A Beginners' Guide to Noise

[Great Bustard]

The easiest way to describe noise is as the photo being randomly "off" from what it "should" be.

A photo that is "too dark", "too bright", "too red", "too blue", etc. isn't "noisy". Instead, a noisy photo would be where a swatch that "should all" be roughly one color at a certain brightness and/or color vary "too much".

The two basic types of noise are luminance noise (where the brightness varies "too much") and color noise (where the color varies "too much").

So, what causes noise? There are two principle sources of noise: the total amount of light that falls on the sensor and the efficiency of the sensor.

If we think of the sensor as a swimming pool, the total amount of light is analogous to the amount of water in the pool, and the sensor efficiency is analogous to amount of contamination that was in the pool before water was ever added.

The weird thing is, that the water itself is contaminated by the light itself. But the contamination isn't as simple as a percent of the water, rather it's the square root of the amount of water. I know that's weird, but that's how it is, and this is called the photon noise (also called the "shot noise"), which is the primary source of noise in a photo.

So, for example, if you have 10,000 gallons of water in the pool, then sqrt 10,000 = 100 gallons are contaminated.

Now here's where it gets a bit weirder still. If the pool had 100 gallons of contaminents before the water was added (this is called the "read noise"), then the total amount of contamination isn't the linear sum of the photon noise and the read noise. That is, the total contamination isn't 100 gallons + 100 gallons = 200 gallons.

Instead, the total noise is root sum square: sqrt (100² + 100²) = 141 gallons. But, if that's a bit too technical, then just think of the total noise as the linear sum -- it's not that important a distinction to make at this point in the game.

The noise in the photo is the analogous to the density of the contaminants in the water. For example, if we have 141 gallons of the 10,000 gallons of water is contaminated, then the contamination is 141 / 10,000 = 1.41%.

This measurement of noise is called the NSR -- Noise to Signal Ratio. Often you will here people refer to the SNR which is simply the reciprocal of this value. For example, an NSR of 1.41% means the SNR = 1 / 1.41% = 71.

OK, moving on. How does the water get into the pool? Well, think of the scene as rain falling down on the pool, and as the shutter as a covering over the pool. The amount we open the cover is analogous to the aperture in the lens, and the length of time we keep the pool uncovered is analogous to the shutter speed.

Thus, how much water gets in the pool depends on how hard it's raining, how wide we open the covering, and how long we keep it open. How much light gets on the sensor depends on how bright the scene is, how wide we open the aperture, and how long our shutter speed is.

The exposure is analogous to the depth of the water in the pool. Clearly, a larger pool will hold more water for a given depth. In the same way, a larger sensor will collect more total light for a given exposure. That's the reason that larger sensor systems usually have less noise.

I say "usually have less noise" because a big pool with lots of contaminents in it before the water began falling may still be more contaminated than a smaller, cleaner, pool would be. In other words, a smaller, more efficient, sensor may outperform a larger, less efficient sensor even for the same exposure.

So, where does ISO fit into all this? In an AE (auto exposure) mode, such as Auto, P, Av, Tv, etc., choosing a higher ISO results in the camera selecting a smaller aperture (higher f-ratio), faster shutter speed, and/or less flash power, any of which will result in less light falling on the sensor.

What about pixel size? Well, the effect of pixel size is a secondary effect on noise, and is a bit more technical. This link:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=35068712

and, even more detailed, this link:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=32064270

cover the topic pretty thoroughly, and this link:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=37714016

provides a pretty clear demonstration of what the previous two links discuss. But, if anyone's read this far, and has questions, I'd be happy to answer them.

I really hope this discussion was helpful to beginners. I sincerely apologize if it was overly technical. Clearly, the issue of noise is more complicated than my post makes it out to be, but I'm trying to keep my post and accurate as possible, at the expense of being more complete.

EDIT: I also apologize for any typos and/or errors that may be in the post. If you feel I made an error anywhere, please don't hesitate to bring it to my attention, so that I can acknowledge it if I agree, or so we can discuss it if I disagree.

 

[nelsonal]

My simplistic model is to imagine a large array of test tubes on your lawn (photosites), with a sprinkler in the middle raining down on them (light exposure). If we measure the levels, noise is the variance between a high and a low reading since in this case it should be even (ie a grey card or something similar). If the difference is 1/10" (roughly 2mm) that's not bad if our scale is from 0-4" (0 to 100mm) and our levels are 2.5 to 2.6" (50-57mm) [low ISO setting] but it's not going to work well if our min max range is 2.4-2.75" [high ISO setting] in other words an ISO setting has a similar effect to changing the measured range of the photosites (but then I always preferred the water model for electricity.

 

[Great Bustard]

My simplistic model is to imagine a large array of test tubes on your lawn (photosites), with a sprinkler in the middle raining down on them (light exposure). If we measure the levels, noise is the variance between a high and a low reading since in this case it should be even (ie a grey card or something similar).

This analogy works well for a flat scene, like a gray card or white wall.

If the difference is 1/10" (roughly 2mm) that's not bad if our scale is from 0-4" (0 to 100mm) and our levels are 2.5 to 2.6" (50-57mm) [low ISO setting] but it's not going to work well if our min max range is 2.4-2.75" [high ISO setting] in other words an ISO setting has a similar effect to changing the measured range of the photosites (but then I always preferred the water model for electricity.

The effect of a higher ISO, as I said in the OP, is to cause the camera to choose a more narrow aperture, shorter shutter speed, and/or use less flash power, all of which result in less light falling on the sensor, and this is why there is more noise at higher ISOs. There may also be an analog gain associated with the ISO, and this analog gain will result in less read noise that will partially offset the higher noise due to less light.

The primary purpose of ISO is to map the recorded signal to the desired output level. So, if the base ISO is 100, then using ISO 1600 results in a 16x multiplier to the signal if no analog amplification is used. If there is, for example, 2x analog amplification, then ISO 1600 would result in an 8x multiplier to the 2x analog amplification.

If there is no analog amplification, then there is no reason to use a higher ISO in the capture, since the multiplier will simply result in lost data for the resulting values that are too large for the camera's bit depth after the multiplication.

So, using a higher ISO is simply an indirect manner to make the camera choose a more narrow aperture (higher f-ratio), faster shutter speed, and/or less flash power in an AE (auto exposure) mode. This will result in less light on the sensor (lower exposure) and thus more noise, but that increased noise may be partially offset if the sensor uses analog gain, which results in greater efficiency.

I hope it's obvious why I didn't discuss this point in more detail in the OP.

 

[Graystar]

...of why you shouldn't be explaining anything to beginners.

Here's a simpler explanation...

"Noise" refers to statistical variations in light as it arrives at the sensor.
These variations are a percentage of the amount of light recorded.
The percentage gets smaller as more light is collected.
Therefore, collecting more light results in less noise.
Increasing ISO results in less light collected, and therefore more noise.

For those who want to know more...

The variation in light is defined as the square root of the amount of light recorded. So if the light collected gives a signal of 100 (don't worry about "100 what?" for now) then the noise is 10, which is 10%. If the light collected gives a signal of 10,000, then the noise is 100, which is 1%.

What if we were to multiply the smaller signal (100) by 100 to get a signal of 10,000? Does the noise become 1? No, the noise is also multiplied by 100. So the noise is 1,000 and is still 10%. The more-light-less-noise process only works when you're actually collecting the light.

When you increase the ISO of your camera, you are telling the camera to first collect less light than is required to properly expose the image, and then multiply the light signals by some factor. The resulting signals are at the same level that a low-ISO image would have provided. That's how you get the same image with less light. However, remember that multiplying the light signals results in more noise than having collected more light to begin with. Therefore, increasing the ISO setting results in images with more noise.

.

 

[Great Bustard]

...did it?

http://forums.dpreview.com/forums/read.asp?forum=1002&message=38953022

I should have known better than to respond to you. Well I won't make the same mistake again.

Funny -- you seem to have a knack for mistakes. Not that I mind, of course -- please, feel free to respond anytime. Just a bit childish to make comments like that, brag about how many people are on your ignore list:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=37942523

Yeah, I have 26 people on ignore already. I wish that ignore function would completely hide the existence of a post by an ignored person. As it stands now, I have to read their subject line...terrible!

and then come back at me when you proudly announced that I was one of them:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=37956141

You're ignored.

In any case, like I said, please feel free to comment anytime. Myself, I don't feel like I need to an "ignore list" to aid me in ignoring anyone, but I suppose if there were so many people I was ignoring, maybe it would come in handy. Still unclear about why anyone would brag about the size of their ignore list or announce to people that they're ignored, but, well, that's just me.

OK, on to business!

...of why you shouldn't be explaining anything to beginners.

Opinion noted.

Here's a simpler explanation...

"Noise" refers to statistical variations in light as it arrives at the sensor.
These variations are a percentage of the amount of light recorded.
The percentage gets smaller as more light is collected.
Therefore, collecting more light results in less noise.
Increasing ISO results in less light collected, and therefore more noise.

It is indeed simpler (and even correct!), but not as complete.

For those who want to know more...

The variation in light is defined as the square root of the amount of light recorded. So if the light collected gives a signal of 100 (don't worry about "100 what?" for now) then the noise is 10, which is 10%. If the light collected gives a signal of 10,000, then the noise is 100, which is 1%.

When do they get to know "100 what"?

What if we were to multiply the smaller signal (100) by 100 to get a signal of 10,000? Does the noise become 1? No, the noise is also multiplied by 100. So the noise is 1,000 and is still 10%. The more-light-less-noise process only works when you're actually collecting the light.

OK, were you ever going to discuss photon noise vs read noise, or is sensor efficiency unimportant? How about luminance vs color noise? Too advanced to even mention for these poor ignorant beginners? Will it make their head explode?

When you increase the ISO of your camera, you are telling the camera to first collect less light than is required to properly expose the image, and then multiply the light signals by some factor. The resulting signals are at the same level that a low-ISO image would have provided. That's how you get the same image with less light. However, remember that multiplying the light signals results in more noise than having collected more light to begin with. Therefore, increasing the ISO setting results in images with more noise.

I'm wondering if anyone might not be wonder why increasing the ISO tells the camera to collect less light, and why anyone would want to do that.

So, it seems that feel a post with 1/3 the text that omits discussing luminance vs color noise, photon vs read noise, why raising the ISO makes the camera collect less light, and without any links to other references is much better for the beginner.

OK, I can see that. I'll chalk it up to a "difference in opinion".

 

[bobn2]

For those who want to know more...

The variation in light is defined as the square root of the amount of light recorded. So if the light collected gives a signal of 100 (don't worry about "100 what?" for now) then the noise is 10, which is 10%. If the light collected gives a signal of 10,000, then the noise is 100, which is 1%.

When do they get to know "100 what"?

The '100 what' is rather important, since the noise (or at least the shot noise) bit comes from the random arrival of photons , so that (or, at least, the derived photoelectrons) is what you need to count. As an example, imaging a grey patch results in 10000 electrons per pixel, and those 10000 electrons result on a 0.1 V signal to the ADC which results in a count of 1000 in the raw file.

The standard deviation of the noise is 100 electrons, the square root of the number collected, which corresponds to 0.001V and a count of 10. If we were mistakenly to think that it was the square root of the sensor output voltage that matters, we'd estimate the noise at 0.3162V, for a count of 3162, or if we decided to take the square root of the count, we'd get 31.62. So, the sentence ' if the light collected gives a signal of 10,000, then the noise is 100, which is 1%' is really rather misleading. You have to be specific about what the 'signal' is, and it only works if it's photoelectrons.
--
Bob

 

[sherwoodpete]

So, it seems that feel a post with 1/3 the text that omits discussing luminance vs color noise, photon vs read noise, why raising the ISO makes the camera collect less light, and without any links to other references is much better for the beginner.

When I'm starting out in any subject, I like to absorb information in small, easily digestible chunks. So there are advantages in a shorter text.

Other topics can be introduced initially by name, and then described using sub-headings.

I was going to say that the problem lies with the limitations of this forum. But not really. It's possible to use such things as

• numbered or bulleted lists
• underlining for key words
• underlining or bold text (or both ) for headings or sub-headings
• italics are available too , for whatever purpose may be suitable

A little attention to layout and breaking up a topic into easily-digested sub-topics would go a long way towards making technical posts comprehensible to experts, and ultimately perhaps even to beginners too.

No offence intended, but I found the opening post of this thread caused my attention to wander very rapidly, it may not be much use to a beginner. Just my opinion, for what it's worth.

Regards,
Peter

 

[Great Bustard]

So, it seems that feel a post with 1/3 the text that omits discussing luminance vs color noise, photon vs read noise, why raising the ISO makes the camera collect less light, and without any links to other references is much better for the beginner.

When I'm starting out in any subject, I like to absorb information in small, easily digestible chunks. So there are advantages in a shorter text.

Other topics can be introduced initially by name, and then described using sub-headings.

I was going to say that the problem lies with the limitations of this forum. But not really. It's possible to use such things as

• numbered or bulleted lists
• underlining for key words
• underlining or bold text (or both ) for headings or sub-headings
• italics are available too , for whatever purpose may be suitable

A little attention to layout and breaking up a topic into easily-digested sub-topics would go a long way towards making technical posts comprehensible to experts, and ultimately perhaps even to beginners too.

Thanks for the feedback! I considered using the bulleting, but decided against it, as I thought it would make it look "too definitive", and I wanted a more informal look since my post left out a lot.

I don't like underlining (looks like links) or bold (shouting), but I usually do use italics quite a bit for emphasis in posts. However, this time, I thought it made it look too technical, as if there was going to be a quiz afterwards.

No offence intended, but I found the opening post of this thread caused my attention to wander very rapidly, it may not be much use to a beginner. Just my opinion, for what it's worth.

Well, that's the bottom line, isn't it? Give me some time, and maybe after a bit more feedback, and I'll post a rewrite. What do you think of a multi-post format with just a few paragraphs each like Daniel Browning's treastice that I linked:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=32064270

One post could be "types of noise", the next, "what is noise", the next "photon and read noise", etc. I actually thought of that idea, too, but decided against it because each post would only be a couple of paragraphs.

Anyway, thanks for the critique!

 

[Steen Bay]

For those who want to know more...

The variation in light is defined as the square root of the amount of light recorded. So if the light collected gives a signal of 100 (don't worry about "100 what?" for now) then the noise is 10, which is 10%. If the light collected gives a signal of 10,000, then the noise is 100, which is 1%.

When do they get to know "100 what"?

The '100 what' is rather important, since the noise (or at least the shot noise) bit comes from the random arrival of photons , so that (or, at least, the derived photoelectrons) is what you need to count. As an example, imaging a grey patch results in 10000 electrons per pixel, and those 10000 electrons result on a 0.1 V signal to the ADC which results in a count of 1000 in the raw file.

The standard deviation of the noise is 100 electrons, the square root of the number collected, which corresponds to 0.001V and a count of 10. If we were mistakenly to think that it was the square root of the sensor output voltage that matters, we'd estimate the noise at 0.3162V, for a count of 3162, or if we decided to take the square root of the count, we'd get 31.62. So, the sentence ' if the light collected gives a signal of 10,000, then the noise is 100, which is 1%' is really rather misleading. You have to be specific about what the 'signal' is, and it only works if it's photoelectrons.

I've often been wondering.. when discussing noise we're almost always assuming a "grey patch" like you did above, in order to simplify things, but that a rather rare scenario in real world shooting. So, how do the colors of the real world affect the amount of visible noise in the image? Let's take for example a blue sky, that often, as far as I can tell, seems to have surprisingly much visible noise, even in a low ISO DSLR image. Does a blue sky in practice have more visible noise than a grey patch with the same 'brightness'?

 

[bobn2]

For those who want to know more...

The variation in light is defined as the square root of the amount of light recorded. So if the light collected gives a signal of 100 (don't worry about "100 what?" for now) then the noise is 10, which is 10%. If the light collected gives a signal of 10,000, then the noise is 100, which is 1%.

When do they get to know "100 what"?

The '100 what' is rather important, since the noise (or at least the shot noise) bit comes from the random arrival of photons , so that (or, at least, the derived photoelectrons) is what you need to count. As an example, imaging a grey patch results in 10000 electrons per pixel, and those 10000 electrons result on a 0.1 V signal to the ADC which results in a count of 1000 in the raw file.

The standard deviation of the noise is 100 electrons, the square root of the number collected, which corresponds to 0.001V and a count of 10. If we were mistakenly to think that it was the square root of the sensor output voltage that matters, we'd estimate the noise at 0.3162V, for a count of 3162, or if we decided to take the square root of the count, we'd get 31.62. So, the sentence ' if the light collected gives a signal of 10,000, then the noise is 100, which is 1%' is really rather misleading. You have to be specific about what the 'signal' is, and it only works if it's photoelectrons.

I've often been wondering.. when discussing noise we're almost always assuming a "grey patch" like you did above, in order to simplify things, but that a rather rare scenario in real world shooting. So, how do the colors of the real world affect the amount of visible noise in the image? Let's take for example a blue sky, that often, as far as I can tell, seems to have surprisingly much visible noise, even in a low ISO DSLR image. Does a blue sky in practice have more visible noise than a grey patch with the same 'brightness'?

Well 'grey' is a figurative term. The point is that there is no noise in a single pixel. There is only noise as a variance from a mean in a set of observations (pixels). All the noise calculations make a tacit assumption of a set which has a fixed, steady mean, and then you measure the variance over the set. Joofa has made the point that the same statistics do not apply to a set where the mean varies across the set. I've been doing some experiments to find out whether that has a practical effect, but it's actually quite tricky to do them properly.

As to colours, what you're doing is separating into three separate sets. with different populations. Imagine we have a flat grey sky, so that the mean sample is 6400 electrons in all of the RGB pixels. We end up with a SNR of 64, or 6 stops, not at all bad. Now ye have a blue sky with, say, a mean of 10000 in the blue but 400 in the green and red. In the blue channel we have an SNR of 100, or 6.6 stops - all hunky dory. In the red channel we have an SNR of 20, or 4.3 stops - not so good. In the green we have an SNR of 28 or 4.8 stops - also not great. Now remember also that on a Bayer sensor the 'green' is the primary 'luminance' channel, and you can see why with some decodings, the sky might look noisy. That is a case where some smart noise reduction does some good. Most NR goes for a high 'luminance' definition and filters the 'chrominance', but with a blue sky (or any other flat area predominantly one colour) the calculation of 'luminance' needs to be done differently.
--
Bob

 

[Steen Bay]

For those who want to know more...

The variation in light is defined as the square root of the amount of light recorded. So if the light collected gives a signal of 100 (don't worry about "100 what?" for now) then the noise is 10, which is 10%. If the light collected gives a signal of 10,000, then the noise is 100, which is 1%.

When do they get to know "100 what"?

The '100 what' is rather important, since the noise (or at least the shot noise) bit comes from the random arrival of photons , so that (or, at least, the derived photoelectrons) is what you need to count. As an example, imaging a grey patch results in 10000 electrons per pixel, and those 10000 electrons result on a 0.1 V signal to the ADC which results in a count of 1000 in the raw file.

The standard deviation of the noise is 100 electrons, the square root of the number collected, which corresponds to 0.001V and a count of 10. If we were mistakenly to think that it was the square root of the sensor output voltage that matters, we'd estimate the noise at 0.3162V, for a count of 3162, or if we decided to take the square root of the count, we'd get 31.62. So, the sentence ' if the light collected gives a signal of 10,000, then the noise is 100, which is 1%' is really rather misleading. You have to be specific about what the 'signal' is, and it only works if it's photoelectrons.

I've often been wondering.. when discussing noise we're almost always assuming a "grey patch" like you did above, in order to simplify things, but that a rather rare scenario in real world shooting. So, how do the colors of the real world affect the amount of visible noise in the image? Let's take for example a blue sky, that often, as far as I can tell, seems to have surprisingly much visible noise, even in a low ISO DSLR image. Does a blue sky in practice have more visible noise than a grey patch with the same 'brightness'?

Well 'grey' is a figurative term. The point is that there is no noise in a single pixel. There is only noise as a variance from a mean in a set of observations (pixels). All the noise calculations make a tacit assumption of a set which has a fixed, steady mean, and then you measure the variance over the set. Joofa has made the point that the same statistics do not apply to a set where the mean varies across the set. I've been doing some experiments to find out whether that has a practical effect, but it's actually quite tricky to do them properly.

As to colours, what you're doing is separating into three separate sets. with different populations. Imagine we have a flat grey sky, so that the mean sample is 6400 electrons in all of the RGB pixels. We end up with a SNR of 64, or 6 stops, not at all bad. Now ye have a blue sky with, say, a mean of 10000 in the blue but 400 in the green and red. In the blue channel we have an SNR of 100, or 6.6 stops - all hunky dory. In the red channel we have an SNR of 20, or 4.3 stops - not so good. In the green we have an SNR of 28 or 4.8 stops - also not great. Now remember also that on a Bayer sensor the 'green' is the primary 'luminance' channel, and you can see why with some decodings, the sky might look noisy. That is a case where some smart noise reduction does some good. Most NR goes for a high 'luminance' definition and filters the 'chrominance', but with a blue sky (or any other flat area predominantly one colour) the calculation of 'luminance' needs to be done differently.

Thanks, that'll do for now, but interesting subject I think, and maybe I'll get back to it later. And yes, a bit of smart noise reduction is probably a good thing. The base ISO (iso80) images from my 5.6x crop 'superzooms' are actually surprisingly good, with lots af 'natural' looking detail and pretty low noise, but without the clever NR/processing it probably wouldn't be a pretty sight.

 

[Michael Firstlight]

This is an excellent tutorial on the subject - much easier to comprehend and apply:

http://ronbigelow.com/articles/noise-1/noise-1.htm

Regards,
Mike

 

[cpw]

I've often been wondering.. when discussing noise we're almost always assuming a "grey patch" like you did above, in order to simplify things, but that a rather rare scenario in real world shooting. So, how do the colors of the real world affect the amount of visible noise in the image? Let's take for example a blue sky, that often, as far as I can tell, seems to have surprisingly much visible noise, even in a low ISO DSLR image. Does a blue sky in practice have more visible noise than a grey patch with the same 'brightness'?

Well 'grey' is a figurative term. The point is that there is no noise in a single pixel. There is only noise as a variance from a mean in a set of observations (pixels). All the noise calculations make a tacit assumption of a set which has a fixed, steady mean, and then you measure the variance over the set. Joofa has made the point that the same statistics do not apply to a set where the mean varies across the set. I've been doing some experiments to find out whether that has a practical effect, but it's actually quite tricky to do them properly.
...
--

Hi Bob and Steen, Well, the effects that Joofa has talked about over the years is true for real world images; which attempts to explain things other than that of uniform grey patches. In order to understand this difference, let us first take a picture of a uniform wall, that might give us an average per pixel signal of the aforementioned 10,000 electrons. Then we might say that, since this is a uniform area, the noise = 100 e, and SNR = 100:1. But then again, how exciting is this, since there is no detail in this patch.

Then we can go up a step in complication in this wall image by now imagining that we impose some detail into the patch. The next level in going to detail would be to imagine that there is a bar pattern of some low spatial frequency, sinusoidally varying intensity pattern that modulates the uniform field. Let's imagine that the amplitude of this sinusoidal pattern is 10% of the backround, or an excursion of 1,000 e about the mean of 10,000 e, as measured by the camera (i.e. after all of the camera's optics has had their MTF influence upon the scene. Then what we see that we're interested in is pulling out this 1,000 e of detail out of the noise created by the 10,000 e average signal, this concept is called Contrast SNR. In this case the Contrast SNR = 1,000 e/100 e = 10:1, which is much lower than the 'usual' SNR of 100:1. This is a different quantity that goes in the numerator, in this next level of complication.

This Contrast SNR is defined as:
SNR(con) = MTF*C*SNR

where C is the contrast of the scene and MTF is from the optics, for any spatial frequency. We can think of SNR as what you'd get at zero spatial frequency (which is the definition of a uniform patch!). So now MTF is involved in this.

If you're interested, the math I got from this is found in the following references:

Janesick, J., "Lux transfer: Complementary metal oxide semiconductors versus charge-coupled devices," Optical Engineering, Vol. 41 No. 6, June 2002.

http://spie.org/x648.html?product_id=725073

Both of which are good light reading!

Chris

 

[bobn2]

This is an excellent tutorial on the subject - much easier to comprehend and apply:

http://ronbigelow.com/articles/noise-1/noise-1.htm

This tutorial propagates some of the usual errors that surround this subject, or at best contains unclear wording that leads people to misunderstandings. The bits to beware of are the ones concerning ISO and noise (not actually incorrect, but could be interpreted as suggesting that it is additional amplification which causes noise at high ISO's) and the effects of pixel size on noise (the usual blooper of considering a pixel by itself without considering how they aggregate to form an image)

--
Bob

 

[bobn2]

I've often been wondering.. when discussing noise we're almost always assuming a "grey patch" like you did above, in order to simplify things, but that a rather rare scenario in real world shooting. So, how do the colors of the real world affect the amount of visible noise in the image? Let's take for example a blue sky, that often, as far as I can tell, seems to have surprisingly much visible noise, even in a low ISO DSLR image. Does a blue sky in practice have more visible noise than a grey patch with the same 'brightness'?

Well 'grey' is a figurative term. The point is that there is no noise in a single pixel. There is only noise as a variance from a mean in a set of observations (pixels). All the noise calculations make a tacit assumption of a set which has a fixed, steady mean, and then you measure the variance over the set. Joofa has made the point that the same statistics do not apply to a set where the mean varies across the set. I've been doing some experiments to find out whether that has a practical effect, but it's actually quite tricky to do them properly.
...
--

Hi Bob and Steen, Well, the effects that Joofa has talked about over the years is true for real world images; which attempts to explain things other than that of uniform grey patches. In order to understand this difference, let us first take a picture of a uniform wall, that might give us an average per pixel signal of the aforementioned 10,000 electrons. Then we might say that, since this is a uniform area, the noise = 100 e, and SNR = 100:1. But then again, how exciting is this, since there is no detail in this patch.

Then we can go up a step in complication in this wall image by now imagining that we impose some detail into the patch. The next level in going to detail would be to imagine that there is a bar pattern of some low spatial frequency, sinusoidally varying intensity pattern that modulates the uniform field. Let's imagine that the amplitude of this sinusoidal pattern is 10% of the backround, or an excursion of 1,000 e about the mean of 10,000 e, as measured by the camera (i.e. after all of the camera's optics has had their MTF influence upon the scene. Then what we see that we're interested in is pulling out this 1,000 e of detail out of the noise created by the 10,000 e average signal, this concept is called Contrast SNR. In this case the Contrast SNR = 1,000 e/100 e = 10:1, which is much lower than the 'usual' SNR of 100:1. This is a different quantity that goes in the numerator, in this next level of complication.

This Contrast SNR is defined as:
SNR(con) = MTF*C*SNR

where C is the contrast of the scene and MTF is from the optics, for any spatial frequency. We can think of SNR as what you'd get at zero spatial frequency (which is the definition of a uniform patch!). So now MTF is involved in this.

If you're interested, the math I got from this is found in the following references:

Janesick, J., "Lux transfer: Complementary metal oxide semiconductors versus charge-coupled devices," Optical Engineering, Vol. 41 No. 6, June 2002.

http://spie.org/x648.html?product_id=725073

Both of which are good light reading!

Chris, thanks, that is the first time I've seen a simple formula for this effect - very useful.

--
Bob

 

[ejmartin]

This is an excellent tutorial on the subject - much easier to comprehend and apply:

http://ronbigelow.com/articles/noise-1/noise-1.htm

This tutorial propagates some of the usual errors that surround this subject, or at best contains unclear wording that leads people to misunderstandings. The bits to beware of are the ones concerning ISO and noise (not actually incorrect, but could be interpreted as suggesting that it is additional amplification which causes noise at high ISO's) and the effects of pixel size on noise (the usual blooper of considering a pixel by itself without considering how they aggregate to form an image)

Yes, I wrote to him (Ron Bigelow) a number of times explaining some of the mistakes and got no response. Some of his other tutorials on post-processing are quite helpful, but I think here he steps out of his area of expertise.

--
emil
--



http://theory.uchicago.edu/~ejm/pix/20d/

 

[Graystar]

...did it?

http://forums.dpreview.com/forums/read.asp?forum=1002&message=38953022

I should have known better than to respond to you. Well I won't make the same mistake again.

Funny -- you seem to have a knack for mistakes. Not that I mind, of course -- please, feel free to respond anytime. Just a bit childish to make comments like that, brag about how many people are on your ignore list:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=37942523

Yeah, I have 26 people on ignore already. I wish that ignore function would completely hide the existence of a post by an ignored person. As it stands now, I have to read their subject line...terrible!

It's up to 68 now! And ignoring those people has contributed greatly to a more pleasant forum experience for me.

You, however, are a special case and your OP illustrates why. While the posts of 60 or so of those people are simply worthless, your posts (and those of a few others) have great potential to induce such as exasperated state of confusion in a beginner that he’s dishearten and discouraged from pursuing the understanding he sought. And as such, must be reviewed from time to time.

Here's a simpler explanation...

"Noise" refers to statistical variations in light as it arrives at the sensor.
These variations are a percentage of the amount of light recorded.
The percentage gets smaller as more light is collected.
Therefore, collecting more light results in less noise.
Increasing ISO results in less light collected, and therefore more noise.

It is indeed simpler (and even correct!), but not as complete.

It provides enough for a beginner to understand why images can have different degrees of noise, and identifies the camera function that will affect that noise, which is generally what most beginners are concerned about (a generalization drawn from reading hundreds upon hundreds of posts from beginners.)

For those who want to know more...

The variation in light is defined as the square root of the amount of light recorded. So if the light collected gives a signal of 100 (don't worry about "100 what?" for now) then the noise is 10, which is 10%. If the light collected gives a signal of 10,000, then the noise is 100, which is 1%.

When do they get to know "100 what"?

When the beginner feels he wants to know...he'll ask.

What if we were to multiply the smaller signal (100) by 100 to get a signal of 10,000? Does the noise become 1? No, the noise is also multiplied by 100. So the noise is 1,000 and is still 10%. The more-light-less-noise process only works when you're actually collecting the light.

OK, were you ever going to discuss photon noise vs read noise, or is sensor efficiency unimportant?

It not important because generally the beginner is concerned with controlling the output of his camera. A discussion of read noise or sensor efficiency does not contribute to that end.

How about luminance vs color noise?

Noise is noise. The beginner can read about that distinction in his software help when he wonders why where are two sliders to reduce noise.

Too advanced to even mention for these poor ignorant beginners? Will it make their head explode?

Did your teachers tell you everything about the subject on the first day of class?

When you increase the ISO of your camera, you are telling the camera to first collect less light than is required to properly expose the image, and then multiply the light signals by some factor. The resulting signals are at the same level that a low-ISO image would have provided. That's how you get the same image with less light. However, remember that multiplying the light signals results in more noise than having collected more light to begin with. Therefore, increasing the ISO setting results in images with more noise.

I'm wondering if anyone might not be wonder why increasing the ISO tells the camera to collect less light, and why anyone would want to do that.

That's good info to know, but it has nothing to do with noise. That's a discussion on applying exposure controls to photography, and beyond the scope of a discussion on noise. If the beginner is wondering, he'll ask.

So, it seems that [you] feel a post with 1/3 the text that omits discussing luminance vs color noise, photon vs read noise, why raising the ISO makes the camera collect less light, and without any links to other references is much better for the beginner.

It addresses most beginners' immediate concerns. You seem to have an issue with letting beginners learn at their own pace.

.

 

[Great Bustard]

I've often been wondering.. when discussing noise we're almost always assuming a "grey patch" like you did above, in order to simplify things, but that a rather rare scenario in real world shooting. So, how do the colors of the real world affect the amount of visible noise in the image? Let's take for example a blue sky, that often, as far as I can tell, seems to have surprisingly much visible noise, even in a low ISO DSLR image. Does a blue sky in practice have more visible noise than a grey patch with the same 'brightness'?

Well 'grey' is a figurative term. The point is that there is no noise in a single pixel. There is only noise as a variance from a mean in a set of observations (pixels). All the noise calculations make a tacit assumption of a set which has a fixed, steady mean, and then you measure the variance over the set. Joofa has made the point that the same statistics do not apply to a set where the mean varies across the set. I've been doing some experiments to find out whether that has a practical effect, but it's actually quite tricky to do them properly.
...
--

Hi Bob and Steen, Well, the effects that Joofa has talked about over the years is true for real world images; which attempts to explain things other than that of uniform grey patches. In order to understand this difference, let us first take a picture of a uniform wall, that might give us an average per pixel signal of the aforementioned 10,000 electrons. Then we might say that, since this is a uniform area, the noise = 100 e, and SNR = 100:1. But then again, how exciting is this, since there is no detail in this patch.

Then we can go up a step in complication in this wall image by now imagining that we impose some detail into the patch. The next level in going to detail would be to imagine that there is a bar pattern of some low spatial frequency, sinusoidally varying intensity pattern that modulates the uniform field. Let's imagine that the amplitude of this sinusoidal pattern is 10% of the backround, or an excursion of 1,000 e about the mean of 10,000 e, as measured by the camera (i.e. after all of the camera's optics has had their MTF influence upon the scene. Then what we see that we're interested in is pulling out this 1,000 e of detail out of the noise created by the 10,000 e average signal, this concept is called Contrast SNR. In this case the Contrast SNR = 1,000 e/100 e = 10:1, which is much lower than the 'usual' SNR of 100:1. This is a different quantity that goes in the numerator, in this next level of complication.

This Contrast SNR is defined as:
SNR(con) = MTF*C*SNR

where C is the contrast of the scene and MTF is from the optics, for any spatial frequency. We can think of SNR as what you'd get at zero spatial frequency (which is the definition of a uniform patch!). So now MTF is involved in this.

Now that is some pretty clever thinking, especially inasumuch as I am a big proponent of saying that lens sharpness contributes a great deal image noise, since a sharper photo can withstand more NR (noise reduction) for a given level of detail.

If you're interested, the math I got from this is found in the following references:

Janesick, J., "Lux transfer: Complementary metal oxide semiconductors versus charge-coupled devices," Optical Engineering, Vol. 41 No. 6, June 2002.

http://spie.org/x648.html?product_id=725073

Both of which are good light reading!

The "Photon Transfer" book looks pretty damned good, actually! I did a quick browse, and it has a derivation of why photon noise follows a Poisson Distribution, something I've long wondered and never new! I may well have to buy that book!

Much thanks!

 

[Great Bustard]

This is an excellent tutorial on the subject - much easier to comprehend and apply:

http://ronbigelow.com/articles/noise-1/noise-1.htm

...I do have to agree that it is a pretty good tutorial, although Bob's criticisms have a great deal of merit, particularly the differentiation between noise at the pixel level, and noise at the image level. In particular, he attributes smaller pixels with greater noise than larger pixels for a given exposure, as opposed to smaller sensors having more noise than smaller sensors for a given exposure (and efficiency).

However, while that error is a biggie, overall I have to say it's an excellent and well written article on noise, and I much appreciate you linking to it! As Dr. Martinec (ejmartin) has said above that he has emailed the author pointing out errors, one would hope that the author would correct those errors, and then I would happily concede that his explanation on noise is superior to my attempt in the OP.

Again, much thanks for that link, but it does need some rewording.

 

[Great Bustard]

...did it?

http://forums.dpreview.com/forums/read.asp?forum=1002&message=38953022

I should have known better than to respond to you. Well I won't make the same mistake again.

Funny -- you seem to have a knack for mistakes. Not that I mind, of course -- please, feel free to respond anytime. Just a bit childish to make comments like that, brag about how many people are on your ignore list:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=37942523

Yeah, I have 26 people on ignore already. I wish that ignore function would completely hide the existence of a post by an ignored person. As it stands now, I have to read their subject line...terrible!

It's up to 68 now! And ignoring those people has contributed greatly to a more pleasant forum experience for me.

Whatever it takes to make the "DPR experience" more pleasant for you.

You, however, are a special case and your OP illustrates why. While the posts of 60 or so of those people are simply worthless, your posts (and those of a few others) have great potential to induce such as exasperated state of confusion in a beginner that he’s dishearten and discouraged from pursuing the understanding he sought. And as such, must be reviewed from time to time.

I'm pleased to have you watching over me. That's not a sarcastic quip, by the way. I always appreciate the opportunity to explain why the naysayers are in error, and I always appreciate when the naysayers find me to be in error, so that I can correct myself.

However, I have to say that I'm likely in much less need of correction than you are, as clearly demonstrated here:

http://forums.dpreview.com/forums/read.asp?forum=1002&message=38953107

I mean, that's a huge slew of errors that you made. In fact, I find it more than a little hypocritical of you to say that I was wrong when I wrote:
http://forums.dpreview.com/forums/read.asp?forum=1002&message=38952700

But don't you think telling them why it increases noise might not be of value? That is, raising the ISO in an AE (auto exposure) mode results in a more narrow aperture (larger f-ratio), faster shutter speed, and/or less light output from the flash, all of which have the effect of putting less light on the sensor, and less light means more noise.

Yet incorporating the whole of what I said in your version of what you thought was a better tutorial on noise for beginners. As for the rest of your post, you'll excuse me if I skip it. But, if I may be so bold, may I suggest reading Michael Firstlight's criticism of my post and his alternative suggestion:

http://forums.dpreview.com/forums/read.asp?forum=1002&message=38962486

as well as the replies to it. A perfect example of a naysayer making a constructive criticism, linking to a resource I did not know about, and giving me (and others) the opportunity to offer some correction of our own.