I was recently reading an interesting article from the ARRL to get a better understanding of the SWR.

The following case is still unclear: A low loss feed-line with an impedance that doesn't match the antenna's impedance and the receiver's impedance.

In this case a large part of the signal should "bounces" back and forth into the feed-line due to the impedance mismatch. At this point the article states that:

The energy bounces back and forth inside the cable until it’s all radiated by the antenna for a lossless transmission line. An important point to realize is that with extremely low loss transmission line, no matter what the SWR, most of the power can get delivered to the antenna.

I understand that during each "bounce", due to the impedance mismatch, some amount of energy will be transmitted and the rest reflected.

I don't understand how this transmitted power could be useful. After "bouncing" back and forth the part of the signal that will be transmitted will probably be out of phase with the signal sent by the transmitter. Even if by any luck the reflected signal and the signal currently sent by the transmitter happened to be in phase, the information conveyed wouldn't be the same.

So to me, even if almost of the power is transmitted, due to the high SWR most of this power should be just noise and don't improve the quality of the transmission in any case. But it's not what the article seems to explain. What did I miss ?


3 Answers 3


For many modulations, the modulation is very slow compared to the propagation delay of the feedline. For example, SSB is typically limited to no more than 4 kHz. That corresponds to a wavelength of of 75 km. As long as the feedline is significantly shorter than this, then the delay due to the feedline is negligible.

It may be easier to understand intuitively in a digital case. For example, PSK31 sends 31.25 symbols per second. 31.25 Hz corresponds to 9600 km, meaning by the time you are sending a bit, the previous bit is 9600 km away.

This breaks down for very high-speed modulations. For example, 8VSB modulation used by HDTV broadcasts transmits 10.76 million symbols per second, equivalent to a 29 meter wavelength. At these speeds, a feedline is long enough to introduce significant delay, which could come out looking like multipath distortion. Then again, TV channels are 6 MHz wide. Most amateur transmissions are several orders of magnitude narrower. Protip: don't operate your commercial TV broadcast station with a poorly matched feedline.

But for most amateur transmissions, we can consider the transmission as a pure sine wave, which is a reasonable approximation given the timescales dictated by the length of the feedline and the modulation. Then, it helps to visualize what a standing wave in the feedline looks like. From Wikipedia:

enter image description here

If we regard the transmitter to be on the left and the antenna to be on the right, then the blue wave represents the transmitted wave, and the red wave the reflected wave.

Now here's the thing to realize: this particular image depicts a complete standing wave on a lossless transmission line. In this case, we are delivering 0 power to the antenna, the VSWR is infinite, and the transmitter's finals are probably about to explode. This doesn't happen in practice because our antennas always accept at least some of the power.

In practice, only some of the power is reflected back, and the result is a partial standing wave. Dan A Russel has a page of great animations, like this one:

enter image description here

In the case of a partial standing wave, we can see that the envelope (outlined by the dotted lines) does not reach 0 amplitude at the nodes. In this case, some power is transferred. This is also a good visualization of VSWR: it is the ratio of the voltage amplitude at the antinodes (envelope maximums) to the voltage amplitude at the nodes (envelope minimums).

Notice also that the reflected wave might arrive back at the transmitter in any phase, depending on the length of the feedline and the antenna's reflection coefficient. It may or may not be in phase, but it will be a constant phase, not just noise. By adding an antenna tuner we can reflect the reflected wave back at the antenna, and adjust the phase of this re-reflected wave to be whatever we want, within the limits of the tuner's capabilities of course.

  • $\begingroup$ That last paragraph of the reference article is particularly confusing, And wrong. The transmission baud rate is 10.76 Mbaud, or 10,760,000 symbols per second. Also, it is a bad example, as there is compression involved, and trellis encoding, which doesn't even make the 31.x Mbits/second rate the effective one. en.wikipedia.org/wiki/8VSB $\endgroup$
    – jcoppens
    Commented Nov 18, 2014 at 18:31
  • $\begingroup$ Of course, this also means that the bit delay would only be some 30 meters, not kilometers. $\endgroup$
    – jcoppens
    Commented Nov 18, 2014 at 18:34
  • $\begingroup$ @jcoppens good catch. Better now? $\endgroup$ Commented Nov 18, 2014 at 21:07
  • $\begingroup$ I still don't really think this helps to explain the problem in a 'clear and concise' way. But then the quoted text doesn't really address the larger problem: it's very probable that with a high SWR, the transmission will get distorted by the output stage which doesn't expect near-100% reflection. So having low loss cable is moot. Though you're right that DTV signals are a good candidate for problems, those signals were designed to cope with them through error protocols (not for high SWR of course) but to cope with reflections (physical ones) $\endgroup$
    – jcoppens
    Commented Nov 19, 2014 at 0:41
  • $\begingroup$ @jcoppens 1) what quoted text? 2) the antenna tuner solves precisely the problem of the output stage not expecting reflection, no? $\endgroup$ Commented Nov 19, 2014 at 13:00

...will probably be out of phase with the signal sent by the transmitter.

I think this might help you visualize what is happening when you have an untuned antenna and a high SWR on the line, tuned with an Antenna Tuner.

SWR stands for Standing Wave Ratio. An SWR of 1:1 means that the peaks and valleys of the wave are equal, which is desired. However, a high SWR means that the peaks and valleys (nodes) have a high ratio. If you could go along the feed line and measure the voltage (or current) at different places on the line, you would find that the voltage varied to the ratio of the SWR. In fact, in early radio (1920, for example) this was how they actually measured SWR. They would have a feed line and go with their measuring equipment to sample the voltages along the line.

But the fact is that the Standing Waves were just that - Standing still so they could be measured. The peaks and valleys stayed in place and didn't move around on the line.

This means that the power magically is in phase with itself. The power does not jitter around like you are thinking, instead, the power organizes itself into standing waves. Really. I know that this goes against your common sense. But this is the fact.

So when we say that the radio energy is bounced back and forth between the tuner and the antenna, it is in standing waves that are organized, and reach the antenna (the original signal and the bounced signals) together. And thus are radiated together. The bounced signals radiate a certain percentage on each bounce, so eventually are all radiated (except for heat loss in the line).

Which is a good reason for using ladder line instead of coax when using a Tuner and untuned antenna. Because coax will exhibit greater resistance and greater heat to the bouncy signals with high SWR. Ladder line is much better in this respect.

One more thing. The time lag between the original signal and the bounced signals is very small. Say you are on 7.268 MHz, and say "CQ". The audio will vary maybe 2 or 3 thousand times. But the radio will vary over 7 million times. Each cycle of audio is changed to 2 or 3 thousand radio "cycles". So if the bouncey radio Standing Waves bounce 5 or even 10 times until completely radiated, all of the bounces will be in approximately the same phase of the audio cycle. So it doesn't really matter the time delay of the bounces.

vry 73, Joe W3TTT


"After "bouncing" back and forth the part of the signal that will be transmitted will probably be out of phase with the signal sent by the transmitter."

This seems to be the crux of the question. The phase shift in the RF carrier caused by the reflected wave delay(s) is irrelevant because the length of the transmission line can vary the phase shift by any amount including 360 degrees if the feedline is one wavelength long. There is no HF amateur mode that I know of that depends on the phase of the RF signal for data transmission. It would be nice to be able to send a data bit with every HF RF cycle but Mother Nature seems to have other ideas.

So the real question is: What effect does the round trips made by the reflected waves on the transmission line have on the (audio) frequency phase of the modulation of the radiated waves?

With each round trip to the tuner and back to the antenna feedpoint, the magnitude of the reflected wave decreases until it has negligible effect. So let's assume after being re-reflected 20 times, the re-reflected wave has no effect. Let's assume that 20 re-reflections take four microseconds at the speed of light in the medium. Does it make sense that if an individual reflected wave is damped and disappears (losses and radiation) after four microseconds that it cannot possibly have much effect on a modulating signal with a cycle time of 1000 microseconds (e.g. 1000 Hz)?

Since we don't depend upon HF/RF phase shifts to convey any information, worrying about HF/RF phase shifts is a waste of time. Since the delay in the re-reflected waves traveling at the speed of light in the medium is negligible compared to the cycle time of our modulating waves, worrying about audio modulation frequency phase shifts is a waste of time at HF/RF (except maybe for software designers).

However, at UHF and higher frequencies, re-reflections can be a problem.


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