten your exposure or introduce something to block some of the rain (an aperture).

The point is simply that light arrives at your sensor just as rain drops arrive in a test tube: as separate packets, randomly over time.

Comment edited 13 seconds after posting

1 upvote

@By rhurani (3 hours ago)

"@mostlyboringphotog understood your question. the answer is NO. no difference, the FF lens just shed light on and around the crop sensor (what goes around is wasted)"

When I think one way I agree with you; then I think the cropped printed image of a FF sensor should still have the SNR of uncropped printed image? If so, the printed image of a "crop lens" that lets in less light than the FF lens will have less SNR than the cropped print from a FF image.
This is more of a thought experiment than if the difference if any would be visible.
It may be that I need to understand "Poisson" distribution better :)
BTW, I do not fret about the noise differences in my photos as I have other issues :(
BTW2: a radio with a larger antenna will sound louder and less noisy than a radio with a small antenna. But if you reduce the volume of a radio with a large antenna to the level of radio with a small antenna, the radio with a larger antenna will still sound clearer, no?

Comment edited 4 times, last edit 14 minutes after posting

1 upvote

I think I know where you're coming from, and you're right, but the issue on the cropped sensor is the focal length with be different even if the same photons are coming through the same lens.

So lets say I take a pic on a FF sensor with a 50mm at 1/100 f2.8 ISO100. Then I take that same 50mm lens, and put it on a crop sensor and take a picture with the same settings.
The SNR of the photons through the lens are the same, but the crop sensor is only seeing a 75mm field of view (the rest of the light is spilling off the edges of the sensor and are ignored). If you crop the 50mm FF image down to the 75mm field of view you'll see basically the same noise level between the two sensors.

The example in the article didn't compare identical lenses, the focal lengths were different so the resulting image would be the same between the FF and MFT image, leading to a bit of confusion.

0 upvotes

Very informative article. Learned alot from it. Time to do some experimenting.

0 upvotes

Excellent article and I am absolutely looking forward to the next part. Thank you!

0 upvotes

In my opinion the writer of this article is missing the real story here. The suggestion that apperture and shutter speeds have influence on the amount of noise is not correct. Making pictures at the suggested shutter speed / apperture combination will show the same amount of noise levels, even if you were to change one of the parameters (so the other changes accordingly). If I change my apperture from f/3.2 to f/8 and the shutter speed changes accordingly (at same ISO setting), I will not have two different images with two different noise levels. However, when using EV compensation and actually over-exposing, I might achieve a cleaner image. However, in a case whereby you would have to raise the ISO setting in order to make exposure to the right possible, your brighter picture may well show a lot more noise than the picture taken with the suggested shutter speed and lower ISO. So, to conclude, it is all about exposure, not about the shutter speed and apperture used.

0 upvotes

the author did not say otherwise, you are both in agreement

5 upvotes

And what if the 'suggested combination' isn't optimal?

Its a change in the overall exposure (defined by the combination of f-number and shutter speed) that changes the noise.

0 upvotes

I've taken shots where increasing the ISO a stop to get the image exposed a stop to the right resulted in less noise in the finished product.

I was really confused why my ISO 100 shots were being beat by an image at ISO 200. It's cool seeing an actual article explain why noise sometimes shows up at ISO 100.

0 upvotes

This is an honest question: Considering the speed of light compared to the shutter speeds we use what kind of shutter speeds can actually show up the randomness of "light packets"? There must be a point where it all averages out pretty well - where is the boundary?

0 upvotes

The speed of light is a non-issue. The issue is how much light falls on the sensor in a given time interval. For the sake of simplicity, let's say you have a 24 MP sensor that is uniformly illuminated with 480 billion green photons during the exposure. If the QE of the sensor is 50%, this means all the pixels under the green filters will record 240 billion photoelectrons -- 10000 photoelectrons/pixel on average.

But each pixel will not record exactly 10000 photoelectrons. The standard deviation of the signal (noise) is sqrt 10000 = 100 photoelectrons/pixel. This gives a per-pixel photon NSR (noise-to-signal ratio) of 100/10000 = 1%.

Let's say we downsample the photo to 6 MP. Then each of the "pixels" in the new image file will have recorded 40000 photoelectrons on average and have a standard deviation of sqrt 40000 = 200 photoelectrons for a per-"pixel" photon NSR of 200/40000 = 0.5%.

So, there is no boundary. More light means less image noise.

Comment edited 1 minute after posting

1 upvote

speed of light has nothing to do with this effect. it is the distribution of photons (light packets) among pixels that matters. if say 100 photons on average fall on each pixel, pixel a may get 99 photons, pixel b gets 105, pixel c gets 96 photon etc- average is 100, max difference between them is 105-100/100= 5%, here you get pixels that differ in brightness by no more that 5%. At 1000 photons the difference will be less noticed

1 upvote

I wasn't thinking when I asked this question - obviously the amount of light falling on the sensor will depend on shutter speed and aperture settings -reflected in the iso needed which will suggest how low the existing light levels are. So even if I have a shutter speed of 1 second, if the 'rain' is just a very light drizzle the 'tubes' will still not collect even amounts of 'drops' right? :) I was obviously mislead by thinking more of the actual shutter speed rather than resulting combined settings.

Comment edited 3 minutes after posting

1 upvote

Exactly :-)

1 upvote

If I may, the attributes that make a great image, a great image, the level of image noise is so far below the noise threshold that a great image would not be detracted because of noise of the imaging sensor size that took the image, IMHO.
Are we not the luckiest photographers, yet?

3 upvotes

Is there an "executive summary" for those of us that are technically challenged? My brief read leads me to believe this is something I could spend a whole lot of time concerned about when in reality I would never know if my concern and new knowledge accomplished anything at all. Why bother? Just do your best and enjoy your pictures.
Alice

1 upvote

Here is it: "Expose To The Right".

2 upvotes

There is. Skim the article, looking at the gold colored section headers and read the one that says:
"How does this affect my photography? Exposing to the right"

Or ... put as much light on the sensor as possible (without clipping highlights).

0 upvotes

...."without clipping highlights that you want to keep detail in." :)

2 upvotes

@Alice0102:

The more light that makes up the photo, the lower the noise and higher the DR of the photo, but the greater the risk of blown highlights.

To maximize the IQ of the photo, all these factors must be balanced according to your aesthetics.

3 upvotes

Actually, that article is an "executive summary" already, written by an author who undoubtedly knows much more.

1 upvote

Expose to the right...more light. Thanks all, my technically challenged brain understands this.
Alice

Comment edited 32 seconds after posting

0 upvotes

Noise is simply the unwanted part of the information that was expected, IMHO.

1 upvote

Is there any difference if one uses FF lens on a crop sensor compared to a crop lens on a crop sensor?
For the same f/stop and shutter speed, FF lens will let in more (total) light thus higher SNR than a crop lens, no?
This is equivalent to cutting out the crop equivalent print off the FF image vs. crop image from a crop lens.
In other words, is it the total light available or the total light captured?

Comment edited 14 minutes after posting

0 upvotes

Unless otherwise specified, comparisons are assumed to be using the same angle of view. So, for example, we would compare 85mm on [Canon] APS-C to 135mm on FF. And, yes, an 85mm lens designed specifically for APS-C would have the same angle of view as an 85mm lens designed for FF when both are mounted on APS-C.

The difference is that 135mm f/2 on FF would put 2.5x as much light on the sensor as 85mm f/2 on APS-C for a given scene and exposure time (and also have a 1.6x more shallow DOF).

It is for this reason that larger formats have lower noise than smaller formats, although the specifics of how different the noise will be will also depend on the sensor design (typically, sensors of the same generation are rather close except at the extremes).

1 upvote

Greetings,
I'm still trying to have someone with a more knowledge than I to chime in and say, if there is a difference between FF image that is cropped after the fact to the crop image size vs. same image taken with a crop lens and crop sensor. Clarifying this apparent paradox, SNR cannot change after the fact vs. crop sensor with a crop lens have less SNR, I think will help to lessen the controversy.

0 upvotes

If you use the same lens and settings on FF and a crop camera, then crop the FF photo to the same framing as the photo from the crop camera, then both photos will be made from the same total amount of light.

The IQ advantage will go to the system that put more pixels on the subject and/or had the sensor with the higher QE* and lower electronic noise**.

*The QE is the Quantum Efficiency, which is the proportion of light falling on the sensor that is recorded. For sensors of the same generation, there is remarkably little variation amongst systems regardless of sensor size, pixel count, or brand.

**The electronic noise is the additional noise added by the sensor and supporting hardware, but is very small compared to the photon noise. Thus, the electronic noise is significant only for portions of the photo made with very little light, such as deep shadows at base ISO, and progressively more of the photo as the ISO rises, becoming "significant" around ISO 3200 in modern cameras.

0 upvotes

GB,
Thanks but it's still not clear to me if you have answered, if the noise of image of FF cropped (after the fact) to the image taken with a crop lens to the crop sensor size is equal or different.

0 upvotes

Let's say we have a FF sensor made with the same pixels as the sensor in a cropped DSLR (e.g. the Canon 5Ds and Canon 7D2) and took a photo of a scene from the same position with the same lens and settings, cropped the 5Ds photo to the same framing as the 7D2 photo, and displayed the photos at the same size.

The resulting photos would be all but identical.

0 upvotes

GB,
FWIW, I appreciate your response above.
Thanks!!

0 upvotes

@mostlyboringphotog understood your question. the answer is NO. no difference, the FF lens just shed light on and around the crop sensor (what goes around is wasted)

0 upvotes

Just great, shot noise is something else to obsess about besides bokeh. No wonder phone cameras are so popular, only the chief obsession is selfie sticks!

6 upvotes

I know right?!

The whole reason I bought a camera with the P for Professional on the dial was so that I could take better pictures without thinking about all this.
;)

0 upvotes

I wouldn't say you have to obsess about it... but now when you look at those low light shots and wonder where all that noise comes from, or all that smudged detail from all the noise reduction to get rid of that noise, you'll know.

And it's just kind of cool to know. :)

Also gives you an understanding of why bigger sensor tend to perform better in lower light, how smaller sensors can catch up with faster lenses, etc. And the scene/sensor/Raw/JPEG diagrams set us up for what we'll talk more about in subsequent sections: how you can't actually always compensate by giving the smaller sensors more light ad infinitum, because of finite pixel capacities.

Basically, this sets the stage for explaining many of the truths you know about (and don't know about), but don't quite exactly know the why behind.

Ultimately, understanding these things can help you make more informed decisions when considering tradeoffs of different systems, as well as tradeoffs in actual shooting/exposure.

1 upvote

In theory and in practice exposing to the right does indeed decrease image noise in low light shots however, it also introduces other factors that may be more detrimental to overall image quality such as lower shutter speeds (motion and camera blur induced softness) and the risk of clipping highlights. My experimentation has led me to conclude that the associated risks outweigh the benefits.

I suppose the entire exercise only proves that the benefits of ETTR are nothing to really get excited about in the real world. In reality, I would prefer the higher shutter speed with lower risk of camera shake induced blur at the cost of just slightly more noise which can easily be dealt with.

Nice article but no revelations here, sorry.

5 upvotes

There are a lot more sources than noise, and I thought you explained them very well in your A7s low light article - namely read noise and amp glow.

0 upvotes

Oh, that's coming next :) I'm just glad that next time I have to write an article like that a7S one, I can just refer to this one instead of trying to stop and explain shot noise in the middle of other, already dense, text!

1 upvote

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