Skip to main content
2 of 3
added 77 characters in body
Marcus Müller
  • 100.1k
  • 5
  • 141
  • 262

Why power losses in fiber-optical are so high (Because atenuation)?

Let´s suppose we have two arbitrary point A and B. Transmition trhought these points are about optical-fiber. We don´t care the wavelenght, distance and and other stuffs.

But now Let´s suppose we have total atenuation of 16 dB in average between these 2 points. According to general decibels formula:

$$ 10\log_{10}\left(\frac{P_\text{out}}{P_\text{in}}\right) = 16\,\text{dB}$$

When we solve the formula and the relationship between Power Out and Power in :

$$10^{-1.6}=\frac{P_\text{out}}{P_\text{in}}=0.025$$

It means to the custumer, just only arrives 0.025 of the power sent from de source (ISP, and so on). That means If we want to the custumers arrive 1 watts, We need to sent at least 40 watts:

$$P_\text{in} = \frac{1\,\text{W}}{0.025}=40\,\text{W}$$

My question is:

Does this really happen? Companies spend hundres of watts so that only a few watts reach customers because of attenuation? Does that make sense?

NIN
  • 423
  • 4
  • 11