|
I played with it a bit and found that converting it to a T from an L
actually reduced the overall gain at 30 deg elevation and did not improve it below that. I ended up with 73' on each side of the center for the T, slightly inverted. In fact, I took the inverted L in one of my files (already constructed) and just added the 2nd wire. I, did, of course shorten the vertical section to the 42' that I have available. It did eliminate the overhead radiation, but did not significantly improve the low angle (which seems impossible), but I'll play some more. So far, it looks like the Inverted L is the better choice, even for more power at lower launch angles than the "T". (It is also easier to construct...for a quick throw it up antenna, you only need a long piece of wire and an insulator at the top, pullling as much wire as you can get to go vertical and stretching out the rest.) Then, the real work comes, putting down a good radial field before the frost. Thanks for the tip. 73 ....hasan, N0AN "Roy Lewallen" wrote in message ... hasan schiers wrote: Now that is interesting, Roy. I was going to put up a 160 m inverted L this summer. I am limited to only being able to go up about 45 feet, so I would need about another 90 feet horizontal. Are you suggesting that it might be a better arrangement to go up the 45' and then put up the top "T"? It might be. You might benefit from the radiation from the horizontal portion of an L, and you might not. But if it's quite low, the radiation will be mostly straight up, and a fair amount will be expended warming up the ground. Neither will occur with a T. If so, roughly how long should the top part of the T be (each side of center) to get me to 160? That's just what antenna modeling programs are for! Dust off your EZNEC and you'll have the answer in minutes. . . . Roy Lewallen, W7EL |
I found the error, I had to fix two conditions that I had changed in the
model: Copper wire (for loss) Ground characteristics Now that both antennas have the same conditions, the T has ever so slightly better gain at 20 degrees than the Inverted L. Not enough to bother with the increased complexity, and the input Z is now down around 5 ohms for the T and 8 ohms for the L. Now, is it worth matching the 8 ohms up to 50 at the feedpoint, or just using the tuner in the shack to take care of it? (coax feed, LMR-400, about 50') ....hasan, N0AN "Roy Lewallen" wrote in message ... John Ferrell wrote: . . . I am a perpetual antenna student! And so are we all. Roy Lewallen, W7EL |
Since you are talking about 50 Ohms I assume you are talking about a
transmission line. If that is the case you should definitely match the feedline to antenna at the antenna feed point. Any attempt to match the feedline with a tuner in the shack only turns the whole feedline into part of the antenna system. By doing that you have lost any good work in building the antenna. Dan hasan schiers wrote: I found the error, I had to fix two conditions that I had changed in the model: Copper wire (for loss) Ground characteristics Now that both antennas have the same conditions, the T has ever so slightly better gain at 20 degrees than the Inverted L. Not enough to bother with the increased complexity, and the input Z is now down around 5 ohms for the T and 8 ohms for the L. Now, is it worth matching the 8 ohms up to 50 at the feedpoint, or just using the tuner in the shack to take care of it? (coax feed, LMR-400, about 50') ...hasan, N0AN "Roy Lewallen" wrote in message ... John Ferrell wrote: . . . I am a perpetual antenna student! And so are we all. Roy Lewallen, W7EL |
dansawyeror wrote:
Since you are talking about 50 Ohms I assume you are talking about a transmission line. If that is the case you should definitely match the feedline to antenna at the antenna feed point. Why is it definite? What is the loss in 50 ft. of LMR-400 at the frequency of interest when the SWR is 50/8 = 6.25:1? Any attempt to match the feedline with a tuner in the shack only turns the whole feedline into part of the antenna system. Simply not true if the currents remain differentially balanced. SWR doesn't cause feedline radiation. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
On Sat, 27 Aug 2005 15:26:25 -0500, "hasan schiers"
wrote: I found the error, I had to fix two conditions that I had changed in the model: Copper wire (for loss) Ground characteristics Now that both antennas have the same conditions, the T has ever so slightly better gain at 20 degrees than the Inverted L. Not enough to bother with the increased complexity, and the input Z is now down around 5 ohms for the T and 8 ohms for the L. Now, is it worth matching the 8 ohms up to 50 at the feedpoint, or just using the tuner in the shack to take care of it? (coax feed, LMR-400, about 50') If you would go he http://www.qsl.net/ac6la/tldetails.html download the program and enter your load Z and 50' of LMR400 @ 3.5 MHz, you would immediately see the answer to your question. If the 8 ohm is real (j=0) then the total loss is all of 0.4 dB and if the line is 50' long, the resulting input Z is easily matched with low loss. |
On Sat, 27 Aug 2005 14:31:06 -0700, dansawyeror
wrote: Since you are talking about 50 Ohms I assume you are talking about a transmission line. If that is the case you should definitely match the feedline to antenna at the antenna feed point. Any attempt to match the feedline with a tuner in the shack only turns the whole feedline into part of the antenna system. By doing that you have lost any good work in building the antenna. The feedline is -always- part of the antenna system. If it troubles you to think about matching in the shack, just think of the transmission line as part of a matching network located at the feedpoint. In other words, the feedpoint network is comprised of 50' of coax and coupla LCs in a box. This network then connects to the transmitter through a non-resonant (flat) coax line. |
Let's take the case of a 50 Ohm line and some mismatched antenna. The result is
a combination other then 50 Ohm with most likely a zero complex component. All a tuner does is match 50 Ohm at the radio to the complex impedance presented to it at the source of the line. That the only place with 50 Ohms and zero inductance in the line - antenna system. The combination of cable and antenna presents something other then R = 50 ohms 0 reactance and the the transmission line see discontinuities. The result is it radiates. Dan Cecil Moore wrote: dansawyeror wrote: Since you are talking about 50 Ohms I assume you are talking about a transmission line. If that is the case you should definitely match the feedline to antenna at the antenna feed point. Why is it definite? What is the loss in 50 ft. of LMR-400 at the frequency of interest when the SWR is 50/8 = 6.25:1? Any attempt to match the feedline with a tuner in the shack only turns the whole feedline into part of the antenna system. Simply not true if the currents remain differentially balanced. SWR doesn't cause feedline radiation. |
I am not sure what you mean by -always-. If you mean there is no such thing as a
perfect coax line then your statement is true but does not add any real value. If you mean by -always- the feedline is a significant component in the antenna system then I would have to disagree. When operated at their design point coax transmission lines do not radiate and are not part of the radiating "antenna system". Coax is designed to work in a specific environment as a transmission line. These transmission lines are designed not to radiate. When transmission lines are operated significantly outside their design range the radiate. Adding a tuner to one end only controls the characteristics at that point. It does not 'clean up' the mismatchs. Dan Wes Stewart wrote: On Sat, 27 Aug 2005 14:31:06 -0700, dansawyeror wrote: Since you are talking about 50 Ohms I assume you are talking about a transmission line. If that is the case you should definitely match the feedline to antenna at the antenna feed point. Any attempt to match the feedline with a tuner in the shack only turns the whole feedline into part of the antenna system. By doing that you have lost any good work in building the antenna. The feedline is -always- part of the antenna system. If it troubles you to think about matching in the shack, just think of the transmission line as part of a matching network located at the feedpoint. In other words, the feedpoint network is comprised of 50' of coax and coupla LCs in a box. This network then connects to the transmitter through a non-resonant (flat) coax line. |
Cecil Moore wrote:
Why is it definite? What is the loss in 50 ft. of LMR-400 at the frequency of interest when the SWR is 50/8 = 6.25:1? Ok, I need teaching here. Why would the loss change? The loss on the line is forced to what happens at the nominal 50 ohms doesn't it? The SWR shouldn't be able to change it unless the voltage limits are hit I would think. I need an explanation of why it wouldn't be so. Thanks. tom K0TAR |
On Sat, 27 Aug 2005 17:49:42 -0700, dansawyeror
wrote: I am not sure what you mean by -always-. If you mean there is no such thing as a perfect coax line then your statement is true but does not add any real value. If you mean by -always- the feedline is a significant component in the antenna system then I would have to disagree. When operated at their design point coax transmission lines do not radiate and are not part of the radiating "antenna system". An antenna tuner or transmatch, if you prefer, doesn't radiate and is part of an antenna system. the fact that a transmission line radiates or not doesn't mean it isn't part of the system. Danny, K6MHE |
dansawyeror wrote:
The combination of cable and antenna presents something other then R = 50 ohms 0 reactance and the the transmission line see discontinuities. The result is it radiates. If the currents are balanced, a 50 ohm transmission line seeing something other than a 50 ohm load does NOT cause it to radiate. If a 50 ohm unbalanced transmission line sees a 50 ohm balanced load and common-mode currents flow on the outside of the coax, it will usually result in radiation from the feedline. Simply knowing the magnitude of the feedpoint impedance doesn't tell us anything about feedline radiation. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Tom Ring wrote:
Cecil Moore wrote: Why is it definite? What is the loss in 50 ft. of LMR-400 at the frequency of interest when the SWR is 50/8 = 6.25:1? Ok, I need teaching here. Why would the loss change? The loss on the line is forced to what happens at the nominal 50 ohms doesn't it? The SWR shouldn't be able to change it unless the voltage limits are hit I would think. I need an explanation of why it wouldn't be so. The loss specified on transmission line charts are usually matched-line losses (SWR=1:1). When the SWR is higher than 1:1, additional losses appear due to the higher SWR. There's a chart in my ARRL Handbook that gives those additional losses. The matched-line loss for 50 ft. of LMR-400 at 3.5 MHz is about 0.1 dB. Wes calculated the total loss at 0.4 dB but I think that must have been for 100 ft. of LMR-400 at 3.5 MHz with an SWR of 6.25:1. The higher SWR causes additional losses because the maximum current and maximum voltage is higher for the same power delivered to the load than for the matched-line case. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
The coil measures about 60 uH. The antenna is elevated about 3 feet on a
short tripod. The radials angle down the tripod legs and then out. The coil is about 4 inchs in diameter, number 12, wound on a fiberglass form. It is centerloaded. I am looking at it accross the yard, it is about 6 inches long. It is would with about a point .5 pitch. Calculations for a 1:1 pitch predict a Q of about 450. Thanks, Dan Dan, I have just run a NEC 2 model of your antenna. I have used the Sommerfeld/Norton ground, with average parameters of: Er = 13, and Sigma = 5 S/m. The top of the antenna is at 18 ft, and the base at 3 ft. I have twelve 40 ft radials, spaced at 30 deg, and within the limitations of NEC 2, placed them at 3" above ground. The first 5 ft of the radials drops from 3 ft to 3" at an angle of 45 deg. The monopole is center loaded with an inductor of Q = 450. The model has 640 (6") segments and takes 3 minutes to run (3.5 - 4.0 MHz in 50 kHz increments). What I notice is that I need 92 uH to resonate at 3.9 MHz. The input impedance is 12 ohms. I used a lumped element model for the inductor. I may try a physical helix later. These data do not seem to agree with your measured results. NEC 2, with the Sommerfeld/Norton ground solution, is supposed to give a reasonable result with wires at 10^(-3) wavelengths above ground (Basic Antenna Modeling, Cebik p. 15-16 Nittany Scientific). Gain and take-off angle are excellent, with max gain of -3 dBi at 28 deg. elevation. The lower 3 dB point (8 deg elevation) gain is -6.6 dBi. The NEC output file indicates an antenna efficiency of 54%. A free space model shows an input impedance of 8 ohms, so your ground losses are not significant (At least with my model). Apart from adding horizontal wires, in "T" or inverted "L" fashion, I doubt any antenna you could put up would match its performance at distances over 500 miles. With 100 ft of LMR 400 the additional loss is about 0.45 dB. I would be very interested to know if anybody has any ideas why my calculations appear to be different from the measurements. Regards, Frank |
On Sat, 27 Aug 2005 17:49:42 -0700, dansawyeror
wrote: I am not sure what you mean by -always-. If you mean there is no such thing as a perfect coax line then your statement is true but does not add any real value. If you mean by -always- the feedline is a significant component in the antenna system then I would have to disagree. When operated at their design point coax transmission lines do not radiate and are not part of the radiating "antenna system". Well, now, we've just morphed "antenna system" to "radiating antenna system." But never mind, nothing's changed, you're still all wet. Of course the feedline is a component in the antenna system regardless of whether it radiates or not. BTW the line can be perfectly matched and radiate or it can be highly mismatched and not radiate. Coax is designed to work in a specific environment as a transmission line. These transmission lines are designed not to radiate. When transmission lines are operated significantly outside their design range the(y) radiate. Squeak squeak---squeak squeak---squeak squeak, there I just screwed a tee connector on the end of a run of 50 ohm coax and on the tee I screwed on two 50 ohm loads. The line is now terminated in 25 ohm making the SWR 2:1. 'Splain to me how well this line, "operating outside its design range", radiates. If that's not "significantly" far enough outside the "design range" then allow me to remove the loads, add two more tees and terminate them with four 50 ohm loads, making the SWR 4:1. How well does the line radiate now? Should I continue to 8:1 or are you convinced? Adding a tuner to one end only controls the characteristics at that point. It does not 'clean up' the mismatchs. It can certainly "clean up" the mismatch at the input to the tuner, which unless I've been deluding myself for 45 years or so, is the point. |
On Sat, 27 Aug 2005 16:53:39 -0700, dansawyeror
wrote: Let's take the case of a 50 Ohm line and some mismatched antenna. The result is a combination other then 50 Ohm with most likely a zero complex component. Surely you don't believe this do you? It is -much- more likely that the impedance is reactive than not. At one (fundamental) frequency the reactance is zero. At every other frequency it is reactive. All a tuner does is match 50 Ohm at the radio to the complex impedance presented to it at the source of the line. Isn't that enough? That the only place with 50 Ohms and zero inductance in the line - antenna system. The combination of cable and antenna presents something other then R = 50 ohms 0 reactance and the the transmission line see discontinuities. The result is it radiates. Oh dear me. |
Thanks Wes...done...btw the antenna is for 160 meters, not 80m, so the loss
is even less. It doesn't look to me like it's worth doing anything more than tuning out the mismatch in the shack. 73 ....hasan, N0AN |
Dan,
I appreciate your reply, but it is filled with misinformation. Consider reading some reference materials. The errors have been discussed in other responses to your post. I don't mean to insult you for trying to help, but the information you provided is so incorrect, that I couldn't let it pass. I hope no one attempts to make use of what you said...it will only compound the error. I asked simply is it worth it? (in terms of loss). Wes had the calculator software that showed the loss is insignificant, so it saves the work/complexity at the base of the antenna. Radiation from the feedline is not a function of mismatch. I'm surprised Reg didn't go apoplectic over that one. The tuner in the shack will do the job nicely. ....hasan, N0AN "dansawyeror" wrote in message ... Since you are talking about 50 Ohms I assume you are talking about a transmission line. If that is the case you should definitely match the feedline to antenna at the antenna feed point. Any attempt to match the feedline with a tuner in the shack only turns the whole feedline into part of the antenna system. By doing that you have lost any good work in building the antenna. Dan hasan schiers wrote: I found the error, I had to fix two conditions that I had changed in the model: Copper wire (for loss) Ground characteristics Now that both antennas have the same conditions, the T has ever so slightly better gain at 20 degrees than the Inverted L. Not enough to bother with the increased complexity, and the input Z is now down around 5 ohms for the T and 8 ohms for the L. Now, is it worth matching the 8 ohms up to 50 at the feedpoint, or just using the tuner in the shack to take care of it? (coax feed, LMR-400, about 50') ...hasan, N0AN "Roy Lewallen" wrote in message ... John Ferrell wrote: . . . I am a perpetual antenna student! And so are we all. Roy Lewallen, W7EL |
Tom Ring wrote:
Cecil Moore wrote: Why is it definite? What is the loss in 50 ft. of LMR-400 at the frequency of interest when the SWR is 50/8 = 6.25:1? Ok, I need teaching here. Why would the loss change? The loss on the line is forced to what happens at the nominal 50 ohms doesn't it? The SWR shouldn't be able to change it unless the voltage limits are hit I would think. I need an explanation of why it wouldn't be so. For the same amount of power delivered to the load, transmission line losses increase with increasing SWR. For a Z0-matched system: (PLoad = Pfor - Pref) As the load mismatch is increased, energy reflected from the mismatched load increases and is re-reflected back toward the load at the tuner Z0-match point. As the load mismatch is increased, more energy is stored in the transmission line during steady-state. This is indirect proof that reflected energy waves actually exist. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
"Cecil Moore" wrote in message
... dansawyeror wrote: The combination of cable and antenna presents something other then R = 50 ohms 0 reactance and the the transmission line see discontinuities. The result is it radiates. If the currents are balanced, a 50 ohm transmission line seeing something other than a 50 ohm load does NOT cause it to radiate. If a 50 ohm unbalanced transmission line sees a 50 ohm balanced load and common-mode currents flow on the outside of the coax, it will usually result in radiation from the feedline. Simply knowing the magnitude of the feedpoint impedance doesn't tell us anything about feedline radiation. -- 73, Cecil http://www.qsl.net/w5dxp Even grossly mismatched open wire transmission line does not radiate significantly. For example a 66 ft length of 3" spaced, open wire line, shorted at one end, radiates only 4 - 5% of the input power. 95% is dissipated in the conductor losses. 73, Frank |
Frank,
Thank you. Is there any way you can forward the saved parameters. This is a screwdriver antenna, I will remeasure the coil and double check. My modeling of the free space antenna showed about 4 Ohms but it was with a much simpler program. It was that program I used to measure Q. Thanks, Dan Frank wrote: The coil measures about 60 uH. The antenna is elevated about 3 feet on a short tripod. The radials angle down the tripod legs and then out. The coil is about 4 inchs in diameter, number 12, wound on a fiberglass form. It is centerloaded. I am looking at it accross the yard, it is about 6 inches long. It is would with about a point .5 pitch. Calculations for a 1:1 pitch predict a Q of about 450. Thanks, Dan Dan, I have just run a NEC 2 model of your antenna. I have used the Sommerfeld/Norton ground, with average parameters of: Er = 13, and Sigma = 5 S/m. The top of the antenna is at 18 ft, and the base at 3 ft. I have twelve 40 ft radials, spaced at 30 deg, and within the limitations of NEC 2, placed them at 3" above ground. The first 5 ft of the radials drops from 3 ft to 3" at an angle of 45 deg. The monopole is center loaded with an inductor of Q = 450. The model has 640 (6") segments and takes 3 minutes to run (3.5 - 4.0 MHz in 50 kHz increments). What I notice is that I need 92 uH to resonate at 3.9 MHz. The input impedance is 12 ohms. I used a lumped element model for the inductor. I may try a physical helix later. These data do not seem to agree with your measured results. NEC 2, with the Sommerfeld/Norton ground solution, is supposed to give a reasonable result with wires at 10^(-3) wavelengths above ground (Basic Antenna Modeling, Cebik p. 15-16 Nittany Scientific). Gain and take-off angle are excellent, with max gain of -3 dBi at 28 deg. elevation. The lower 3 dB point (8 deg elevation) gain is -6.6 dBi. The NEC output file indicates an antenna efficiency of 54%. A free space model shows an input impedance of 8 ohms, so your ground losses are not significant (At least with my model). Apart from adding horizontal wires, in "T" or inverted "L" fashion, I doubt any antenna you could put up would match its performance at distances over 500 miles. With 100 ft of LMR 400 the additional loss is about 0.45 dB. I would be very interested to know if anybody has any ideas why my calculations appear to be different from the measurements. Regards, Frank |
On Sun, 28 Aug 2005 07:02:03 -0500, "hasan schiers"
wrote: Thanks Wes...done...btw the antenna is for 160 meters, not 80m, Sorry, I was just thinking about the subject of the thread, not reading what you wrote. so the loss is even less. It doesn't look to me like it's worth doing anything more than tuning out the mismatch in the shack. Exactly. As I said earlier, if it makes it easier on the someone's (not you, you get it) conscience or ego they can think of the transmission line as a bunch of distributed L and Cs (with a little R thrown in) that are part of a tuner. When they get this "tuner" to show 50 +j0, then they can call the length of line between the "tuner" and the transmitter, the "transmission line." In your case, you have a low-loss line, so the "R" part is low and the system is efficient. |
Frank,
Thank you. Is there any way you can forward the saved parameters. This is a screwdriver antenna, I will remeasure the coil and double check. My modeling of the free space antenna showed about 4 Ohms but it was with a much simpler program. It was that program I used to measure Q. Thanks, Dan Hi Dan, I have run the program to determine the precise resonance. The parameters are as follows: Inductor 89.3 uH, and resonant at 3.92 MHz. I can send you a zipped NEC output text file. It is about 190 kB. Also the NEC code I used. You can plug in the appropriate data into an Excel spread sheet. If you need any specific graphical output I can it as a JPEG file. I used the default input source of 1 V peak, which accounts for the low power values in the output file. The E-field data is far field, normalized to 1 meter. The inductor is described as a lumped element, complex impedance, of 4.9 + j2200 ohms. I arrived at this value based on your Q of 450, and just played around with the imaginary value to achieve resonance within the 75 meter band. Let me know if I can send the above information to the address shown in your posting. Regards, Frank |
Wes,
As a starter, look at this site: http://www.cbtricks.com/~ab7if/coax/coax.htm When a transmission line is terminated in it's characteristic impedance there is no voltage or current reflection from the line. The electromagnetic fields continue to flow into the termination as if the line were infinitely long. When a mismatch of impedance occurs, reflected waves will be produced and they will interact with the incident waves. The total voltage and current on the line are no longer the result of a single traveling wave from the source to the load. Instead, it is the algebraic sum of two waves traveling in opposite directions. This interaction results in what is known as standing waves. The waves remain in fixed positions along the line while they vary in amplitude and polarity. A wave of any shape can be transmitted along the line without any change of waveshape or magnitude. Looking at the gif below, we see a line driven with a sine wave generator, terminated with a short circuit to maximize the reflection. My first claim is a tuner at the source does not materially improve what is happening in the coax. That is a tuner does not recreate the condition above where the coax is functioning as a properly matched and terminated transmission line. All the tuner does is match the impedance at the coax source back to some known, usually 50 Ohm, value. My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself. Dan Wes Stewart wrote: On Sat, 27 Aug 2005 16:53:39 -0700, dansawyeror wrote: Let's take the case of a 50 Ohm line and some mismatched antenna. The result is a combination other then 50 Ohm with most likely a zero complex component. Surely you don't believe this do you? It is -much- more likely that the impedance is reactive than not. At one (fundamental) frequency the reactance is zero. At every other frequency it is reactive. All a tuner does is match 50 Ohm at the radio to the complex impedance presented to it at the source of the line. Isn't that enough? That the only place with 50 Ohms and zero inductance in the line - antenna system. The combination of cable and antenna presents something other then R = 50 ohms 0 reactance and the the transmission line see discontinuities. The result is it radiates. Oh dear me. |
Wes,
As a starter, look at this site: http://www.cbtricks.com/~ab7if/coax/coax.htm When a transmission line is terminated in it's characteristic impedance there is no voltage or current reflection from the line. The electromagnetic fields continue to flow into the termination as if the line were infinitely long. When a mismatch of impedance occurs, reflected waves will be produced and they will interact with the incident waves. The total voltage and current on the line are no longer the result of a single traveling wave from the source to the load. Instead, it is the algebraic sum of two waves traveling in opposite directions. This interaction results in what is known as standing waves. The waves remain in fixed positions along the line while they vary in amplitude and polarity. A wave of any shape can be transmitted along the line without any change of waveshape or magnitude. Looking at the gif below, we see a line driven with a sine wave generator, terminated with a short circuit to maximize the reflection. My first claim is a tuner at the source does not materially improve what is happening in the coax. That is a tuner does not recreate the condition above where the coax is functioning as a properly matched and terminated transmission line. All the tuner does is match the impedance at the coax source back to some known, usually 50 Ohm, value. My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself. Dan Wes Stewart wrote: On Sat, 27 Aug 2005 16:53:39 -0700, dansawyeror wrote: Let's take the case of a 50 Ohm line and some mismatched antenna. The result is a combination other then 50 Ohm with most likely a zero complex component. Surely you don't believe this do you? It is -much- more likely that the impedance is reactive than not. At one (fundamental) frequency the reactance is zero. At every other frequency it is reactive. All a tuner does is match 50 Ohm at the radio to the complex impedance presented to it at the source of the line. Isn't that enough? That the only place with 50 Ohms and zero inductance in the line - antenna system. The combination of cable and antenna presents something other then R = 50 ohms 0 reactance and the the transmission line see discontinuities. The result is it radiates. Oh dear me. |
On Sun, 28 Aug 2005 09:38:52 -0700, dansawyeror
wrote: My first claim is a tuner at the source does not materially improve what is happening in the coax. That is a tuner does not recreate the condition above where the coax is functioning as a properly matched and terminated transmission line. All the tuner does is match the impedance at the coax source back to some known, usually 50 Ohm, value. Hi Dan, As true as that may be, the results run the gamut from trivial to considerable as has been already discussed in this thread. My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself. This mismatch could arise for any number of reasons, and not all contribute to radiation from the coax. Wes has already demonstrated a deliberate mismatch at the end of a cable that exhibits absolutely no radiation from the coax. This is because he has contrived to contain the fields from emerging and coupling to the outside of the coax shield. You should be aware that the shield does support currents on the inside and outside that are wholly unaffected by each other - except at the drive point where the two conduction paths are joined. When you drive a dipole with a coax, the exterior conductive path of the shield (a separate circuit from the interior conductive path of that same shield) is in parallel with one arm of the dipole. This means you have a third radiator that has a length and termination that is undefined. It is THAT radiator that both causes a higher SWR AND radiation that is not a normal condition for an otherwise tuned antenna. Given that the length of the line's external conductive path, and its termination is largely undefined (unless you take great care to both measure and characterize such issues), the occurrence of mismatch and radiation is highly variable. Thus, anecdotal accounts of antennas being poor or good when they are driven by a simple coax are suspect (barring the reporter also supplying the conditions of the external path). To eliminate the effects of this third path, a choke is added to the drive point. The purpose of the choke is to add impedance to this path to reduce Common Mode current. Common Mode current is the current that flows due to an unbalanced system (the unanticipated third radiator does that in spades). It flows in two wire transmission lines too when the unbalance occurs for other reasons (and those are plentiful as well). 73's Richard Clark, KB7QHC |
On Sun, 28 Aug 2005 09:39:24 -0700, dansawyeror
wrote: Wes, As a starter, look at this site: http://www.cbtricks.com/~ab7if/coax/coax.htm snip My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself. Dan Well, Dan, responding to your second claim first, consider either of two situations, 1) center-fed dipole, fed with open wire line, or 2) a center fed dipole, fed with coax with an efficient choke balun. In either of these situations an impedance mismatch between the feedline and the dipole will NOT cause radiation from the feedline. And taking a look at the web site you referenced above, the writer is professing to clear some misconceptions concerning transmission line technique. However, although he does present some straight dope, he is also further spreading some misconceptions concerning SWR. As he stated, there is a lot of misunderstanding concerning the effect of line length on the amplitude of the standing wave, but he continued the incorrect information on the subject, rather than presenting a correction. The fact is, Dan, that with lossless line the SWR is NOT affected by the line length--it remains constant along the entire length of the line. And further, the ONLY affect of line length on SWR is line attenuation, which causes the SWR at the input to be less than that at the load. If you believe everything you read in that reference, you've been duped. Walt, W2DU |
dansawyeror wrote:
My first claim is a tuner at the source does not materially improve what is happening in the coax. That is a tuner does not recreate the condition above where the coax is functioning as a properly matched and terminated transmission line. All the tuner does is match the impedance at the coax source back to some known, usually 50 Ohm, value. No matter what the voltages and currents are, if they are balanced, the transmission line won't radiate (much). If the SWR is 100:1 and the currents are balanced, the transmission won't radiate (much). If the SWR is 1:1 and the currents are unbalanced, the feedline is likely to radiate. My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself. Please understand it is not impedance mismatches that cause radiation from the feedline. It is unbalance in the feedline currents that causes feedline radiation. Current imbalance and impedance mismatches are not necessarily related. Current imbalance in a matched system can cause feedline radiation. Impedance mismatches can exist with negligible feedline radiation. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
On Sun, 28 Aug 2005 09:38:52 -0700, dansawyeror
wrote: Wes, As a starter, look at this site: http://www.cbtricks.com/~ab7if/coax/coax.htm When a transmission line is terminated in it's characteristic impedance there is no voltage or current reflection from the line. The electromagnetic fields continue to flow into the termination as if the line were infinitely long. When a mismatch of impedance occurs, reflected waves will be produced and they will interact with the incident waves. The total voltage and current on the line are no longer the result of a single traveling wave from the source to the load. Instead, it is the algebraic sum of two waves traveling in opposite directions. This interaction results in what is known as standing waves. The waves remain in fixed positions along the line while they vary in amplitude and polarity. A wave of any shape can be transmitted along the line without any change of waveshape or magnitude. Looking at the gif below, we see a line driven with a sine wave generator, terminated with a short circuit to maximize the reflection. I am certainly not the sharpest guy in this forum, however, I have been a ham for almost 47 years and I've been working with antennas from the beginning. I retired after 33+ years in the aerospace business where a good deal of my work involved rf design, measurements, failure analysis and the writing of specifications for rf/microwave devices and assemblies. I regret having to even bring this up, but it seems that you're willing accept something written on a CB radio site as fact while ignoring anything you hear in this forum from professionals in the field. Why is that? My first claim is a tuner at the source does not materially improve what is happening in the coax. Please show me the place where I claimed anything different. That is a tuner does not recreate the condition above where the coax is functioning as a properly matched and terminated transmission line. All the tuner does is match the impedance at the coax source back to some known, usually 50 Ohm, value. Please show me the place where I claimed anything different. My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself. This I've already proved to be wrong by example. Let me try one last time. Since antennas are reciprocal, you don't even need a transmitter. Connect any length of coax to your receiver input. Terminate the far end with a dummy load (50 ohm resistor). Tune around the bands. What do you hear? Nothing, if the coax is any good and the receiver and dummy load are well shielded. So I suppose this proves one of your claims, a matched line doesn't radiate. Remove the dummy load and replace it with a short circuit. What do you hear now on this totally mismatched line that by your reckoning should radiate like crazy? |
Frank,
My e-mail address is above. Thank you for all your help. I will try this. Dan Frank wrote: Frank, Thank you. Is there any way you can forward the saved parameters. This is a screwdriver antenna, I will remeasure the coil and double check. My modeling of the free space antenna showed about 4 Ohms but it was with a much simpler program. It was that program I used to measure Q. Thanks, Dan Hi Dan, I have run the program to determine the precise resonance. The parameters are as follows: Inductor 89.3 uH, and resonant at 3.92 MHz. I can send you a zipped NEC output text file. It is about 190 kB. Also the NEC code I used. You can plug in the appropriate data into an Excel spread sheet. If you need any specific graphical output I can it as a JPEG file. I used the default input source of 1 V peak, which accounts for the low power values in the output file. The E-field data is far field, normalized to 1 meter. The inductor is described as a lumped element, complex impedance, of 4.9 + j2200 ohms. I arrived at this value based on your Q of 450, and just played around with the imaginary value to achieve resonance within the 75 meter band. Let me know if I can send the above information to the address shown in your posting. Regards, Frank |
Somone wrote:
"My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself." Responses already show this is untrue. Radiation from the external coax surface comes from launching a signal on that surface. Good coax does not let signals penetrate its shield. A mismatch between a transmission line and its attached antenna affects both transmitting and receiving from the antenna, but does not launch signals on the outside of the coax. A mismatched transmitting antenna does not accept all available power incident upon it and reflects a portion back toward the sender depending on how bad the mismatch is. A mismatched receiving antenna has a source resistance (radiation resistance) and may also have reactance. A conjugate match is needed for maximum power transfer to the feedline. The mismatched antenna will either not extract all the power available to it in the passing wave or else reradiate more than 50%, (with full extraction, the minimum possible reradiation is with a perfectly matched antenna). Consider a short circuit across the antenna feedpoint. 100% of the energy extracted by the antenna is reradiated. Consider an open circuit at the antenna feedpoint. Little if any power is extracted from the wave sweeping the receiving antenna. The most power is received by a receiving antenna when its radiation resistance is matched to the Zo of the feedline. In this case, 50% is the best possible received carrier power in the receiver input. Nobody tells the antenna it is a receiving antenna. It is a conductor carrying a current, never mind where it came from, so it is going to radiate. When matched resistances are involved in source (radiation resistance) and load (Zo matched), the power is split 50-50 between source and load. The radiation resistance, is the source resistance for the receiver load, and it represents the reradiation from the reeiving atenna. 50% of the received power accepted by the load is the best possible performance. Mismatch means less. Either less power accepted by the antenna or more power reradiated by by the antenna. A transmatch can make the feedline appear as a matching load at the antenna junction for receiving. If matched for both transmitting and receiving, all available power will be transmitted and received. Best regards, Richard Harrison, KB5WZI |
Richard,
Simply create a model of a coax feed to a poorly tuned dipole. It is quite easy to create a situation where a significant portion or the power is radiated from the coax feedline. It is really quite simple: Coax is designed as an unbalanced transmission line from a source of a known impedance (nominally 50 Ohms for most common coax) to a similar known destination impedance. Under those circumstances there in trivial leakage. However in the case where the 'antenna' at the other end is poorly matched it is possible to have significant radiation from the coax. Again look at a poorly tuned dipole. The whole idea of coax is to use it the way it was intended. Then it works well. Dan Richard Harrison wrote: Somone wrote: "My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself." Responses already show this is untrue. Radiation from the external coax surface comes from launching a signal on that surface. Good coax does not let signals penetrate its shield. A mismatch between a transmission line and its attached antenna affects both transmitting and receiving from the antenna, but does not launch signals on the outside of the coax. A mismatched transmitting antenna does not accept all available power incident upon it and reflects a portion back toward the sender depending on how bad the mismatch is. A mismatched receiving antenna has a source resistance (radiation resistance) and may also have reactance. A conjugate match is needed for maximum power transfer to the feedline. The mismatched antenna will either not extract all the power available to it in the passing wave or else reradiate more than 50%, (with full extraction, the minimum possible reradiation is with a perfectly matched antenna). Consider a short circuit across the antenna feedpoint. 100% of the energy extracted by the antenna is reradiated. Consider an open circuit at the antenna feedpoint. Little if any power is extracted from the wave sweeping the receiving antenna. The most power is received by a receiving antenna when its radiation resistance is matched to the Zo of the feedline. In this case, 50% is the best possible received carrier power in the receiver input. Nobody tells the antenna it is a receiving antenna. It is a conductor carrying a current, never mind where it came from, so it is going to radiate. When matched resistances are involved in source (radiation resistance) and load (Zo matched), the power is split 50-50 between source and load. The radiation resistance, is the source resistance for the receiver load, and it represents the reradiation from the reeiving atenna. 50% of the received power accepted by the load is the best possible performance. Mismatch means less. Either less power accepted by the antenna or more power reradiated by by the antenna. A transmatch can make the feedline appear as a matching load at the antenna junction for receiving. If matched for both transmitting and receiving, all available power will be transmitted and received. Best regards, Richard Harrison, KB5WZI |
Cecil,
To this point I do not know where the radiation comes from. I suspect it is from the antenna. The easiest case is to model a poorly balanced dipole directly feed by coax. It radiates significant power back down the shield. After the response of the majority of posters to this concept I plan to find out more. Dan Cecil Moore wrote: dansawyeror wrote: My first claim is a tuner at the source does not materially improve what is happening in the coax. That is a tuner does not recreate the condition above where the coax is functioning as a properly matched and terminated transmission line. All the tuner does is match the impedance at the coax source back to some known, usually 50 Ohm, value. No matter what the voltages and currents are, if they are balanced, the transmission line won't radiate (much). If the SWR is 100:1 and the currents are balanced, the transmission won't radiate (much). If the SWR is 1:1 and the currents are unbalanced, the feedline is likely to radiate. My second claim is when the mismatch condition at the coax destination, i.e. antenna that may result in significant radiation from the coax itself. Please understand it is not impedance mismatches that cause radiation from the feedline. It is unbalance in the feedline currents that causes feedline radiation. Current imbalance and impedance mismatches are not necessarily related. Current imbalance in a matched system can cause feedline radiation. Impedance mismatches can exist with negligible feedline radiation. |
Richard,
Thank you for this well constructed reply. This mismatch could arise for any number of reasons, and not all contribute to radiation from the coax. Wes has already demonstrated a deliberate mismatch at the end of a cable that exhibits absolutely no radiation from the coax. This is because he has contrived to contain the fields from emerging and coupling to the outside of the coax shield. You should be aware that the shield does support currents on the inside and outside that are wholly unaffected by each other - except at the drive point where the two conduction paths are joined. When you drive a dipole with a coax, the exterior conductive path of the shield (a separate circuit from the interior conductive path of that same shield) is in parallel with one arm of the dipole. This means you have a third radiator that has a length and termination that is undefined. In the case where a mistuned dipole is being driven directly from coax there is radiation from the coax feed. This can only happen from current in the shield. Is this what you are referring to in the second paragraph? Thanks, Dan |
On Mon, 29 Aug 2005 21:12:58 -0700, dansawyeror
wrote: When you drive a dipole with a coax, the exterior conductive path of the shield (a separate circuit from the interior conductive path of that same shield) is in parallel with one arm of the dipole. This means you have a third radiator that has a length and termination that is undefined. In the case where a mistuned dipole is being driven directly from coax there is radiation from the coax feed. This can only happen from current in the shield. Is this what you are referring to in the second paragraph? Hi Dan, Hmmm, The dipole is mistuned by the third conductor, the coax's shield's exterior; otherwise, the dipole would be suitably matched (this is the presumption, of course). The source of the current on the coax's shield's exterior comes from the excitation voltage seen across the dipole drive point (to which the shield is common to one of the arms). The arm of the dipole that is not attached to the shield, sees both its opposite arm, and the undefined length of the shield's exterior path. This additional load both unbalances, and mismatches. It is the unbalance that gives rise to the Common Mode current, the mismatch simply comes for free. Of course, you could fall into the condition where the dipole would not normally be tuned, but through luck and happenstance, the addition of the third radiator creates a match - this is strictly opportunistic and sometimes the source for glowing reports of an otherwise horrible antenna design. And this is the genesis of favorable accolades for many of the mythic antennas that go by initials: CFA, EH, and so on down the line. The "inventors" have simply contrived to tune the driveline to their "inventions." Their aversion to discussing driveline isolation is a hallmark of their "science." Their insistence that choking the driveline is unnecessary or an impediment to the design's utility, is further evidence of a generous thumb on the scale of proof. The addition of the choke gives its Z to snub this Common Mode current. As both interior paths (that of the line's center wire, and the interior of the shield) driving the dipole pass through the same loops, their magnetic fields are unperturbed and see no additional impedance. However, the "return" path of the shield exterior sees these loops alone, and thus the Z is inserted into series with it. If you think in terms of the W2DU style BalUn, the interior current/magnetic lines both transit THROUGH the beads, whereas the exterior shield current/magnetic lines CUT the beads - hence the choking action is more apparent in this configuration. 73's Richard Clark, KB7QHC |
Dan wrote:
"Simply create a model of a coax feed to a poorly tuned dipole." Dan has something there. The "ARRL Antenna Book" has an explanation of "Commonn-Mode Transmission Line Currents" on page 26-16 in my 19th edition. Launch of coax radiation is shown in Fig 24 on the same page. This picture may be worth 1000 words. Best regards, Richard Harrison, KB5WZI |
dansawyeror wrote:
To this point I do not know where the radiation comes from. I suspect it is from the antenna. The easiest case is to model a poorly balanced dipole directly feed by coax. It radiates significant power back down the shield. Actually, a perfectly balanced dipole can also radiate from the coax shield because coax is unbalanced. Take a look at www.w2du.com and www.eznec.com for information on baluns and chokes which tend to reduce feedline radiation. It appears to me that you may be confusing "balance" with "matching". Those words have very different definitions. A balanced 50 ohm antenna fed with coax and having an SWR of 1:1 is matched but unbalanced and is likely to radiate from the feedline. A balanced 50 ohm antenna fed with 600 ohm balanced line and having an SWR of 12:1 is unmatched but balanced and is not likely to radiate much from the feedline. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
dansawyeror wrote:
Simply create a model of a coax feed to a poorly tuned dipole. Please define, "poorly tuned", "mismatched", and "unbalanced". -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
On Mon, 29 Aug 2005 20:55:08 -0700, dansawyeror
wrote: I think I've got this guy figured out; it's Fractenna come back to haunt us. |
On Tue, 30 Aug 2005 07:13:14 -0700, Wes Stewart
wrote: On Mon, 29 Aug 2005 20:55:08 -0700, dansawyeror wrote: I think I've got this guy figured out; it's Fractenna come back to haunt us. No Wes, I think it's a new guy getting his kicks from reading that new technical best seller, "Good Buddy for Dummies." |
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