Optical fiber attenuation (${\alpha }_{\mathrm{dB}}$)

Optical fiber attenuation is the reduction of transmitted optical power along an optical fiber. Define a positive loss in decibels by comparing optical power $P(0)$ already launched into the guide with surviving optical power $P(z)$:The factor ten applies to a power ratio; the factor twenty applies to a wave amplitude ratio only when power is proportional to its square under the same conditions.^[ref]

For a uniform illustrative model with constant loss coefficient ${\alpha }_{\mathrm{dB}}$ in $dB/km$ and distance $z$ in kilometres,For ${\alpha }_{\mathrm{dB}}=0.2dB/km$ over $10\mathrm{km}$, the loss is $2\mathrm{dB}$ andThus about $63.1\%$ of the launched optical power survives and $36.9\%$ is lost. This is an illustrative coefficient, not a property of all optical fibers, and is independent of the earlier illustrative refractive indices.

The exponential decay follows from assuming each equal distance interval transmits the same fraction. For two such intervals, power ratios multiply; their logarithms add, so the losses in decibels add. Two $5\mathrm{km}$ sections each transmit $10^{-0.1}\simeq 0.794328$; together they transmit $(0.794328{)}^2\simeq 0.630957$. The loss curve is straight in decibels and curved on a linear power scale.

Wave absorption and scattering of light can remove guided optical power even in a straight fibre, while bends can let it escape. Entrance losses, joints and connectors contribute additional losses; they belong separately from a uniform distance coefficient. This is why a measured end-to-end power ratio need not equal the straight-fibre model.^[ref]

An optical fiber cutback measurement estimates distributed optical fiber attenuation by comparing a long guide with the short input portion left after cutting the output end back. Keep the launch untouched. For the following original functional model, fix wavelength, source output, input alignment and collected output cone; subtract a dark reading and keep the photodetector in its linear range. Renewed end-face quality and changed normal mode populations can still spoil the comparison. The diagram describes a measurement principle, not Kao's historical short-glass apparatus.

Let $L_{\ell }>L_s$ be lengths in kilometres, $P_{\ell },P_s$ the corresponding positive dark-subtracted readings, and $C$ the same launch and detection factor. A uniform guide obeysHere ${\alpha }_{\mathrm{dB}}$ and its estimate are numerical values in $dB/km$. Dividing eliminates $C$; taking a logarithm proves the formula. A changed $C$ instead adds an error $10{\log }_{10}(C_s/C_{\ell })/(L_{\ell }-L_s)$. Reconnecting the launch between readings defeats this cancellation.

For original chosen values $L_s=0.1\mathrm{km}$, $L_{\ell }=2.1\mathrm{km}$, $P_s=100\mu \mathrm{W}$ and ${\alpha }_{\mathrm{dB}}=0.5dB/km$, the predicted long reading is $P_{\ell }=10010^{-0.1}\simeq 79.4328\mu \mathrm{W}$. The ratio gives $1\mathrm{dB}$ for the removed $2\mathrm{km}$ and hence $0.5dB/km$.

Suppose an unremoved additive detector background contributes $b=5\mu \mathrm{W}$ to both readings. Then the apparent coefficient isThis is a bias of an estimator: a common additive background pushes the ratio toward one, whereas a common multiplicative factor cancels. For independent small zero-mean errors with standard deviations ${\sigma }_s,{\sigma }_{\ell }$ on the dark-subtracted readings, a first-order Taylor expansion givesChoosing both standard deviations as $1\mu \mathrm{W}$ gives ${\sigma }_{\hat{\alpha }}\simeq 0.03491dB/km$. A shared estimated dark offset introduces covariance, so it requires a different uncertainty calculation. Near zero signal, the logarithm and this linear approximation become unreliable.