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Suppose I have an RF source with 50 Ohm output impedance. First lets imagine the RF source outputs a low frequency 10 kHz signal (electrical wavelength is about 20 km). Suppose I have a 50 ohm coax that is a few meters long. Now I plug this cable into a high impedance input oscilloscope. Suppose I read 1 V peak-to-peak. If I now 50 ohm terminate the oscilloscope (by putting a 50 ohm resistor in parallel with the scope input or using an internal 50 ohm mode) then I know the signal reading on the oscilloscope will drop to 0.5 V peak-to-peak.

I understand that in the first case (high impedance input) that the signal reflects off of the scope and the sum of the incident and reflected wave causes the measured voltage to double relative to the 50 ohm terminated case in which there is no reflection. There are no cable length effects because the cable length is much less than an optical wavelenth so it is the same circuit analysis as if the signal were DC.

Now, suppose the signal frequency is increased to 100 MHz and we consider the high input impedance scope. Now the electrical wavelength is about 2 m so cable length matters. Here is my question, given that the source is 50 ohm terminated, will the amplitude that I measure at the scope depend on the length of the cable? My intuition seems to say it should be I can't work out why. It seems to me that the first signal will hit the scope and then reflect and this will cause the signal to double relative to the 50 ohm terminated case. however, the reflected signal travels back to the source where it is dissipated in the 50 ohm terminator which is there. So the wave doesn't bounce back multiple times. Without these multiple bounces I don't see how you can get these sort of interference affects at the oscilloscope.

So my question in a nutshell:

  • If you output a high frequency signal from a 50 ohm output impedance generator and put it through a long cable, will the signal amplitude that you measure on a high-impedance oscilloscope depend on the length of the cable and why or why not?
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  • \$\begingroup\$ Note that it's a good thing to either use ordinary 50 (or 75) ohm cable properly terminated at the scope, or to use a proper O-scope test lead (which uses resistance wire and compensation to insure no reflections). \$\endgroup\$ – TimWescott Aug 11 at 20:36
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Here is my question, given that the source is 50 ohm terminated, will the amplitude that I measure at the scope depend on the length of the cable? ... It seems to me that the first signal will hit the scope and then reflect and this will cause the signal to double relative to the 50 ohm terminated case. however, the reflected signal travels back to the source where it is dissipated in the 50 ohm terminator which is there. So the wave doesn't bounce back multiple times. Without these multiple bounces I don't see how you can get these sort of interference affects at the oscilloscope.

This is correct. In this scenario you'll measure double the signal when the scope has a high-impedance termination, compared to if it had 50 ohm termination. This result won't depend on the length of the cable.

In the real world, the cable doesn't have exactly 50 ohm characteristic impedance and the source won't have exactly 50 ohm output impedance. Also, the cable won't be lossless. So you will see some variation in the measured amplitude, depending on the length of the cable.

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  • \$\begingroup\$ Great, thank you for the information. This makes sense to me. Especially the part that non-idealities in the source or characteristic impedance will lead to reflections that can affect the measured signal (I had been observing this in the lab and was wondering if I was misunderstanding scope termination or seeing a non-ideality in the transmission lines or signal generator.) \$\endgroup\$ – Jagerber48 Aug 11 at 19:32
  • \$\begingroup\$ It's a "voltage divider". 50-ohms in the cable, 50-ohms at the termination. I know it seems odd to apply Ohms Law to this, but it works \$\endgroup\$ – Kyle B Aug 11 at 19:41
  • \$\begingroup\$ @KyleB, that only gives the amplitude of the initial reflection. If you want the final value after multiple reflections have settled out you need to consider the source impedance and termination impedance. \$\endgroup\$ – The Photon Aug 11 at 20:12
  • \$\begingroup\$ @ThePhoton Right, but if I read correctly there's matching source termination too , so there shouldn't be reflections (of significance)???? \$\endgroup\$ – Kyle B Aug 11 at 22:02
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If the cable impedance, the signal generator impedance, and the terminator impedance are all the same, then there will not be any reflections.

Reflections occur at places where the impedance changes.

If you use a long cable, but with impedances matched at all components (and a terminator at the oscilloscope end,) then there will be no reflections. There will be losses in the cable, so with a really long cable you will measure an amplitude lower than the half you would expect from a signal generator driving a terminator. It won't be from reflections, though.

If you don't terminate the oscilloscope end, then you will see reflections. The voltage you measure will depend on the length of the cable and the wavelength of the signal in the cable. It can vary between double the signal generator output level to zero.

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