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.

  • $\begingroup$ The flow of energy in an electromagnetic field is governed, in the theory, by the Poynting Vector, which has only one direction at any point. In a coax cable it points at all times from transmitter to antenna. along the length of the cable, in the dielectric between the inner and outer conductors. It never points the other way. In a lossy cable it is slightly divergent , transferring energy into the conductors to be dispersed as heat. $\endgroup$ Oct 21, 2015 at 11:26
  • $\begingroup$ But if you calculate the voltage and current in the transmission line as separate forward and reflected waves, don't you get two associated Poynting vector fields, pointing in opposite directions? And if you add those vector fields together, don't you get exactly the same result? $\endgroup$ Oct 26, 2015 at 19:56

4 Answers 4


I'm afraid that it is not correct that there is no energy transfer back to the source after an initial period.

Wherever you have a mismatch on a transmission line, there is a reflection (at least partial) back to the source. A percentage of power is "actually" reflected back. It can be separated from the transmitted power using a suitable device like a hybrid. When the reflected wave comes back to the source, it can be absorbed if it is terminated at Zo (eg 50 ohms). A typical transmitter does not provide a 50 ohm load to returning power, however, so there is a (partial) reflection of energy back in the forward direction. There is an infinite series of partial reflections back and forth, fading away as power is transferred to the load or lost in the transmission line. In the end, a fraction of the transmitter's power is ultimately transferred to the antenna. The fraction not transferred ends up partly as heat due to transmission line loss and partly as heat dissipation in the transmitter.

It is useful, as you suggest, to consider the "time domain" situation, such as when you transmit a very short pulse (much shorter than the propagation time of the transmission line). You can "watch" the reflection at the far end and the series of "internal" reflections in the transmission line. In the end (neglecting the wide bandwidth required), you should see just the same power transfer and loss as you would with CW waves. (A sine wave can be thought of as a sum of multiple short pulses if you like.)

  • 1
    $\begingroup$ "There is an infinite series of partial reflections back and forth..." -- Surely not infinite. If there were an infinite series of reflections then all antennas that ever existed in the past are still reflecting power back and forth however small. Of course, the energy reflection ultimately stops but that means there are not an infinite number. $\endgroup$
    – K7PEH
    Oct 20, 2015 at 15:58
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    $\begingroup$ Mathematically, it's an infinite series of reflections, but the amplitude (and power) falls off pretty fast and it gets down into the thermal noise level before long. If you have a 10% reflection amplitude, the reflected power is reduced by at least 20 dB per "bounce." So in practice you don't need to worry about ancient information in your transmission line. $\endgroup$ Oct 20, 2015 at 17:46
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    $\begingroup$ The mathematical model is merely a model -- it is not physical reality. Off hand, I can't think of any real physical process that is infinite. My graduate work was Mathematical Physics so I do understand notions of infinite series. $\endgroup$
    – K7PEH
    Oct 20, 2015 at 18:17
  • $\begingroup$ Thanks to those who commented. I agree that there is theoretically an infinite series of reflections, but, as Martin says, they fade out, just as charging a capacitor takes a infinite time to charge up from a constant voltage source, but we assume, for all practical purposes, that it is fully charged after a few time constants. $\endgroup$ Oct 21, 2015 at 10:57

It's true that if the transmitter is sending a pure sine wave which began infinitely long ago, and will continue on forever, then we can take the mismatched load impedance, rotate it the appropriate amount according to the length of the transmission line, and arrive at some impedance that the transmitter sees. And, from the transmitter's perspective, this long transmission line plus the mismatched load is indistinguishable from a lumped impedance right at its output.

For example, this is exactly true:


simulate this circuit – Schematic created using CircuitLab

This is a simplification we can make in some circumstances. The problem with your statement is you neglect the conditions:

  • it must be a pure sine wave
  • that began forever ago
  • and continues forever into the future

Of course, such a signal contains zero information, and real communications are never like that. Real transmissions modulate a carrier over time. Or we can say the same thing in the frequency domain: real transmissions aren't just one frequency.

As such, we can't reduce the entire system to just one impedance that the transmitter sees. For a complete analysis we must think about the frequency response of the system over the entire range of frequencies present in the signal. We can depict the frequency as a Bode plot which shows the changes in magnitude and phase present in a system.

For a feedline and an antenna, we want the magnitude to be unity, or as close as possible across the entire channel. Less than unity would be a loss, and more than unity would violate energy conservation.

But magnitude is only half the frequency domian: we must also consider phase. If we have a transmission line, all frequencies will be delayed by some constant amount of time. Higher frequencies will advance more in phase during this time than low frequencies, so the phase plot will be a straight line, although perhaps slanted. We say such a frequency response has linear phase.

Since we care only that the phase response is linear, but any amount of delay is acceptable, it is common to differentiate the phase response with respect to frequency to arrive at the group delay. If the phase response is linear, the group delay will be flat.

And group delay has a intuitive explanation: it is the delay introduced to the amplitude envelopes of the frequency components of the signal. And it makes sense that we want the group delay to be flat: our signal consists of many frequency components, and we want them all delayed by the same amount so they arrive at the receiver at the same time.

So what happens to the group delay if the load is not matched to the transmission line? It stops being flat, meaning some of the frequency components of the signal arrive after others. This interferes with demodulation of the signal.

For SSB and CW this is rarely a problem: these modulations work over a very narrow channel. Over this small range of frequencies it is hard to get enough group delay variation to significantly degrade intelligibility. But for high-speed modulations such as used in TV broadcasting or Wi-Fi, this can very much be a problem. Even commercial FM broadcasters work to minimize variations in group delay.

So, a mismatched load can be a real problem. Now getting back to your original statement:

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.

As it turns out, all this explanation of frequency response and group delay is just saying that there are reflections, in different terms. The feedline and antenna are a linear time-invariant (LTI) system. If I were to transmit an impulse and measure the reflections that came back over time, I would know the impulse response of the system. And through the mathematics of Fourier analysis and the properties of all LTI systems, I can calculate the frequency response knowing just the impulse response.

So, it does not matter if we think about reflections in the time domain, or variations in group delay in the frequency domain. They can mathematically be demonstrated to be the same thing. So you can not say that there is no reflection. And, you can not simplify a mismatched load to a simple impedance, though that may be a valid approximation under some circumstances.

  • $\begingroup$ Thank you Phil for that comprehensive and detailed study. I cannot argue with your analysis. It is true that in the maths we can work with reflections, but it's the assumption that they are real, energy flowing both ways at once, that buzzes in my bonnet. I don't agree that you need an infinite length of time, the whole thing settles down to a steady state very soon after switch on, and switching it off doesn't affect what is happening now. By the way it is the initial transient, which radiates as any changing EM field does, that causes the "click" in a radio when you switch the lights on. $\endgroup$ Oct 21, 2015 at 16:06
  • $\begingroup$ @HarryWeston No, you really do need an infinite amount of time. With each iteration of reflections the current state approaches the steady state, but it never gets fully there unless you wait forever. It's not unlike charging a capacitor through a resistor. At some point you can say it's "close enough" for some purpose and neglect any discrepancies. Anyway, I'm still not sure why you insist on saying the reflected power is not real, since we can see its effects everywhere. It's as real as anything that can be predicted by physics can be. $\endgroup$ Oct 22, 2015 at 2:13
  • $\begingroup$ Thanks again Phil Frost. It is true that in theory there would be reflections for ever, but that is getting too far into pedantry, and not necessary to consider in order to understand what is going on. The capacitor example is apt, soon the diminishing change in voltage is lost in the thermal noise anyway. I would be interested to know what effects you would say can only be explained by assuming that energy does flow separately back from load to source, and when you say "reflected power " do you mean "reflected energy"? $\endgroup$ Oct 22, 2015 at 14:33

I think your description is fairly reasonable, with one exception: I would change it to say that

  1. the net energy transfer is forward, not reverse, and
  2. don't try argue the point of whether there is a “reverse flow” at all.

Energy is not a substance, it's a quantity. I would say that there is no meaningful difference between “energy only moves from the source to the load” and “the forward flow is slightly larger than the reverse”. They are simply two different models of the exact same situation. You can't put a label on a piece of energy and say where that energy ended up.

  • $\begingroup$ Of course net energy is transferred to an antenna, that's the whole point of a transmitter. And energy is real enough, it burns my face in the sun: true not a substance, at any rate since we abolished phlogiston. $\endgroup$ Oct 21, 2015 at 10:51
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    $\begingroup$ @HarryWeston Sorry, I screwed up that sentence totally. Fixed to say what I meant. $\endgroup$
    – Kevin Reid AG6YO
    Oct 21, 2015 at 14:48
  • $\begingroup$ Thanks Kevin, I think I can now agree with what you say. But, about not arguing that there is no reverse flow, well the very assumption that there is shows a complete lack of understanding of the theory. It stems from the solution of the Telegraphers' Equations to give the mathematical situation of two wave travelling in opposition. So seductive that it is taken as a description of the physical world, rather than just the very useful, and correct, theoretical analysis. $\endgroup$ Oct 21, 2015 at 15:48
  • $\begingroup$ @HarryWeston A correct analysis is any one that accurately (or accurately enough) predicts the outcome of an experiment. There can be more than one. It sounds like your issues are more with the philosophy of physics than any particular technical issue. $\endgroup$ Oct 22, 2015 at 14:21

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.

If there were no more reflection, then the transmitter could not distinguish between a mismatched load and an infinitely long transmission line. But of course this is not the case, so I think your statement is demonstrably wrong.

You can say that the transmitter is putting energy into the system at some rate (the transmit power), and the antenna and losses are taking it out at exactly the same rate. But that's just an obvious consequence of the law of conservation of energy.

You can also say that if you had a lossless transmission line, and a transmitter capable of ideal operation into any impedance without damage, that all the transmit power ends up in the antenna eventually. And for moderate mismatches and operation on HF where feedline losses are typically low, this is approximately true in practice, also.

You can also say that it is sometimes a valid engineering simplification to not concern ourselves with the forward versus reflected power, and instead just think of the impedance at the load, transformed by rotation according to the length of the transmission line. The operative word here is sometimes.

But you can't say "there is no more reflection". As Martin Ewing points out, we can at any time extract the reflected power with a hybrid. We can measure it with an SWR meter. We can transmit an impulse and see the reflection in the time domain. So it's demonstrably happening.

Perhaps this is the issue: "reflection" is inherently a time domain concept. So if we are thinking about the "steady state", we are thinking about sinusoids and the frequency domain. There can't be "reflections" in the frequency domain, because a reflection requires a concept of time and the frequency domain has none. So is there a frequency domain equivalent of reflection in the time domain? There is: in a mismatched system there will be variations in group delay with respect to frequency.

Is that a problem? It depends. See What happens in a low loss feed-line with a high SWR

  • $\begingroup$ I'm afraid that this answer goes awry from the first paragraph. Of course the transmitter cannot distinguish between a mismatched load and an infinitely long transmission line, it can only work into the impedance presents to it by the feeder and its load, assuming. $\endgroup$ Oct 21, 2015 at 10:54
  • $\begingroup$ @HarryWeston sometimes you can make that simplification, if the reflections are insignificant to your problem. Read the whole answer, and I point out no less than four ways the transmitter can make the distinction. $\endgroup$ Oct 21, 2015 at 12:21
  • $\begingroup$ @HarryWeston I'll try another answer now that I know specifically what your beef is. $\endgroup$ Oct 21, 2015 at 12:30
  • $\begingroup$ Sorry Phil, I was a bit hasty with that dismissal, however there are other issues I would disagree with: We can of course use a narrow pulse and see reflections. and it is a moot point whether they still exist invisibly later on. I suppose what I am really saying is that, after the transient, you cannot distinguish any reflected energy. The directional couplers in SWR meters do a clever trick on the phases of current and voltage to enable the calculation of various parameters like reflection coefficient. They are not a proof that reflected energy travels back towards the transmitter. $\endgroup$ Oct 21, 2015 at 15:40
  • $\begingroup$ @Harry Weston, there is no fundamental difference between a wave travelling left-to-right and a wave travelling right-to-left. They are both transferring energy. $\endgroup$
    – Chu
    Oct 24, 2015 at 0:18

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