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An antenna question--43 ft vertical
In message , Wayne
writes As a young man I was given a problem of solving poor antenna performance on an aircraft band fixed station antenna. The SWR at the transmitter was close to 1:1, but the antenna didn't work well. I climbed up on the tower and found that the coax had never been connected to the antenna. That was with about 400 feet of coax at 120 MHz. A length of coax, which has (say) at least 10dB loss at the frequency of interest, can indeed make a superb SWR dummy load! -- Ian |
An antenna question--43 ft vertical
On 7/2/2015 3:24 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message ... On 7/2/2015 1:56 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... You are correct in that if a 75 ohm bridge is used, the indicated SWR would be 1:1, because everything from that point on is 75 ohms. However, the mismatch (and reflection) occurs on the transmitter side of the bridge, not the antenna side. So the bridge will never see it. But an accurate bridge will show lower power output due to the mismatch. A mismatch is a mismatch, no matter where in the system it occurs. And any mismatch will cause less than 100% power to be transferred. The rest is reflected. Just look at the specs of any amateur transceiver. They show an impedance of 50 ohms. So a load of 50 ohms provides for maximum power transfer; any other impedance causes a mismatch. -- The real impedance of the transmitter is not 50 ohms. It is whatever the device is used in the final stage and the poewr level. For a 100 watt transmitter it is in the thousand ohm range and for solid state devices it is very low. The matching circuit is often fixed to be 50 ohms,but could be made for most any impedance. The older tube circuits were adjustable by the user for a range of somewhat bleow 50 ohms to around 200 ohms. Could be more or less depending on the design. Incorrect. The real impedance of the transmitter is 50 ohms. The impedance of the final stage may be above or below that, and is matched to the 50 ohm standard. What happens before the match is unimportant. The only important part of the discussion is the 50 ohm output. The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the transmitter. Whatever power comes out of the transmitter will make it to the antenna minus the loss of the coax, but not additional loss due to swr. The power comming out of a 50 ohm transmitter will be less due to mismatch, but not because of swr of the antenna system which is 1:1. Incorrect. The connection between the 50 ohm transmitter and the 75 ohm coax is also part of the antenna system. The system starts at the transmitter output (actually the output of the final stage - but since this is converted to the 50 ohm standard, you can effectively consider the output of the matching network to be the start of the antenna system), not the coax. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
Wayne wrote:
wrote in message ... Wayne wrote: snip In my own particular case, an automatic remotely tuned ATU would be a pain to install/maintain. This part I do not understand at all. At the antenna end is a box with a connector for the feed line and a connector for the antenna. There is nothing to maintain there. If you get an ATU that gets it's power through the coax, you put the power injector in line with the feed line in the shack. There is nothing to maintain there either and you do not need to run any extra wires out to the antenna. The problem is with my own particular case. The antenna is a whip mounted in the middle of a metal roof. At my age, I shouldn't be wandering around on or climbing such a roof. Once installed, any failure would require a trip to the roof. The ATU would be exposed to extreme temperature and sunlight that might eventually induce failures. The age part I can understand; an inverted, cheap plastic trash can will provide more than adequate protection against the elements. -- Jim Pennino |
An antenna question--43 ft vertical
"Jerry Stuckle" wrote in message ... The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the transmitter. Whatever power comes out of the transmitter will make it to the antenna minus the loss of the coax, but not additional loss due to swr. The power comming out of a 50 ohm transmitter will be less due to mismatch, but not because of swr of the antenna system which is 1:1. Incorrect. The connection between the 50 ohm transmitter and the 75 ohm coax is also part of the antenna system. The system starts at the transmitter output (actually the output of the final stage - but since this is converted to the 50 ohm standard, you can effectively consider the output of the matching network to be the start of the antenna system), not the coax. -- So with my Icom 746 with a built in tuner, where does the system start ? Within a small range 3 or 4 to 1, the internal SWR meter will show a 1:1 match, an external SWR meter will show a differant SWR if there is a mismatch. If I hook up a good 75 ohm load and 75 ohm coax I will have no SWR by definition . A 50 ohm bridge will show 1.5:1 but the internal tuner/bridge will show 1:1. Which SWR meter is correct ? |
An antenna question--43 ft vertical
On 7/2/2015 3:17 PM, Wayne wrote:
wrote in message ... Wayne wrote: snip In my own particular case, an automatic remotely tuned ATU would be a pain to install/maintain. This part I do not understand at all. At the antenna end is a box with a connector for the feed line and a connector for the antenna. There is nothing to maintain there. If you get an ATU that gets it's power through the coax, you put the power injector in line with the feed line in the shack. There is nothing to maintain there either and you do not need to run any extra wires out to the antenna. The problem is with my own particular case. The antenna is a whip mounted in the middle of a metal roof. At my age, I shouldn't be wandering around on or climbing such a roof. Once installed, any failure would require a trip to the roof. The ATU would be exposed to extreme temperature and sunlight that might eventually induce failures. A friend of mine who is past 70 has had a TV antenna preamp on his roof for some 50 years. He has been up there to check it at least once when he couldn't get the digital TV signals as well any more. It was not the problem. If a unit is constructed well, it should live a rich, full life on the roof, protected from the abuse it might receive in your shack. -- Rick |
An antenna question--43 ft vertical
On 7/2/2015 3:52 PM, Wayne wrote:
"rickman" wrote in message ... On 7/2/2015 12:18 PM, Wayne wrote: "John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) # Transferred where? The match at the transmitter output only matches the # output to the line. There are still reflections from the mismatch at # the antenna. These reflections result in extra losses in the line as # well as power delivered back into the transmitter output stage # (especially with a perfect impedance match). Well, I put a few (unrealistic) qualifiers into my question: a transmitter with a a 1063 ohm output (not 50), and a lossless RG-8. Thus, the back and forth reflections would not have attenuation. And the transmitter and load are conjugately matched for maximum power transfer. Your quoting style is very confusing. If you use with a space at the front of lines you are quoting it will show up the same as everyone else's quotes. Why will the reflections not have losses? Every load that is not an infinite impedance will absorb some of the signal that would be reflected. That applies to the transmitter output as well as the antenna, no? A matched impedance does not mean no losses. It means the maximum transfer of power. These are not at all the same thing. # But I don't see anyone taking wavelength vs. feed line length into # account. If the wavelength is long compared to the feed line I believe # a lot of the "bad" stuff goes away. But then I am used to the digital # transmission line where we aren't really concerned with delivering # power, rather keeping a clean waveform of our (relatively) square waves. # So I guess a short feed line doesn't solve the SWR problems... or does # it? The attenuation at a given high SWR depends upon the the matched feedline loss, as reflections encounter that loss with every forward or backward trip. Thus feedline length/attenuation should be considered. As a young man I was given a problem of solving poor antenna performance on an aircraft band fixed station antenna. The SWR at the transmitter was close to 1:1, but the antenna didn't work well. I climbed up on the tower and found that the coax had never been connected to the antenna. That was with about 400 feet of coax at 120 MHz. So how was the SWR 1:1? -- Rick |
An antenna question--43 ft vertical
In message , rickman
writes On 7/2/2015 3:52 PM, Wayne wrote: "rickman" wrote in message ... On 7/2/2015 12:18 PM, Wayne wrote: "John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) # Transferred where? The match at the transmitter output only matches the # output to the line. There are still reflections from the mismatch at # the antenna. These reflections result in extra losses in the line as # well as power delivered back into the transmitter output stage # (especially with a perfect impedance match). Well, I put a few (unrealistic) qualifiers into my question: a transmitter with a a 1063 ohm output (not 50), and a lossless RG-8. Thus, the back and forth reflections would not have attenuation. And the transmitter and load are conjugately matched for maximum power transfer. Your quoting style is very confusing. If you use with a space at the front of lines you are quoting it will show up the same as everyone else's quotes. Why will the reflections not have losses? Every load that is not an infinite impedance will absorb some of the signal that would be reflected. That applies to the transmitter output as well as the antenna, no? A matched impedance does not mean no losses. It means the maximum transfer of power. These are not at all the same thing. # But I don't see anyone taking wavelength vs. feed line length into # account. If the wavelength is long compared to the feed line I believe # a lot of the "bad" stuff goes away. But then I am used to the digital # transmission line where we aren't really concerned with delivering # power, rather keeping a clean waveform of our (relatively) square waves. # So I guess a short feed line doesn't solve the SWR problems... or does # it? The attenuation at a given high SWR depends upon the the matched feedline loss, as reflections encounter that loss with every forward or backward trip. Thus feedline length/attenuation should be considered. As a young man I was given a problem of solving poor antenna performance on an aircraft band fixed station antenna. The SWR at the transmitter was close to 1:1, but the antenna didn't work well. I climbed up on the tower and found that the coax had never been connected to the antenna. That was with about 400 feet of coax at 120 MHz. So how was the SWR 1:1? It's probably the go-and-return loss of (say) 20dB's worth of coax (at 120MHz), with the far end open (or short) circuit. That would give you an RLR of 40dB (an SWR of 1.02), which probably far exceeds the capability of an SWR meter to read. http://bit.ly/1T8TwcF http://cgi.www.telestrian.co.uk/cgi-bin/www.telestrian.co.uk/vswr.pl -- Ian |
An antenna question--43 ft vertical
On 7/2/2015 6:08 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message ... The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the transmitter. Whatever power comes out of the transmitter will make it to the antenna minus the loss of the coax, but not additional loss due to swr. The power comming out of a 50 ohm transmitter will be less due to mismatch, but not because of swr of the antenna system which is 1:1. Incorrect. The connection between the 50 ohm transmitter and the 75 ohm coax is also part of the antenna system. The system starts at the transmitter output (actually the output of the final stage - but since this is converted to the 50 ohm standard, you can effectively consider the output of the matching network to be the start of the antenna system), not the coax. -- So with my Icom 746 with a built in tuner, where does the system start ? Within a small range 3 or 4 to 1, the internal SWR meter will show a 1:1 match, an external SWR meter will show a differant SWR if there is a mismatch. If I hook up a good 75 ohm load and 75 ohm coax I will have no SWR by definition . A 50 ohm bridge will show 1.5:1 but the internal tuner/bridge will show 1:1. Which SWR meter is correct ? A built in tuner is an entirely different story. The output from the transmitter is not necessarily 50 homes at this point. It's a shame they've dumbed down the exams so much that you don't need to know anything to hold a ticket any more. Back when I got mine, even the General Class was tougher than the Extra Class today - and it was administered by the FCC, with no public question pool or cheat sheets. You had to actually know something other than just memorizing a few answers. And at the time, the Amateur Extra was harder than the First Class Radiotelephone. Nowadays, an 8-year-old can pass the Amateur Extra with virtually no knowledge of math or electronics at all - just the ability to memorize a few answers. Learn something about electronics, AC power transfer and other theory. Than maybe we can converse with a modicum of intelligence. As it is, I'm tired of trying to teach an idiot who refuses to learn. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On Thu, 02 Jul 2015 14:55:53 -0400, Jerry Stuckle
wrote: It's easy enough to demonstrate that you're wrong. 1. Setup your favorite HF xmitter and attach a Tee connector to the antenna connector, your favorite VSWR meter, a length of 50 ohm coax, and a 50 ohm load. 2. Transmit and convince yourself that the VSWR is 1:1. Make sure the transmitter is not into ALC. 3. Now, take another 50 ohm dummy load and connect it to the Tee connector. The transmitter now sees 25 ohms, so the PA stage has half the normal gain. You may need to increase the drive level to obtain the same RF power as before. 4. Measure the VSWR again. It should also be 1:1. Looking back towards the transmitter, the 50 ohm coax cable sees a 2:1 mismatch of 25 ohms (the alleged 50 ohms from radio in parallel with another 50 ohms from the extra dummy load). I know this part works because I've demonstrated it twice to the local non-believers. Now, we go into uncharted territory and do it again with a 75 ohm coax and a 75 ohm dummy load at the far end (antenna end) of the coax. Same procedure. 1. Check the VSWR and it should be 1.5:1. 2. Connect a 50 ohm dummy load to the Tee connector, and measure the VSWR again. It should still be 1.5:1. Looking back towards the transmitter, the 75 ohm coax cable sees the same 2:1 mismatch of 25 ohms. If you want to go further, I think it can be demonstrated that almost any number of extra dummy loads at the Tee connector will still produce the same 1.5:1 VSWR. I'll try it on the bench, but I have other plans for the holiday weekend. If I find time, and manage to get all the junk off my workbench, I'll give try it. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
On 7/2/2015 6:53 PM, Ian Jackson wrote:
In message , rickman writes On 7/2/2015 3:52 PM, Wayne wrote: The attenuation at a given high SWR depends upon the the matched feedline loss, as reflections encounter that loss with every forward or backward trip. Thus feedline length/attenuation should be considered. As a young man I was given a problem of solving poor antenna performance on an aircraft band fixed station antenna. The SWR at the transmitter was close to 1:1, but the antenna didn't work well. I climbed up on the tower and found that the coax had never been connected to the antenna. That was with about 400 feet of coax at 120 MHz. So how was the SWR 1:1? It's probably the go-and-return loss of (say) 20dB's worth of coax (at 120MHz), with the far end open (or short) circuit. That would give you an RLR of 40dB (an SWR of 1.02), which probably far exceeds the capability of an SWR meter to read. http://bit.ly/1T8TwcF http://cgi.www.telestrian.co.uk/cgi-bin/www.telestrian.co.uk/vswr.pl I like the comment that the antenna "didn't work well". Lol. I wonder how much better it worked when connected. The feed line would still have a lot of loss one way. I wonder why the power amp wasn't closer to the antenna. Reminds me of a time I was working a job installing TV antennas and one was up the side of the ridge near here. In the "old days" we would unplug the TV to prevent shocks from a "hot" chassis set. This installation also had a wall jack for the antenna connection. The guy on the roof told me to plug the TV in and check the signal while the turned the antenna. The image was a little snowy, but not bad. No matter how they turned the antenna the image didn't get much better. I looked behind the set and saw the three foot twin lead from the set laying on the floor. I plugged it in and the picture was *perfect*! I was amazed a short piece of lead in could receive as good a signal. -- Rick |
An antenna question--43 ft vertical
On 7/2/2015 11:18 AM, Wayne wrote:
"John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? No. As I said, one watt is applied at the source end. This is condition defined by the example and has nothing to do with source matching. The 32% loss is due to transmission line loss. The mismatch at the load end causes the high SWR which increases the line loss due to high current at some point in the line as well as increased voltage at other point(s). The impedance of the 1063+J0 load is transformed to 54+J192 ohms at the source. However, at an electrical quarter wave away from the antenna, the impedance is about 2.8+j0 ohms. So that point is a relative hot spot in the line. If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) With a lossless transmission line and one watt applied to one end the other end will have one watt available. The only place the power can go is into the antenna. To put one watt into 1063 ohms will require .0306 amps and 32.6V at the feed point. Does this make sense? |
An antenna question--43 ft vertical
"Jerry Stuckle" wrote in message ... On 7/2/2015 6:08 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... It's a shame they've dumbed down the exams so much that you don't need to know anything to hold a ticket any more. Back when I got mine, even the General Class was tougher than the Extra Class today - and it was administered by the FCC, with no public question pool or cheat sheets. You had to actually know something other than just memorizing a few answers. And at the time, the Amateur Extra was harder than the First Class Radiotelephone. I had my First Phone at the age of 22 back in 1972. Passed the 2nd and first the same day on the first try. And back at you on trying to teach someone that will not learn. Seems that several on here think you are wrong most of the time. |
An antenna question--43 ft vertical
"rickman" wrote in message ... On 7/2/2015 3:52 PM, Wayne wrote: "rickman" wrote in message ... On 7/2/2015 12:18 PM, Wayne wrote: "John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) # Transferred where? The match at the transmitter output only matches the # output to the line. There are still reflections from the mismatch at # the antenna. These reflections result in extra losses in the line as # well as power delivered back into the transmitter output stage # (especially with a perfect impedance match). Well, I put a few (unrealistic) qualifiers into my question: a transmitter with a a 1063 ohm output (not 50), and a lossless RG-8. Thus, the back and forth reflections would not have attenuation. And the transmitter and load are conjugately matched for maximum power transfer. # Your quoting style is very confusing. If you use with a space at the # front of lines you are quoting it will show up the same as everyone # else's quotes. It's a problem with my newsreader not doing the proper job. #Why will the reflections not have losses? Because the assumption I posed was for a lossless line. In that case, with a conjugate match on both ends, wouldn't there be maximum power transmission regardless of the SWR? ......just a question I'm posing to the group. With no line losses, and a conjugate match, is the SWR of any consequence? A matched impedance does not mean no losses. It means the maximum transfer of power. These are not at all the same thing. # But I don't see anyone taking wavelength vs. feed line length into # account. If the wavelength is long compared to the feed line I believe # a lot of the "bad" stuff goes away. But then I am used to the digital # transmission line where we aren't really concerned with delivering # power, rather keeping a clean waveform of our (relatively) square waves. # So I guess a short feed line doesn't solve the SWR problems... or does # it? The attenuation at a given high SWR depends upon the the matched feedline loss, as reflections encounter that loss with every forward or backward trip. Thus feedline length/attenuation should be considered. As a young man I was given a problem of solving poor antenna performance on an aircraft band fixed station antenna. The SWR at the transmitter was close to 1:1, but the antenna didn't work well. I climbed up on the tower and found that the coax had never been connected to the antenna. That was with about 400 feet of coax at 120 MHz. # So how was the SWR 1:1? It was a long run of coax at 120 MHz. The reflected wave was was attenuated considerably by the time it returned to the source. |
An antenna question--43 ft vertical
On 7/2/2015 8:53 PM, Wayne wrote:
"rickman" wrote in message ... On 7/2/2015 3:52 PM, Wayne wrote: "rickman" wrote in message ... On 7/2/2015 12:18 PM, Wayne wrote: "John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) # Transferred where? The match at the transmitter output only matches the # output to the line. There are still reflections from the mismatch at # the antenna. These reflections result in extra losses in the line as # well as power delivered back into the transmitter output stage # (especially with a perfect impedance match). Well, I put a few (unrealistic) qualifiers into my question: a transmitter with a a 1063 ohm output (not 50), and a lossless RG-8. Thus, the back and forth reflections would not have attenuation. And the transmitter and load are conjugately matched for maximum power transfer. # Your quoting style is very confusing. If you use with a space at the # front of lines you are quoting it will show up the same as everyone # else's quotes. It's a problem with my newsreader not doing the proper job. #Why will the reflections not have losses? Because the assumption I posed was for a lossless line. In that case, with a conjugate match on both ends, wouldn't there be maximum power transmission regardless of the SWR? You aren't grasping the issue. Losses are *not* only in the transmission line. When a reflected wave returns to the transmitter output, it is not reflected 100%. If the output and transmission line are matched exactly, 50% of the reflected wave reaching the output will be reflected and 50% will be dissipated in the output stage. Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? -- Rick |
An antenna question--43 ft vertical
On 7/2/2015 7:37 PM, John S wrote:
On 7/2/2015 11:18 AM, Wayne wrote: "John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? No. As I said, one watt is applied at the source end. This is condition defined by the example and has nothing to do with source matching. The 32% loss is due to transmission line loss. The mismatch at the load end causes the high SWR which increases the line loss due to high current at some point in the line as well as increased voltage at other point(s). The impedance of the 1063+J0 load is transformed to 54+J192 ohms at the source. However, at an electrical quarter wave away from the antenna, the impedance is about 2.8+j0 ohms. So that point is a relative hot spot in the line. Correction: 0.5+j0 at 1/4 wavelength away from the antenna. If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) With a lossless transmission line and one watt applied to one end the other end will have one watt available. The only place the power can go is into the antenna. To put one watt into 1063 ohms will require .0306 amps and 32.6V at the feed point. Does this make sense? |
An antenna question--43 ft vertical
On 7/2/2015 8:51 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message ... On 7/2/2015 6:08 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... It's a shame they've dumbed down the exams so much that you don't need to know anything to hold a ticket any more. Back when I got mine, even the General Class was tougher than the Extra Class today - and it was administered by the FCC, with no public question pool or cheat sheets. You had to actually know something other than just memorizing a few answers. And at the time, the Amateur Extra was harder than the First Class Radiotelephone. I had my First Phone at the age of 22 back in 1972. Passed the 2nd and first the same day on the first try. And back at you on trying to teach someone that will not learn. Seems that several on here think you are wrong most of the time. I've got you there. First phone in 1970 at age 18 - before I started college. Amateur I also passed Second and First the same day. My Amateur Extra came 9 months later, but only because I had to hold a General for two years before I could take the Extra exam. Did you ever use your first phone? I was an engineer for one broadcast station and chief engineer for another. I also repaired everything from $40 cb sets to multi-million dollar mainframe computers. And even did some digital design work back on the 70's. But it's quite obvious from your updates that you have no idea what you're talking about. You're an appliance operator with no understanding of what's going on underneath the covers. And yes, I know there are some idiotic trolls here who don't think I know what I'm talking about. My EE professors would disagree with them. -- ================== Remove the "x" from my email address Jerry Stuckle ================== |
An antenna question--43 ft vertical
On 7/2/2015 7:56 PM, Jeff Liebermann wrote:
On Thu, 02 Jul 2015 14:55:53 -0400, Jerry Stuckle wrote: It's easy enough to demonstrate that you're wrong. 1. Setup your favorite HF xmitter and attach a Tee connector to the antenna connector, your favorite VSWR meter, a length of 50 ohm coax, and a 50 ohm load. 2. Transmit and convince yourself that the VSWR is 1:1. Make sure the transmitter is not into ALC. 3. Now, take another 50 ohm dummy load and connect it to the Tee connector. The transmitter now sees 25 ohms, so the PA stage has half the normal gain. You may need to increase the drive level to obtain the same RF power as before. 4. Measure the VSWR again. It should also be 1:1. Of course it will be. You have 50 ohms on one end. But you're not measuring what the TRANSMITTER sees. The fact that you may have to increase drive level indicates the circumstances have changed. Looking back towards the transmitter, the 50 ohm coax cable sees a 2:1 mismatch of 25 ohms (the alleged 50 ohms from radio in parallel with another 50 ohms from the extra dummy load). I know this part works because I've demonstrated it twice to the local non-believers. Sure. But you don't have power going from one resistor to the transmitter and the other resister, so your measurement is meaningless - and you are as full of crap as you normally are. Stick your VSWR meter between the power source (the transmitter) and BOTH loads (i.e. before the T). You will see a 2:1 SWR. Now, we go into uncharted territory and do it again with a 75 ohm coax and a 75 ohm dummy load at the far end (antenna end) of the coax. Same procedure. 1. Check the VSWR and it should be 1.5:1. 2. Connect a 50 ohm dummy load to the Tee connector, and measure the VSWR again. It should still be 1.5:1. Looking back towards the transmitter, the 75 ohm coax cable sees the same 2:1 mismatch of 25 ohms. If you want to go further, I think it can be demonstrated that almost any number of extra dummy loads at the Tee connector will still produce the same 1.5:1 VSWR. Again, the dummy load is not producing any power, so adding something to the other end of the T will have no effect. So once again your "test" is meaningless and you are full of crap. Now in this case if you connect the SWR bridge before the T, you will show a 1.5:1 with one 75 ohm load, and a 1.3:1 SWR with two 75 ohm loads (37.5 ohms). SWR measurements are only valid when the VSWR meter is connected between the power generator (transmitter) and the total load (one or both dummy loads). Connecting between one leg of the T and the load only shows VSWR for that leg - but not the entire system. I'll try it on the bench, but I have other plans for the holiday weekend. If I find time, and manage to get all the junk off my workbench, I'll give try it. Go ahead - continue to mae a fool of yourself. You're real good at it. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/3/2015 2:50 AM, Jeff wrote:
Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? Yes, it must. For example with an external ATU that provides a conjugate match it is clearly the case that if a 1:1 VSWR is achieved then no reflected power reaches the TX. (as shown on an SWR meter between the Tx and ATU.) I am very certain that this assumption is not correct. I wish I had the math to back me up. The only total reflection I am aware of is an open circuit which of course absorbs no power at all. Here is a point. The VSWR only shows no power being sent back to the txmt output. That does not mean no power is absorbed from the reflected wave by the matching circuit. I believe the example you gave was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent to a resistor? Resistors dissipate power don't they? -- Rick |
An antenna question--43 ft vertical
In message , rickman
writes On 7/3/2015 2:50 AM, Jeff wrote: Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? Yes, it must. For example with an external ATU that provides a conjugate match it is clearly the case that if a 1:1 VSWR is achieved then no reflected power reaches the TX. (as shown on an SWR meter between the Tx and ATU.) I am very certain that this assumption is not correct. I wish I had the math to back me up. The only total reflection I am aware of is an open circuit which of course absorbs no power at all. Here is a point. The VSWR only shows no power being sent back to the txmt output. That does not mean no power is absorbed from the reflected wave by the matching circuit. I believe the example you gave was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent to a resistor? Resistors dissipate power don't they? I have to admit that I am, to some extent, confused. Maybe it helps to look at the situation from the point of view that the matching circuit doesn't 'know' that there is a reflected wave. All it sees is the impedance looking into the sending end of the coax - and this is whatever is on the antenna end, transformed by the length of coax. The load the matching unit sees could be replaced with the same physical values of L, C and R, so there IS nowhere for a reflected wave to exist. Provided the TX sees a 50 ohm load when looking into the input of the matcher, there will be no theoretical losses. However, a real-life matcher WILL have loss, and so will the coax. Also, the coax will have a loss greater than when it is matched, mainly because of the 'I -squared R' (literal) hot-spots. -- Ian |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , rickman writes On 7/3/2015 2:50 AM, Jeff wrote: Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? Yes, it must. For example with an external ATU that provides a conjugate match it is clearly the case that if a 1:1 VSWR is achieved then no reflected power reaches the TX. (as shown on an SWR meter between the Tx and ATU.) I am very certain that this assumption is not correct. I wish I had the math to back me up. The only total reflection I am aware of is an open circuit which of course absorbs no power at all. Here is a point. The VSWR only shows no power being sent back to the txmt output. That does not mean no power is absorbed from the reflected wave by the matching circuit. I believe the example you gave was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent to a resistor? Resistors dissipate power don't they? I have to admit that I am, to some extent, confused. Maybe it helps to look at the situation from the point of view that the matching circuit doesn't 'know' that there is a reflected wave. All it sees is the impedance looking into the sending end of the coax - and this is whatever is on the antenna end, transformed by the length of coax. The load the matching unit sees could be replaced with the same physical values of L, C and R, so there IS nowhere for a reflected wave to exist. Provided the TX sees a 50 ohm load when looking into the input of the matcher, there will be no theoretical losses. However, a real-life matcher WILL have loss, and so will the coax. Also, the coax will have a loss greater than when it is matched, mainly because of the 'I -squared R' (literal) hot-spots. Surely it *is* the reflected wave that mediates the transformation of the aerial impedance to what is seen at the transmitter end? The transmitter sees the vector sum of all the waves traversing the transmission line at that point. Or else how would it "know" what was happening at the other end? -- Roger Hayter |
An antenna question--43 ft vertical
Jeff wrote:
On 03/07/2015 08:29, rickman wrote: On 7/3/2015 2:50 AM, Jeff wrote: Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? Yes, it must. For example with an external ATU that provides a conjugate match it is clearly the case that if a 1:1 VSWR is achieved then no reflected power reaches the TX. (as shown on an SWR meter between the Tx and ATU.) I am very certain that this assumption is not correct. I wish I had the math to back me up. The only total reflection I am aware of is an open circuit which of course absorbs no power at all. Here is a point. The VSWR only shows no power being sent back to the txmt output. That does not mean no power is absorbed from the reflected wave by the matching circuit. I believe the example you gave was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent to a resistor? Resistors dissipate power don't they? The point can be easily proved with a lossy feeder, the lossier the better. If your assumption is correct then the power delivered to the antenna would be the Tx power less the cable loss less the reflected power at the antenna mismatch, however it is the case that can be measured and seen on computer simulation that all of the power is delivered to the antenna except that which is dissipated in the cable loss (for the multiple reflections). Also it can been seen that with perfect components in the matching circuit no power is dissipated there. The 1063 ohms that you refer to is not resistive so with perfect Ls & Cs no power will be dissipated in it. In the real world the Cs and Ls in the matching unit will have some loss associated with them but that is a different story. Jeff While conjugate matching is the way to transfer the maximum power from a voltage (or current) generator to a load, it is not the way power amplifiers are set up. The transmitter normally does not present a match to signals from the aerial, hence the re-reflection. -- Roger Hayter |
An antenna question--43 ft vertical
In message , Roger Hayter
writes Ian Jackson wrote: In message , rickman writes On 7/3/2015 2:50 AM, Jeff wrote: Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? Yes, it must. For example with an external ATU that provides a conjugate match it is clearly the case that if a 1:1 VSWR is achieved then no reflected power reaches the TX. (as shown on an SWR meter between the Tx and ATU.) I am very certain that this assumption is not correct. I wish I had the math to back me up. The only total reflection I am aware of is an open circuit which of course absorbs no power at all. Here is a point. The VSWR only shows no power being sent back to the txmt output. That does not mean no power is absorbed from the reflected wave by the matching circuit. I believe the example you gave was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent to a resistor? Resistors dissipate power don't they? I have to admit that I am, to some extent, confused. Maybe it helps to look at the situation from the point of view that the matching circuit doesn't 'know' that there is a reflected wave. All it sees is the impedance looking into the sending end of the coax - and this is whatever is on the antenna end, transformed by the length of coax. The load the matching unit sees could be replaced with the same physical values of L, C and R, so there IS nowhere for a reflected wave to exist. Provided the TX sees a 50 ohm load when looking into the input of the matcher, there will be no theoretical losses. However, a real-life matcher WILL have loss, and so will the coax. Also, the coax will have a loss greater than when it is matched, mainly because of the 'I -squared R' (literal) hot-spots. Surely it *is* the reflected wave that mediates the transformation of the aerial impedance to what is seen at the transmitter end? The transmitter sees the vector sum of all the waves traversing the transmission line at that point. Or else how would it "know" what was happening at the other end? I guess that until reflections are received back from the far end of the coax, the transmitter will see the 50 ohms Zo (surge impedance) of the coax. But once things have settled worn, the transmitter neither knows nor cares what's at the far end. All it knows is that the load presented to it isn't what it ought to be. But insert a matcher, and it will be as happy as Larry. The system will work fine, but will suffer the penalty of the additional SWR losses on the coax, and those of the matcher. Provide these are not unacceptable, the benefit is that all the matching can be done in the comfort of shack. -- Ian |
An antenna question--43 ft vertical
On 6/29/2015 10:48 AM, Wayne wrote:
As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. |
An antenna question--43 ft vertical
In message , John S
writes On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. A fixed-tuned TX will still need a matcher. At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. A fixed-tuned TX will probably be reasonably happy with a direct connection - although maybe even happier with a series capacitor of -J22 ohms. There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. The question is really whether the losses with the 4:1 transformer, plus those of any matcher at the TX end, exceed those when there is no transformer (but with higher loss on the coax), plus a matcher. Put another way, for short feeder lengths, is it better to use the transformer? -- Ian |
An antenna question--43 ft vertical
In message , Ian Jackson
writes although maybe even happier with a series capacitor of -J22 ohms. Sorry - somebody obviously swapped the '2' and '3' keys. -- Ian |
An antenna question--43 ft vertical
"John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. Thanks John. Yes, we have strayed from the original question, but I have found the discussion stimulating. Perhaps a new thread should be started to address those subjects. If I use EZNEC to model the 43 footer over perfect ground with a 3 inch diameter radiator, I get impedances in the same ball park as you list. If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna would see as a feedline, if a 4:1 unun had 50 ohm coax on the other side), the SWR plot becomes interesting. The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR getting below 2.5:1 around 29 MHz. Is that a valid approach? |
An antenna question--43 ft vertical
On 7/3/2015 10:17 AM, Ian Jackson wrote:
In message , John S writes On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. A fixed-tuned TX will still need a matcher. That was not part of the original question(s). At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. A fixed-tuned TX will probably be reasonably happy with a direct connection - although maybe even happier with a series capacitor of -J22 ohms. That was not part of the original question(s). There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. The question is really whether the losses with the 4:1 transformer, plus those of any matcher at the TX end, exceed those when there is no transformer (but with higher loss on the coax), plus a matcher. Put another way, for short feeder lengths, is it better to use the transformer? That was not the question he asked. Please re-read the OP. I was trying to address his original question(s) as best as I could. In addition I also said that there were "several disclaimers I could include" which may involve your personal concerns. I did not want to muddy the waters. I think I answered Wayne's question(s), but I will wait to hear from him to see if that is so. |
An antenna question--43 ft vertical
On 7/3/2015 10:37 AM, Wayne wrote:
"John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. Thanks John. Yes, we have strayed from the original question, but I have found the discussion stimulating. Perhaps a new thread should be started to address those subjects. If I use EZNEC to model the 43 footer over perfect ground with a 3 inch diameter radiator, I get impedances in the same ball park as you list. If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna would see as a feedline, if a 4:1 unun had 50 ohm coax on the other side), the SWR plot becomes interesting. The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR getting below 2.5:1 around 29 MHz. Is that a valid approach? I have not done what you have done, but it sounds correct. I'll try to verify what you have done when time permits. I really think you know what you are doing. Don't forget that EZNEC can use transmission lines, transformers, inductors, capacitors, resistors and other stuff to help in your analysis. Although the true answers come from the physical implementation, it is very helpful to use EZNEC to gain insight into the situation. And, I think you know that as well. |
An antenna question--43 ft vertical
On 7/3/2015 10:37 AM, Wayne wrote:
"John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. Thanks John. Yes, we have strayed from the original question, but I have found the discussion stimulating. Indeed! So have I. Perhaps a new thread should be started to address those subjects. Please start one if you feel compelled. If I use EZNEC to model the 43 footer over perfect ground with a 3 inch diameter radiator, I get impedances in the same ball park as you list. Ha! I used 1.5 inches. I will re-do. If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna would see as a feedline, if a 4:1 unun had 50 ohm coax on the other side), the SWR plot becomes interesting. I've never done that. I will explore this set-up. The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR getting below 2.5:1 around 29 MHz. Are we still considering a 10MHz to 30Mhz frequency sweep? Is that a valid approach? You might be ahead of me on this. |
An antenna question--43 ft vertical
"John S" wrote in message ... On 7/3/2015 10:37 AM, Wayne wrote: "John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. Thanks John. Yes, we have strayed from the original question, but I have found the discussion stimulating. Indeed! So have I. Perhaps a new thread should be started to address those subjects. Please start one if you feel compelled. If I use EZNEC to model the 43 footer over perfect ground with a 3 inch diameter radiator, I get impedances in the same ball park as you list. Ha! I used 1.5 inches. I will re-do. If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna would see as a feedline, if a 4:1 unun had 50 ohm coax on the other side), the SWR plot becomes interesting. I've never done that. I will explore this set-up. The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR getting below 2.5:1 around 29 MHz. Are we still considering a 10MHz to 30Mhz frequency sweep? Well, I have been running the SWR across 4 to 30 MHz, but mainly looking at 10 MHz and above. As for EZNEC and transmission lines, I have never done that, but plan to when I can. I don't follow how to do it. In the few cases I wanted the info for a single frequency, I just used a Smith chart. This thread has given me a lot to consider in improving my whip setup, but details of the possibilities would run the thread off in the weeds :) |
An antenna question--43 ft vertical
In message , John S
writes On 7/3/2015 10:17 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. A fixed-tuned TX will still need a matcher. That was not part of the original question(s). At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. A fixed-tuned TX will probably be reasonably happy with a direct connection - although maybe even happier with a series capacitor of -J22 ohms. That was not part of the original question(s). There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. The question is really whether the losses with the 4:1 transformer, plus those of any matcher at the TX end, exceed those when there is no transformer (but with higher loss on the coax), plus a matcher. Put another way, for short feeder lengths, is it better to use the transformer? That was not the question he asked. Please re-read the OP. I was trying to address his original question(s) as best as I could. In addition I also said that there were "several disclaimers I could include" which may involve your personal concerns. I did not want to muddy the waters. I think I answered Wayne's question(s), but I will wait to hear from him to see if that is so. You have certainly answered "Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical?" (ie to reduce the horrendous mismatch). However, don't you think there's any virtue in wondering whether, in the circumstances described (with the relatively short feeder), it will be any better than a direct connection to the antenna, and to do all the matching at the TX end? Also, would you use a transformer if there was hardly any feeder at all, or (in an extreme case) if the antenna was fed directly from the TX? I'm not advocating anything - only wondering. -- Ian |
An antenna question--43 ft vertical
"rickman" wrote in message ... On 7/2/2015 8:53 PM, Wayne wrote: "rickman" wrote in message ... On 7/2/2015 3:52 PM, Wayne wrote: Why will the reflections not have losses? Because the assumption I posed was for a lossless line. In that case, with a conjugate match on both ends, wouldn't there be maximum power transmission regardless of the SWR? You aren't grasping the issue. Losses are *not* only in the transmission line. When a reflected wave returns to the transmitter output, it is not reflected 100%. If the output and transmission line are matched exactly, 50% of the reflected wave reaching the output will be reflected and 50% will be dissipated in the output stage. I don't think I've ever heard that anywhere before. Could you elaborate? Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? Well, yes. Minus losses in matching networks and transmission lines. In examples with lossless lines and lossless matching networks, wouldn't it be 100%. |
An antenna question--43 ft vertical
On 7/3/2015 1:34 PM, Ian Jackson wrote:
In message , John S writes On 7/3/2015 10:17 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? I think we strayed off the path to answering your original question. The short answer is that you are correct and there is no magic involved. A bit longer answer is: A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30 meters. Using a 1:4 transformer at the feed point will reduce that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line. A fixed-tuned TX will still need a matcher. That was not part of the original question(s). At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4 transformer will reduce that to 37 + J 33 ohms. A fixed-tuned TX will probably be reasonably happy with a direct connection - although maybe even happier with a series capacitor of -J22 ohms. That was not part of the original question(s). There are several disclaimers I could include, but I think you understand that the answers cannot be exact with the info presented. I hope this helps. The question is really whether the losses with the 4:1 transformer, plus those of any matcher at the TX end, exceed those when there is no transformer (but with higher loss on the coax), plus a matcher. Put another way, for short feeder lengths, is it better to use the transformer? That was not the question he asked. Please re-read the OP. I was trying to address his original question(s) as best as I could. In addition I also said that there were "several disclaimers I could include" which may involve your personal concerns. I did not want to muddy the waters. I think I answered Wayne's question(s), but I will wait to hear from him to see if that is so. You have certainly answered "Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical?" (ie to reduce the horrendous mismatch). However, don't you think there's any virtue in wondering whether, in the circumstances described (with the relatively short feeder), it will be any better than a direct connection to the antenna, and to do all the matching at the TX end? Also, would you use a transformer if there was hardly any feeder at all, or (in an extreme case) if the antenna was fed directly from the TX? I'm not advocating anything - only wondering. Yes, I agree that there is virtue in fully examining all the possibilities. I may or may not use a transformer with a direct connection to the antenna. It depends on my source's capabilities. Of course, anything added to improve a match also causes a bit of loss. It's all tradeoffs, as you well know. |
An antenna question--43 ft vertical
Okay. This data set is for a 43' carbon steel antenna on a perfect
ground fed with 25' of RG8A/U. No transformer. Freq R X SWR 4.000 2.52 -12.74 21.140 5.000 20.16 20.82 2.977 6.000 28.78 -30.81 2.586 7.000 8.04 -5.76 6.307 8.000 5.68 12.25 9.337 9.000 6.08 30.66 11.344 10.000 8.95 56.02 12.703 11.000 20.21 104.84 13.685 12.000 137.06 277.80 14.297 13.000 126.21 -267.32 14.174 14.000 19.15 -95.40 12.417 15.000 11.09 -43.81 8.067 16.000 17.90 -11.07 2.949 17.000 56.99 -15.98 1.384 18.000 20.29 -31.49 3.568 19.000 9.29 -12.21 5.711 20.000 6.95 3.56 7.236 21.000 6.99 19.13 8.217 22.000 9.03 38.15 8.829 23.000 15.83 67.71 9.161 24.000 48.70 132.94 9.150 25.000 418.45 49.46 8.488 26.000 60.45 -119.25 6.589 27.000 28.69 -45.41 3.465 28.000 46.51 -9.99 1.245 29.000 62.03 -48.79 2.397 30.000 21.38 -41.26 4.116 This data set is the same except with a 1:4 transformer at the antenna. Freq R X SWR 4.000 4.07 19.59 14.187 5.000 25.57 74.28 6.633 6.000 153.39 -83.56 4.058 7.000 25.51 -38.52 3.333 8.000 15.35 -6.11 3.312 9.000 16.11 18.43 3.567 10.000 26.59 49.05 3.970 11.000 87.00 102.27 4.497 12.000 180.75 -114.05 5.136 13.000 28.87 -72.84 5.814 14.000 11.18 -31.02 6.256 15.000 8.52 -4.23 5.914 16.000 13.30 23.05 4.607 17.000 50.10 61.48 3.196 18.000 107.24 -32.07 2.383 19.000 36.37 -30.20 2.135 20.000 22.85 -5.54 2.222 21.000 22.77 16.08 2.474 22.000 33.85 41.69 2.827 23.000 85.10 73.92 3.268 24.000 165.89 -60.03 3.790 25.000 39.38 -70.12 4.323 26.000 15.47 -31.46 4.605 27.000 12.03 -4.86 4.197 28.000 18.81 20.71 3.175 29.000 54.98 43.57 2.257 30.000 85.50 -14.78 1.786 Is this of any help? |
An antenna question--43 ft vertical
On 7/4/2015 1:34 AM, John S wrote:
Okay. This data set is for a 43' carbon steel antenna on a perfect ground fed with 25' of RG8A/U. No transformer. I made a mistake. I used 3" copper pipe. I will re-do with 3" carbon steel pipe. No transformer: Freq R X SWR 4.000 2.61 -12.68 20.382 5.000 20.92 20.65 2.867 6.000 28.18 -30.18 2.600 7.000 8.03 -5.57 6.304 8.000 5.70 12.35 9.317 9.000 6.11 30.75 11.314 10.000 9.00 56.13 12.662 11.000 20.37 105.05 13.626 12.000 139.02 278.20 14.204 13.000 125.69 -265.07 14.021 14.000 19.40 -94.97 12.182 15.000 11.41 -43.56 7.809 16.000 18.65 -11.02 2.830 17.000 56.19 -17.88 1.426 18.000 19.92 -31.06 3.599 19.000 9.26 -12.02 5.720 20.000 6.96 3.68 7.228 21.000 7.02 19.23 8.195 22.000 9.09 38.27 8.792 23.000 15.99 67.89 9.102 24.000 49.53 133.22 9.057 25.000 412.93 41.45 8.343 26.000 60.97 -117.33 6.399 27.000 29.70 -44.72 3.324 28.000 48.31 -10.74 1.247 29.000 60.26 -49.68 2.445 30.000 21.05 -40.86 4.142 1:4 transformer: Freq R X SWR 4.000 4.21 19.69 13.745 5.000 26.48 74.45 6.450 6.000 149.09 -82.20 3.972 7.000 25.55 -37.90 3.288 8.000 15.45 -5.81 3.285 9.000 16.25 18.68 3.549 10.000 26.91 49.36 3.954 11.000 88.45 102.31 4.478 12.000 177.47 -114.21 5.105 13.000 28.81 -72.20 5.756 14.000 11.31 -30.71 6.152 15.000 8.73 -3.99 5.767 16.000 13.78 23.29 4.469 17.000 51.75 60.81 3.109 18.000 104.23 -32.25 2.336 19.000 36.12 -29.39 2.111 20.000 22.93 -5.09 2.210 21.000 22.96 16.44 2.467 22.000 34.32 42.05 2.819 23.000 86.65 73.65 3.255 24.000 162.78 -61.65 3.764 25.000 39.08 -69.10 4.270 26.000 15.65 -30.95 4.510 27.000 12.38 -4.51 4.075 28.000 19.52 20.93 3.075 29.000 56.33 42.38 2.197 30.000 83.40 -15.44 1.755 |
An antenna question--43 ft vertical
"John S" wrote in message ... On 7/4/2015 1:34 AM, John S wrote: Okay. This data set is for a 43' carbon steel antenna on a perfect ground fed with 25' of RG8A/U. No transformer. I made a mistake. I used 3" copper pipe. I will re-do with 3" carbon steel pipe. No transformer: Freq R X SWR 4.000 2.61 -12.68 20.382 5.000 20.92 20.65 2.867 6.000 28.18 -30.18 2.600 7.000 8.03 -5.57 6.304 8.000 5.70 12.35 9.317 9.000 6.11 30.75 11.314 10.000 9.00 56.13 12.662 11.000 20.37 105.05 13.626 12.000 139.02 278.20 14.204 13.000 125.69 -265.07 14.021 14.000 19.40 -94.97 12.182 15.000 11.41 -43.56 7.809 16.000 18.65 -11.02 2.830 17.000 56.19 -17.88 1.426 18.000 19.92 -31.06 3.599 19.000 9.26 -12.02 5.720 20.000 6.96 3.68 7.228 21.000 7.02 19.23 8.195 22.000 9.09 38.27 8.792 23.000 15.99 67.89 9.102 24.000 49.53 133.22 9.057 25.000 412.93 41.45 8.343 26.000 60.97 -117.33 6.399 27.000 29.70 -44.72 3.324 28.000 48.31 -10.74 1.247 29.000 60.26 -49.68 2.445 30.000 21.05 -40.86 4.142 1:4 transformer: Freq R X SWR 4.000 4.21 19.69 13.745 5.000 26.48 74.45 6.450 6.000 149.09 -82.20 3.972 7.000 25.55 -37.90 3.288 8.000 15.45 -5.81 3.285 9.000 16.25 18.68 3.549 10.000 26.91 49.36 3.954 11.000 88.45 102.31 4.478 12.000 177.47 -114.21 5.105 13.000 28.81 -72.20 5.756 14.000 11.31 -30.71 6.152 15.000 8.73 -3.99 5.767 16.000 13.78 23.29 4.469 17.000 51.75 60.81 3.109 18.000 104.23 -32.25 2.336 19.000 36.12 -29.39 2.111 20.000 22.93 -5.09 2.210 21.000 22.96 16.44 2.467 22.000 34.32 42.05 2.819 23.000 86.65 73.65 3.255 24.000 162.78 -61.65 3.764 25.000 39.08 -69.10 4.270 26.000 15.65 -30.95 4.510 27.000 12.38 -4.51 4.075 28.000 19.52 20.93 3.075 29.000 56.33 42.38 2.197 30.000 83.40 -15.44 1.755 Yes, very interesting. Throw in 75 feet of cable, and things get "better". The lowest SWR is about 2:1 at 19 MHz. It is about 6:1 on 20 meters. |
An antenna question--43 ft vertical
On 7/4/2015 10:55 AM, Wayne wrote:
Yes, very interesting. Throw in 75 feet of cable, and things get "better". The lowest SWR is about 2:1 at 19 MHz. It is about 6:1 on 20 meters. Of course. More loss in the cable makes it "better" (but, you know that). |
An antenna question--43 ft vertical
On 7/4/2015 10:55 AM, Wayne wrote:
By the way, Wayne... Are you aware of a companion Excel application for EZNEC called AutoEZ? You can run many test cases in a few seconds using it. You can find it on the EZNEC site. It is how I generated the data I posted. |
An antenna question--43 ft vertical
"John S" wrote in message ... On 7/4/2015 10:55 AM, Wayne wrote: By the way, Wayne... Are you aware of a companion Excel application for EZNEC called AutoEZ? You can run many test cases in a few seconds using it. You can find it on the EZNEC site. It is how I generated the data I posted. Thanks, I'll look for that. I run the old wood burning version 3.0. |
An antenna question--43 ft vertical
On 7/3/2015 3:27 PM, Wayne wrote:
"rickman" wrote in message ... On 7/2/2015 8:53 PM, Wayne wrote: "rickman" wrote in message ... On 7/2/2015 3:52 PM, Wayne wrote: Why will the reflections not have losses? Because the assumption I posed was for a lossless line. In that case, with a conjugate match on both ends, wouldn't there be maximum power transmission regardless of the SWR? You aren't grasping the issue. Losses are *not* only in the transmission line. When a reflected wave returns to the transmitter output, it is not reflected 100%. If the output and transmission line are matched exactly, 50% of the reflected wave reaching the output will be reflected and 50% will be dissipated in the output stage. I don't think I've ever heard that anywhere before. Could you elaborate? I'm not so sure now. I think I mentioned before that I learned about transmission lines in the digital context where source and loads are largely resistive. Resistance dissipates power. So when matched the source dissipates as much power as delivered to the load (or transmission line). Likewise, matched impedance will not reflect power, but rather it is all absorbed. That is what happens at the antenna for sure. But I'm not clear about what this conjugate network is really. If it is purely reactive, then it will not have losses other than the parasitics. I have to admit I am not fluent in the complex math of networks. So off hand an impedance of 1063 -j0 says to me resistive. The imaginary part implies phase shifting, no? With that term being 0 doesn't that say the capacitive and inductive parts cancel out leaving only resistance? If you can, please explain how I am wrong. Are you suggesting that the conjugate match will reflect back to the antenna 100% of the original reflected wave from the antenna? Well, yes. Minus losses in matching networks and transmission lines. In examples with lossless lines and lossless matching networks, wouldn't it be 100%. I don't get how the matching network will reflect the wave from the antenna 100%. Is that something you can explain? -- Rick |
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