I am trying to bring my knowledge of electronics to the next level. So I am learning about transmission lines.

I perfectly understand the case with an open end. Current flows to the end of the line, can't flow any further, and charges all pile up. Incoming charge carriers pile up, like cars on the motorway and rise the pressure, while flow grinds to zero.

Now, I have problem understanding a line end with a short circuit. Basically a transmission line, like any other wire, consist of an infinite number of short circuits. By adding a short circuit to the end of the T.L. basically, we haven't changed anything. There is no interruption in the continuity. Wave is traveling down the T.L., when it comes to the short circuit, there is just another piece of wire. Nothing should happen, there should be no reflection.

Now, I am obviously wrong, because oscilloscope will show reflection in the short circuit case. Can somebody explain to me where am I wrong?

DROBNJAK:
Now, I have problem understanding a line end with a short circuit. Basically a transmission line, like any other wire, consist of an infinite number of short circuits. By adding a short circuit to the end of the T.L. basically, we haven't changed anything. There is no interruption in the continuity. Wave is traveling down the T.L., when it comes to the short circuit, there is just another piece of wire. Nothing should happen, there should be no reflection.

Now, I am obviously wrong, because oscilloscope will show reflection in the short circuit case. Can somebody explain to me where am I wrong?

My study of T.Ls was over 40 years ago but I don't recall an infinite number of short circuits. Coax is a T.L and is not short circuit.

I do recall points along the line as having zero volts due to the wave traveling along it.

At one quarter wave length the voltage is maximum, at half a wave length the voltage is zero so a short will have no effect.

DeviceNet (CAN BUS) Trunk lines and Drops require a 120 ohm Termination Resistor at each end of the line so if you are MISSING one it will measure 120 ohms instead of 60 ohms. It prevents Relected Wave which occurs when the line is lect unterminated.

The reflection occurs at selected frequencies based on the line length to the "fault" TDR or time-domain-reflectometry is often used to identify cable faults.
I once mucked about trying to do a bodged extension on a TV aerial circuit and the only way I could get it to work correctly was by fitting a short-circuit coax plug into a particular aerial sockets - all other sockets then produced reasonable signals. I presumed this established a zero node point

That short circuit thing is pretty misguided. The capacitance on the line makes it look like a short circuit for a very brief period of time, but as soon as the line charges up, it goes away. The open and short are simply two special cases of the termination having zero resistance. You can make one appear to be the other by simply changing the length of the line.

Yeah, it seems I can't find a descent explanation.

What I just found is, in the case of AC energy along the wire is not carried by electrons, but by EM wave in the dielectric. Try to bend your head around that?

DROBNJAK:
I am trying to bring my knowledge of electronics to the next level. So I am learning about transmission lines.

I perfectly understand the case with an open end. Current flows to the end of the line, can't flow any further, and charges all pile up. Incoming charge carriers pile up, like cars on the motorway and rise the pressure, while flow grinds to zero.

Now, I have problem understanding a line end with a short circuit. Basically a transmission line, like any other wire, consist of an infinite number of short circuits. By adding a short circuit to the end of the T.L. basically, we haven't changed anything. There is no interruption in the continuity. Wave is traveling down the T.L., when it comes to the short circuit, there is just another piece of wire. Nothing should happen, there should be no reflection.

Consider first the perfectly terminated case - the step wave reaches the resistor and
the current flows through the resistor creating a voltage across the resistor that perfectly
matches the voltage on the line - nothing further changes.

By having a short circuit the edge reachs the end of the line and the voltage it brings
is immediately cancelled due to the zero ohm resistor at the end - so there is a voltage
step the other direction generated, which means a new edge starts to flow the other way,
with inverted voltage (cancelling the incoming) and double the current (because the
wave is going the other way).

In the open circuit case the wave reaches the end and the current edge is immediately
countered by its exact opposite (the sum must be zero at the open end), causing a
voltage step to be generated that adds to the incoming voltage, which then propagates
back (leaving no current flow in its wake).

Really though the way to understand TLs is to study the one-dimensional wave equation,
we are talking about waves in a medium, and the open / closed / terminated ends
correspond to different refractive indices.

MarkT:
By having a short circuit the edge reachs the end of the line and the voltage it brings
is immediately cancelled due to the zero ohm resistor at the end - so there is a voltage
step the other direction generated, which means a new edge starts to flow the other way,
with inverted voltage (cancelling the incoming) and double the current (because the
wave is going the other way).

Thanks. 2 Qs:

in the above description for short circuit case, is the reflected wave coming back along the positive or negative wire? With open circuit case, there is only one wire and the reflected wave is coming back along the positive wire, because negative wire is not even connected.

I watched one video that explained that energy is moved as EM wave through dielectric (insulation), because electrons are just crawling at 2 m/sec. Is that true?

The wave travels along the transmission line as a whole, its an electromagnetic
wave involving changing electric and magnetic fields - the voltage across the
wires and currents in them are associated with these fields. Because the charge
carriers are trapped in the metal wires they constrain the fields to follow the line.

I am not sure what you mean by "constrained", but the speed of propagation is set by the dielectric constant, yes. It is directly analogous to the index of refraction.

However, there are some very complex transmission lines like microstrip and co-planar waveguide that have effective dielectric constants based on the interaction of the EM fields and the multiple dielectrics involved.

KeithRB:
I am not sure what you mean by "constrained", but the speed of propagation is set by the dielectric constant, yes. It is directly analogous to the index of refraction.

Thanks. That explains the main point.

When high frequency transmission lines are in a question, energy is transmitted mostly by EM waves in a proximity of the conductor. So all these reflections are much easier to understand when we are talking about fields, as opposed to charge carriers.

The charge carriers can't leap out of the wires into the dielectric - this is
the main boundary condition for the wave equation.

For coax line the speed of propagation is determined by the dielectric,
for twisted pair its more complex as part of the field is in air, part in dielectric,
so you actually get a complex mix of two propagation speeds, but most of the
field is in dielectric material I think.

For waveguide the speed is affected by the mode as the waves can zig-zag
down the guide in some modes.