# Standing Waves

Coaxial cables, AMUs and Standing Waves, an overview.

This article is in non-mathematical terms. Tee-crossers and I-dotters will complain that it does not go into all the whys and wherefores. I make no apologies for this. However, if you subsequently go on to study further, you will not have to unlearn anything you will read here,
Imagine an RF signal is applied to a 50ohm co-axial cable, by a “source”, usually the 50 ohm RF output of a transmitter. The front edge of the signal rushes down the 50 ohm line, usually at 2/3rds the Speed of Light. (Velocity Factor = 0.66) When this front reaches the Antenna Matching Unit, some of it may be reflected back towards the source, where it will again be reflected back towards the load end. This to and fro between increasingly mixed up waves, bangs about briefly until it fades out and everything settles down to a steady state. It is important to realize that now, after this first brief flutter, provided that the match is perfect, no energy is transferred back from the load to the source, i.e there is no more reflection of energy once settled down into the steady state.
If you want a 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 depends entirely on the impedance of the load. If the load impedance matches the line impedance, there will be no reflection, so that RF voltages and currents measured at any point on the line will be exactly the same.
What happens after the start up transient has settled down to the steady state? Since RF is just a very rapid alternation, there is a variation of RF voltages and currents along the cable and an outflow of power into the load.

However, if there is a mismatch, the measured voltages at any point (it is much more difficult to measure the currents), will show a rhythmic variation in amplitude, i.e if the line is long enough, with equal values of this voltage every half wavelength back along the line. This is the standing wave, which occurs at the same frequency as the applied RF, but spaced out equally along the line.  This is not a wave in time, like AC voltages and currents. The peaks and troughs of the Standing Wave are at fixed points along the line, the size of these varying, from a slight ripple if the match is almost exact, to the extreme case where the voltage troughs are zero and the current peaks are very high or vice-versa, i.e the voltage peaks are very high and the current very low.
It is important to realize that these stationary peaks and troughs are not steady values, but are varying in amplitude. The voltage standing wave naturally has a corresponding current standing wave within the line, with current troughs being where the voltage peaks are, and vice-versa.
As a comparison, imagine a child’s swing: at the very end of every swing the seat is momentarily stationary, (zero current I) but has high potential energy (high Voltage V) During the swing, the kinetic energy from the height (V) is converted to speed (I), and so on to the end of the swing, where I becomes V again. Like the stationary standing wave in the coaxial line, the swing oscillates up and down but the swing is not going anywhere.
At any point on the line the ratio of voltage to current, taking account of the V/I phase differences there, gives the impedance of the line at that particular point. This is important because the impedance varies from point to point along the line, and, this being a static condition on the line: the troughs and peaks do not move along the line, but fixed and rising and falling in amplitude. This can have serious consequences for the electronics at the input end of the line.

It is worth repeating here that the steady state depends entirely on the impedance at the load end, a consequence of this being that the standing waves are “anchored” there. What the situation is, at the source end, depends on the length of the line, i.e. how many wavelengths or parts of a wavelength exist, from there to the load. Then, it is the ratio of voltage to current, at the source end of the line that determines the load Z on the transmitter.
It is worth repeating here that what the transmitter sees as its load, depends on the length of the line, the RF frequency and the load Z at the far end.
Two special cases of mismatch at the load end are:
If the load is short-circuited, the voltage there is zero, i.e. the voltage standing wave at that point is in a trough, but the current standing wave is a maximum. This means that a large current flows in and out of whatever is causing the short circuit and generating heat there.

Where no current can flow because of an open circuit there, the situation there must then be a voltage peak and a current trough.
For load impedances between these two extremes there is both current and voltage entering the load, so power is transferred.
Consider now the transmitter as the RF source into a mismatched line. The phase of the standing wave at the source end is determined by the length of the line and the Z at the far end. In the worst case, where the line is an exact number of half wavelengths long and with a short circuit at the load end, there will be a voltage trough there, (0v) but a high current peak. Looking back at the transmitter end, it too will be at a point where there is a virtual short circuit as its load and damage to the PA output transistors is very likely.

If the load end is open circuit and the line is an odd number of quarter wavelengths long, (imagine the load end being moved to the first current trough a quarter wavelength closer to the transmitter, i.e. the line is cut to a shorter length) the same situation occurs. The transmitter will again see a short circuit as its load. How does this happen? No current flowing out of the far end, yet a very heavy current at the transmitter end? The current is just flowing back and forth between the troughs and the peaks as they oscillate, and in the end getting nowhere, just like the child’s swing.
If the match is very poor and the transmitter end sees a voltage  peak, this voltage will be much higher than that being generated by the transmitter, and can easily damage the output transistors by exceeding their maximum voltage rating!
We can see from the above that the input impedance Zin of a length of coaxial cable depends on four factors:,
Its characteristic impedance Z0, determined by its physical construction and determines what Z the AMU must give it.
The ACTUAL load impedance, Zload  i.e the Z that the AMU is actually presenting to it.
Its electrical length, taking into account its velocity factor.
The wavelength/frequency of the RF.
Most formulae relating to SWR & Z are complex, but two simple cases are often seen: Zin = Zload, which can only apply if the line is an exact whole number of half wavelengths long.
Also Z02 = Zin x Zload  which only applies to quarter or multiples of quarter wavelength lines. You will need to read published articles on applications of quarter wave stubs and impedance transformers which have different formulae.
Please note, that the above is a reworking of an article by an unknown amateur and, I hope, a tad more understandable. G3JKX