Why don't the traveling waves on a dipole go back down the transmission line ?
Why does the standing wave on a dipole stop at the feed point ?
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$\begingroup$ Short answer: they either radiate out the dipole, or they do go back down the feedline. And it doesn't stop there -- it could even go back into the radio. This is what causes SWR. $\endgroup$– user10489Aug 7 at 14:57
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1$\begingroup$ “This is what causes SWR.” That’s … inaccurate as stated. $\endgroup$– Joshua Nozzi W4JLNAug 9 at 1:03
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$\begingroup$ @JoshuaNozziW4JLN it's what causes the reactive part of the antenna impedance to appear at the feedpoint, so nearly there :) $\endgroup$– hobbs - KC2GAug 10 at 15:18
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1$\begingroup$ It's what causes the resistive part of the antenna impedance to appear at the feedpoint also. The feedpoint impedance need not be reactive to have an SWR higher than 1:1. $\endgroup$– Phil Frost - W8IIAug 10 at 21:01
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1$\begingroup$ @user10489 but you do have an SWR if there's no standing wave. It's 1:1. $\endgroup$– Phil Frost - W8IIAug 11 at 18:57
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If you fed a dipole not with a sine wave but with an impulse, you'd see a wave travel down the feedline, to the feedpoint, and then "ring" several times in the dipole at the dipole's resonant frequency. The ringing doesn't continue forever: on each oscillation this wave loses some energy to radiation. If the feedline is terminated on the non-antenna end then the feedline looks like a resistor, and some energy is lost there also.
Waves in the dipole certainly do go back down the transmission line. They must, by reciprocity. Also, if they did not then the impedance looking into the feedpoint would have be the same regardless of the length of the dipole because there would be nothing "coming back" to yield any information about what the impedance should be. Obviously we know this is false.
You can consider the standing waves to be the sum (by superposition) of infinitely many such impulses. At any given time, all these impulses are traveling all over. The more recent ones are still quite strong; the ones that happened a while ago are weaker because they've made more trips around the dipole and have lost more energy. This infinite but diminishing series of impusles is a geometric series and it sums to a finite value.
That finite value is the standing wave. The standing wave does not "stop" anywhere. It can't stop because it is by definition not traveling, and so having never "gone" anywhere it can't stop going.
And if you like, you can consider the standing wave as the sum of a geometric series of impulses, each of these impulses being associated with traveling waves that do indeed travel around the dipole and the feedline an infinite amount of times, approaching but never reaching zero amplitude. If and only if the dipole is matched to the feedline the reflections from all these impulses sums to zero on the feedline, the VSWR is 1:1 and we see no standing waves on the feedline.
answered Aug 6 at 21:14