# Relation between nepers and decibels

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 2 7 - 46 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 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}} 