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I have posted this question in math.stackexchange.com, [can be found here], which I originally thought it is a math question. I now believe this would fit more to the Signal Processing community. For convenience, I am reposting the question in below. I would appreciate any help.

I am trying to understand what it means by $0.01\log_{10}$, which was published in the following paper at page 7 in IEEE.

Berrier, Keith L., Danny C. Sorensen, and Dirar S. Khoury. "Solving the inverse problem of electrocardiography using a Duncan and Horn formulation of the Kalman filter." IEEE Transactions on biomedical engineering 51.3 (2004): 507-515.

I cannot convert $0.01\log_{10}$ to a number since I expected to see something like $0.01\log_{10}^3$ to be able to say it is equal to 0.0048. The authors uses this expression twice in the paper.

Could someone kindly help me in understanding what the authors meant by $0.01\log_{10}$?

The following is a snapshot of the paragraph of the paper in which $0.01\log_{10}$ was used.

enter image description here

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Differences in log scales correspond to ratios of the underlying values. Saying you are within .01 on a $\log_{10}$ scale means the true value is within $10^{-.01}$ and $10^{.01}$ of the stated value, ratio wise. That calculates out to 0.9772 and 1.0233. Since the value is so close to 1, an interval on the ratio scale is close to an interval on a difference scale, so you could also say the results are within about 2.3%.

I haven't read the paper.

Hope this helps.

Ced


Followup:

So, on a whim I looked up "centibel" and found https://en.wiktionary.org/wiki/centibel. A "cB" if you will. Not too many other references though.

I should have also looked at the OP's math link. The correct answer was already there.

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It looks to me that they are specifying a tolerance of within $1/1000^{\mathrm{th}}$ of a dB.

If they have experimentally determined a parameter $P$, which in dB is $10 \log_{10}{P}$, then saying they are within $0.01 \log_{10}$ looks like one thousandth of a dB to me.

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  • $\begingroup$ I think you went the wrong way on the decimal point. $$ 1 dB = 0.1 \log_{10} $$ So the result is 1/10 dB. $\endgroup$ – Cedron Dawg Apr 21 '18 at 1:19
  • $\begingroup$ Hmm. I would call $10 \log_{10}$ within 1 dB ("on the scale of a dB"), $1 \log_{10}$ within 0.1 dB, etc. Although I will concede, claiming to know something more precisely than a 10th of a dB is silly, if a log scale is the correct way to present the data. Either way, it's a really odd way to specify the error in an emperically determined value. $\endgroup$ – Andy Walls Apr 21 '18 at 3:49
  • $\begingroup$ Yeah, I stated my equation poorly. It should say "1 dB is 0.1 on a Log (Bel) scale" and not implied the multiplication. I think "within 0.1 dB" would be better than "within 0.01 Bel" which is better than "within .01 Log10". If a log scale isn't inherent, I would think "within 2.3%" is a lot clearer. $\endgroup$ – Cedron Dawg Apr 21 '18 at 4:12

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