Is this text correct?
I use it when helping people to study for their Licence, and get a lot of indignant scorn for my assertion "It is important to realise, that, after the first brief flutter, no energy is transferred back from the load to the source, that is there is no more reflection of energy when it settles down to the steady state." I hope that here I can get endorsement or at least a fully thought out rebuttal.
Standing waves: reflections and impedances of a radio frequency feeder.
This is an overview in non-mathematical terms. Eye-crossers and tea-dotters will complain that I have been too ready to say “always”, and have not gone into all the ifs and buts. I don't apologise for this: you will not have to unlearn anything you read here if you then go on to study the subject more deeply.
When an RF signal is applied to a transmission line, such as a co-ax cable, by a "source", usually the output stage of a transmitter, the front of it rushes down the line at the speed of light (somewhat slower than in a vacuum) until it reaches a point where the impedance changes, typically at a connection to an Antenna Matching Unit, the "load". Unless the match of the AMU to the line is exact, the wave front, or part of it, is reflected back towards the source, where it will again be reflected towards the load end. This to and fro', between more and more mixed up waves, bangs about for a brief moment until it fades out and everything settles to a steady state, with RF energy being transferred from the source to the load. Since RF is just very rapid alternation, there is an RF variation of voltage along the cable, and a corresponding current in and out of each end.
It is important to realise, that, after the first brief flutter, no energy is transferred back from the load to the source, that is there is no more reflection of energy when it settles down to the steady state. However in the mathematical analysis of the situation, it is convenient and common to express things in terms of forward and reflected waves as though they exist as separate entities.
What the resulting steady state is depends (all else being equal) entirely on the impedance of the load. If the load impedance matches the line impedance, happy, happy state, there is no reflection at switch on, and the voltage and current measured at any point on the line is the same.
From now on I am talking about the steady state after the start up transient has settled down.
When there is a mismatch, measuring the voltage (it's more difficult to measure the current) at any point will show a rhythmic variation, if the line is long enough, with equal values of voltage at distances equal to the wavelength of the RF in the line. This is the standing wave. It is a wave in space, along the line, not a wave in time like voltage and current. It has peaks and troughs at fixed points along the line, and the height of these can vary, from a slight ripple if the match to the load is almost exact, to an extreme where the voltage troughs are zero and the peaks are very high. I will come back to this again later.
It is important to realise that the peaks are not a steady value, but a wildly oscillating voltage, and the “peak” is a measure of its amplitude, r.m.s. or whatever. Also a reminder that all this happens in the standing wave at the frequency of the applied RF.
As well as the voltage standing wave there is a current standing wave within the line, with its troughs where the voltage peaks are, and vice versa. Compare the standing wave with a child's swing: at the ends (momentarily stationary = zero current I) the swing is at its height (high Voltage V); as it swings past the bottom the height (V) converts to speed (I), and so on. Just like the stationary wave in the line, the swing is going nowhere, but it is full of speed and energy. See this Standing Wave diagram
At any point on the line the ratio of voltage to current at that point, taking account of the phase differences, gives the impedance of the line at that particular point. This is important: the impedance varies from point to point along the line, and this is a static condition: the troughs and peaks do not move; and it has serious consequences especially for the transmitter end.
It's worth repeating that the steady state depends (all else being equal) entirely on the impedance of the load. A first consequence of this is that the standing wave is "anchored" at the load end. What happens at the source end then depends on the length of the line, that is how many wavelengths of the standing wave, or parts of a wavelength, it carries, working back from the load. Then, it is the ratio of voltage to current, at the source end that determines the load on the transmitter, and it's also worth repeating that what the transmitter sees as the load on its own output terminals, at its end of the line, depends on the length of the line, and the load on the far end, and the frequency, and nothing else.
Two special cases of mismatch at the load end are a short circuit and an open circuit. In the short circuit case there can be no voltage across that end, it is at a trough of the voltage standing wave, and at the peak of the current standing wave, so there is a large current in and out of whatever it is that forms the short circuit. At an open circuit end no there can be no current, and the situation is reversed, it's a voltage peak and a (zero) current trough. For a load impedance between these extremes there is both a current in the load, and a voltage across it.
Considering now the transmitter at the source end of a mismatched line, once again the phase of the standing wave at this end is determined by the length of the line. In the worst case, on a line that is a number of half wavelengths long, with a short circuit at the end of it, then the load end is at a voltage trough, zero voltage in this case, and a current peak. Looking back at the transmitter, it too is at a point where it has a short circuit as its immediate load, with possible damage to its output amplifier transistors.
It is strange, but true, that if the load end is open circuit, and the line is an odd number of quarter wavelengths long, (in the diagram imagine the end being moved a quarter wavelength closer to the transmitter, to the first current trough, i.e. the line is cut to a shorter length) the same situation will occur, with the transmitter again seeing a short circuit as its load. How can this happen? With no current out of the far end, yet a very heavy current at the transmitter end? Well, that current is just back and forth between the troughs and the peaks as they oscillate, and in the end getting nowhere, like the child's swing.
If the transmitter is at a voltage peak, this again can have serious consequences: the instantaneous peak voltage is much higher than that generated within the transmitter, and can damage the output transistors by exceeding their maximum voltage rating.
We see from the above that the input impedance Zin of a length of coax depends on four factors: its characteristic impedance, Z0, which decides the match to the load, and is determined by its construction, and is independent of the RF frequency; its load impedance, Zload and its length in terms of wavelength, the latter both at the frequency of the RF, the fourth factor.
The formula relating these is complex, but two simple cases are often used: Zin = Zload if the line is a whole number of wavelengths long; and another formula commonly quoted is Z0² = Zin × Zload but note that this last formula applies only to quarter (or multiples of a quarter) wavelength lines. See articles on quarter wave stubs and impedance transformers for applications of these formulae.
Many thanks to everybody who has commented on this, it has been a lively discussion. Most of the comments are on my assertion about (un)real reflected energy. Please note the paragraph "This is an overview ...". I did not want to go too deeply into it, I really intended it to give some background about how a transmitter is affected by the load on the end of a feeder, at an elementary and easily digested level. I don't yet see any need to change anything.