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In order to determine the characteristic impedance of a coaxial cable, I put a resistance on the one end and connected the other end of the cable to a signal generator and an oscilloscope.

This worked fine when using a sine signal as I just had to find the resistance, for which the returning signal was (almost) gone. However, I also wanted to try this with a square signal. The result was first a positive sharp peak decreasing similar to an exponential function followed by a negative peak of the same form.

Why does this happen? My first thought was a differentiator since the sine signal kept its form and the sides of the square signal resulted in sharp peaks with the constant middle part being changed to $0$ voltage. After some reading I found out, that an RL-circuit can be looked at as an differentiator. So my assumption is that the coaxial cable's inductance makes the experiment a RL-circuit. However, where do you draw the inductor into the circuit? My first guess would be to just add it in series with R:

enter image description here

Where V is the oscilloscope and the red parts are the coaxial cable. This also produces the next question: I am measuring the voltage over R and the inductor while for the differentiator the voltage is taken over the inductor only:

enter image description here

I only made a sketch of the seen image on the oscilloscope:

enter image description here

The coaxial cable was $40\; \mathrm{m}$ long.

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    $\begingroup$ First, what is meant by "wave resistance"? Do you mean characteristic impedance? Second, can you share an image of the scope waveform you saw when you measured the square wave? Third, was your scope set for 50 ohm or 1 Megohm input resistance? Fourth what was the rise-time of the square wave stimulus and how long was the cable? $\endgroup$
    – The Photon
    May 3, 2021 at 14:25
  • $\begingroup$ A photo of your physical set-up might also help us work out what happened. $\endgroup$
    – The Photon
    May 3, 2021 at 14:29
  • $\begingroup$ Yes, i mean the characteristic impedance. I only have a sketch of the waveform i saw, sadly. The input resistance was 1 Megohm, but I used a 5 Ohm resistance parallel to the oscilloscope to not have a reflection there. The rise time was 20 ns and the cable length was 40 m. I also do not have any photos of the experiment. I had two connector boxes, one had one end of the cable and the resistance connected to it. The other one had the 50 Ohm resistance, the function generator, the oscilloscope aswell as the second end of the cable connected. $\endgroup$
    – Kekks
    May 3, 2021 at 18:24

2 Answers 2

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An high-pass filter is characteristic of a capacitance in series with a signal. An unexpected high-pass filter is often diagnostic of a broken cable. For example, imagine the connector at one end of the cable gets separated from the ground sheath: that little gap is a capacitance interrupting your signal return path. This kind of problem is easy to cause by bumping the cable with your butt, and easy to fix by replacing the connector.

You say you have your function generator and oscilloscope close to each other, and your 40m cable going to a terminating resistor. If your termination were correct, the oscilloscope would see the signal produced by the function generator, and the part of the signal going down the transmission line would be swallowed by the resistor without evidence. You sketch a figure and give a rise time of 20 ns, so it looks like a time between start-of-rise and start-of-fall of about 70 ns, and a time between start-of-rise and the positive spike of a little over 200 ns.

One model which reproduces this signal is that the connection between the far end of the cable and its terminating resistor is bad. Instead of the echo being swallowed, it’s differentiated by a capacitive connection and returned to you. A good estimate for signal speed in coaxial cable is $c/2$ or $2c/3$, so without taking your horizontal time scale too seriously it’s the right order of magnitude delay for an 80m round trip.

If your oscilloscope has two inputs, you might connect the near end of the cable on channel 1 and the far end on channel 2, to measure this delay time directly (including the time for the echo to return to channel 1, depending on how you terminate the scope connection at channel 2). This will also give you an excuse to touch all of the connectors and look for anything loose or weird.

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  • $\begingroup$ You are right, I didn't properly scale the horizontal axis as i wanted the image to stay compact and just show the shape of the signals. The used signal (start-of-rise to start-of-fall) was $100 \; \mathrm{ns}$, the distanse between start-of-rise to the first spike was $420 \; \mathrm{ns}$ which results in a speed of $2c/3$. This was also the case with the sine wave and with a different resistor. What I found weird, that I was able to completely terminate the returning signal when it was a sine signal but as soon as I changed to a square signal, I was not able to fully terminate it. $\endgroup$
    – Kekks
    May 4, 2021 at 13:47
  • $\begingroup$ For a continuous sine wave, both the differentiator and the delayed echo contribute a phase shift, and the outgoing and reflected/shifted signals will interfere at the oscilloscope. You can’t distinguish between the function generator’s “clean” output versus the sum of that output with a phase-shifted signal at the same frequency, unless you do clever things to connect the scope trigger to the function generator’s timing signals. If you sweep the sine wave’s frequency, you’ll see the amplitude at the scope change as the echo goes between constructive and destructive interference. $\endgroup$
    – rob ♦
    May 4, 2021 at 20:17
  • $\begingroup$ I used the signal generator's trigger signal to trigger the oscilloscope, so the signals I saw should be once the "clean" output aswell as the reflected signal, shouldn't it? $\endgroup$
    – Kekks
    May 4, 2021 at 20:42
  • $\begingroup$ If you are triggering from the function generator, and you have echoes returning from the cable, then disconnecting the cable from the scope should change the amplitude and phase at the scope as the echo stops interfering. $\endgroup$
    – rob ♦
    May 4, 2021 at 22:14
  • $\begingroup$ So without having the ability to go back and check the cable / the connectors, the most likely reason is that one of these was broken resulting in a capacitor (in series). Otherwhise I would have had an integrator due to the parallel capacity of the coaxial cable? $\endgroup$
    – Kekks
    May 7, 2021 at 8:31
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The dominant characteristic of a piece of coax is its parallel capacitance per foot of length. This effect is significant and must be included in a model of the cable along with its series inductance (small) and series DC resistance (very small) per foot.

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  • $\begingroup$ Looking at the circuit diagram of a coaxial cable, the capacitor would be drawn between the two red lines parallel to the resistor in the above circuit diagram. I know that capacitor in row with a resistance acts as a differentiator, but I don't see how this should work for a parallel capacitor. $\endgroup$
    – Kekks
    May 4, 2021 at 13:34
  • $\begingroup$ I'm sorry, I don't understand your overall objective. If you are trying to accurately model the cable, you'll need to include this effect- then look at the system response. $\endgroup$
    – niels nielsen
    May 4, 2021 at 16:06
  • $\begingroup$ I am trying to understand why the coaxial cable with a resistor at the end acts like a differentiator or why it results in signals that look like it acts as a differentiator. (I added the capacitor, the resistance and the inductor to the circuit above) $\endgroup$
    – Kekks
    May 4, 2021 at 20:53
  • $\begingroup$ Since you said the capacitor is the dominant characteristics. Can I look at the right part of the circuit as an RC-circuit with the voltage being measured over the capacitor? In an RC-circuit switching on a constant voltage creates the a voltage over the capacitor of the form $1-\exp(-\alpha t)$ which would fit the second signal seen in the sketch $\endgroup$
    – Kekks
    May 4, 2021 at 21:05
  • $\begingroup$ Note that parallel capacitance, as is present in an intact cable, works as a low-pass filter or an integrator. $\endgroup$
    – rob ♦
    May 4, 2021 at 22:16

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