Relation between units of attenuation

In summary, the coefficient of loss, lambda, is a dimensionless number that is equal to 4343 log10e units of attenuation per kilometre.
  • #1
ParoxysmX
21
0

Homework Statement



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

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

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

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?
 
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  • #2
Hi ParoxysmX! :smile:
ParoxysmX said:
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?

I don't understand your question. :redface:

Does it help if I point out that Pout/Pin is a dimensionless number, and therefore so is log(Pout/Pin) ?
Also, is attenuation per metre just an arbitrary value of loss per metre?

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 :wink:
 
  • #3
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
TSny said:
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.
Where does [itex]\lambda[/itex] 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.
 
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  • #5
##\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.
 
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  • #6
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
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
ParoxysmX said:
… I'm still not sure where 4343 comes from …

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? :redface: see
log-4f.gif
)
 
  • #9
tiny-tim said:
oooh, do people still have log tables?

Yes, they do.
 

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  • #10
tiny-tim said:
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? :redface: see
log-4f.gif
)

Ah yes this makes sense, thanks tiny-tim!
 

1. What is the definition of attenuation?

Attenuation is the gradual loss of intensity or strength of a signal as it travels through a medium.

2. What units are used to measure attenuation?

The most commonly used units to measure attenuation are decibels (dB) and nepers (Np).

3. How are decibels and nepers related?

Decibels and nepers are logarithmic units that are used to express the ratio between two quantities, such as the input and output power of a signal. They are related by the formula: 1 Np = 8.686 dB.

4. What is the difference between attenuation and loss?

Attenuation refers to the reduction in signal strength as it travels through a medium, while loss refers to the amount of signal that is lost during transmission. Attenuation is a natural phenomenon, while loss can be caused by external factors such as interference or impedance mismatch.

5. How does the medium affect the attenuation of a signal?

The medium through which a signal travels can affect its attenuation. For example, different materials have different attenuation coefficients, which determine how much a signal will be weakened as it passes through that material. Additionally, factors such as distance, temperature, and frequency can also impact the attenuation of a signal.

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