In one line
Attenuation is the gradual loss of optical power as light travels along a fibre, and the attenuation coefficient gives this loss in decibels per kilometre.
Explained simply
Even a very clear glass of water looks greenish when you look through a long length of it. Over kilometres, glass also absorbs and scatters a little light. Attenuation tells us how fast the light gets weaker, which decides how far a signal can go before it needs a repeater or amplifier.
How it works
- Measure input power \(P_{in}\) at the start and output power \(P_{out}\) after length \(L\).
- Loss in decibels: \(\text{Loss} = 10\log_{10}(P_{in}/P_{out})\) dB.
- Divide by length in km to get the attenuation coefficient \(\alpha\) in dB/km.
- To find output power: \(P_{out} = P_{in}\times10^{-\alpha L/10}\).
- Silica fibres have low-loss "windows" around 850 nm, 1310 nm and 1550 nm; the lowest loss, about 0.2 dB/km, is near 1550 nm.
Key formula
\(\alpha = \dfrac{10}{L}\log_{10}\left(\dfrac{P_{in}}{P_{out}}\right)\) dB/km. \(\alpha\): attenuation coefficient (dB/km), \(L\): fibre length (km), \(P_{in}, P_{out}\): input and output power (W or mW, same unit for both).
Quick example
A 2 km fibre receives 1 mW and delivers 0.5 mW. Loss \(= 10\log_{10}2 = 3.01\) dB, so \(\alpha = 3.01/2 \approx 1.5\) dB/km. Second case: \(\alpha = 0.2\) dB/km, \(L = 50\) km, \(P_{in} = 1\) mW. Total loss = 10 dB, so \(P_{out} = 1\times10^{-1} = 0.1\) mW.
Common mistakes
- Swapping \(P_{in}\) and \(P_{out}\), which gives a negative loss.
- Using natural log instead of \(\log_{10}\).
- Forgetting to convert length to km when \(\alpha\) is in dB/km.
How MU asks it
Asked as "Define attenuation and attenuation coefficient" (3 marks) and frequently as a numerical (3 to 5 marks): find \(\alpha\), output power or the length for a given loss. Write the formula, substitute with units, and state the answer in dB/km or mW.