Optic fibers atenuation problem

[Jalo]

Homework Statement

In a given communication system the attenuation of a optic fiber is 0.5 dB/km, existing an extra increment of 1dB every 10km due to the junctions of the fiber. Find the maximum length that the fiber can have, knowing that the debited power by the emmiting laser, $P_{0}$, is 1.5mW and that the minimum power at the end of fiber,$P_{0}$, is 2μW.

Homework Equations

$P_{t}=P_{0}e^{-\alpha L}$ , where $\alpha$ is the attenuation constant.
$\alpha_{dB}=-\frac{10}{L}log(\frac{P_{t}}{P_{0}})$ , where $\alpha_{dB}$ is the attenuation given in units of dB/km.

The Attempt at a Solution

I solved this problem by assuming first that the total distance was:
$L = 10n + \delta$, where n is a positive natural number.
Basically for each interval of 10km I used an attenuation of $\alpha_{dB}=0.6$, and for the last kilometers, $\delta<10km$, I used an attenuation of $\alpha_{dB}=0.5$.

First I found the value of $n$:
$\alpha_{dB}=-\frac{10}{L}log(\frac{P_{t}}{P_{0}})$
$n = -\frac{1}{\alpha_{dB}}log(\frac{P_{t}}{P_{0}})$

Using $\alpha_{dB}=0.6$ and the values of $P_{t}$ and $P_{0}$ I got the value
$n=4.79$.


To find the remaining length of the optic fiber I tried to find the power at the end of 40km. This was easily achieved using the equation presented earlier.

$P_{t}=P_{0}e^{-n*\alpha_{dB}}$
$P_{t}=1.36*10^{-4}$

Finally I tried to find the remaining length using the equation again, but using the power at the 40km mark.

$\delta = -\frac{10}{\alpha_{dB}}log(\frac{P_{t}}{P_{0}})$
$-\frac{10}{0.5}log(\frac{2*10^{-6}}{1.35*10^{-4}}) = 36.65km$

This is obviously wrong since the result for delta had to be less that 10km.

If someone could give me a hint on where I went wrong I'd appreciate.
Thanks!

 

[gneill]

Since dB losses are additive, why not begin by finding the total end-to-end dB loss? Then you can play with the combinations of 0.5dB/km and 1dB/10km to sum up to that.