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Dec 26, 2020 at 14:08 vote accept Yoomo
Dec 20, 2020 at 12:55 comment added Andy aka At the instant the pulse is applied, the potential divider formed by the source impedance and the cable's characteristic impedance is created. It's instant so, if you measure after 1 ns or 10 ns or 100 ns or 1 us, you will get the same result (providing the cable is long enough so that far end reflections are avoided during the measurement period).
Dec 20, 2020 at 12:40 comment added Yoomo I am a bit confused. When I use a long square wave pulse like above the impedance will be a real value and can be calculated like this Vcable=1 volt⋅(100/150)?
Dec 20, 2020 at 12:29 comment added Andy aka @Yoomo use a pulse and pretty much after applying that pulse (via a resistor) look at the magnitude at the cable end where you apply the pulse. If the cable is short then you might get a reflection coming back too quickly and it screws up the measurement of course.
Dec 20, 2020 at 12:26 comment added Yoomo So with your formula I can calculate the impedance of the cable, when I am using either a pulse or a DC voltage?
Dec 20, 2020 at 12:24 comment added Andy aka @Yoomo If you are using pulses then this is going to be the case. The characteristic impedance will be \$\sqrt{L/C}\$ and that will be resistive above 1 Mohm.
Dec 20, 2020 at 12:22 comment added Yoomo But this assumes that the impedance of the cable is a real value?
Dec 20, 2020 at 10:53 history answered Andy aka CC BY-SA 4.0