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# Signal Attenuation in an Optical Fiber

Signal attenuation within the optical fibers is usually expressed in the logarithmic unit of the decibel.

Decibel is defined by the input and output power ratio of the signal for a specific operating wavelength.

The overall signal attenuation is defined by the number of dB, which is expressed as given below,

$\large&space;dB=10log_{10}\frac{P_{in}}{P_{out}}$

$\large&space;\frac{P_{in}}{P_{out}}=10^{\frac{dB}{10}}$

In optical fiber communications, the attenuation unit is in decibels per unit length (i.e. dB/Km)

$\large&space;\alpha&space;_{dB}=\frac{10}{L}log_{10}\frac{P_{in}}{P_{out}}$

$\large&space;\alpha&space;_{dB}$ is the signal attenuation per unit length. L is a length of the optical fiber.

### Example:

When the mean optical power launched into an 8 km length of fiber is 120 μW. the mean optical power at the fiber output is 3 μW.

You have to determine;

1. The overall signal attenuation through the fiber assuming there are no connectors.
2. The signal attenuation per kilometer for the fiber
3. The overall signal attenuation for a 10km optical link using the same fiber with splices at 1 km intervals, each giving an attenuation of 1dB.
4. The numerical input/output power ratio in the (3).

#### Solution:

##### (1) Part

The overall signal attenuation in decibels through the fiber is

$\large&space;dB=10log_{10}\frac{P_{in}}{P_{out}}$

Given, input power = 120 μW; output power= 3μW; length of an optical fiber=8 km

so putting all these values in above formula, we get

$\large&space;=10log_{10}\frac{120\mu&space;W}{3\mu&space;W}$

$\large&space;=10log&space;_{10}40&space;=16.0&space;dB$

Overall signal attenuation is 16.0dB for this optical fiber. Joining the two fiber cables together is known by the connectors and splicers. In this part, there are no connectors. So, no signal loss due to the connectors.

##### (2) Part

The signal attenuation per kilometer for the same fiber may be obtained by the formula;

$\large&space;\alpha&space;_{dB}=\frac{dB}{L}=\frac{10}{L}log_{10}\frac{P_{in}}{P_{out}}$

In the first part, we found overall signal attenuation dB=16.0 dB

so now divide it by the L i.e. 8 Km. we will get attenuation per unit length, and it will be

$\large&space;\alpha&space;_{dB}$=2.0 dB/km

##### (3) Part

To understand this part of the attenuation problem, we have to observe this picture carefully. Here the total length of optical fiber is 10 km and at every kilometer, there is a splice. So, count the total number of splices, are these will be nine or more. Just check how these are nine? Because this is the only important part of this problem.

The optical fiber has nine splices with an attenuation of 1 dB. So, loss due to all nine splices will be 9 dB. Hence the overall signal attenuation for the optical fiber is

= 20 dB + 9 dB = 29dB

From the second part, we have seen the attenuation per unit kilometer is 2 dB. Here, the optical fiber is the same and length is 10 km is given. By this way, the total attenuation will be 20 dB. Also, 9 dB due to splices that is additional.

##### (4)Part

The input/output ratio in case of (3). We know the formula

$\large&space;\frac{P_{in}}{P_{out}}=&space;10^{\frac{dB}{10}}$

Put the value of dB that is 29 in the above formula,

$\large&space;=10^{\frac{29}{10}}=10^{2.9}=794.3$

This is the result.

### QUIZ:

1. What are the two main causes of losses in signals of an optical fiber?

2. What is the unit of Loss in an optical fiber?

3. How signal transmitted through the optical fiber?

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