I need to do twisted pair cable measurements. I have differential signal generator and two channel osciloscope (Analog discovery 2) for measurements. Cable is around 300m length, but unknown impedance. I need to measure cable at 1kHz frequency. Please advice how to do cable attenuation measurements?
I am planning to generate differential sine signal, and measure input and output voltage of the cable with differential input scope.
1. But the issue is that do I need to load this cable?
2. If yes what impedance as impedance is unknown?
3. If load is required how can I separate how much loss will be on load and than what is cable loss?
Thank you in advance
Electrical length of your cable is very short compared to the wavelength on 1 kHz so transmission line effects should negligible. But we still have resistive losses due to finite conductivity and skin effect so the voltage you are going to see at the far end depends on impedance of the generator, resistive cable losses and load impedance at the far end. I suggest you to review the actual application for what kind of impedances can be expected at the transmitting and receiving ends and use those values to mimic the actual use case. If it is ethernet, the impedance is 100 ohms. In traditional telephone world, everything is 600 ohms.
Thank you very much. I wondering how then I can calculate attenuation of this cable. In case if I measure only voltage at the input and output, the higher impedance of driver and load, the lower voltage drop I receive. Please find the LTSpice model I attached.
Using analyzer I get L=0.20079mH, while capacitance is 15.760nF. In this case I calculate Z to be 112Ohm. It LTSpice model I use 56R resistors at all RL parameters as presented in circuit.
I believe that cable attenuation should be single number, but it looks like it depends on loading impedance. What would be your thoughts?
That's correct. Cable attenuation is specified with termination to nominal impedance so it is a single number. Referring to spice circuit, find out the voltage at cable input (after the generator internal resistace) and output (over the load resistor). 20*log10 (outV / inV) gives you the cable loss in dB. In physical test set-up, you can measure those voltages with an oscilloscope.
Thank you, this is what I am doing, but I was going for Power comparison at input and output. In this case I still assume that cable loss should be lets say 1dB for any kind of load, but it looks like that it is depending on loading.
Please find figure with different load resistance.
Why the smallest attenuation is not when I matched cable impedance to input driver and load?
As I noted earlier, the electrical length of the cable is very short compared to the wavelength at 1 kHz so one could very well neglect the transmission line effects and consider the circuit as generator impedance, cable resistance and load resistance in series.
Now think the extremes: load resistance infinite and load resistance short circuited. In the first case, voltages at the ends of the cable are equal so no cable loss but power at the load is lim outV^2/R, R->inf is also zero. In the second case outV = 0 and load current finite so in both cases you have no power at the load. If you look only at the voltage ratios at cable ends, you get zero attenuation in the first case and infinite attenuation in the second. So indeed we have to specify the loading condition when defining the cable attenuation.
Natural selection for the load impedance is by the maximum power transfer theorem as the main motivation is to transfer some energy to the load and that happens with conjugate match, i.e. impedance real parts equal and imaginary parts cancelling. Maximising the power at load does not necessarily mean the minimisation on the cable losses at the same time.
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