Relation between nepers and decibels

**Problems in Exploration Seismology and their Solutions**
Series	Geophysical References Series
Title	Problems in Exploration Seismology and their Solutions
Author	Lloyd P. Geldart and Robert E. Sheriff
Chapter	2
Pages	7 - 46
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

Problem 2.17

The natural logarithm of the ratio of two amplitudes is measured in nepers. Show that one neper = 8.68 dB.

Background

By definition, if $E_{1}$ and $E_{2}$ are energies, $\log _{10}\left(E_{2}/E_{1}\right)$ is the value of the ratio in bels. One bel = 10 decibels (dB), and energy is proportional to (amplitude) $^{2}$ , so

{\begin{aligned}\mathrm {dB} =10\log _{10}\left(E_{2}/E_{1}\right)=20\log _{10}\left(A_{2}/A_{1}\right).\end{aligned}}

(2.17a)

Solution

Let $N={}$ value measured in nepers, dB = same value in decibels. Then,

${\begin{aligned}N=\ln \left(A_{2}/A_{1}\right)=\left(\log _{e}10\right)\log _{10}\left(A_{2}/A_{1}\right)=2.3026\log _{10}\left(A_{2}/A_{1}\right).\end{aligned}}$

Since $\mathrm {dB} =20\log _{10}\left(A_{2}/A_{1}\right)$ .

${\begin{aligned}N=\left(20/2.3026\right)\ \mathrm {dB} =8.686\ \mathrm {dB} .\end{aligned}}$

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