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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!

Elias
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1 Answers1

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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

$a= 0.737624 dB/km * 20 km = 14.7525 dB$.

Dimitris U.
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