Calculation of snr

"David J Taylor" wrote:
Floyd L. Davidson wrote:
"David J Taylor"
[]
However, dB are actually a ratio, and can be used to measure other
things than power (as I'm sure you will know).

That is what gets people into big trouble trying to use
dB.

dB is *defined* as a power ratio. It only makes sense
when the right side of the equation is a formula that
equates to a *power* ratio.

I have seen the dB used for voltage levels (typically relative to 1
microvolt or 1 millivolt - dBuV or dBmV), in digital audio (relative to

Power levels.

full scale - dBFS), and in bandwidth (dBHz). In all of these, it is used

Power levels. (And note that dBHz is not bandwidth, it
is a power ratio, commonly used for carrier to noise.)

as a ratio, with an implication of power, although the voltage
measurements may not be at the same impedance.

The voltage measurement is in fact defined at a specified impedance.

I completely agree that you need to be very careful!

But your post is not. Decibels are relative to _power_,
every time.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

Kennedy McEwen wrote:
In article , Floyd L. Davidson
writes
Kennedy McEwen wrote:
In article , Floyd L. Davidson
writes
Kennedy McEwen wrote:
In article , Floyd L. Davidson
writes
ransley wrote:
On May 22, 5:28*am, Marc Wossner wrote:

SNR = 20 log (Signal RMS / Noise RMS)

dB has no particular attachment to sound, any more than
it does to light. 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.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Read what I wrote very carefully Kennedy; it should be
very apparent that right from the start I know exactly
what you are referring to.

I have, and I think you will realise that, from the start I have
considered what you wrote to be ambiguous at best or indeed, if your
words are interpreted in a logical manner but not necessarily the manner
you meant, completely wrong.

"Its a ratio of two numbers" followed by "the numbers" (ie. the SAME
numbers) "represent signal power, regardless..."

That is completely wrong. "The numbers" do NOT represent signal power,
they are voltages. The ratio (signal / noise) is a VOLTAGE ratio.

A decibel is a power ratio, by definition.

The ratio (signal^2 / noise^2) IS a power ratio,

End of discussion.

....

It is a *power* ratio. That is true regardless of what
the two numbers actually represent.

No it isn't!

It is a power ratio by definition.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

Floyd L. Davidson wrote:
[]
But your post is not. Decibels are relative to _power_,
every time.

No, decibels are a ratio every time. It's quite possible to have a
voltage amplifier where the input impedance is very high, so that while
the voltage gain may be 20dB, the power gain is huge. Another example, a
photodetector working as a current source, into a near zero-resistance
load.

As I said, you need to be careful in such circumstances.

David

In article , Floyd L. Davidson
writes
Kennedy McEwen wrote:
In article , Floyd L. Davidson
writes
Kennedy McEwen wrote:
In article , Floyd L. Davidson
writes
Kennedy McEwen wrote:
In article , Floyd L. Davidson
writes
ransley wrote:
On May 22, 5:28*am, Marc Wossner wrote:

SNR = 20 log (Signal RMS / Noise RMS)

dB has no particular attachment to sound, any more than
it does to light. 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.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Read what I wrote very carefully Kennedy; it should be
very apparent that right from the start I know exactly
what you are referring to.

I have, and I think you will realise that, from the start I have
considered what you wrote to be ambiguous at best or indeed, if your
words are interpreted in a logical manner but not necessarily the manner
you meant, completely wrong.

"Its a ratio of two numbers" followed by "the numbers" (ie. the SAME
numbers) "represent signal power, regardless..."

That is completely wrong. "The numbers" do NOT represent signal power,
they are voltages. The ratio (signal / noise) is a VOLTAGE ratio.

A decibel is a power ratio, by definition.

The DECIBEL is, but NOT the ratio of the numbers you referred to.

The ratio (signal^2 / noise^2) IS a power ratio,

End of discussion.

...

It is a *power* ratio. That is true regardless of what
the two numbers actually represent.

No it isn't!

It is a power ratio by definition.

Only if you follow the definition. If you don't, such as by using
numbers without regard to their meaning, exactly as you suggested, then
you are not following the definition and therefore end up with something
that is NOT a power ratio.
--
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)

"David J Taylor" wrote:
Floyd L. Davidson wrote:
[]
But your post is not. Decibels are relative to _power_,
every time.

No, decibels are a ratio every time. It's quite possible to have a
voltage amplifier where the input impedance is very high, so that while
the voltage gain may be 20dB, the power gain is huge. Another example, a
photodetector working as a current source, into a near zero-resistance
load.

As I said, you need to be careful in such circumstances.

David, you still don't get it: a decibel is a power
ratio EVERY TIME. That is by definition.

You continue to use the term carelessly.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

Floyd L. Davidson wrote:
[]
David, you still don't get it: a decibel is a power
ratio EVERY TIME. That is by definition.

You continue to use the term carelessly.

I'm not saying that I use the term that way, simply that it is common
practice.

David

Kennedy McEwen 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.

numbers without regard to their meaning, exactly as you suggested, then

I was talking about *decibels*, and was very precise.

you are not following the definition and therefore end up with something
that is NOT a power ratio.

That is what you continue to boost. So does David J Taylor.

But it does not apply to decibels.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

[A complimentary Cc of this posting was sent to
Marc Wossner
], who wrote in article :
As this highly theoretical discussion is not so easy to follow for me,

As it is not for everybody. The recipe is very simple: DO NOT USE dB
in non-linear context; and, more specific, DO NOT USE dB in context of
light intensity.

The common usage of dB [one imported from measuring voltage in the
universe of resistive loads] (as 6dB = 2x-change) contradicts the
formal definition in context of light intensity (which is 3db =
2x-change).

What is boils down is that dB in such a context are too confusing.
Use values with logs, or use log-base-2 ("steps").

I=B4d 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
=3D 20 log (Signal RMS / Noise RMS) correct or should it be SNR =3D 10 log
(Signal RMS / Noise RMS)?

Just forget about this question...

Hope this helps,
Ilya

"David J Taylor" wrote:
Floyd L. Davidson wrote:
[]
David, you still don't get it: a decibel is a power
ratio EVERY TIME. That is by definition.

You continue to use the term carelessly.

I'm not saying that I use the term that way, simply that it is common
practice.

David, you listed a number of what you thought were
examples of it not being a power ratio (dBV and dBHz are
two that I remember), and when that was pointed out as
an error, you gave "examples" that did not mention
decibels and that did not even apparently involve
decibels!

As I noted to start with, what we are discussing *is*
the most common _error_ made in using the term. Keep
trying all you like, but it is indeed in error if it is
not a power ratio, and you are confused when you think it
correct and not a power ratio.

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)

Ilya Zakharevich wrote:
[A complimentary Cc of this posting was sent to
Marc Wossner
], who wrote in article :
As this highly theoretical discussion is not so easy to follow for me,

As it is not for everybody. The recipe is very simple: DO NOT USE dB
in non-linear context; and, more specific, DO NOT USE dB in context of
light intensity.

And do use it for the analog output of a light sensor!

The common usage of dB [one imported from measuring voltage in the
universe of resistive loads] (as 6dB = 2x-change) contradicts the
formal definition in context of light intensity (which is 3db =
2x-change).

What is boils down is that dB in such a context are too confusing.
Use values with logs, or use log-base-2 ("steps").

Such as dB!

I=B4d 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
=3D 20 log (Signal RMS / Noise RMS) correct or should it be SNR =3D 10 log
(Signal RMS / Noise RMS)?

Just forget about this question...

The digital values are derived from the analog voltages
generated by the sensor. To get a power ratio for dB
one can either square the digital value, or multiply the
resulting logarithm value by 2.

The basic formula is:

dB = 10 log ( Signal_power / Noise_power )

Given a voltage measurement the power can be derived with
either of these methods:

dB = 10 * log ( Signal_voltage ^ 2 / Noise_voltage ^ 2 )
dB = 2 * 10 * log ( Signal_voltage / Noise_voltage )

--
Floyd L. Davidson http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska)