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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 |
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! =----- |
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 |
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 |
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 |
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! =----- |
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? |
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! =----- |
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 |
Tom Bruhns wrote:
"What`s the reactance at the anti-resonance point? Is highest SWR at anti-resonance or at maximum reactance or at some point between?" I don`t quite understand the questions but that`s no inhibition. The anti-resonance point must be at the drivepoint of an end-fed 1/2-wave element because anti-resonance is the frequency at which the impedance of the system is very high. If the system is anti-resonant, reactance is zero. Highest SWR comes from greatest mismatch. It gets no worse that from an open or short circuit. The original question concerned 1/4-wave and 1/2-wave resonances. SWR is highest at the reflection point because it is a creature of the reflected wave`s reaction with the incident wave, and nowhere can the reflected wave be stronger than at the tip of the antenna. The open circuit at the tip has some capacitance, but the only effect of this reactance is to shift the standing wave pattern along the antenna according to transmission line analogy. Unlike a transmission line, the impedance seen by a wave traveling in either direction along an antenna varies from one end to the other. As usual, I hope to learn something from this thread. Best regards, Richard Harrison, KB5WZI |
Reg Edwards wrote:
Cec, & Co. Zo of an antenna wire seems to be an important parameter in your discussions. Why do you say that? The only thing I have dealt with is a constant feedpoint impedance at any one frequency. -- 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! =----- |
Reg Edwards wrote:
Cec, & Co. Zo of an antenna wire seems to be an important parameter in your discussions. Here's an example of what I have been talking about. The feedpoint impedance of my 130 ft. dipole (according to EZNEC) is: 60+j0 ohms at 3.63 MHz - resonance 2513+j2314 at 6.575 MHz - maximum reactance 5000+j0 at 7.118 MHz - anti-resonance Assuming the antenna is fed with Z0 = 550 ohm open-wire line, the SWR ranges from 8.5:1 to 9.2:1 and the impedance at a current maximum point on the ladder-line ranges from 60 ohms to 65 ohms. The feedpoint impedances from 3.63 MHz (resonance) to 7.118 MHz (anti-resonance) describe an imperfect semi-circle on a Smith Chart normalized for 550 ohms. If one plots the feedpoint impedances for all the different bands on a Smith Chart normalized for SQRT(Rresonant * Ranti-resonant), it will resemble an imperfect SWR circle (or imperfect spiral). By feeding my dipole only at the current-maximum points, I achieve a near-perfect match on all amateur HF bands without an antenna tuner. At least for my multi-band dipole, it appears that the anti-resonant feedpoint impedance is about 100 times the resonant feedpoint impedance. The feedpoint impedance at the maximum reactance point is about 5000/2+j5000/2, i.e. the R is about half the anti-resonant resistance and the Xmax is about half the anti-resonant resistance times 'j'. These are my rules-of-thumb for my dipole. -- 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! =----- |
"Cecil Moore" wrote in message ... 5000+j0 at 7.118 MHz - anti-resonance making up terms again cecil?? I have never heard of 'anti-resonance' and don't see it in my references here... i would say it is just resonant at the second harmonic, just because the impedance is high doesn't mean it is 'anti' anything, or anything else special. it is just plain resonant at 7118. |
making up terms again cecil?? I have never heard of 'anti-resonance' and
don't see it in my references here I have heard the term used before. I typed anti resonance into Google and got 296,000 hits. I don't think the term was made up. 73 Gary N4AST |
"JGBOYLES" wrote in message ... making up terms again cecil?? I have never heard of 'anti-resonance' and don't see it in my references here I have heard the term used before. I typed anti resonance into Google and got 296,000 hits. I don't think the term was made up. 73 Gary N4AST it would appear from a quick check that many of those 'anti-resonance' references are mechanical (preventing resonance), quantum mechanical(some kind of spin resonance), optical(something about lithography masks), or at best electro-mechanical (lots of piezo-electric references in there), i did see one that was explaining electrical resonance as mechanical analogies that used that term. and, oh yes, there was another discussion group that had latched onto it as an alternate term for a 'parallel resonant' circuit. |
Dave wrote:
"Cecil Moore" wrote: 5000+j0 at 7.118 MHz - anti-resonance making up terms again cecil?? I have never heard of 'anti-resonance' and don't see it in my references here... i would say it is just resonant at the second harmonic, just because the impedance is high doesn't mean it is 'anti' anything, or anything else special. it is just plain resonant at 7118. There's lots of thing of which I am ignorant but I don't make postings about those things. The first four books I picked up had information on anti-resonance. Are you familiar with books? :-) A web search for "anti-resonance" turned up 700 hits. Perhaps you don't need a book. _Radio_Handbook_, 20th edition, by Bill Orr: "In radio circuits, parallel resonance (MORE CORRECTLY TERMED ANTIRESONANCE) is more frequently encountered than series resonance; ..." emphasis mine _The_ARRL_Handbook_, 17th edition, page 14-5: "A capacitor connected across a cavity provides an ANTI-RESONANT notch below the resonant frequency (f0),..." emphasis mine _Electronic_Fundamentals_and_Applications_, John D. Ryder, page 287 under section 9-13. Parallel-Resonant Circuits: "The total admittance of the parallel-resonant (SOMETIMES CALLED ANTIRESONANT) circuit is ..." emphasis mine _IEEE_Dictionary_: "antiresonant frequency - Usually in reference to a crystal unit or the parallel combination of a capacitor and inductor." Since the resonant feedpoint impedance of a 1/2WL dipole acts somewhat like a series resonant circuit, at Texas A&M we called that the resonant frequency. Since the resonant feedpoint impedance of a one-wavelength dipole acts somewhat like a parallel resonant circuit, at Texas A&M we called that the anti-resonant frequency. Differentiating between the resonant frequency and anti-resonant frequency gives one additional information about the impedance to be matched. "Resonant" = low, "Anti-resonant" = high. It's a useful concept. You really should try it. -- 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! =----- |
Dave wrote:
... and, oh yes, there was another discussion group that had latched onto it as an alternate term for a 'parallel resonant' circuit. According to the IEEE Dictionary, that's exactly what it is. And the feedpoint of a one-wavelength center-fed dipole acts a lot like a parallel resonant circuit. Somehow, saying that the resonant feedpoint impedance of your dipole is 6000 ohms leaves something to be desired. -- 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! =----- |
Cecil Moore wrote in message ...
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. I'm sorry, Cecil, but you lost me there. For any given SWR circle, there are only two (complex conjugate) points at which reactance arcs are tangent. Why would we think that the point of max reactance on the antenna impedance curve will necessarily be at the point of tangency? The antenna impedance arc of the simple dipole I modelled indeed does not lie tangent to the max reactance arc at the same point as the SWR circle that's tangent that reactance arc. In any event, I don't see that this tells us anything about _why_ the dipole shows max reactance at that particular frequency. |
Nope! Cecil's not making up terms.
You just missed a point in your education. Dave wrote: "Cecil Moore" wrote in message ... 5000+j0 at 7.118 MHz - anti-resonance making up terms again cecil?? I have never heard of 'anti-resonance' and don't see it in my references here... i would say it is just resonant at the second harmonic, just because the impedance is high doesn't mean it is 'anti' anything, or anything else special. it is just plain resonant at 7118. |
Tom Bruhns wrote:
I'm sorry, Cecil, but you lost me there. For any given SWR circle, there are only two (complex conjugate) points at which reactance arcs are tangent. Why would we think that the point of max reactance on the antenna impedance curve will necessarily be at the point of tangency? Note: I am talking about the frequencies between the 1/2WL resonant point and the one-wavelength (anti)resonant point for a fixed dipole. There will exist a maximum reactance point between those two frequencies. By definition of the bi-linear transformation rules involving the Smith Chart, the maximum reactance point will be located at the point where the SWR circle is tangent to the reactance arc. It simply cannot be located anywhere else. It's an obvious geometrical thing, Tom. The SWR circle is centered at the center of the Smith Chart. The reactance arc (circle) is centered somewhere else outside of the Smith Chart. Where these two circles are tangent, the reactance is at a maximum, by definition. If the two circles are not tangent and not touching, then that cannot possibly be the maximum reactance point. If the two circles are not tangent and intersect at two points, then that cannot possibly be the maximum reactance point. In the latter case, the maximum reactance point lies between those two intersection points. The antenna impedance arc of the simple dipole I modelled indeed does not lie tangent to the max reactance arc at the same point as the SWR circle that's tangent that reactance arc. Sorry, you did something wrong or don't understand what I am saying. It is impossible for the maximum reactance point not to be tangent to the reactance arc at the maximum reactance point. On the inductive top part of the Smith Chart, between 1/2WL and 1WL, if the circles intersect at more than one point, you are not at the maximum reactance point. If the circles intersect at one and only one point, they are tangent, by definition, and you are at the maximum reactance point. If they don't intersect at all, you are not at the maximum reactance point. In any event, I don't see that this tells us anything about _why_ the dipole shows max reactance at that particular frequency. Because it's an obvious geometrical thing, Tom. It simply cannot be any other frequency and can be proved with relatively simple geometry. EXAMPLE: 1/2WL resonant feedpoint impedance is 50+j0 ohms. One-wavelength (anti)resonant feedpoint impedance is 5000 ohms. Maximum reactance point has a feedpoint impedance of 2500+j2500 ohms. The SWR circle (at the frequency of maximum reactance) will pass through the 2500+j2500 ohm point. Do you disagree? The reactance arc (at the frequency of maximum reactance) will pass through the 2500+j2500 ohm point. Do you disagree? ERGO: The SWR circle will be tangent to the reactance arc at the 2500+j2500 ohm point no matter what Z0 is being used. Do you disagree? -- 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! =----- |
I agree that it's an obvious geometrical thing...and it could be that
I'm simply not understanding what you're trying to communicate. But...do a simulation of the "dipole1.ez" that ships with EZNec, after first changing the number of segments to 31 so it doesn't complain. Find the frequency between 300MHz and 600MHz which maximized the reactance. I believe you will find it at about 494MHz, and the reactance is +j866.0 ohms. At that frequency, the resistive part is 1085 ohms. Put that on a Smith chart whose reference resistance is 50 ohms. Draw a constant-SWR circle (35.542; magnitude of rho = .9453) through that point. Note that there are two constant-reactance arcs tangent to that SWR circle, and note that they are NOT tangent at the antenna's impedance point. In fact, it's reasonably easy to calculate them as +/-j887.8. Now change the reference impedance to 2000 ohms, and note that the constant SWR circle which passes through the antenna max-reactance impedance point is nowhere near the point that SWR circle is tangent to a reactance arc. The SWR for 1085+j866 referred to 2000 ohms is 2.296, magnitude of rho = .3932. The constant SWR circle in this case is tangent to reactance arcs of +/-j1860.28. I certainly agree that the point where a reactance arc is tangent with an SWR circle is the maximum reactance on that SWR circle, but that's not necessarily (and generally is NOT) also a point on the curve representing antenna feedpoint impedance versus frequency. -- Did I miss something fundamental here? Are we not discussing how the antenna feedpoint impedance changes with frequency, and specifically the frequency between half and full wave resonances at which the antenna feedpoint reactance is maximum? And still, all this does NOTHING to tell us WHY the antenna's reactance reaches maximum at that particular frequency. Again, maybe I missed it, but I didn't see anything in your swr circles and reactance arcs explanation that even mentioned frequency. Example: Cecil Moore wrote in message ... Tom Bruhns wrote: I'm sorry, Cecil, but you lost me there. For any given SWR circle, there are only two (complex conjugate) points at which reactance arcs are tangent. Why would we think that the point of max reactance on the antenna impedance curve will necessarily be at the point of tangency? Note: I am talking about the frequencies between the 1/2WL resonant point and the one-wavelength (anti)resonant point for a fixed dipole. There will exist a maximum reactance point between those two frequencies. By definition of the bi-linear transformation rules involving the Smith Chart, the maximum reactance point will be located at the point where the SWR circle is tangent to the reactance arc. It simply cannot be located anywhere else. It's an obvious geometrical thing, Tom. The SWR circle is centered at the center of the Smith Chart. The reactance arc (circle) is centered somewhere else outside of the Smith Chart. Where these two circles are tangent, the reactance is at a maximum, by definition. If the two circles are not tangent and not touching, then that cannot possibly be the maximum reactance point. If the two circles are not tangent and intersect at two points, then that cannot possibly be the maximum reactance point. In the latter case, the maximum reactance point lies between those two intersection points. The antenna impedance arc of the simple dipole I modelled indeed does not lie tangent to the max reactance arc at the same point as the SWR circle that's tangent that reactance arc. Sorry, you did something wrong or don't understand what I am saying. It is impossible for the maximum reactance point not to be tangent to the reactance arc at the maximum reactance point. On the inductive top part of the Smith Chart, between 1/2WL and 1WL, if the circles intersect at more than one point, you are not at the maximum reactance point. If the circles intersect at one and only one point, they are tangent, by definition, and you are at the maximum reactance point. If they don't intersect at all, you are not at the maximum reactance point. In any event, I don't see that this tells us anything about _why_ the dipole shows max reactance at that particular frequency. Because it's an obvious geometrical thing, Tom. It simply cannot be any other frequency and can be proved with relatively simple geometry. EXAMPLE: 1/2WL resonant feedpoint impedance is 50+j0 ohms. One-wavelength (anti)resonant feedpoint impedance is 5000 ohms. Maximum reactance point has a feedpoint impedance of 2500+j2500 ohms. The SWR circle (at the frequency of maximum reactance) will pass through the 2500+j2500 ohm point. Do you disagree? The reactance arc (at the frequency of maximum reactance) will pass through the 2500+j2500 ohm point. Do you disagree? Certainly the +j2500 ohm reactance arc will pass through 2500+j2500. But that reactance arc is not tangent to the SWR circle that passes through that point, necessarily. Draw the chart with Zref set to 10,000 ohms, and you'll instantly see this. Draw the chart with Zref=50 ohms, and it will be difficult to see, but in fact if you go through the calcs, I believe you'll find a slight discrepancy even there! The point of tangency is not exactly where R=X, though it's close for high SWR. But try it for a low SWR circle, and it will be visually obvious on the Smith chart. For example, on a 50 ohm chart, SWR=1.63 will be tangent to +/-j25 ohm reactance arcs, but at roughly 55 ohms resistive, no where near 25 ohms resistive. ERGO: The SWR circle will be tangent to the reactance arc at the 2500+j2500 ohm point no matter what Z0 is being used. Do you disagree? Sure do. See numerical examples above. |
Tom Bruhns wrote:
I certainly agree that the point where a reactance arc is tangent with an SWR circle is the maximum reactance on that SWR circle, but that's not necessarily (and generally is NOT) also a point on the curve representing antenna feedpoint impedance versus frequency. -- Did I miss something fundamental here? I suspect you missed that my input to the discussion assumes a thin wire at HF frequencies. Try an HF dipole with a thin wire and see what you get. It will be approximately the same curve as Fig. 10, page 2-10, ARRL Antenna Book, 15th edition. Since this is the graph of an end-fed antenna, the dipole response can be obtained by multiplying the frequencies by 2. Everything I have said has been extrapolated from that graph in the ARRL Antenna Book which is, as I said before, an "imperfect circle or imperfect spiral". Are we not discussing how the antenna feedpoint impedance changes with frequency, and specifically the frequency between half and full wave resonances at which the antenna feedpoint reactance is maximum? Yes, that's true. Now try it with a thin wire on HF. I believe what you will find is that at the maximum reactance point, the resistance is approximately half of the one-wavelength (anti)resonant value. At the maximum reactance point, the resistance and reactance are approximately equal. Since the maximum reactance value lies between two points of pure resistance, doesn't it make sense that it might be approximately where the resistance is half of the maximum value of resistance? Maybe you should just tell us why you disagree with Fig. 10 in the ARRL Antenna Book, 15th edition, page 2-10. If you want to see a dipole feedpoint impedance Vs frequency, it is illustrated in Figs. 2-5, Page 2-3,4, ARRL Antenna Book CD, version 2.0. Unfortunately, they plotted reactance on a linear scale and resistance on a log scale and thus messed up the shape of the "imperfect circle or imperfect spiral". Even with that, one can see that the point of maximum reactance approximately equals the resistance at that point and is approximately half the value of the maximum resistance. -- 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! =----- |
Tom Bruhns wrote, (Snip) And still, all this does NOTHING to tell us WHY the antenna's reactance reaches maximum at that particular frequency. Again, maybe I missed it, but I didn't see anything in your swr circles and reactance arcs explanation that even mentioned frequency. In Constantine Balanis book _Antenna Theory Analysis and Design_ second edition Section 8.5 page 411, there are a couple of charts of resistance and reactance that illustrate the original question. On the facing page (page 410) equations are given for R and X at current maximum points on a dipole. If these equations are correct it's not hard to believe that their respective maxima don't coincide since the equations are different. If anyone wants to understand where the equations come from he can read chapter eight in all its gory entirety. It's not simple and requires more than just a familiarity with a Smith chart. Cecil should read before he thinks. 73, Tom Donaly, KA6RUH |
Cecil Moore wrote in message ...
.... The SWR circle (at the frequency of maximum reactance) will pass through the 2500+j2500 ohm point. Do you disagree? The reactance arc (at the frequency of maximum reactance) will pass through the 2500+j2500 ohm point. Do you disagree? ERGO: The SWR circle will be tangent to the reactance arc at the 2500+j2500 ohm point no matter what Z0 is being used. Do you disagree? that specific example, using a reference impedance of 50 ohms... magnitude of rho for 2500.00+j2500.00 = .9801999804 magnitued of rho for 2500.25+j2500.00 = .9801999801 magnitude of rho for 2500.50+j2500.00 = .9801999800 magnitude of rho for 2500.75+j2500.00 = .9801999801 magnitude of rho for 2501.00+j2500.00 = .9801999804 The point of tangency to an SWR circle for the +j2500.00 arc is therefore NOT at 2500.00+j2500.00. It will be much more obvious if you recalculate things at a higher reference impedance, say 1000 ohms. Then following the +j2500 reactance arc, mag(rho) reaches a minimum when Z is about 2692+j2500. And in any event, I think it highly unlikely that the antenna's resistance when the reactance peaks at 2500 ohms will be exactly 2500 ohms as well. |
Tdonaly wrote:
Cecil should read before he thinks. Tom, why do you require 6 decimal point precision out of a ballpark rule-of-thumb estimate? The maximum reactance point of a thin-wire dipole is in the neighborhood of the maximum reactance point of the SWR circle as proved by the graphs in the ARRL Antenna Book CD. The maximum reactance is approximately equal to 1/2 the maximum resistance. The resistance at the maximum reactance point is approximately equal to 1/2 the maximum resistance. If someone carries those ballpark concepts around in his head, he will be reasonably close to reality. It's all cut-and-try after that. Exactly what agenda forces you to sacrifice your reputation trying to find some unusual esoteric exception to my statements? (Never mind, I already know the answer.) -- 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! =----- |
Tom Bruhns wrote:
that specific example, using a reference impedance of 50 ohms... magnitude of rho for 2500.00+j2500.00 = .9801999804 magnitued of rho for 2500.25+j2500.00 = .9801999801 magnitude of rho for 2500.50+j2500.00 = .9801999800 magnitude of rho for 2500.75+j2500.00 = .9801999801 magnitude of rho for 2501.00+j2500.00 = .9801999804 The point of tangency to an SWR circle for the +j2500.00 arc is therefore NOT at 2500.00+j2500.00. Good Lord, Tom, do you have to stoop so low as to argue over 0.0000000003? I said it was a ballpark rule-of-thumb estimate to start with. Just what agenda are you following that requires greater than 0.0000000003 accuracy out of a ballpark rule-of- thumb estimate? No need to answer - I already know the answer. -- 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! =----- |
On Fri, 26 Mar 2004 16:29:59 -0600, Cecil Moore
wrote: No need to answer - I already know the answer. On Fri, 26 Mar 2004 16:20:24 -0600, Cecil Moore wrote: (Never mind, I already know the answer.) Who would've thougth otherwise? Foolish boys. |
Cecil Moore wrote:
Good Lord, Tom, do you have to stoop so low as to argue over 0.0000000003? I said it was a ballpark rule-of-thumb estimate to start with. Just what agenda are you following that requires greater than 0.0000000003 accuracy out of a ballpark rule-of- thumb estimate? No need to answer - I already know the answer. In case anyone has forgotten, here's what I said in an earlier posting: At least for my multi-band dipole, it appears that the anti-resonant feedpoint impedance is about 100 times the resonant feedpoint impedance. The feedpoint impedance at the maximum reactance point is about 5000/2+j5000/2, i.e. the R is about half the anti-resonant resistance and the Xmax is about half the anti-resonant resistance times 'j'. These are my rules-of-thumb for my dipole. Does everyone understand the meaning of "appears", "about", and "rules-of-thumb"? In my world, 20% accuracy out of a rule-of-thumb is pretty good. Does anyone else (besides Tom) have rules-of-thumb that achieve 0.0000000003 accuracy? -- 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! =----- |
Cecil wrote,
Tdonaly wrote: Cecil should read before he thinks. Tom, why do you require 6 decimal point precision out of a ballpark rule-of-thumb estimate? The maximum reactance point of a thin-wire dipole is in the neighborhood of the maximum reactance point of the SWR circle as proved by the graphs in the ARRL Antenna Book CD. The maximum reactance is approximately equal to 1/2 the maximum resistance. The resistance at the maximum reactance point is approximately equal to 1/2 the maximum resistance. If someone carries those ballpark concepts around in his head, he will be reasonably close to reality. It's all cut-and-try after that. Cecil, I know you have the Balanis book and that you even took a course from the great man, himself. You can crack the book and look at the curves as well as anyone can. You can also read the rest of the explanation, and, with a lot of work, puzzle it out and improve your understanding so you don't have to rely on rules of thumb that get you into arguments like this. Exactly what agenda forces you to sacrifice your reputation trying to find some unusual esoteric exception to my statements? (Never mind, I already know the answer.) I don't have a reputation. If I did it wouldn't mean anything to me anyway. Balanis is hardly an "unusual esoteric exception." If you think I have an "agenda" you're letting your paranoia get the better of you. (No, I'm not looking for a job on Kerry's campaign committee.) -- 73, Cecil http://www.qsl.net/w5dxp 73, Tom Donaly, KA6RUH |
On Fri, 26 Mar 2004 17:31:28 -0600, Cecil Moore
wrote: Does anyone else (besides Tom) have rules-of-thumb that achieve 0.0000000003 accuracy? E = I ยท R |
Tdonaly wrote:
... so you don't have to rely on rules of thumb that get you into arguments like this. Nice try at obfuscation, Tom, but I have previously identified my postings as only rules of thumb. You attempted to hold my rules of thumb to 0.00000000003 accuracy. Doesn't that make you feel the least bit silly? That's stretching things pretty far to try to prove that anyone who believes in reflected waves is crazy. Have you figured out how standing waves can occur without the existence of reflected waves yet? I've been holding my breath for that proof you promised. If my rules of thumb are within 20% accuracy, I consider that pretty good. And here I repeat my rules of thumb. The ratio of the resonant feedpoint impedance to the antiresonant impedance of a dipole is about 100 to 1. The maximum reactance point between those two frequencies is about Rmax/2+jRmax/2. If you can't prove that rule of thumb is less than 20% accurate, you have no argument. -- 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! =----- |
Cecil Moore wrote in message ...
Yes, that's true. Now try it with a thin wire on HF. I believe what you will find is that at the maximum reactance point, the resistance is approximately half of the one-wavelength (anti)resonant value. At the maximum reactance point, the resistance and reactance are approximately equal. Since the maximum reactance value lies between two points of pure resistance, doesn't it make sense that it might be approximately where the resistance is half of the maximum value of resistance? That's pretty approximate, Cecil. From EZNec for a 10 meter 1mm diam wire, I got about 2570+j2277 at 27.00MHz. The resistive and reactive parts differ by over ten percent. Be that as it may, if you evaluate the SWR tangency thing for a high reference impedance, you will see that it's a GROSS error, not a tiny one. You said it wouldn't change with changes in reference impedance, but it does, and in a major way. Yes, the error is tiny for your assumed 2500+j2500 evaluated against Z0 of 50 ohms, but the error is huge if you evaluate against Z0 of, say, 2000 ohms. And we're still back to not having said anything about _why_ the reactance peaks at the frequency it does relative to the half- and full-wave resonances. Cheers, Tom |
Tom Bruhns wrote:
That's pretty approximate, Cecil. From EZNec for a 10 meter 1mm diam wire, I got about 2570+j2277 at 27.00MHz. The resistive and reactive parts differ by over ten percent. For an approximation that I carry around in my head, 20% accuracy is good enough for me. Be that as it may, if you evaluate the SWR tangency thing for a high reference impedance, you will see that it's a GROSS error, not a tiny one. You said it wouldn't change with changes in reference impedance, but it does, and in a major way. Actually, what I said is that the maximum reactance point on an SWR circle doesn't depend upon Z0 and proved it with equations. Xmax/Z01 = X1 normalized for Z01. Xmax/Z02 = X2 normalized for Z02. X1 and X2 are different but Xmax is the same value no matter what the Z0. Yes, the error is tiny for your assumed 2500+j2500 evaluated against Z0 of 50 ohms, but the error is huge if you evaluate against Z0 of, say, 2000 ohms. That wouldn't be a logical thing to do. The highest Z0 commonly available to hams is around 600 ohms. The higher the SWR, the more accurate is this approximation. Conversely, the lower the SWR the more inaccurate is this approximation. Since I encounter SWR's in the general range of 10-25, it works pretty well for me. And we're still back to not having said anything about _why_ the reactance peaks at the frequency it does relative to the half- and full-wave resonances. _Why_ am I sitting at my computer right now? Because I'm not somewhere else? :-) The reactance seems to peak about 85% of the way between the 1/2WL frequency and the one-wavelength frequency. QED folks seem to be satisfied with just an equation which doesn't ask or answer, _why?_. They say, "That's just the way it is." -- 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! =----- |
Cecil wrote,
Tdonaly wrote: ... so you don't have to rely on rules of thumb that get you into arguments like this. Nice try at obfuscation, Tom, but I have previously identified my postings as only rules of thumb. You attempted to hold my rules of thumb to 0.00000000003 accuracy. Doesn't that make you feel the least bit silly? That's stretching things pretty far to try to prove that anyone who believes in reflected waves is crazy. Have you figured out how standing waves can occur without the existence of reflected waves yet? I've been holding my breath for that proof you promised. I think you've got your Toms mixed up, Cecil. I haven't said anything about 0.000...003 accuracy. The rest of your post you'll have to take up with Tom Bruhns, although I think your comments on whether or not I "believe in" reflected waves is unmitigated balderdash. I'm going to quit posting on this, Cecil. Your replies are irrational and it's clear that the strain of dealing with two Toms at once is too hard on your head. 73, Tom Donaly, KA6RUH |
Tdonaly wrote:
... I think your comments on whether or not I "believe in" reflected waves is unmitigated balderdash. Have you changed your mind from last time? -- 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! =----- |
Tdonaly wrote:
I think you've got your Toms mixed up, Cecil. I haven't said anything about 0.000...003 accuracy. I didn't say you did, Tom. That "you" I used was plural. Maybe I should have said "y'all"? -- 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! =----- |
On Sat, 27 Mar 2004 12:10:36 -0600, Cecil Moore
wrote: Tdonaly wrote: I think you've got your Toms mixed up, Cecil. I haven't said anything about 0.000...003 accuracy. I didn't say you did, Tom. That "you" I used was plural. Maybe I should have said "y'all"? On Fri, 26 Mar 2004 19:14:03 -0600, Cecil Moore wrote: Nice try at obfuscation, The plural "you" is inclusive, which is still a miss-attribution. This compounds the error. |
Cecil Moore wrote in message ...
For an approximation that I carry around in my head, 20% accuracy is good enough for me. Ah, we started out with exact geometric relationships that defined precise points, and now we're down to 20% accuracy being OK. I got it. Thanks. Cheers, Tom |
Richard Clark wrote:
The plural "you" is inclusive, which is still a miss-attribution. This compounds the error. Not necessarily, Richard. I can say to an Oklahoma State basketball player, "You won your last game", even though he sat on the bench the entire game and didn't score any points. All members of the same team are guilty by association if not by actions. -- 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! =----- |
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