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In message , Wayne
writes "Frank Turner-Smith G3VKI" wrote in message ... On 30/10/14 14:04, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes On 30/10/14 08:47, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes In a full wave dipole the voltage at both ends will always be in phase, Are you sure? Think on't! so I would expect to see a very high impedance at the feed point. Correct. As you point out, matching the full-wave could be difficult and very lossy. Double zepp? OK, what did I miss? In a full wave dipole, at the instant the voltage at one end is peak positive, the voltage at the other end will also be peak positive. Similarly, at the feed point, both legs would be at peak negative and no current would flow in the feeder, hence the high impedance. There would be a current flowing in each leg of the dipole, but the currents would be in anti-phase. Where have I got it wrong? Do I need another drink? Maybe I need a drink too. However, all dipoles/doublets have to fed 'push-pull', so when one leg goes +ve, the other leg goes -ve. The voltage at all points along the antenna that are equidistant from the feedpoint will be in antiphase, so if the feedpoint is in the centre, the voltages at the ends will be in antiphase. [Or is my thinking seriously muddled?] # Looks like I owe you a pint. You've described the situation where a TX # is feeding the dipole. I was trying to visualise the RX conditions, but # it reciprocates. One of us has to be wrong, and I strongly suspect it's # me. Time for a drink. With drinking involved, I must throw in my 2 cents. I'd go with Frank....for full wave assume positive peak at one end, negative peak in the middle, and positive peak at the other end. (or vice versa) But, I suppose I should think about it a little more.....Laphroig would help ![]() See: http://tinyurl.com/q8nxqep ten rows of images down, second from left: This shows the amplitude and the polarity of the voltage and current for a halfwave dipole. [Lots of diagrams only show the amplitude.] You will see that the polarities on each leg are +ve and -ve. For a fullwave, just imagine it continuing on for another halfwave each side. -- Ian |
#2
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![]() "Ian Jackson" wrote in message ... In message , Wayne writes "Frank Turner-Smith G3VKI" wrote in message ... On 30/10/14 14:04, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes On 30/10/14 08:47, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes In a full wave dipole the voltage at both ends will always be in phase, Are you sure? Think on't! so I would expect to see a very high impedance at the feed point. Correct. As you point out, matching the full-wave could be difficult and very lossy. Double zepp? OK, what did I miss? In a full wave dipole, at the instant the voltage at one end is peak positive, the voltage at the other end will also be peak positive. Similarly, at the feed point, both legs would be at peak negative and no current would flow in the feeder, hence the high impedance. There would be a current flowing in each leg of the dipole, but the currents would be in anti-phase. Where have I got it wrong? Do I need another drink? Maybe I need a drink too. However, all dipoles/doublets have to fed 'push-pull', so when one leg goes +ve, the other leg goes -ve. The voltage at all points along the antenna that are equidistant from the feedpoint will be in antiphase, so if the feedpoint is in the centre, the voltages at the ends will be in antiphase. [Or is my thinking seriously muddled?] # Looks like I owe you a pint. You've described the situation where a TX # is feeding the dipole. I was trying to visualise the RX conditions, but # it reciprocates. One of us has to be wrong, and I strongly suspect it's # me. Time for a drink. With drinking involved, I must throw in my 2 cents. I'd go with Frank....for full wave assume positive peak at one end, negative peak in the middle, and positive peak at the other end. (or vice versa) But, I suppose I should think about it a little more.....Laphroig would help ![]() # See: # http://tinyurl.com/q8nxqep # ten rows of images down, second from left: # This shows the amplitude and the polarity of the voltage and current for # a halfwave dipole. [Lots of diagrams only show the amplitude.] You will # see that the polarities on each leg are +ve and -ve. For a fullwave, # just imagine it continuing on for another halfwave each side. # -- # Ian Isn't that figure for a full wave?... lambda |
#3
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In message , Wayne
writes "Ian Jackson" wrote in message ... In message , Wayne writes "Frank Turner-Smith G3VKI" wrote in message ... On 30/10/14 14:04, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes On 30/10/14 08:47, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes In a full wave dipole the voltage at both ends will always be in phase, Are you sure? Think on't! so I would expect to see a very high impedance at the feed point. Correct. As you point out, matching the full-wave could be difficult and very lossy. Double zepp? OK, what did I miss? In a full wave dipole, at the instant the voltage at one end is peak positive, the voltage at the other end will also be peak positive. Similarly, at the feed point, both legs would be at peak negative and no current would flow in the feeder, hence the high impedance. There would be a current flowing in each leg of the dipole, but the currents would be in anti-phase. Where have I got it wrong? Do I need another drink? Maybe I need a drink too. However, all dipoles/doublets have to fed 'push-pull', so when one leg goes +ve, the other leg goes -ve. The voltage at all points along the antenna that are equidistant from the feedpoint will be in antiphase, so if the feedpoint is in the centre, the voltages at the ends will be in antiphase. [Or is my thinking seriously muddled?] # Looks like I owe you a pint. You've described the situation where a TX # is feeding the dipole. I was trying to visualise the RX conditions, but # it reciprocates. One of us has to be wrong, and I strongly suspect it's # me. Time for a drink. With drinking involved, I must throw in my 2 cents. I'd go with Frank....for full wave assume positive peak at one end, negative peak in the middle, and positive peak at the other end. (or vice versa) But, I suppose I should think about it a little more.....Laphroig would help ![]() # See: # http://tinyurl.com/q8nxqep # ten rows of images down, second from left: # This shows the amplitude and the polarity of the voltage and current for # a halfwave dipole. [Lots of diagrams only show the amplitude.] You will # see that the polarities on each leg are +ve and -ve. For a fullwave, # just imagine it continuing on for another halfwave each side. # -- # Ian Isn't that figure for a full wave?... lambda Maybe you're looking at the wrong one. I've had another look, and it's now 9 down, far left. It's the one with the thick black dipole, entitled "Halfwave Dipole Antenna (Hertz)". Ah, I've found the source, here (Fig 1): http://www.digikey.com/en/articles/t...standing-anten na-specifications-and-operation -- Ian |
#4
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On 30/10/14 20:44, Ian Jackson wrote:
In message , Wayne writes "Ian Jackson" wrote in message ... In message , Wayne writes "Frank Turner-Smith G3VKI" wrote in message ... On 30/10/14 14:04, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes On 30/10/14 08:47, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes In a full wave dipole the voltage at both ends will always be in phase, Are you sure? Think on't! so I would expect to see a very high impedance at the feed point. Correct. As you point out, matching the full-wave could be difficult and very lossy. Double zepp? OK, what did I miss? In a full wave dipole, at the instant the voltage at one end is peak positive, the voltage at the other end will also be peak positive. Similarly, at the feed point, both legs would be at peak negative and no current would flow in the feeder, hence the high impedance. There would be a current flowing in each leg of the dipole, but the currents would be in anti-phase. Where have I got it wrong? Do I need another drink? Maybe I need a drink too. However, all dipoles/doublets have to fed 'push-pull', so when one leg goes +ve, the other leg goes -ve. The voltage at all points along the antenna that are equidistant from the feedpoint will be in antiphase, so if the feedpoint is in the centre, the voltages at the ends will be in antiphase. [Or is my thinking seriously muddled?] # Looks like I owe you a pint. You've described the situation where a TX # is feeding the dipole. I was trying to visualise the RX conditions, but # it reciprocates. One of us has to be wrong, and I strongly suspect it's # me. Time for a drink. With drinking involved, I must throw in my 2 cents. I'd go with Frank....for full wave assume positive peak at one end, negative peak in the middle, and positive peak at the other end. (or vice versa) But, I suppose I should think about it a little more.....Laphroig would help ![]() # See: # http://tinyurl.com/q8nxqep # ten rows of images down, second from left: # This shows the amplitude and the polarity of the voltage and current for # a halfwave dipole. [Lots of diagrams only show the amplitude.] You will # see that the polarities on each leg are +ve and -ve. For a fullwave, # just imagine it continuing on for another halfwave each side. # -- # Ian Isn't that figure for a full wave?... lambda Maybe you're looking at the wrong one. I've had another look, and it's now 9 down, far left. It's the one with the thick black dipole, entitled "Halfwave Dipole Antenna (Hertz)". Ah, I've found the source, here (Fig 1): http://www.digikey.com/en/articles/t...standing-anten na-specifications-and-operation The original proposal in this thread was that long antennas performed better than short ones. If that was true you'd get a good 600MHz UHF TV picture using a 132ft end fed longwire. I've not tried it, but it doesn't seem very likely. -- ;-) .. 73 de Frank Turner-Smith G3VKI - mine's a pint. .. http://turner-smith.co.uk .. Ubuntu 12.04 Thunderbirds are go. |
#5
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In message , Frank Turner-Smith G3VKI
writes On 30/10/14 20:44, Ian Jackson wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , Wayne writes "Frank Turner-Smith G3VKI" wrote in message ... On 30/10/14 14:04, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes On 30/10/14 08:47, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes In a full wave dipole the voltage at both ends will always be in phase, Are you sure? Think on't! so I would expect to see a very high impedance at the feed point. Correct. As you point out, matching the full-wave could be difficult and very lossy. Double zepp? OK, what did I miss? In a full wave dipole, at the instant the voltage at one end is peak positive, the voltage at the other end will also be peak positive. Similarly, at the feed point, both legs would be at peak negative and no current would flow in the feeder, hence the high impedance. There would be a current flowing in each leg of the dipole, but the currents would be in anti-phase. Where have I got it wrong? Do I need another drink? Maybe I need a drink too. However, all dipoles/doublets have to fed 'push-pull', so when one leg goes +ve, the other leg goes -ve. The voltage at all points along the antenna that are equidistant from the feedpoint will be in antiphase, so if the feedpoint is in the centre, the voltages at the ends will be in antiphase. [Or is my thinking seriously muddled?] # Looks like I owe you a pint. You've described the situation where a TX # is feeding the dipole. I was trying to visualise the RX conditions, but # it reciprocates. One of us has to be wrong, and I strongly suspect it's # me. Time for a drink. With drinking involved, I must throw in my 2 cents. I'd go with Frank....for full wave assume positive peak at one end, negative peak in the middle, and positive peak at the other end. (or vice versa) But, I suppose I should think about it a little more.....Laphroig would help ![]() # See: # http://tinyurl.com/q8nxqep # ten rows of images down, second from left: # This shows the amplitude and the polarity of the voltage and current for # a halfwave dipole. [Lots of diagrams only show the amplitude.] You will # see that the polarities on each leg are +ve and -ve. For a fullwave, # just imagine it continuing on for another halfwave each side. # -- # Ian Isn't that figure for a full wave?... lambda Maybe you're looking at the wrong one. I've had another look, and it's now 9 down, far left. It's the one with the thick black dipole, entitled "Halfwave Dipole Antenna (Hertz)". Ah, I've found the source, here (Fig 1): http://www.digikey.com/en/articles/t...standing-anten na-specifications-and-operation The original proposal in this thread was that long antennas performed better than short ones. If that was true you'd get a good 600MHz UHF TV picture using a 132ft end fed longwire. I've not tried it, but it doesn't seem very likely. A 132' endfed will have one hell of a gain on 600MHz - but it will be almost straight off the ends. -- Ian |
#6
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![]() "Ian Jackson" wrote in message ... The original proposal in this thread was that long antennas performed better than short ones. If that was true you'd get a good 600MHz UHF TV picture using a 132ft end fed longwire. I've not tried it, but it doesn't seem very likely. A 132' endfed will have one hell of a gain on 600MHz - but it will be almost straight off the ends. -- Ian I know the gain will be off the end of the wire, but still wonder if an antenna that long (in wavelengths) will actually work or will it be too long and the gain does not meet the expectations or if programs like NEC will predict it or fall apart. --- This email is free from viruses and malware because avast! Antivirus protection is active. http://www.avast.com |
#7
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In message , Ralph
Mowery writes "Ian Jackson" wrote in message ... The original proposal in this thread was that long antennas performed better than short ones. If that was true you'd get a good 600MHz UHF TV picture using a 132ft end fed longwire. I've not tried it, but it doesn't seem very likely. A 132' endfed will have one hell of a gain on 600MHz - but it will be almost straight off the ends. -- Ian I know the gain will be off the end of the wire, but still wonder if an antenna that long (in wavelengths) will actually work or will it be too long and the gain does not meet the expectations or if programs like NEC will predict it or fall apart. On that point, you'll have to ask the experts! -- Ian |
#8
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Ralph Mowery wrote:
"Ian Jackson" wrote in message ... The original proposal in this thread was that long antennas performed better than short ones. If that was true you'd get a good 600MHz UHF TV picture using a 132ft end fed longwire. I've not tried it, but it doesn't seem very likely. A 132' endfed will have one hell of a gain on 600MHz - but it will be almost straight off the ends. -- Ian I know the gain will be off the end of the wire, but still wonder if an antenna that long (in wavelengths) will actually work or will it be too long and the gain does not meet the expectations or if programs like NEC will predict it or fall apart. EZNEC handles it just fine. I modeled a 120' (I had forgotten the exact number posted) long wire at 6' over real ground at 600 Mhz: Impedance: 55-j308 Max vertical gain: 21 dBi at 4 degrees Horizontal gain: two 21 dBi lobes at +/- 4 degrees Front/back: 9.5 dB LOTS of little lobes... -- Jim Pennino |
#9
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![]() "Ian Jackson" wrote in message ... In message , Wayne writes "Ian Jackson" wrote in message ... In message , Wayne writes "Frank Turner-Smith G3VKI" wrote in message ... On 30/10/14 14:04, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes On 30/10/14 08:47, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes In a full wave dipole the voltage at both ends will always be in phase, Are you sure? Think on't! so I would expect to see a very high impedance at the feed point. Correct. As you point out, matching the full-wave could be difficult and very lossy. Double zepp? OK, what did I miss? In a full wave dipole, at the instant the voltage at one end is peak positive, the voltage at the other end will also be peak positive. Similarly, at the feed point, both legs would be at peak negative and no current would flow in the feeder, hence the high impedance. There would be a current flowing in each leg of the dipole, but the currents would be in anti-phase. Where have I got it wrong? Do I need another drink? Maybe I need a drink too. However, all dipoles/doublets have to fed 'push-pull', so when one leg goes +ve, the other leg goes -ve. The voltage at all points along the antenna that are equidistant from the feedpoint will be in antiphase, so if the feedpoint is in the centre, the voltages at the ends will be in antiphase. [Or is my thinking seriously muddled?] # Looks like I owe you a pint. You've described the situation where a TX # is feeding the dipole. I was trying to visualise the RX conditions, but # it reciprocates. One of us has to be wrong, and I strongly suspect it's # me. Time for a drink. With drinking involved, I must throw in my 2 cents. I'd go with Frank....for full wave assume positive peak at one end, negative peak in the middle, and positive peak at the other end. (or vice versa) But, I suppose I should think about it a little more.....Laphroig would help ![]() # See: # http://tinyurl.com/q8nxqep # ten rows of images down, second from left: # This shows the amplitude and the polarity of the voltage and current for # a halfwave dipole. [Lots of diagrams only show the amplitude.] You will # see that the polarities on each leg are +ve and -ve. For a fullwave, # just imagine it continuing on for another halfwave each side. # -- # Ian Isn't that figure for a full wave?... lambda # Maybe you're looking at the wrong one. I've had another look, and it's # now 9 down, far left. It's the one with the thick black dipole, entitled # "Halfwave Dipole Antenna (Hertz)". Ah, I've found the source, here (Fig # 1): # http://www.digikey.com/en/articles/t...standing-anten # na-specifications-and-operation # -- # Ian OK, but I'm losing touch with what the point is. The figures referenced both times show voltage peaks of opposite phase at the ends of half wave dipoles, thus voltage peaks of the same phase at the end of full wave dipoles. I'm assuming we have agreement on that. And with that, it currently is time here in CA for Lagavulin 16. |
#10
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In message , Wayne
writes "Ian Jackson" wrote in message ... In message , Wayne writes "Ian Jackson" wrote in message ... In message , Wayne writes "Frank Turner-Smith G3VKI" wrote in message ... On 30/10/14 14:04, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes On 30/10/14 08:47, Ian Jackson wrote: In message , Frank Turner-Smith G3VKI writes In a full wave dipole the voltage at both ends will always be in Are you sure? Think on't! so I would expect to see a very high impedance at the feed point. Correct. As you point out, matching the full-wave could be difficult and very lossy. Double zepp? OK, what did I miss? In a full wave dipole, at the instant the voltage at one end is peak positive, the voltage at the other end will also be peak positive. Similarly, at the feed point, both legs would be at peak negative and no current would flow in the feeder, hence the high impedance. There would be a current flowing in each leg of the dipole, but the currents would be in anti-phase. Where have I got it wrong? Do I need another drink? Maybe I need a drink too. However, all dipoles/doublets have to fed 'push-pull', so when one leg goes +ve, the other leg goes -ve. The voltage at all points along the antenna that are equidistant from the feedpoint will be in antiphase, so if the feedpoint is in the centre, the voltages at the ends will be in antiphase. [Or is my thinking seriously muddled?] # Looks like I owe you a pint. You've described the situation where a TX # is feeding the dipole. I was trying to visualise the RX conditions, but # it reciprocates. One of us has to be wrong, and I strongly suspect it's # me. Time for a drink. With drinking involved, I must throw in my 2 cents. I'd go with Frank....for full wave assume positive peak at one end, negative peak in the middle, and positive peak at the other end. (or vice versa) But, I suppose I should think about it a little more.....Laphroig would help ![]() # See: # http://tinyurl.com/q8nxqep # ten rows of images down, second from left: # This shows the amplitude and the polarity of the voltage and current for # a halfwave dipole. [Lots of diagrams only show the amplitude.] You will # see that the polarities on each leg are +ve and -ve. For a fullwave, # just imagine it continuing on for another halfwave each side. # -- # Ian Isn't that figure for a full wave?... lambda # Maybe you're looking at the wrong one. I've had another look, and it's # now 9 down, far left. It's the one with the thick black dipole, entitled # "Halfwave Dipole Antenna (Hertz)". Ah, I've found the source, here (Fig # 1): # http://www.digikey.com/en/articles/t...standing-anten # na-specifications-and-operation # -- # Ian OK, but I'm losing touch with what the point is. The figures referenced both times show voltage peaks of opposite phase at the ends of half wave dipoles, thus voltage peaks of the same phase at the end of full wave dipoles. I'm assuming we have agreement on that. You assume wrongly. A centre-fed fullwave is also fed 'push-pull', ie in antiphase - except that it's a high voltage feed instead of high current. Just draw a diagram similar to the halfwave, - again showing both the voltage amplitude and polarity. The lines you draw on each side for the amplitudes are copies of each other - not mirror images. However, it seems to a different matter if you feed the fullwave off-centre, a quarterwave from one end. At least on my drawing, the voltages at the ends ARE in phase - so I guess the radiation pattern will be different from the centre-fed. And with that, it currently is time here in CA for Lagavulin 16. No such luck here. FWIW, it's nearly midday, and I'm having a (rather late) strong 'coffee-bag' coffee, with two teaspoons of maple syrup and one teaspoon (heaped) of dark drinking chocolate powder. -- Ian |
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