This is waveform that I got on the osciloscopewhen experimenting in the lab. A pulse was sent through the coaxial cable was used with a capacitor at the load. I have my own idea why is this happening but I am not quite sure.

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  • \$\begingroup\$ What is your own idea? \$\endgroup\$ – Chu Nov 18 '18 at 9:39

When you send the pulse down the transmission line (t-line), initially the source of the pulse cannot know what load is at the far end so, in order to initially pass current, it uses the t-line's characteristic impedance (50 ohm, 75 ohm etc.) and produces a current proportionate with the voltage and that resistance (i.e. ohms law).

So, for a period of time you have a voltage and current "front" hurtling down the t-line and, if the terminating impedance was matching the characteristic impedance, the power associated with the two "fronts" would be totally dissipated by the load.

But your load is a capacitor and, it is receiving real power AND that power cannot be dissipated because the load is a capacitor hence, it is reflected back to the source (because that power has to travel somewhere). If the source has a driving impedance that matches the t-line's characteristic impedance, then the power is dissipated back at the source and you get the waveform you see.

If the source has a zero ohm driving impedance then the power is once again reflected back down the t-line to the load-end and, once again, it cannot be dissipated by the capacitor and things continue back and forth.

  • \$\begingroup\$ I like the "hurtling down" expression. \$\endgroup\$ – Michael Karas Nov 18 '18 at 12:07
  • \$\begingroup\$ Thanks for your answer ! You know when the reflected voltage is upside down because I was thinking that the capacitor acts is kind of like an short circuit because the reflected voltage is flipped, thus Zload probably decreased? After that there is also a bump for the voltage, I don't get why. \$\endgroup\$ – Neon Star Nov 18 '18 at 13:50
  • \$\begingroup\$ After the initial negative (indicating an initial short, as you say), it doesn’t get easier to intuitively analyse for a capacitor. All I can say from your picture is that the capacitor absorbs about half of the power and then starts to look more like an open circuit (compared to the impedance of the coax) and so the voltage rises from negative to a positive value (as it would with an open ended t-line). \$\endgroup\$ – Andy aka Nov 18 '18 at 15:20

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