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Cable loss for coax is mostly given in loss per 100 feet, express in dB. Say I have cable which is 20 dB loss per 100 feet at frequency x. If I use 50 feet (1/2 the length,) would my loss be:

20 dB / 2 = 10 dB

or

2 x the power = 20 dB - 3 dB = 17 dB

KevinHJ
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    20dB loss / 100 ft is huge... am guessing the numbers are for convenience and not measured or spec'd? – webmarc Mar 24 '22 at 15:36

2 Answers2

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It's the first one, if you use 50 feet (1/2 the length) then the loss is 20 dB / 2 = 10 dB.

similarly, if you used 200 feet, the loss would be double… 20dB + 20dB = 40dB loss.

there are several online calculators also available, here's a link to one.

webmarc
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  • Thanks. That is how I thought it would be, however I read sites that said the cable length was inversely proportional to the power. Thus, 2 x cable length would be 1/2 power (according to some), thus, a -3dB difference (if that were correct.) – KevinHJ Mar 29 '22 at 00:33
  • Marking your answer as accepted. – KevinHJ Mar 29 '22 at 00:33
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Your answer is correct.

Simply calculate the loss per unit length and multiply by the length you are using.

If the reference length is 100 ft and the loss at the reference length is 20 dB, then the loss per unit length is:

$$20\text{ dB}/100\text{ ft} = 0.2\text{ db/ft}$$

The loss at any length $d$ ft will be:

$$L(d) = 0.2\times d$$

At 50 ft, $d=50$, so $L = 10\text{ dB}$.

AG5CI
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