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Calculation of snr

 
 
David J Taylor
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      05-26-2008
Floyd L. Davidson wrote:
[]
>> Not at all. David is discussing the misuse of the term decibel.

>
> His examples of what it is are a misuse. He continues
> to claim that such things as dBHz and dBv are not a
> power ratio. Now he even claims that weighted values,
> such as dBrnC are somehow significant to the discussion.


Excuse me, I have never talked about dBrnC - you introduced that! Please
don't misquote me.

dBV are voltages, and only become powers when an impedance is considered.
dBHz are bandwidths, even though they are based on a noise-equivalent
ratio.

David


 
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ASAAR
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      05-26-2008
On Mon, 26 May 2008 05:21:50 -0800, Floyd L. Davidson emerged from
his bunker to write:

>> Being out of touch and confused are just some of the side effects
>> of seeing the world only through Uncle Floyd's personal lense!

>
> 1) My point is that the standard definition of the
> term decibel determines correct usage.
>
> 2) Your point is that gratuitous personal insults are
> your one and only contribution to the discussion.


How bizarre. The gratuitous insult king is seemingly offended by
an insult more humorous and less mean that the ones that are his
stock in trade . . . If you could only understand the real reason
why I replied as I did (and learn from it), the discussions in this,
and in other threads would have a greater signal to noise ratio!


> Readers can draw their own conclusions about side
> effects and confusion...


They have, and they are legion, whether you recognize them or not.

 
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Kennedy McEwen
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      05-26-2008
In article <(E-Mail Removed)>, Floyd L. Davidson
<(E-Mail Removed)> writes
>Kennedy McEwen <(E-Mail Removed)> wrote:
>>In article <(E-Mail Removed)>, Floyd L. Davidson
>><(E-Mail Removed)> writes
>>>Kennedy McEwen <(E-Mail Removed)> wrote:
>>>>>It is a power ratio by definition.
>>>>>
>>>>Only if you follow the definition. If you don't, such as by using
>>>
>>>If you don't, you are not talking about decibels, but
>>>something else.
>>>

>>Correct.

>
>Then stick to talking about decibels, rather than something
>else.
>

Whilst you would like to focus the discussion on decibels, it is
actually the calculation of those decibels that are the discussion - and
have been since Marc made his original post. Irrespective of the claim
you made in your original post in this thread, and have subsequently
repeated, one cannot discuss the calculation of those decibels WITHOUT
reference to the initial parameters available.
>
>The equation, because it produces a figure in decibels,
>has to represent power, regardless of what the signal
>actually is.
>

Only if it is a correct equation. Using the wrong input parameters, as
you suggested can be done, does not result in decibels.

Its a bit like saying distance = velocity x time, but it doesn't matter
whether your variables are velocity and time or not, its still distance!
Not if your variables are mass and acceleration its not!

>How precise could I have been!
>

Much more. You could have stated that this equation was only correct
when using variable which were proportional to the square root of power,
instead of stating that it didn't matter at all.

>You still don't seem to understand the distinction that
>is necessary for precision when discussing how decibels
>are used.
>

Since we are discussing your lack of precision, I suggest you review
where the problem lies!

>>David is discussing the misuse of the term decibel.

>
>His examples of what it is are a misuse. He continues
>to claim that such things as dBHz and dBv are not a
>power ratio. Now he even claims that weighted values,
>such as dBrnC are somehow significant to the discussion.
>
>That is confusion, not "discussing the misuse"!
>

You seem to be confused about what he is discussing. Come to think of
it, you seem to be confused about what anyone else is discussing!

>>Your continued confusion of the two issues is barely matched by your
>>false assurance to readers that the same equation results in the power
>>ratio in decibels "regardless of what the signal actually is". Precisely
>>this false belief results in the misapplication of the originally quoted
>>equation not only in engineering examination papers but in practical
>>engineering.

>
>If you cannot correctly summarize what I've said, it is
>invalid to attempt to draw conclusions from *your* invalid
>summary.
>

I have not summarised what you said, I quoted it directly! Go back and
READ what you wrote!

>I've said that all of the various equations result in a
>power ratio


No you didn't. At the point where you replied, only one equation had
been tabled and you entered to "explain" that "It's a ratio of two
numbers, as shown in the formula above. The numbers represent signal
power, regardless of what the signal actually is."

NO MENTION AT ALL OF ANY DIFFERENT EQUATION BEING REQUIRED IF THE SIGNAL
IS NOT PROPORTIONAL TO THE SQUARE ROOT OF VOLTAGE.

That actually came from my response, pointing out that your final
sentence was the cause of the most common mistake in using the dB.

>, not that they are the exact same equation
>for different signals. The *result* (a power ratio) is
>the same. It *has* to be that, because that is the
>definition of decibel.
>

Ignoring the topic while perpetually repeating the definition does
nothing to further your case since nobody is arguing about the
definition or what a decibel is, but how the available parameters are
used to calculate it.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's ****ed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)
 
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Kennedy McEwen
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      05-26-2008
In article <(E-Mail Removed)>, Floyd L. Davidson
<(E-Mail Removed)> writes
>
>The basic problem with the deleted discussion is the
>question of exactly *what* SNR it is you want to
>measure. Kennedy is trying to measure the SNR of the
>light that lands on the sensor. I doubt that a typical
>photographer cares about that


Au contraire, I think that is precisely what the typical photographer
cares about: just how well any camera interprets the light it receives.

There are a million and more ways to design a camera or sensor and each
can have a different output format. It is how that format relates to
the input, the light, that is important to the photographer, not the
power dissipation into a fixed resistive load buried somewhere inside
the electronics.

Sure, if you are designing the sensor interface circuit then knowing the
peak power ratio that the sensor produces is important, but that should
be totally transparent to the end user.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's ****ed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)
 
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David J Taylor
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      05-26-2008
John O'Flaherty wrote:
[]
> dBV are ratios of voltage squared, so they aren't voltages on two
> counts: they are dimensionless, and the voltage arguments have been
> squared by the factor of 20 applied to the logarithm. The purpose of
> the unit is to express a ratio of two different powers at the same
> point in a circuit in terms of voltage.
> Suppose you have a circuit where you now measure a signal of 100 V,
> but originally you measured 1 V. The second signal is
> 20*log10(100/1) = 40 dBV.
> Now calculate the power
> 10*log10(100^2/r / 1^2/r) = 40 dB
> The numerical equality shows that dBV measures power; the only
> difference between dB and dBV is that the latter assumes a reference
> power resulting from a level of 1 V.
> While dBV can be used to express a voltage level, it will do so
> indirectly by referring to the developed power.


They are a power (ratio) only if the impedances are the same. I can say
40dBV, and that's 100V rms irrespective of what the impedance is, or what
the power is. dBV is most certainly /not/ dimensionless.

Yes, of course, if you want to know the power gain, you need to know the
impedances and, yes, if the impedances are the same (such as in a 50-ohm
system), the relative voltage levels and the relative power levels will be
the same - a 20dB amplifier providing ten times voltage gain and one
hundred times power gain.

Cheers,
David


 
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Marc Wossner
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      05-26-2008
On 26 Mai, 15:15, Kennedy McEwen <(E-Mail Removed)> wrote:
> In article
> <(E-Mail Removed)>,
> Marc Wossner <(E-Mail Removed)> writes
>
> >As this highly theoretical discussion is not so easy to follow for me,
> >IŽd like to bring it back to something that I can work with: If I
> >chose the numerator of snr to be *the range available in the image
> >file -256 for 8-bit or *4096 for 12-bit - and the denominator to be
> >the standard deviation as the equivalent to RMS noise, is the form SNR
> >= 20 log (Signal RMS / Noise RMS) correct or should it be SNR = 10 log
> >(Signal RMS / Noise RMS)?

>
> Sadly, Mark, despite Floyds assurance that it doesn't matter what the
> signal is, neither of these equations are generically suitable because
> the image data has been processed from the sensor output according to a
> response curve.
>
> To estimate the SNR in dB you must use signals which can be converted to
> power by a fixed formula - it doesn't have to be linear, there can be
> exponential terms, but it must apply across all ranges. *Typically the
> response curve used to produce the image data is an "S-type" curve with
> different responses used in the lower values from those in the peaks.
> That means that your noise terms will be artificially high, while the
> signal terms will be artificially low - and hence the estimated SNR will
> be very low indeed.
>
> What you really need is the raw data from the sensor. *Some
> manufacturers do provide this directly however others apply significant
> processing to the raw data from some of their cameras. *For example,
> Nikon, are known to apply chroma noise reduction to many of their
> camera's raw outputs.
>
> If you can get hold of the raw data and convert it to a linear output
> (free from response curves) then the first step is to baseline in. *This
> is necessary because "no light" does not necessarily result in "0" in
> the raw data file - there may be significant dark current in the sensor,
> more likely though, there can be significant voltage offsets between the
> sensor output and the analogue to digital convertor.
>
> At this point you can crudely approximate the SNR by using the 10xlog
> version. *The sensor converts individual photons to electrons with a
> response known as the quantum efficiency. *Each photon has an energy
> which is determined by its wavelength (E = hc/w) and so the light power
> can be determined from the number of monochromatic photons arriving at
> the sensor over a period of time (P=E/t). *Similarly the sensor outputs
> a certain number of electrons per second to create a photocurrent, which
> is then proportionally converted to volts by the load and hence into
> numbers by the analogue to digital convertor.
>
> The significant point here (actually the underlying reason for my
> extended discussion with Floyd) is that the voltage output by the
> sensor, and hence the corrected numbers in the raw data file, are
> linearly proportional to the incident monochromatic light power on the
> sensor. *Consequently, even though the numbers represent voltages
> directly, they are linearly proportional to power input to the camera
> and that is what determines the correct formula to use. *(Floyds
> statement about disregarding what the signal represents is exceedingly
> bad advice.)
>
> As Ilya says in his post however, the common usage in this context
> actually contradicts the formal definition - primarily because it is so
> common to make Floyd's error. *Hence it is best to avoid this approach,
> or at least be prepared to apply corrective factors of 2 to the data you
> wish to compare your results with.
> --
> Kennedy
> Yes, Socrates himself is particularly missed;
> A lovely little thinker, but a bugger when he's ****ed.
> Python Philosophers * * * * (replace 'nospam' with 'kennedym' when replying)



Thanks a lot for this clean summary Kennedy!
Now the context is even clearer to me than in Ilyas post.
But if db should be avoided in this context, how should snr be defined
to be comparable?

Best regards!
Marc Wossner
 
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Peter Irwin
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      05-27-2008
Floyd L. Davidson <(E-Mail Removed)> wrote:
>
> People that work with signal transmission, analysis or
> processing are necessarily comfortable using decibels,
> but most photographers do not have that sort of
> background, and on initial contact are confused by it.
>

Photographers have been using log10 since before there
were any electronics. If you use 6dB to indicate the
same change that photographers know as a log exposure difference
of 0.3 then you have to clearly explain why you are doing so.
Yes, it may make sense in the specifications for the amplifier
and ADC, but photographers are rarely concerned with the analogue
electrical signal levels in their digital cameras. Photographers
are concerned with the light that goes in to the camera and the
recording of that light which comes out.

Peter.
--
http://www.velocityreviews.com/forums/(E-Mail Removed)


 
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David J Taylor
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      05-27-2008
John O'Flaherty wrote:
[]

I don't agree that dBV is dimensionless - yes, it is a ratio of voltages
but you could not equate dBV with dBm or dBA. dBV defines a voltage
level. As with other uses of dB you can think of it as the ratio of a
power at a particular impedance to the power from a one volt signal, hence
20 * log (V2/V1), but dBV is defining a voltage level and not a sound
pressure level or power level or a bandwidth.

I am slightly concerned that trying to use dB as a single measure of
camera performance may be fraught with danger unless the effects of MTF,
at least, are taken into account. I would be happy with using voltage
decibels to measure SNR, but with the OP mentioning things like 12-bit raw
and 8-bit JPEG does bring up the whole issue of the non-linear processing
in the camera (to get the 8-bit JPEG), and of the effects of the Poisson
noise in the basic photon stream.

It's a complex subject!

Cheers,
David


 
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Kennedy McEwen
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      05-27-2008
In article <(E-Mail Removed)>, Floyd L. Davidson
<(E-Mail Removed)> writes
>Kennedy McEwen <(E-Mail Removed)> wrote:
>>In article <(E-Mail Removed)>, Floyd L. Davidson
>><(E-Mail Removed)> writes
>>>
>>>The basic problem with the deleted discussion is the
>>>question of exactly *what* SNR it is you want to
>>>measure. Kennedy is trying to measure the SNR of the
>>>light that lands on the sensor. I doubt that a typical
>>>photographer cares about that

>>
>>Au contraire, I think that is precisely what the typical photographer
>>cares about: just how well any camera interprets the light it receives.

>
>Why, for example, did the OP ask about the SNR of the
>digital image values (specifically mentioning 8-bit and
>12-bit values)?
>

Because that was the data available to him.
>
>You were referring to the design of a system to measure
>noise in the light.
>

No, I was explaining how to measure the signal to noise ratio that the
camera is capable of measuring.

>If we are designing cameras, not light noise meters, we
>are interested in the noise within the system as a
>separate entity from the noise in the light source.
>

Not at all. We are interested in how well the camera interprets the
light it receives: the ratio of the highest light power it can receive
without saturating compared to the lowest light power it can
meaningfully produce a signal from.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's ****ed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)
 
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Marc Wossner
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Posts: n/a
 
      05-27-2008
On 27 Mai, 03:23, (E-Mail Removed) (Floyd L. Davidson) wrote:
> Marc Wossner <(E-Mail Removed)> wrote:
> >On 26 Mai, 15:15, Kennedy McEwen <(E-Mail Removed)> wrote:
> >> In article
> >> <(E-Mail Removed)>,
> >> Marc Wossner <(E-Mail Removed)> writes

>
> >> >As this highly theoretical discussion is not so easy to follow for me,
> >> >IŽd like to bring it back to something that I can work with: If I
> >> >chose the numerator of snr to be *the range available in the image
> >> >file -256 for 8-bit or *4096 for 12-bit - and the denominator to be
> >> >the standard deviation as the equivalent to RMS noise, is the form SNR
> >> >= 20 log (Signal RMS / Noise RMS) correct or should it be SNR = 10 log
> >> >(Signal RMS / Noise RMS)?

>
> >> Sadly, Mark, despite Floyds assurance that it doesn't matter what the
> >> signal is, neither of these equations are generically suitable because
> >> the image data has been processed from the sensor output according to a
> >> response curve.

>
> >> To estimate the SNR in dB you must use signals which can be converted to
> >> power by a fixed formula - it doesn't have to be linear, there can be
> >> exponential terms, but it must apply across all ranges. *Typically the
> >> response curve used to produce the image data is an "S-type" curve with
> >> different responses used in the lower values from those in the peaks.
> >> That means that your noise terms will be artificially high, while the
> >> signal terms will be artificially low - and hence the estimated SNR will
> >> be very low indeed.

>
> >> What you really need is the raw data from the sensor. *Some
> >> manufacturers do provide this directly however others apply significant
> >> processing to the raw data from some of their cameras. *For example,
> >> Nikon, are known to apply chroma noise reduction to many of their
> >> camera's raw outputs.

>
> >> If you can get hold of the raw data and convert it to a linear output
> >> (free from response curves) then the first step is to baseline in. *This
> >> is necessary because "no light" does not necessarily result in "0" in
> >> the raw data file - there may be significant dark current in the sensor,
> >> more likely though, there can be significant voltage offsets between the
> >> sensor output and the analogue to digital convertor.

>
> >> At this point you can crudely approximate the SNR by using the 10xlog
> >> version. *The sensor converts individual photons to electrons with a
> >> response known as the quantum efficiency. *Each photon has an energy
> >> which is determined by its wavelength (E = hc/w) and so the light power
> >> can be determined from the number of monochromatic photons arriving at
> >> the sensor over a period of time (P=E/t). *Similarly the sensor outputs
> >> a certain number of electrons per second to create a photocurrent, which
> >> is then proportionally converted to volts by the load and hence into
> >> numbers by the analogue to digital convertor.

>
> >> The significant point here (actually the underlying reason for my
> >> extended discussion with Floyd) is that the voltage output by the
> >> sensor, and hence the corrected numbers in the raw data file, are
> >> linearly proportional to the incident monochromatic light power on the
> >> sensor. *Consequently, even though the numbers represent voltages
> >> directly, they are linearly proportional to power input to the camera
> >> and that is what determines the correct formula to use. *(Floyds
> >> statement about disregarding what the signal represents is exceedingly
> >> bad advice.)

>
> >> As Ilya says in his post however, the common usage in this context
> >> actually contradicts the formal definition - primarily because it is so
> >> common to make Floyd's error. *Hence it is best to avoid this approach,
> >> or at least be prepared to apply corrective factors of 2 to the data you
> >> wish to compare your results with.
> >> --
> >> Kennedy
> >> Yes, Socrates himself is particularly missed;
> >> A lovely little thinker, but a bugger when he's ****ed.
> >> Python Philosophers * * * * (replace 'nospam' with 'kennedym' when replying)

>
> >Thanks a lot for this clean summary Kennedy!
> >Now the context is even clearer to me than in Ilyas post.

>
> Like pond water, after somebody stirs the mud with a
> stick?



I meant that it made sense to me and my limited knowledge but due to
that limited knowledge I cant figure out if its also true.


> >But if db should be avoided in this context, how should snr be defined
> >to be comparable?

>
> The same way that virtually all engineers do it. *Using
> dB! *Because it is comparable, and it makes good sense.
>
> Note that "comparable" may have a meaning here that is
> or is not obvious. *There is comparable between
> different systems, which is probably what you were
> thinking. *


Yes, how to compare signal quality/noise between different cameras.

> But another much more significant point is
> that within a single system the values need to be
> compared between component parts. *Gain, Dynamic Range,
> and SNR values at each point along the signal path only
> have clear and obvious meaning in relation to each other
> when they are expressed in dB.


This makes sense to me either.

> Note that if you dig up specifications for CCD's, ADC,
> signal amplifiers, or other signal path components in
> digital cameras (or for digital cameras themselves) it
> is typical to specify SNR, Gain, and Dynamic Range
> in terms of dB's.
>
> People that work with signal transmission, analysis or
> processing are necessarily comfortable using decibels,
> but most photographers do not have that sort of
> background, and on initial contact are confused by it.


Absolutely: Totally confused now!

Best regards!
Marc Wossner

 
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