I have the "book Data communications and networking" and I am stuck on this question:
Tabulate the attenuation (in db) of a 2.6/9.5 mm coaxial cable for the indicated frequencies and distances.
Distance | dB at 1KHz | dB at 10 KHz | dB at 100 KHz|
1km
10km
15lm
20km
I have no idea how this should be calculated, there were no formulas in the chapter.
All help appreciated!
As you can see in the recommendation of 2.6/9.5 mm coaxial cable in this link 1 there is an approximated formula for the attenuation coefficient a. As is written there:
"The following equation, in which α is expressed in dB/km and f in MHz, gives an approximation of the attenuation coefficient from 1 MHz onwards":
$α=0.01 +2.3\sqrt {f} +0.003f$
Which means that the attenuation coefficient depends on the channel frequency. (This formula may provide a small error for frequencies less than 1 MHz).
For example, for a frequency equal to 100 kHz:
$a= 0.737624 dB/km$.
and for a length equal to 20 km, the total attenuation in dB is
$a= 0.737624 dB/km * 20 km = 14.7525 dB$.