Talk:Signal-to-noise ratio

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Information theory has been around since 1948; telecommunications much longer, and usenet only since the '80s. So "signal-to-noise ratio" cannot have originated in usenet. 131.183.81.100 01:58 Mar 4, 2003 (UTC)

1 What does this mean?
2 What's better?
3 Associated Page may crash your PC
4 Digital SNR derivation
- 4.1 A million different equivalent equations
5 SNR = 8 sigma ... what does the sigma mean?

[edit] What does this mean?

In the article, this was placed without context:

In Audio Coding there are terms like
- NMT
- TMN
- SMR

These links don't really go anywhere or appear to be much use, so I removed them. --Twinxor 04:07, 12 May 2005 (UTC)

[edit] What's better?

What's better? A high or a low SNR? Maybe someoine can give the formula for calculating SNR? Thanks, --Abdull 10:01, 21 May 2005 (UTC)

Generally, a high SNR is desirable since the signal is defined to be something of interest. I'm hard pressed to think of an example when one would desire a very low SNR (no signal, only noise). (The best I can do is if you interpret the channel as an encrypted one where the signal is the unencrypted data....then you would want the "viewer" to see as much "noise" as possible so they can't "see" the unencrypted data.) But, generally, the higher the SNR the better.

As for the formula, it can vary slightly. Generally:

$SNR = \frac{P_{signal}}{P_{noise}}$

which is the ratio of powers of the signal and the noise. Cburnett 20:42, May 24, 2005 (UTC)

[edit] Associated Page may crash your PC

Modern PC, plenty of memory using Internet Explorer 6.0. PC crashed about where the second formula is.

Can edit the page OK - but don't know enough to work out why I got a blue screen of death. This occurred on a colleagues PC, so is repeatable (older PC but same vsn of IE).

Help?

[edit] Digital SNR derivation

As far as I know, 6*n dB is an approximation. Also, the formula listed contradicts itself:

$\mathrm{SNR} = 10(2n-1) \cdot \log_{10}(2)+10 \cdot 3 = 6.02 \cdot n + 26.989$

not $6.02 \cdot n + 1.763$

This PDF has a different derivation:

$\mathrm{SNR} = 2^n ( {\sqrt {3} / \sqrt{2}} )$

$\mathrm{SNR_{dB}} = 20 \log_{10} \left ( 2^n ( {\sqrt {3} / \sqrt{2}} ) \right )$

$\mathrm{SNR_{dB}} = 20 \log_{10} ( 2) \cdot n + 20 \log_{10} ( \sqrt {3} / \sqrt{2})$

SNR dB = 6.021 n + 1.763

— Omegatron 15:13, August 30, 2005 (UTC)

Thanks for fixing the formula, User:212.119.9.186!

With more precision is 6.020599 n + 1.760913, if it matters. — Omegatron 15:49, 10 October 2005 (UTC)

Also, $10(2n-1) \cdot \log_{10}(2)+10\log_{10}(3) = 20 \log_{10} ( 2) \cdot n + 20 \log_{10} ( \sqrt {3} / \sqrt{2})$ — Omegatron 01:20, 11 October 2005 (UTC)

Ok, the 6.02n formula corresponds to a noise level caused by a uniform fluctuation between two quantization levels. I think another way of saying this is that the sampled value is assumed to be uniformly distributed across the possible range of values.

The 6.02n+1.761 formula maybe only applies to the ratio of a full-scale sine wave to the error signal, as hinted at in the article? So the signal is assumed to be a sine wave instead of being unknown but uniformly distributed. I will try to do the math later.

Is there another for a Gaussian distribution? — Omegatron 17:32, 30 November 2005 (UTC)

As I've explained in the article, the noise is signal-dependent, so SNR for a system requires that you come up with a model of what that signal is going to look like first. — Omegatron 20:25, 12 March 2006 (UTC)

RMS signal = $A \over \sqrt{2}$ or $FSR/2 \over \sqrt{2}$ where FSR is the full-scale range of the converter

RMS noise = $q \over \sqrt{12}$ where q is the LSB

[edit] A million different equivalent equations

$10(2n-1) \cdot \log_{10}(2)+10\log_{10}(3)$
$20 \log_{10}(2^{n-1} \sqrt{6})$ [1]
$20 \log_{10} \left ( {2^n \sqrt{12} \over 2 \sqrt{2}} \right )$ [2]
$20 \log_{10} ( 2) \cdot n + 20 \log_{10} ( \sqrt {3} / \sqrt{2})$ [3]
$2n\cdot10\log_{10}(2)+10\log_{10}(3/2)$ Totally forgot we have a derivation over at Talk:Quantization_noise#Thesis — Omegatron 02:35, 13 March 2006 (UTC)

[edit] SNR = 8 sigma ... what does the sigma mean?

Nasa describes the performance of a CCD in a space-based optical telescope as having SNR = 8 Sigma

at this page: http://kepler.nasa.gov/sci/basis/character.html

I'm sure someone can explain to me if this just means SNR is 10^8? Or what does it mean?

-Kevin

I'm not exactly sure, but sigma often refers to standard deviation, so my guess would be that this means that the SNR is equivalent to (1/) 8 standard deviations in a gaussian distribution or 6x10^-16 or 150dB. This is of course unfounded conjecture, use at your own risk (this corresponds to 25bits, which seems dubious)

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