Communication Basics #27 Worked Example: Fiber Link Budget

Instructor: Dr. Süleyman Burak ÇELİK

Second final preparation video in the Communication Basics series. Classic fiber optic link budget — in light mode, step by step, with a proper math proof for dB arithmetic this time. A laser transmitter outputs +10 dBm. Its output passes through a 1 dB connector, then 100 km of fiber at 0.3 dB/km, then another 1 dB connector into the receiver. The receiver sensitivity threshold is −60 dBm. We answer five parts: (a) find the received optical power, (b) compute the power margin, (c) find the maximum fiber length that keeps a 14 dB margin, (d) check whether a 300 km link operates, (e) evaluate whether 50 km would be a good design. Before the arithmetic, we do something this series has been missing — a full derivation of why dBm ± dB = dBm. Starting from the linear equation P_out = P_in × G, we take 10 log of both sides and use log(a·b) = log(a) + log(b). That turns multiplication into addition in the dB world, and shows why the 1 mW reference in dBm stays put when we add dimensionless dB values. We also explain why mixing linear and dB in the same expression is nonsense. With the math nailed down, the answers are clean: received power = −22 dBm, margin = 38 dB (extremely healthy), max length = 180 km for a 14 dB margin, 300 km fails by 22 dB (needs amplifiers), 50 km is over-specified and wasteful. We close with three takeaways. Topics covered: - dB math derivation — why log(×) = + (full proof) - dBm as absolute, dB as ratio, unit algebra that keeps dBm - warning: never mix linear power with dB (10 mW − 14 dB is nonsense) - fiber link budget chain: laser → connector → fiber → connector → Rx - received power calculation and power margin - maximum-length solve for a target margin (180 km) - under-range and over-spec design cases (300 km, 50 km) - engineering principles: margin, hardware matching Part of the AcEdumy Communication Basics series.