>
> I have struggled with this and have seen it both ways.
>
> Take an example of 100 feet of some coax at 5 db loss.
> Is the loss at 200 feet 8 db or is it 10 db ????
>
10dB
> And is the loss at 50 feet 2 db or 2.5 db ????
>
2.5dB
dB/100ft is a strange type of unit but since the loss has an exponential
dependence on length, it makes sense.
P(x) = P_o * 10^{-a*x}
where a is an attenuation constant per unit length in the same units of x...
(not quite in dB yet, it turns out to be in Bels per unit length, but hang
on...)
divide both sides by P_o and take the log (base 10) of both sides, and you
get
log(P(x)/P_o) = -a*x
multiply both sides by -10 :
-10*log(P(x)/P_o) = (10*a) * x
The left hand side is the familiar dB power difference at some distance x
from the transmitter
The right hand side, (10*a) * the distance
so now 10*a is dB loss per unit length.
Long story short, If you double the length, you double the ****dB loss****.
You don't double the loss. It goes much steeper than that, so you double
the dB number.
If you look in a book with a first-principles calculation of a transmission
line, you'll probably see the power attenuation as
P(x) = P_o * e^{-bx}... with b in units of Nepers per unit length, same
units as x. A neper is like a dB but on a natural log scale rather than a
log base 10 scale.
73
Dan
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