signal attenuation thru cabling

[tom]

If one signal is 3 dB less than another, it's half as strong. Some types of
coax are supposed to attenuate signals they carry by up 8 dB per 100 feet.
If a cable was rated as attenuating a signal by 3 dB over a given length,
and a source put 100watts into one end, does that mean that HALF of the
power is getting absorbed by the cable, such that there's only 50watts
reaching the load at the other end?

 

[PeteS]

Attenuation of 3dB would indeed mean 50% of the input energy is
absorbed (in terms of power, which dB always refers to - that nifty
20log10 V1/V2 is derived from the power equation).

One key thing here though, is that normally the cable attenuation is
given at some frequency, with load and source impedances. Copper cables
are low pass filters, so their attenuation increases with frequency
(although we have some tricks to balance that out in a system).

Cheers

PeteS

 

[tom]

I guess that means the higher the frequncy the more it pays to use better
feedlines.
Thanks, that's usefull info.

 

[tlbs]

If the signal you are transmitting through the cable is of a particular
frequency (i.e carrier, narrow band), then you should be able to
calculate the exact loss through a particular length of that cable by
looking on a nomograph or by using the manufacturers formula. Cable
type, length, and frequency will yield a loss in either the nomograph
or formula. Note that these losses usually assume the cable is
terminated into its characteristic impedance.

If you are trying to transmit wideband information (i.e. DC - 100 MHz)
through the cable, and you want a flat frequency response, you need to
use a cable equalizer at one end or the other. In my work 2 decades
ago, sometimes 100:1 (40dB) equalizers were necessary.