3 ms·

what's the dimensional analysis on this equation? if we assume blue light, that's 6.66 * 10^14 Hz * log(...? / ...?) = x Tbps

a1369209993 5y ago
For the photon quantization regime, I think we have a photon energy scaled as `hbar THz`, so a noise floor as `hbar THz^2` (1/2 photon / cycle is about 23uW at 666THz [450nm], or 19uW at 600THz [500nm]). Your Tbps is then `log(power / noise floor) * frequency` aka `log(power/0.5 hbar frequency^2) * frequency`. The effect of noise floor (and consequently quantization-regime frequency) on data rate isn't really cooperative with dimensional analysis on account of the logarithm.
- roflc0ptic 5y ago
  thanks!
  - a1369209993 5y ago
    Also, to optimise data rate at a given power budget, you want to choose freq such that `d rate/d freq` is zero, which (assuming I didn't screw up the math) goes like: 0 = d/df rate = d/df (log(power/0.5 hbar freq^2) * freq) = d/df log(power/0.5 hbar freq^2) * freq + log(power/0.5 hbar freq^2) * d/df freq = 1/(ln2 power/0.5 hbar freq^2) * d/dx (power/0.5 hbar freq^2) * freq + log(power/0.5 hbar freq^2) = 0.5 hbar freq^2 / ln2 power * power / 0.5 hbar * -2 freq^-3 * freq + log(power/0.5 hbar freq^2) = 0.5 hbar freq^2 / ln2 / 0.5 hbar * -2 freq^-3 * freq + log(power/0.5 hbar freq^2) = freq^2 / ln2 * -2 freq^-2 + log(power/0.5 hbar freq^2) = log(power/0.5 hbar freq^2) - 2/ln2 so log(power / 0.5 hbar freq^2) = 2/ln2 power / 0.5 hbar freq^2 = 2^(2/ln2) freq^2 0.5 hbar / power = 2^(-2/ln2) freq^2 = 2^(1-2/ln2) power / hbar freq = 2^(1/2-1/ln2) sqrt(power / hbar) So (again assuming I got that math right), the optimal single photon frequency to use (in the quantum noise floor regime) scales with the square root of your power budget. However, at one milliwatt it's 1600THz or 187nm, and only increases from there, so for practical purposes you probably just want to use the bluest light your fiber can carry, and worry more about classical noise.

a1369209993 5y ago

For the photon quantization regime, I think we have a photon energy scaled as `hbar THz`, so a noise floor as `hbar THz^2` (1/2 photon / cycle is about 23uW at 666THz [450nm], or 19uW at 600THz [500nm]). Your Tbps is then `log(power / noise floor) * frequency` aka `log(power/0.5 hbar frequency^2) * frequency`. The effect of noise floor (and consequently quantization-regime frequency) on data rate isn't really cooperative with dimensional analysis on account of the logarithm.

roflc0ptic 5y ago
thanks!
- a1369209993 5y ago
  Also, to optimise data rate at a given power budget, you want to choose freq such that `d rate/d freq` is zero, which (assuming I didn't screw up the math) goes like: 0 = d/df rate = d/df (log(power/0.5 hbar freq^2) * freq) = d/df log(power/0.5 hbar freq^2) * freq + log(power/0.5 hbar freq^2) * d/df freq = 1/(ln2 power/0.5 hbar freq^2) * d/dx (power/0.5 hbar freq^2) * freq + log(power/0.5 hbar freq^2) = 0.5 hbar freq^2 / ln2 power * power / 0.5 hbar * -2 freq^-3 * freq + log(power/0.5 hbar freq^2) = 0.5 hbar freq^2 / ln2 / 0.5 hbar * -2 freq^-3 * freq + log(power/0.5 hbar freq^2) = freq^2 / ln2 * -2 freq^-2 + log(power/0.5 hbar freq^2) = log(power/0.5 hbar freq^2) - 2/ln2 so log(power / 0.5 hbar freq^2) = 2/ln2 power / 0.5 hbar freq^2 = 2^(2/ln2) freq^2 0.5 hbar / power = 2^(-2/ln2) freq^2 = 2^(1-2/ln2) power / hbar freq = 2^(1/2-1/ln2) sqrt(power / hbar) So (again assuming I got that math right), the optimal single photon frequency to use (in the quantum noise floor regime) scales with the square root of your power budget. However, at one milliwatt it's 1600THz or 187nm, and only increases from there, so for practical purposes you probably just want to use the bluest light your fiber can carry, and worry more about classical noise.