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Induced signal?
Here's a question overflowing from eHam.net and it's
not a trick question. Assume that the radiating portion of a 40m vertical is made out of 33 feet of RG-213 and the braid is the radiator. The center conductor of the coax is left floating at both ends. How much RF voltage and/or current will be induced in that center wire when using the outside braid as the radiator for 100 watt operation? How much of an EM field can exist inside the coax braid? -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
In article ,
Cecil Moore wrote: Here's a question overflowing from eHam.net and it's not a trick question. Assume that the radiating portion of a 40m vertical is made out of 33 feet of RG-213 and the braid is the radiator. The center conductor of the coax is left floating at both ends. How much RF voltage and/or current will be induced in that center wire when using the outside braid as the radiator for 100 watt operation? How much of an EM field can exist inside the coax braid? No RF Current, as the cenbter conductor is "Floating"...... There maybe some voltage buildup , but it will not have current flow untill the center conductor makes a circuit with something...... |
Induced signal?
You wrote:
No RF Current, as the cenbter conductor is "Floating"...... There maybe some voltage buildup , but it will not have current flow untill the center conductor makes a circuit with something...... Well, that's essentially what I assumed. But W8JI disagrees so now I am not sure my assumption was correct. -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
The center conductor isn't entirely floating.
If the shield were closed at both ends, there would be no fields in the coax, but the coax center conductor at each end can have current induced on it from the outside world. I think you'd go about describing the coupling as being via "fringing fields" at the ends if you were to think of a free end of coax with fields in it radiating into space, I'm thinking of the reciprocal behavior... But this is a behavior where I'm imagining a differential-mode current existing to start with... I think the answer is highly influenced by the exact details of the end of the coax and where it is with respect to other objects... the coupling is very light and mostly to the shield at each end... and I think this makes it not entirely unlike just putting the shield wire and the center conductor wire in parallel in space with the ends tied together. I think current flows in the center conductor in phase with the current in the shield and there's very little differential mode current if any. Certainly I'm wrong, I'm going to think on it some more and try to figure out how wrong... What current flows in the center conductor if you short it to the shield with one wire at each end? What about if you do it with a solid metal cap ? 73, Dan |
Induced signal?
Cecil Moore wrote:
Here's a question overflowing from eHam.net and it's not a trick question. Assume that the radiating portion of a 40m vertical is made out of 33 feet of RG-213 and the braid is the radiator. The center conductor of the coax is left floating at both ends. How much RF voltage and/or current will be induced in that center wire when using the outside braid as the radiator for 100 watt operation? How much of an EM field can exist inside the coax braid? From my EM311 days: If I integrate the fields inside a short conducting cylinder, since there is no charge within the surface integral, there is no field within the cylinder. So the volume integral, ?E.dL = 0 = No voltage. But, this isn't the case! Now, if the cylinder is a 1/4 wavelength with distributed L and C ... ??? Since the cylinder is long compared to a wavelength, the distributed capacitance will couple a voltage to the inner conductor. The terminal impedance is open circuited [High Z] so no current flows. Conclusion: a standing wave exists on the inner conductor. It is caused by the distributed capacitance and the magnitude of the standing wave on the cylinder. Been away from EM for almost 50 years. I've probably forgotten too much. |
Induced signal?
"Dave" wrote in message ... Cecil Moore wrote: Here's a question overflowing from eHam.net and it's not a trick question. Assume that the radiating portion of a 40m vertical is made out of 33 feet of RG-213 and the braid is the radiator. The center conductor of the coax is left floating at both ends. How much RF voltage and/or current will be induced in that center wire when using the outside braid as the radiator for 100 watt operation? How much of an EM field can exist inside the coax braid? From my EM311 days: If I integrate the fields inside a short conducting cylinder, since there is no charge within the surface integral, there is no field within the cylinder. So the volume integral, ?E.dL = 0 = No voltage. But, this isn't the case! Now, if the cylinder is a 1/4 wavelength with distributed L and C ... ??? Since the cylinder is long compared to a wavelength, the distributed capacitance will couple a voltage to the inner conductor. The terminal impedance is open circuited [High Z] so no current flows. ah, but it isn't! remember, 1/4 wave long (more or less) makes the open at one end look like a short at the other end. there is also that distributed capacitance all along the length between the inner side of the shield and the center conductor... and also, the driving voltage at the ends are about 90 degrees out of phase, so there could be some non-trivial currents in the center conductor. Conclusion: a standing wave exists on the inner conductor. It is caused by the distributed capacitance and the magnitude of the standing wave on the cylinder. Been away from EM for almost 50 years. I've probably forgotten too much. |
Induced signal?
Dave wrote:
remember, 1/4 wave long (more or less) makes the open at one end look like a short at the other end. There is an open at both ends. Does that make it look like a short at both ends? -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
Cecil Moore wrote:
Dave wrote: remember, 1/4 wave long (more or less) makes the open at one end look like a short at the other end. There is an open at both ends. Does that make it look like a short at both ends? If it looked like a short at both ends, it would look like an open at both ends. |
Induced signal?
Built a skeleton piece of giant coax in EZNEC, been playing with it as
a monopole over MININEC ground. I'm not going to claim much since it's probably wildly inapplicable, but I will claim that the answer to this question ain't trivial and depends on the details of the ends. I find a current in the center conductor, It's very much bigger when the center conductor is slightly extended past the ends of the "shield" than when it's slightly inside. The shield is eight wires arranged in a regular octogon with the tops and bottoms tied together. It shows more or less normal fat monopole behavior when used alone. (33 foot element resonant around 6.4 MHz) A fatter wire used for the center conductor has greater current than a thinner one. A very thin wire exhibits almost no current. The peak current in the center conductor is about 3% of what's flowing in the shield for a 1 foot "diameter" (the circumscribed circle around the octagon) shield and a 6 inch diameter inner conductor. The current is very low at either end of the center conductor and peaks about 2/3 from the top. This dumb model is full of holes of many kinds, I'm sure, but it does seem to show nontrivial effects of changing things a little bit near the ends of the coax (slight extension of center conductor outside of/retraction of center conductor into the shield) Dan |
Induced signal?
Cecil Moore wrote: Here's a question overflowing from eHam.net and it's not a trick question. Assume that the radiating portion of a 40m vertical is made out of 33 feet of RG-213 and the braid is the radiator. The center conductor of the coax is left floating at both ends. How much RF voltage and/or current will be induced in that center wire when using the outside braid as the radiator for 100 watt operation? How much of an EM field can exist inside the coax braid? I know this will shock regular group users, but he has taken things WAY out of context in the question above! The actual problem is this: A fellow placed a relay at the top of a half square antenna to change directions by switching from one flat top and drop wire to another. This is a VOLTAGE fed antenna at the ground. The vertical wires at the antenna ends have to be an electrical 1/4 wl long on the OUTSIDE for the system to work properly. Cecil suggested he simply run the relay wires up inside a "shield" to the relay, and the shield would prevent the relay control wires from affecting the very high feed impedance at the base. The shield could be used as the actual vertical antenna lead. Now I know to many people the problem is obvious. The problem is the IMPEDANCE of the open stub formed at the bottom of the vertical sleeve by the inner wire that has to go to a control system of some type and the outer sleeve. That impedance has to be many ten's of kilo ohms so the shunting impedance is high compared to the impedance of the sleeve. Full RF voltage of the feedpoint is also across the gap where the center wires leave the shield. In order for the shield to have some meaningful effect on the system other than simply running the wires down in parallel with the fed wire, the impedance between the inner wire and shield must be VERY high at the bottom. It can of course be a SHORT at the top, since the relay just sits up there in the air with only the contacts making a connection, so the top is easy to handle with some bypass caps. What Cecil totally misses is he formed what is in effect the electrical equivalent of a sleeve balun. The velocity factor of the transmission line forming this stub has to be the SAME as the outside of the sleeve so the INSIDE is 1/4 wl long electrical, and the the loss has to be very low. Otherwise the common mode impedance of the relay wires exiting the shield will not be several times higher than the antenna feed impedance, which is several k-ohms. I've seen antenna manufacturers make the same mistake Cecil just made, and assume that running a cable down the center of a "hot" mast that is part of an antenna means the wires have zero current and zero effect since they are inside the shield, but anyone with any understanding of how the system works would catch the flaws in this idea right away. The flaw is the differential IMPEDANCE between the shield and the shell forming an antenna has to be several times the common mode impedance of the shell or the system won't be worth a flip. Without that high impedance, the inner wire might as well just run down the outside of the sleeve and a couple good HV high impedance RF chokes be used to supply relay control voltage. As a matter of fact at AM BC stations, when using two way or RPU antennas on the hot base insulated towers, I never bothered with running the cables INSIDE the tower. I went up 1/4 wl above the base, and bonded the cables to the tower. I spaced the cables a foot or so off the tower face on large insulators, so it formed an open 1/4 wl very low loss stub. This made the differential mode impedance of the open stub end at the ground very high, and allowed the cables to be brought away from the hot tower base without interacting a large amount with the tower base impedance. This is a very simple common system that is often used in antennas (often in BC systems) , and once in a while used incorrectly by Hams and Ham antenna manufacturers (like Gap and MFJ and a few manufacturers of Ham log periodics). Cecil will catch on with help I'm sure, I just don't have time to walk him through the problem step by step. 73 Tom |
Induced signal?
wrote:
The actual problem is this: Seems logical to discuss the simpler coax example before introducing an example that is more complicated and harder to understand. A fellow placed a relay at the top of a half square antenna to change directions by switching from one flat top and drop wire to another. This is a VOLTAGE fed antenna at the ground. The vertical wires at the antenna ends have to be an electrical 1/4 wl long on the OUTSIDE for the system to work properly. Cecil suggested he simply run the relay wires up inside a "shield" to the relay, and the shield would prevent the relay control wires from affecting the very high feed impedance at the base. The shield could be used as the actual vertical antenna lead. I did not use the word "shield" or "sleeve". I said he could run the control wires up inside a hollow 1/4WL radiator feeding the rest of the half-square. Note that the two original antenna leads were completely removed. The tubing becomes the radiator. Here's the diagram of what I suggested where "RFC" is an RF choke with RF bypass caps at A-B and C-D. FP is the half-square feedpoint. FP 1/4 WL radiator 3/4WL ======================tubing====================== ===+---wire A--RFC----------------------wire-----------------RFC--C--relay B--RFC----------------------wire-----------------RFC--D--coil ======================tubing====================== === It is my contention that the RFCs located just inside the tubing at both ends will prevent this configuration from acting like a stub and that there will be little RF EM energy inside the tubing. How many functional stubs has anyone seen with two RF chokes in the conducting path? Now I know to many people the problem is obvious. The problem is the IMPEDANCE of the open stub formed at the bottom of the vertical sleeve by the inner wire that has to go to a control system of some type and the outer sleeve. All DC circuits isolated by RF chokes inside the tubing and bypass caps across points A-B and C-D. That impedance has to be many ten's of kilo ohms so the shunting impedance is high compared to the impedance of the sleeve. Full RF voltage of the feedpoint is also across the gap where the center wires leave the shield. That's where the RF chokes are located inside the tubing. what happens when RF voltage encounters an RF choke? In order for the shield to have some meaningful effect on the system other than simply running the wires down in parallel with the fed wire, the impedance between the inner wire and shield must be VERY high at the bottom. It can of course be a SHORT at the top, since the relay just sits up there in the air with only the contacts making a connection, so the top is easy to handle with some bypass caps. But that wasn't the configuration I suggested. What Cecil totally misses is he formed what is in effect the electrical equivalent of a sleeve balun. Please explain how a sleeve balun functions with two RF chokes installed in the conducting path. The velocity factor of the transmission line forming this stub has to be the SAME as the outside of the sleeve so the INSIDE is 1/4 wl long electrical, and the the loss has to be very low. Otherwise the common mode impedance of the relay wires exiting the shield will not be several times higher than the antenna feed impedance, which is several k-ohms. The impedance of the RF chokes is also pretty high. I've seen antenna manufacturers make the same mistake Cecil just made, and assume that running a cable down the center of a "hot" mast that is part of an antenna means the wires have zero current and zero effect since they are inside the shield, but anyone with any understanding of how the system works would catch the flaws in this idea right away. Don't forget the RF chokes inside the tubing and bypass caps where the wires enter and exit the tubing. The flaw is the differential IMPEDANCE between the shield and the shell forming an antenna has to be several times the common mode impedance of the shell or the system won't be worth a flip. Without that high impedance, the inner wire might as well just run down the outside of the sleeve and a couple good HV high impedance RF chokes be used to supply relay control voltage. RF chokes at each end present a pretty high series impedance. The RF current at each end is virtually zero and the wires are a non- resonant length. As a matter of fact at AM BC stations, when using two way or RPU antennas on the hot base insulated towers, I never bothered with running the cables INSIDE the tower. We weren't talking about a tower. We were talking about solid tubing made from copper or aluminum. Cecil will catch on with help I'm sure, I just don't have time to walk him through the problem step by step. I'm willing to learn but you cannot simply assert something that seems to violate the laws of physics and then say you don't have time to explain it or furnish a reference. Please explain how a stub can be functional with two RF chokes in the conductive path. I know you have a bunch of followers who consider your word to be gospel and depend upon nothing except faith for their belief in you, but I am not one of them. -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
I reran my EZNEC coax model with a cage with an octagon loop every foot
in height... This is a pretty good shield at 40m, I think. The holes are .002x.007 wavelength. The difference is apparent. Not much confidence that this is modeling the problem... but maybe it's better: Ran it with wire inside/outside the shield. Did #2 wire this time to simulate thin wire down the center of a pipe. Before, the inside/outside comparison yielded almost identical results. Now it doesn't... Wire 6 inches outside shield: Wire No. 26: Segment Conn Magnitude (A.) Phase (Deg.) 1 Open .00356 86.70 2 .00776 95.54 3 .01039 103.14 4 .01208 109.03 5 .01323 117.85 6 .01373 126.69 7 .01297 133.85 8 .01159 144.14 9 Open .00684 154.69 Wire centered in shield: Wire No. 26: Segment Conn Magnitude (A.) Phase (Deg.) 1 Open .00201 53.19 2 .00298 65.98 3 .00324 71.00 4 .00509 45.58 5 .00387 78.85 6 .00401 118.35 7 .00308 63.54 8 .00347 137.70 9 Open .00503 164.46 Anyway, the point that the original cage was a bad model is taken. An additional approximation toward a full shield changes things a great deal. Should anyone want to take a look: http://www.n3ox.net/cage_coax.ez Dan |
Induced signal?
Oh, the laws of physics don't preclude RF from getting in the ends of a
piece of coax, by the way. There is no minimum cutoff frequency for the TEM mode in coaxial waveguide. There is in hollow waveguide with no center conductor. You still need to be able to couple to the ends, and a floating center conductor is not the best way to couple energy in. However, there's no fundamental physical reason why currents *won't* flow on the center conductor in an open-ended piece of coax. Dan |
Induced signal?
I should add that sticking the wire even a little bit (six inches) out
the ends of the skeleton shield increases the current on the center conductor... I expect that the situation with a long wire exiting the bottom will couple MUCH more energy into the center conductor. So, in the context of control wires up an antenna element, the wires coming away from the antenna and a load to represent a choke should be included. I'd also like to refine the shield mesh but I ran out of segments ! Dan |
Induced signal?
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Induced signal?
wrote:
You still need to be able to couple to the ends, and a floating center conductor is not the best way to couple energy in. However, there's no fundamental physical reason why currents *won't* flow on the center conductor in an open-ended piece of coax. How about when there's two RF chokes in series? -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
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Induced signal?
On 11 Jul 2006 07:46:49 -0700, "
wrote: Should anyone want to take a look: http://www.n3ox.net/cage_coax.ez Hi Dan, Thanx for the work. 73's Richard Clark, KB7QHC |
Induced signal?
I think it's just a matter of degree. The more RF chokes there are,
the less current will flow. It would be a straightforward matter to add more wire to the model and include loads for bypass caps and chokes. In either case (wire inside or outside of the shield) good decoupling where the wire lead exits is going to be important. The model so far may suggest that being inside the shield is better than being outside... but without that lead trailing away from the antenna some distance, it's not time to conclude much about the relay+half square problem. The 40m coax monopole answer would seem to be "yes, there's current on the center conductor, but it's small and coupled in through the ends" If you're using the standard EZNEC, you're going to have to knock another section off the top of the cage... if you've got EZNEC+ then just add away... I may try it when I get home... knock another section off and try a control wire... Might have to send off my money to Roy and go for EZNEC+... I know I can get around segment limitations with other programs but I do like EZNEC. This is probably the ninth time since I got the program a few months ago that I've hit the segment limit ... i like meshing things... Gives me a question about the (EZ)NEC limitations... should I be watching out for fine 2D meshes? It seems to work OK in this case... the base impedance of the meshed monopole and the current distribution viewed on the segments all makes sense, and it seems to me that there's not much reason to doubt that the currents are calculated correctly in the mesh as long as it's not coarse with respect to a wavelength... any caveats in this regard? 73, Dan |
Induced signal?
Cecil Moore wrote: wrote: I expect that the situation with a long wire exiting the bottom will couple MUCH more energy into the center conductor. My suggested solution over on eHam.net included RF chokes and RF bypass caps at each end of the tubing. -- 73, Cecil http://www.qsl.net/w5dxp Cecil, You put great faith in passing you Mensa exam. With that in mind, what possible difference does the shield make once the inner conductor is bypassed and choked at each end? In case you can't understand, the answer is NO difference. Without the chokes and bypasses, your idea won't work. With the chokes and bypasses, the idea isn't needed. The relay wire can run right down the mast with no ill effects. It's so basic and simple, even a Mensa member can follow it! 73 Tom |
Induced signal?
I should have said "the more RF chokes there are, the less current will
flow in the center conductor of the case where the wire is inside the shield" by the way. If the RF chokes and wire are inside the shield, it will keep the current off the wire inside the shield. However, where the wire exits the shield, currents can still be induced on it... this is what allows current to flow on the unchoked center conductor, right? So why wouldn't it flow in the other direction. So you need chokes some distance outside the shield to isolate the rest of the wire from the antenna. If you put the wire outside the shield and put chokes at the top and bottom, some distance away from the antenna, then a very large current will be induced on the wire between the chokes compared to the current you get on the wire inside the shield. However, this doesn't immediately translate into more current on the control wire trailing away. It is just saying something about the wire between the top and bottom chokes. What you're trying to do is decouple the control wire going away from the antenna from the antenna. Decoupling the control wire that runs up inside or outside of the antenna isn't very important as long as current doesn't flow on the control wire to the shack. Whether or not inside or outside of the tower/tube helps depends on a lot of factors and should be answered with some sort of calculation... Seems that the various claims are testable... I'll post anything I come up with. Dan |
Induced signal?
I modeled the original question Cecil asked on r.r.a.a. (current on a
coax center conductor of a monopole made of coax) because I was curious and because my wetware simulation for the amount of RF coupled into the end of a piece of coaxial cable wasn't up to the task. I didn't start with the eHam question in mind. I don't know offhand how much the coupling between the center conductor and shield is at the ends of the coax, so I tried some stuff with the only electromagnetic code I have lying around my house. The relay question is not very interesting, it's just an engineering one.. throw good chokes at it. Done. I agree. I wanted to know what the current was in the center conductor of a piece of RG-8 being used as a monopole. The answer to that question is not trivially calculable, though it might be trivial in the sense of being trivia... who cares? I did. Dan |
Induced signal?
wrote:
You put great faith in passing you Mensa exam. This member of MENSA is having difficulty in parsing your statement. What does, "you Mensa exam", actually mean? Does the pronoun, "you", address the Mensa exam? Wouldn't it be better to address the Mensa exam as an "it" rather than as a person? Your numerous deviations from the accepted laws of physics and accepted English language construction have me confused. With that in mind, what possible difference does the shield make once the inner conductor is bypassed and choked at each end? What I said is that, contrary to your strange assertions, is that the RF chokes and RF bypass capacitors block the stub function upon which your entire argument rests. The result of my suggestion is *NOT* a stub function as you have so stubbornly insisted. So I ask you once again: Please prove that a stub with two RF chokes in the conductive path is actually functional as a stub. If you cannot, your entire argument falls apart. -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
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Induced signal?
wrote:
If the RF chokes and wire are inside the shield, it will keep the current off the wire inside the shield. I hope W8JI is reading this. He insists that it will still function as a stub and such an approach is "useless". However, where the wire exits the shield, currents can still be induced on it... this is what allows current to flow on the unchoked center conductor, right? So why wouldn't it flow in the other direction. So you need chokes some distance outside the shield to isolate the rest of the wire from the antenna. Just outside the shield is a relay with an RF bypass cap across its coil. -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
Cecil Moore wrote: wrote: If the RF chokes and wire are inside the shield, it will keep the current off the wire inside the shield. I hope W8JI is reading this. He insists that it will still function as a stub and such an approach is "useless". However, where the wire exits the shield, currents can still be induced on it... this is what allows current to flow on the unchoked center conductor, right? So why wouldn't it flow in the other direction. So you need chokes some distance outside the shield to isolate the rest of the wire from the antenna. Just outside the shield is a relay with an RF bypass cap across its coil. -- 73, Cecil http://www.qsl.net/w5dxp Turn in your mensa membership card Cecil. You flunked basic grade-K problem solving logic. :-) |
Induced signal?
wrote:
I wanted to know what the current was in the center conductor of a piece of RG-8 being used as a monopole. The answer to that question is not trivially calculable, though it might be trivial in the sense of being trivia... who cares? I did. I have used that same circuit to switch a top hat in and out. How much RF power did I lose by taking that route? -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
Cecil Moore wrote: wrote: I wanted to know what the current was in the center conductor of a piece of RG-8 being used as a monopole. The answer to that question is not trivially calculable, though it might be trivial in the sense of being trivia... who cares? I did. I have used that same circuit to switch a top hat in and out. How much RF power did I lose by taking that route? That is NOTHING like the problem the ZL on eHam had. |
Induced signal?
wrote:
Cecil Moore wrote: I have used that same circuit to switch a top hat in and out. How much RF power did I lose by taking that route? That is NOTHING like the problem the ZL on eHam had. Clutching at straws, are we? It's exactly like the ZL problem except the relay is used to switch a top hat instead of a wire. -- 73, Cecil http://www.qsl.net/w5dxp |
Induced signal?
On 11 Jul 2006 12:21:06 -0700, "
wrote: I wanted to know what the current was in the center conductor of a piece of RG-8 being used as a monopole. The answer to that question is not trivially calculable, though it might be trivial in the sense of being trivia... who cares? I did. Dan Hi Dan, All the issues you focus on are quite particular and specific, and perhaps too much so. I find them equally interesting and you ask reasonable questions. Unfortunately, you question above suffers from every problem imaginable for measuring, and any report you get without considerable qualification is probably sheer fancy. It suffers from a version of Heisenberg's problem of disturbing what you attempt to measure, and invalidating everything in the process. Another way of stating this problem, with more practicality, is that you have to put a wire into the coax to add your meter to make the measurement. This then brings that wire's own contribution, which, as you've noted, can raise the stakes considerably. I, too, have played with a variant of your model (I can push the mesh finer and have worked with a 16 sided coax model). However, instead of driving the line like a monopole, I simply plunked a source into the wire skeleton of the "shield." But to return to your own published model, I've played with the length of wire 26 after discovering it emerged from both ends of the coax. I don't put much credit to Cecil's invention of topics, so I am unaware if this wire length meets some criteria (even if it did, I would still suspect the detail would have been spurious). Be that as it may. After truncating the wire 26 so that it did not come anywhere closer to the mouth of the coax (either end) than 5 feet, "induced" currents plummeted like a rock. 73's Richard Clark, KB7QHC |
Induced signal?
Richard, You hit this nail on the head, for su
All the issues you focus on are quite particular and specific, and perhaps too much so. I find them equally interesting and you ask reasonable questions. Unfortunately, you question above suffers from every problem imaginable for measuring, and any report you get without considerable qualification is probably sheer fancy. It suffers from a version of Heisenberg's problem of disturbing what you attempt to measure, and invalidating everything in the process. I don't have much trouble with the antennas I actually build. The aluminum and wire I sling around tend to behave. I do have some vested interest in overspecific modeling in the sense that since I'm deploying all my antennas from the balcony of my apartment (with its overhanging roof), and don't have much space to build, adjustments and pruning are very hard to do. Some performance gains can certainly be had by adjustment... but in my situation now it's so much easier to push bits around in EZNEC. This one, though, was just pure impractical theoretical curiosity. That, and a little bit of being concerned that people seemed to "know" the answer, that the answer was obvious, and, well, I didn't think it was. The current in real experiment would be, as you pointed out, pretty much impossible to measure without disturbing it. But as I see it, from a practical standpoint, in a real system where you're using the shield of a piece of coax as an antenna , the current in the center conductor never matters. The existence of the current matters, at least as a learning experience. It matters in the sense that some seem to think that there couldn't be a current, just because a piece of coax is "shielded". Or because the center conductor is "floating". It's precisely the fact that the center conductor is "floating" that makes the (probably) weak coupling at the ends important. Thanks for posting your results. What you describe certainly jives with the fuzzy idea that I've got of what's doing the coupling ... or at least where it's happening. Of course, the tube without center conductor is certainly a waveguide well below cutoff, so we've got solid footing saying a short wire inside a long conductive tube a small fraction of a wavelength in diameter will have little current on it... and the model also exhibits this. I'm done with this one unless I find a leadless current meter (small, battery powered with an A/D converter and optical fiber out?) and a big piece of copper pipe laying around! 73, Dan |
Induced signal?
Cecil Moore wrote: wrote: Cecil Moore wrote: I have used that same circuit to switch a top hat in and out. How much RF power did I lose by taking that route? That is NOTHING like the problem the ZL on eHam had. Clutching at straws, are we? It's exactly like the ZL problem except the relay is used to switch a top hat instead of a wire. -- 73, Cecil http://www.qsl.net/w5dxp Cecil, In your alleged system, you carried RF through the coax and used the coax as a swixcthing stub. In the other fellows system, he had a high impedance feed and had to get a relay conductor past the high impedance feed without affecting the system. They are TOTALLY different. Even a card carrying mensa member can't be that out of touch with a simple system like this. 73 Tom |
Induced signal?
wrote:
In your alleged system, you carried RF through the coax and used the coax as a swixcthing stub. You seem to be confused by my actual system in my yard Vs the hypothetical piece of coax I introduced as a simplified example for discussion purposes. Forget the coax example. In my yard, I have run two wires up a piece of 1/4WL aluminum tubing to a relay which switched a top hat in and out just as the other fellow wanted to switch a wire in and out. The only thing that I would have to add to turn that system into switchable half-squares, like the other fellow has, would be two 3/4WL runs of wire. I suspect the SGC-230 autotuner at the base would match the new feedpoint impedance just fine. I've never seen an impedance it couldn't match. In the other fellows system, he had a high impedance feed and had to get a relay conductor past the high impedance feed without affecting the system. The two physical configurations being discussed are *identical*, being 1/4WL of tubing through which two DC wires are routed. They are TOTALLY different. Once again, you make an outrageous exclusive statement completely divorced from reality. If any one thing about the two antennas is not different, your statement is completely false. One wonders what compels you to make such strange assertions. -- 73, Cecil http://www.qsl.net/w5dxp |
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