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#1
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Antenna Reactance Question
In my dipole/vertical modeling and analyzer measurements, as frequency
is increased past quarter-wave resonance, I've noticed with interest that both reactance and resistance peak at different times, as they increase with frequency toward the half-wave point. Resistance peaks at approximately the half-wave point as expected, but inductive reactance always peaks a little earlier (lower frequency). This is also indicated in the ARRL Antenna Book in Figures 3 through 5 on pages 2-3 and 2-4 (it's the bulging on the right side of each of the curves). For a given antenna of particular length, adding inductive or capacitive reactance changes the magnitude of the reactance peak, but not the frequency at which it occurs. Changing the thickness of the radiating elements changes (lowers) the frequency at which the reactance peak occurs, but it also changes (lowers) the frequency at which resistance peaks, and the difference in these two freqencies stays approximately the same. Why does the reactance peak occur slightly earlier than half-wavelength? Can it be mathmatically predicted/explained? Any help would be appreciated. Al, WA4GKQ |
#2
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alhearn wrote:
Why does the reactance peak occur slightly earlier than half-wavelength? Since the monopole is purely resistive around 1/4WL and around 1/2WL, i.e. the reactance is zero at those two points, it is simply impossible for it to be be any other way. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#3
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Al, WA4GKO wrote:
"Why does the reactance peak earlier than half-wavelength? Can it be mathematically prtedicted / explained?" At resonance, phase shifts abruptly from leading to lagging or vice versa as resonance is crossed. The resonant length depends on wave velocity along the antenna wire and is a function of how thin the wire is. Resonant length is also shortened by increased capacitive effect caused by the voltage pile up at the open-circuit wire end. There are formulas involving wire length to periphery ratio but these values have been already calculated and plotted on convenient charts. My 19th edition of the ARRL Antenna Book has Fig 8 on page 2-5 which gives values from 94% to 98% of free-space wavelength for a range of 1/2-wavelength to conductor diameter ratios. Often the antenna is shortened to 95% for "end effects" to avoid formulas and charts and in most cases this is good enough. Best regards, Richard Harrison, KB5WZI |
#4
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Alhearn If it is "understanding" that you want, I'd suggest you plot the antenna's impedance on a Smith Chart. A plot of an antenna's impedance with varying frequency will describe a continuous curve. Beyond that, any additional comments would probably just make this post more confusing. I'd bet you can answer your own question after looking at a Smith Chart. Note the chart identifies R+/-jX. So the reactance added in series will move an impedance along lines of constant R. The path of the impedance plot resulting from shunt X can also be identified, but maybe thats for another time for discussion Jerry "alhearn" wrote in message om... In my dipole/vertical modeling and analyzer measurements, as frequency is increased past quarter-wave resonance, I've noticed with interest that both reactance and resistance peak at different times, as they increase with frequency toward the half-wave point. Resistance peaks at approximately the half-wave point as expected, but inductive reactance always peaks a little earlier (lower frequency). This is also indicated in the ARRL Antenna Book in Figures 3 through 5 on pages 2-3 and 2-4 (it's the bulging on the right side of each of the curves). For a given antenna of particular length, adding inductive or capacitive reactance changes the magnitude of the reactance peak, but not the frequency at which it occurs. Changing the thickness of the radiating elements changes (lowers) the frequency at which the reactance peak occurs, but it also changes (lowers) the frequency at which resistance peaks, and the difference in these two freqencies stays approximately the same. Why does the reactance peak occur slightly earlier than half-wavelength? Can it be mathmatically predicted/explained? Any help would be appreciated. Al, WA4GKQ |
#6
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Cecil Moore wrote ..
Since the monopole is purely resistive around 1/4WL and around 1/2WL, i.e. the reactance is zero at those two points, it is simply impossible for it to be be any other way. That's very true, but between the 1/4WL and 1/2WL zero-reactance points, reactance increases with frequency, peaks, and then begins to decrease to cross zero again at the 1/2WL point. This peak is NOT half-way between 1/4WL and 1/2WL, but skewed heavily toward the 1/2WL point. My question is what determines where that peak occurs? Can the reactance peak be somehow adjusted (frequency position, not magnitude) for tuning purposes? Magnitude is easily adjusted by simply adding inductive or capacitive reactance, which doesn't change the shape or peak position of the reactance curve, but simply moves the curve up or down to change the points at which it crosses zero (resonance frequency). Controlling the reactance peak position could do the same. The answer may be simple and common knowledge, but I haven't found it. Al |
#7
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alhearn wrote:
My question is what determines where that peak occurs? Mathematically, it will be where the SWR circle is tangent to the reactance arc. The higher the SWR, the higher the maximum possible reactance and the closer it is to the anti-resonance point. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#8
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Cecil Moore wrote in message ...
alhearn wrote: My question is what determines where that peak occurs? Mathematically, it will be where the SWR circle is tangent to the reactance arc. That is not in general true; do you have reason to believe it's true for the impedance of a dipole? Consider what happens when you change the reference impedance for the SWR measurement. The higher the SWR, the higher the maximum possible reactance and the closer it is to the anti-resonance point. What's the reactance at the anti-resonance point? Is highest SWR at anti-resonance, or at maximum reactance, or at some point between? |
#9
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Tom Bruhns wrote:
Cecil Moore wrote: alhearn wrote: My question is what determines where that peak occurs? Mathematically, it will be where the SWR circle is tangent to the reactance arc. That is not in general true; do you have reason to believe it's true for the impedance of a dipole? When the dipole is at maximum feedpoint reactance, can the SWR be calculated? Of course. Will that point, when plotted on a Smith Chart lie on the SWR circle? Of course. Will it also lie on a reactance arc? of course. Will the SWR circle be tangent to that reactance arc? Of course. Will the un-normalized value of the maximum feedpoint reactance be constant? Of course, assuming nothing changes except Z0. Consider what happens when you change the reference impedance for the SWR measurement. It doesn't matter. The Smith Chart reference changes and therefore the SWR changes but *the value of the antenna feedpoint impedance stays the same*. The new SWR circle is still tangent to the reactance arc at the same value of un-normalized reactance even though the normalized reactance value has changed. Xmax/Z01 is different from Xmax/Z02 but Xmax has not changed (assuming Z0 is the only change). The higher the SWR, the higher the maximum possible reactance and the closer it is to the anti-resonance point. What's the reactance at the anti-resonance point? Always zero, by definition. Examples: A full-wave center-fed dipole or a half-wave end-fed monopole. ("Anti-resonant" is what we called such antennas at Texas A&M 50 years ago.) Is highest SWR at anti-resonance, or at maximum reactance, or at some point between? If the Z0 of your transmission line is equal to the feedpoint impedance at anti-resonance, the lowest SWR will occur at anti-resonance. If the Z0 of your transmission line is equal to the feedpoint impedance at maximum reactance, the lowest SWR will occur at maximum reactance. :-) Depends upon the Z0 of the transmission line. If you choose a Z0 that is the square root of the resonant impedance times the anti-resonant impedance, the SWR at resonance and anti-resonance will be the same. For a dipole that Z0 value is usually between 450 ohms and 600 ohms. That's why anti-resonance is no problem for ladder-line/open-wire. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#10
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Cec, & Co.
Zo of an antenna wire seems to be an important parameter in your discussions. Knowledge of its value would appear to be essential before continuing with calculations. Otherwise nobody will get nowhere. So how is the value of Zo obtained (without bringing Terman et al into it)? ---- Reg, G4FGQ |
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