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Suppose we have two arbitrary points A and B. Transmission between these points is over optical-fiber. We don´t care the wavelength, distance and other stuff.

Now suppose we have total attenuation of 16 dB in average between these 2 points. According to general decibel 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 customer, only 0.025 of the power sent from the source (ISP, and so on) arrives. That means if we want to the customers to receive 1 watt, we need to sent at least 40 watts:

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

Does this really happen? Do companies spend hundreds of watts so that only a few watts reach customers because of attenuation? Does that make sense?

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    \$\begingroup\$ Attenuation in optical fibers is actually very low at typical telecommunications wavelengths. SMF28 for example is less than 0.2 dB per kilometer, so 16dB of attenuation would be 80 km long. The cable from your ISP is likely much shorter such that your ISP can ignore attenuation in the fibers themselves. \$\endgroup\$
    – user1850479
    Commented May 23, 2021 at 18:26
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    \$\begingroup\$ Imagine looking through a pane of glass of the purest quality that's 80 km thick. How much daylight do you think you'll see? \$\endgroup\$
    – Transistor
    Commented May 23, 2021 at 18:28
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    \$\begingroup\$ You can't really say the losses are high until you calculate it against copper for comparison. There's a reason we use fiber optic for long distance connections. Also, your question is super flawed in the sense that you want to ignore distance and then start by randomly picking an attenuation out of thin air (or at least that's the way it seems, maybe you actually got the 16dB from somewhere). \$\endgroup\$
    – DKNguyen
    Commented May 23, 2021 at 18:48
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    \$\begingroup\$ also 16 dB?! Pfffft, that's practically perfect conditions. Real-world systems end-mile systems might realistically see much more, and still work: the receiver doesn't depend on much power being delivered. In fact, in passive optical networks, as often used by Fiber-to-the-home/-premises/-office... solutions, energy is split amongst many subscribers, often leading to far less than 1/16 of the power left for each individual, even if there was no attenuation at all. \$\endgroup\$
    – Marcus Müller
    Commented May 23, 2021 at 18:53
  • \$\begingroup\$ @MarcusMüller I chuckled at the anachronism with the term "practically perfect conditions" \$\endgroup\$
    – DKNguyen
    Commented May 23, 2021 at 18:54

3 Answers 3

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Fiber optics are used to transport data, not to transport energy.

Compare to radio transmission: 100 dB is common. So the transmitter pumps kilowatts into the air, the receivers receive only a minute fraction. Useless? As power transmission, yes, but that is not the goal.

For fibre optic transmission, an input power of (10's of) milliwatts is normal. The cost of that amount of power is neglectible. The receivers can still reproduce the data, that's what it is all about.

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  • \$\begingroup\$ With kilo-watts emphasized by hyphen the answer was a way more convincing \$\endgroup\$
    – V.V.T
    Commented May 24, 2021 at 3:11
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Your calculations are correct regarding how much power is lost.

However, your expectations about transmitted and received power are quite absurd.

Transmitted power is usually given in milliwatts, and received power that is still usable is given in microwatts.

So if you transmit 1 milliwatt, and 25 microwatts end up in receiver, that is more in the ballpark of power levels involved.

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Does this really happen? Do companies spend hundreds of watts so that only a few watts reach customers because of attenuation? Does that make sense?

No, it does not happen, and it does not make sense.

Where data is being transmitted, the power levels of the light itself is insignificant compared to the power levels of the data processing and signal conditioning at the transmitter and receiver.

In the industrial applications that I know of where power is being transmitted, it is inside a box, and the attenuation within the fiber is negligible from the standpoint of the cost of the power being lost -- it is only important from the standpoint that the fiber must be kept cool enough so that it does not get damaged.

If you did want to transmit power over "long" distances -- well, there's that 0.2dB/km figure in the comments. Theoretically one could transmit power over fiber -- but the conversion into light energy at the transmit end would be inefficient, and the conversion into electrical or any other form of energy at the receive end would be inefficient, and the size of the fiber bundle you'd need to transmit, say, the 24kW that a house needs would be pretty big and probably more expensive than the copper wire you'd need to just transmit that much electricity.

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