# Relation between units of attenuation

1. Sep 27, 2013

### ParoxysmX

1. The problem statement, all variables and given/known data

It is customary to express fibre loss in units of dB/km:

$\alpha(dB/km) = \frac{10log(P_{in}/P_{out})}{L(km)}$

where $P_{in}$ is the power entering the fibre and $P_{out}$ is the power leaving the fibre. Show that $\alpha(dB/km)$ = 4343$\alpha$($m^{-1}$)

3. The attempt at a solution

Dividing through by decibels -should- give attenuation per kilometre, then converting to metres would give the right dimension, but how can you come up with a coefficient without any numbers to work with? Also, is attenuation per metre just an arbitrary value of loss per metre?

2. Sep 27, 2013

### tiny-tim

Hi ParoxysmX!

Does it help if I point out that Pout/Pin is a dimensionless number, and therefore so is log(Pout/Pin) ?
I think the idea behind it is that if the attenuation is constant, that is the same as saying that the power "does not remember its history" …

P(x)= ekx,

so P(L+x) = P(L)P(x) …
the power loss over distance x is always P(x), no matter at what distance L you start

3. Sep 27, 2013

### TSny

I think there might be some confusion with the notation. I believe that what you want to show is that

if $P_{out} = P_{in}e^{-\lambda L}$, then $\alpha(dB/km) = 4343\lambda(m^{-1})$.

$L$ is measured in meters in the exponential expression so that $\lambda$ has the units of m-1.

4. Sep 28, 2013

### ParoxysmX

Where does $\lambda$ come into it? Is that simply the wavelength of the light entering the fibre? My question sheet definitely says $\alpha(dB/km) = 4343\alpha(m^{-1})$.

e; I see that $P_{out} = P_{in}e^{-\lambda L}$ is the same as the version we were taught where $P_{out} = P_{in}e^{-\alpha L}$. But then, I'm still not sure where 4343 comes from without having data to work with.

Last edited: Sep 29, 2013
5. Sep 29, 2013

### TSny

$\lambda$ is just the notation I picked for the coefficient of $L$ in the argument of the exponential. Apparently, your question sheet uses the same notation $\alpha$ for two different quantities. I find this a little confusing, but I guess it is standard practice. Including the units is then essential in order to distinquish them.

Last edited: Sep 29, 2013
6. Sep 30, 2013

### ParoxysmX

Yeah ok, so this essentially comes down to a conversion between bases. $\αlpha(dB/km)$ should equal $2303 \αlpha(m−1)$ if you account for the difference between metres and kilometres, shouldn't it?

7. Sep 30, 2013

### TSny

In your formula for $\alpha(dB/km)$ you need to know what base you are using for the log. I suspect it is base 10. Likewise, for $\alpha(m^{-1})$ you need to know if it is defined using $P_{out} = P_{in}e^{-\alpha L}$ or $P_{out} = P_{in}10^{-\alpha L}$. I suspect it is the first way.

8. Sep 30, 2013

### tiny-tim

log10e = 0.4343

(so 100.4343 = e)

(and if you look at the top line of your log tables, you'll see it starts 0043 0086 etc, which is short for 004343 …… oooh, do people still have log tables? see )

9. Sep 30, 2013

### TSny

Yes, they do.

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10. Sep 30, 2013

### ParoxysmX

Ah yes this makes sense, thanks tiny-tim!