# Open termination, signal propagation, transmission lines

I'm inexperienced with transmission line theory so I have a question about implications of signal travel time. Most of what I've read so far has been on lossless transmission lines, not sure how relevant that is.

Suppose you have a dc source connected to some transmission line with an open termination: Here the wave travels down the transmission line and is reflected back at the end. Depending on the impedance matching between R1 and the line there might be more back and forth reflections but eventually the voltage at either end of the line will stabilize at the value of the source voltage and no more current flows.

Now the same thing but with an ac source: In this case the end of the line will always oscillate since the line is being driven with a changing velocity, so it can't settle to a single value. Because of the presence of the transmission line, this diagram implies a source driving a load whose physical dimensions are comparable to wavelength of waves launched by the source, meaning that the voltage and current varies with position over the line. Also, the propagation speed of the waves is independent of the length of the line or the frequency of the source. As an example say the waves propagate at $$\c\$$ and that it takes the waves 100 ps to make a one-way trip from the source end to the open termination end. So the load is approximately 3 cm long and the source wavelength is on the order of 3 cm at most, so the frequency is approximately at least 10 GHz.

So now suppose you use the same ac source to drive a load whose physical dimensions are much smaller than the source wavelength: This could happen either because the load circuit was made smaller or because the frequency of the source was reduced, thus making the wavelength much bigger than the dimension of the load. Considering the second case, say the load is still about 3 cm long, but the source wavelength is now increased to 300 cm, or a frequency of 100 MHz. However, changing the source frequency hasn't changed the wave propagation time, so it still takes 100 ps for any electrical signal (or any information at all since the wave travels at $$\c\$$) to reach the open termination and then another 100 ps for any "response" information to come back to the source.

Lumped circuit theory says that the voltage at the open circuit will follow the oscillating source (instantaneously) and there will be 0 current in the branch at all time. In reality the open termination needs at least 100 ps delay between a "new value" being presented at its input (due to changing ac source) and that value having any effect on the open terminated end. Since the load is driven by a constantly changing ac source it seems that there should always be a nonzero current going "into" the load. This seems incorrect so I think I'm missing something.

I think the answer might be that the period of the new source is 10 ns, which is 100 times longer than the one-way travel time of an electrical signal through the load (or there are 50 reflections from open end that make it back to source in one period). Each returning reflection "tells" the source to decrease the amount of current going into the load since the reflection has information about the open circuit. So I think that the low frequency source input changes very slowly relative to how quickly reflections make it back and reduce the amount of incident current. So if you hook up a low frequency source to this kind of load, although the current is not actually zero it is very small since reflections make it back to the source and "dampen" the incoming current at a rate much faster than the input current increases/decreases.

Is this an accurate description?

the propagation speed of the waves is independent of the length of the line or the frequency of the source

The phase velocity, which is probably what you mean by propagation speed, is independent of the length of the transmission line, but varies with the frequency of the source.

Lumped circuit theory says that the voltage at the open circuit will follow the oscillating source (instantaneously) and there will be 0 current in the branch at all time.

I think the answer might be that the period of the new source is 10 ns, which is 100 times longer than the one-way travel time of an electrical signal through the load (or there are 50 reflections from open end that make it back to source in one period). Each returning reflection "tells" the source to decrease the amount of current going into the load since the reflection has information about the open circuit.

When the AC signal is first applied, there is a back and forth of reflections, as you imagine. However, over time, "transmission line" consisting of R4 and a ground somewhere approaches an "oscillation mode". Each point in the resistor/transmission line, will have a sinusoidal instantaneous voltage. The voltage at any other point on the "transmission line" will also have an sinusoidal instantaneous voltage that will generally differ from that of the first point in phase and amplitude.

Yes

In reality the open termination needs at least 100 ps delay between a "new value" being presented at its input (due to changing ac source) and that value having any effect on the open terminated end.

There will be a delay, but again, it is frequency dependent.

Since the load is driven by a constantly changing ac source it seems that there should always be a nonzero current going "into" the load.

I take "the load" to be the R4 in your schematic, and I take "a nonzero current" to mean a nonzero rms current, rather than an instantaneous current which is never zero. With those assumptions, yes, there will be an extremely small AC current into R4 using transmission line theory, rather than "instantaneous" lumped-element theory. The transmission line theory will be correct, there will be some current, but it will be so small that the lumped element model is nearly perfectly correct.

This seems incorrect so I think I'm missing something.

Your reasoning is correct. What you may be missing is that the current will be so small that it hardly differs from the lumped-element prediction.

So if you hook up a low frequency source to this kind of load, although the current is not actually zero it is very small...

Yes

since reflections make it back to the source and "dampen" the incoming current at a rate much faster than the input current increases/decreases.

Yes, when AC is first applied. As time passes, reflections "smooth out", and an oscillation mode is "excited" (or "entered") as described above.