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

Let´s supposeSuppose we have two arbitrary pointpoints A and B. Transmition trhoughtTransmission between these points are aboutis over optical-fiber. We don´t care the wavelenghtwavelength, distance and and and other stuffsstuff.

But now Let´sNow suppose we have total atenuationattenuation of 16 dB in average between these 2 points. According to general decibelsdecibel 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 Outpower out and Powerpower in  :

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

It means to the custumercustomer, just only arrives 0.025 of the power sent from dethe source (ISP, and so on) arrives. That means Ifif we want to the custumers arrivecustomers to receive 1 wattswatt, Wewe 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 Do companies spend hundreshundreds of watts so that only a few watts reach customers because of attenuation? Does that make sense?

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?

Why are power losses in fiber-optical cables so high?

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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Marcus Müller
Marcus Müller
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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:

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

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

enter image description here$$10^{-1.6}=\frac{P_\text{out}}{P_\text{in}}=0.025$$

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:

enter image description here$$P_\text{in} = \frac{1\,\text{W}}{0.025}=40\,\text{W}$$

My question is:

ReallyDoes this really happen  ? Companies spent Companies spend hundres of watts so that only a few watts reach customers because of attenuation? (because atenuation ). Does Itthat make sense  ?

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:

enter image description here

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

enter image description here

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:

enter image description here

My question is:

Really this happen  ? Companies spent hundres of watts so that only a few watts reach customers ? (because atenuation ). Does It make sense  ?

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?

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NIN
NIN
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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:

enter image description here

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

enter image description here

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:

enter image description here

My question is:

Really this happen ? Companies spent hundres of watts so that only a few watts reach customers ? (because atenuation ). Does It make sense ?