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Im trying to understand exactly how a signal from a coaxial cable is interpreted. From my understanding based on disscussing with class mates and a couple of hours in a school laboratory my best guess is:

The cable working as a capasitor loads up to a degree where the change in voltage can be interpeted as a bit (0 or 1), alternating in a pattern that then gets interpeted by a reciever as data.

Also about the RC constant in a coax, is it correct that if I have a low RC, it would take a shorter time to put/remove a load on the cable, thus increasing the bitrate?

Im not studying electronics specifically, its just a part of a basic physics course im taking.

Thanks in advance!

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    \$\begingroup\$ What sort of signal? It's more usual to analyse it as a sine wave, even if it's a digital signal. It's not a single RC constant but a concept called "impedance". \$\endgroup\$
    – pjc50
    May 22, 2016 at 9:48
  • \$\begingroup\$ You're trying to understand coax but then put forward a theory that doesn't make sense leaving the reader with no idea where you are coming from and no immediate answer to supply. \$\endgroup\$
    – Andy aka
    May 22, 2016 at 9:50
  • \$\begingroup\$ LC constant is more signifigant than RC constant, that is hat defines the characteristic impedance of the coax. \$\endgroup\$ May 22, 2016 at 10:46
  • \$\begingroup\$ Sorry if I cant be more specific. In an assignment we had at school we let a function generator generate a signal through a coax cable and we looked at the wave through an occiloscope (square wave). At high frequencies the square wave took the form of a sinus wave and from my understanding it had to do with the RC constant in the cable, the capacitor (cable) didnt have enough time to reach full load. So im just wondering if the wave's geometry is what the reciever interprets as ones and zeros in for example an old network card or such. \$\endgroup\$ May 22, 2016 at 12:24
  • \$\begingroup\$ A bit of reading around transmission lines might prove useful - en.wikipedia.org/wiki/Transmission_line \$\endgroup\$ May 22, 2016 at 14:36

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Either you were not listening carefuly or your teacher was cheating. If the source impedance matches the load impedance (complex conjugate \$Z_{load}=Z^*_{source}\$) and they are connected with a coax of same characteristics impedance (\$Z_c=\sqrt{\frac{L}{C}}\$, resistance of wire and conductance of the isolator are ommited, since negligible), then there is no distortion, however the signal would be delayed, but not distorted. What you saw, wasn't a distortion due to capacitance of the coax, plausible cause is that signal generator reached its max capability in terms of frequency output. The generator was unable to drive a square pulse at such high frequency, that's it.

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  • \$\begingroup\$ If the output impedance of the signal generator is at ideal theoretical value of zero ohm, no matter what the characteristics impedance of the coax, the signal at the end will be the pristine square wave? \$\endgroup\$ May 22, 2016 at 18:12
  • \$\begingroup\$ @soosaisteven No, because of impedance mismatch. \$\endgroup\$ May 22, 2016 at 18:19
  • \$\begingroup\$ Zin=Zout called impedance matching, right? But how do you plan to put zero ohm at the load end and expect signal from the dead short? \$\endgroup\$ May 22, 2016 at 18:22
  • \$\begingroup\$ @soosaisteven What useful load has 0 ohm? All impedances must match, source, cable, load. If the source has 0 ohm, then it doesn't match, nobody said that you have to put a short at the end. \$\endgroup\$ May 22, 2016 at 18:31
  • \$\begingroup\$ But how do you do impedance matching if the source output impedance is very near to zero? I think impedance matching is needed only when the source has non zero output impedance. When the source has extremely low output impedance, it should be able to drive any cable of any impedance with no signal degradation. \$\endgroup\$ May 22, 2016 at 18:37

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