Transmission line reflections from a short termination

[bson]

The explanation for reflections into an unterminated open is rather obvious and intuitive: a signal reaching the end of an unterminated cable has nowhere to go but back, so that's where it goes.

But apply the same logic to a short and the same "nowhere to go" argument fails - obviously it has somewhere to go!  In fact, it has a totally unrestricted path to ground.  So why would it reflect?

TL;DR: in the case of a short the transmission line is effectively twice as long, doubled on itself, and the signal comes back on the ground return.  Yes?  No?  Maybe?

My conjecture is that the short makes the transmission line effectively twice as long, folded in half, and the reflection is the ground return.  Observing signal over ground this ground reflection appears as an inverted version of the outbound signal.  This would also suggest that a proper load termination perfectly balances the signal and ground returns so they fully cancel.  In other words, the signal has no where to go but to get dissipated in the load.  If the ground return is also the chassis ground on the transmitter this returning ground signal will continue onto the chassis and down the mains earth ground.  Since ground is low impedance and is never really "terminated", this reflection can travel quite a ways, spreading across a potentially large portion of the mains network, and only eventually gets consumed by earth coupling and resistive losses.  This, in turn, would suggest under-terminated antenna cables are a Bad Thing as they will radiate somewhere.

Is my guess anywhere near correct?  We're talking power so technically speaking of course the power out can be said to reflect.  So does the voltage, but understanding it comes back on the ground return actually seems kind of important. Naturally, the source has to conjugate match the load so the outgoing I/V phase can be canceled by the load.  (At least at a single frequency, as in the case of an antenna tuner.)  Well, source plus transmission line if the latter isn't a multiple of a half wavelength.

I got thinking about this as I was measuring an RG-8 direct burial antenna feed cable.  I used an AWG to send a 10MHz single cycle burst (just a single sinusoidal cycle) down the cable, and T connected it to my scope, set to 1M input impedance.  Using a 20ft unterminated cable I could see the pulse go out, then return 35.5ns later, for an effective V_P of 0.748ft/ns.  That's about 76% VF, actually a bit shy of the advertised 84%.  (But it does seem very low loss, well under the -0.2dB/100ft spec.)  When I ran it down the main feedline (with the 20' section attached) into an open I measured ~154ns round trip for a length of ~57.5ft.  However, into a short the round trip is 10% longer!  Not much, but easily measured.  This is what led me to the speculate that in the case of a short the reflection is the ground return (foil plus braid shield), which then has to be slower...  (I know I can also hook up the open feedline as a stub filter and measure the notch center on my VNA.  I just wanted to eyeball it in the time domain, for additional reflections and whatnot.)

KM6VKX

[rfeecs]

The explanation for reflections into an unterminated open is rather obvious and intuitive: a signal reaching the end of an unterminated cable has nowhere to go but back, so that's where it goes.

But apply the same logic to a short and the same "nowhere to go" argument fails - obviously it has somewhere to go!  In fact, it has a totally unrestricted path to ground.  So why would it reflect?

TL;DR: in the case of a short the transmission line is effectively twice as long, doubled on itself, and the signal comes back on the ground return.  Yes?  No?  Maybe?

No.  Maybe watch at least the first 6 minutes of this video:

[bson]

I don't see anything in the video that disagrees with what I wrote?

[rfeecs]

I don't see anything in the video that disagrees with what I wrote?

I guess I didn't understand your reasoning.  I thought you were saying a reflection from a short takes twice as long as a reflection from an open, and this is from the ground return.

My long wordy explanation is that for the transmission line you can have two waves that can exist on the line.  Each wave travels in opposite directions.  For an ideal line, the current in the ground return is equal and opposite to the current in the other conductor at each point along the line.  The other conditions that must be satisfied are the voltage at each point is the sum of the voltage of each wave; the current at each point is the difference of the current of each wave; the ratio of voltage to current at each point is the impedance of the line, or at the ends it is the impedance of the source or load.

So when you have an open, the current at the end of the line is zero.  This requires the reflected wave to be equal to the incident wave.

When you have a short, the voltage at the end of the line is zero.  This requires that the reflected wave be the negative of the incident wave.

Other than that, the time required for open and short is the same, and nothing to do with the ground return.

I'm talking about ideal transmission lines.  You may be talking about an unbalanced line where the outer conductor of the coax has some additional current flowing on it and balanced by current through the earth ground.  I guess that is what is going over my head.

Anyway, it looks like you are doing a TDR measurement and as usual w2aew has a great video about that that you have probably already seen.  It shows that the timing when changing the load from matched to a short to a high impedance doesn't affect the timing:

https://youtu.be/Il_eju4D_TM

[bson]

Alright, I devised an experiment to determine if the short reflection returns on the signal lead or on the ground shield:

AWG -> CH1 -> CH3 -> Cable

CH1 and CH3 on the scope are apart by about 2ft of RG-58.  So the pulse from the AWG will go out, pass the 1M CH1, pass the 1M CH3, and go out on the 20ft RG-8 cable.  It reflects at the end of the cable, coming back, where it passes CH3, then CH1, before going back to the AWG (and then reflecting a bit more).  The cable is 20ft.

Now, for an open termination it's clear it's going to come back the way it went out:



So, here's the experiment.  If it returns on the signal conductor it's going to look the same, only 180˚ inverted.
If, however, it returns on the shield, then CH3 with its very low ground impedance is going to split it, and it will be strongly attenuated.



But it's not attenuated!  No split, hence it passed the 1M CH3 input, not the low impedance scope ground, or it would have been split.

So... the short reflection does return on the center conductor!!!

So how then can it be slower on a short reflection... although looking at the screenshots above they're the same.  Hmm.

And then we're also back to square 1... the energy does have somewhere to go, so why is it reflecting?!  I have to surmise this is because the conductor and shield are closely coupled by the dielectric, and this is why it behaves the way it does.

[David Hess]

One way or other the signal reflects back.  The difference between an open and a short is whether the signal is inverted; with an open the signal doubles after the reflection and with a short, it becomes, um, shorted.

[T3sl4co1l]

Right, it stays inside the cable as long as the shield is the outermost conductor, i.e., you aren't connecting anything backwards.

The current or voltage doesn't "go" anywhere, it's the field between the conductors that's going.  And if it's not absorbed, then it comes back around, one way or the other.

Tim

[AG6QR]

And then we're also back to square 1... the energy does have somewhere to go, so why is it reflecting?!  I have to surmise this is because the conductor and shield are closely coupled by the dielectric, and this is why it behaves the way it does.

What do you mean, the energy has somewhere to go?  A short cannot dissipate energy.  Power is current times voltage.  When voltage is zero, power is zero, and energy is zero.

Only a resistance can dissipate energy.  When the resistance of the termination equals the characteristic impedance of the transmission line, that's when no reflection happens.  At any other impedance of the termination, there will be a reflection.

Of course, there are many ways of looking at the issue, and as long as your explanation doesn't violate the physics, you'll get the right answer.  The way it made sense to me was to consider the distributed capacitance and inductance of the transmission line.  For the sake of illustration, model it as a hundred discrete capacitors across and inductors along the line.  As an impulse travels down the line, the source generator has to charge up each capacitor, one at a time, in turn, and the inductors limit the speed at which this can happen.  When the impulse encounters the short at the end, the short can now discharge each capacitor in turn, starting with the capacitor closest to the short.  The short is effectively an ideal voltage source, but it's just one that's permanently set to zero volts.  Other than the fact that it's stuck at zero volts, it behaves the same as any other voltage source feeding a transmission line.