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#1
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Jim Kelley wrote:
That angle is the 'phase' of the current standing wave as a function of position, not to be confused with the phase of the current with respect to voltage. Roger? Since any reference to voltage on an antenna seems to be verboten, I have avoided any such reference. Otherwise, here is a quote from Kraus: "It is generally assumed that the current distribution of an infinitesimally thin antenna is sinusoidal, and that the phase is *constant* over a 1/2WL interval, changing abruptly by 180 degrees between intervals." Don't you just love the phrase, "It is generally assumed ..."? He doesn't say the current is sinusoidal. He doesn't say the phase is constant over a 1/2WL interval. For that general assumption to be true, the reflected current would have to equal the forward current on a standing-wave antenna. But we know it doesn't. However, this implies that the reflected current arriving back at the feedpoint is not extremely/severely attenuated. Here's a cute ballpark analysis sure to drive the gurus crazy. Assume the Z0 of a traveling-wave dipole is 600 ohms. The ratio of forward voltage to forward current is 600 ohms. The ratio of reflected voltage to reflected current is 600 ohms. Assume the feedpoint current is one amp and the feedpoint voltage is 50 volts. Assume the forward current and the reflected current are in phase at the feedpoint. Assume the forward voltage and reflected voltage are 180 degrees out of phase at the feedpoint. This is enough information to solve for the ratio of forward current to reflected current at the feedpoint. Assuming the net current is one amp at the feedpoint, I get 0.542 amps for the forward current and 0.458 amps for the reflected current, i.e. the reflected current is 85% of the value of the forward current. Remember, that is a ballpark estimate. That means the current only decreases by 15% in its round trip to the end of the antenna and back. Same for the voltage. Oops, I mentioned voltage - sorry. But please note that the power loss is a lot higher than 15% since both the current and voltage are reduced by the same 15%. The argument seems to occur due to the ignoring of the component waves on a standing wave antenna. Such is the steady-state model seduction attended by its sacred cows. Who wants to join me in a beer bust and barbecue? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#2
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Cecil Moore wrote:
Jim Kelley wrote: That angle is the 'phase' of the current standing wave as a function of position, not to be confused with the phase of the current with respect to voltage. Roger? Since any reference to voltage on an antenna seems to be verboten, I have avoided any such reference. I can understand how at DC, references to current in a dipole might be verboten. :-) I wonder if voltage on a dipole could be roughly likened to transverse velocity at various points along a whip. Don't you just love the phrase, "It is generally assumed ..."? I guess it allows: 'I'm not responsible should it turn out not to be a good assumption'. :-) For that general assumption to be true, the reflected current would have to equal the forward current on a standing-wave antenna. But we know it doesn't. However, this implies that the reflected current arriving back at the feedpoint is not extremely/severely attenuated. Seems to me it just implies that current at the end of a dipole isn't really zero. Assume the Z0 of a traveling-wave dipole is 600 ohms. The ratio of forward voltage to forward current is 600 ohms. The ratio of reflected voltage to reflected current is 600 ohms. Assume the feedpoint current is one amp and the feedpoint voltage is 50 volts. Assume the forward current and the reflected current are in phase at the feedpoint. Assume the forward voltage and reflected voltage are 180 degrees out of phase at the feedpoint. This is enough information to solve for the ratio of forward current to reflected current at the feedpoint. Assuming the net current is one amp at the feedpoint, I get 0.542 amps for the forward current and 0.458 amps for the reflected current, i.e. the reflected current is 85% of the value of the forward current. Remember, that is a ballpark estimate. 50 volts difference across a 600 ohm impedance, over 2, plus and minus half an amp. That means the current only decreases by 15% in its round trip to the end of the antenna and back. I think it may actually make many round trips. There may be multiple reflections. The argument seems to occur due to the ignoring of the component waves on a standing wave antenna. Such is the steady-state model seduction attended by its sacred cows. Well, you'll either have to write out everything as a series, which is a lot of busy work, or use the steady state equivalent. So what does all this say about reflectivity at the end of the dipole? ;-) 73, Jim AC6XG |
#3
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Jim Kelley wrote:
Seems to me it just implies that current at the end of a dipole isn't really zero. The net current is very close to zero because the forward and reflected currents are very nearly equal and 180 degrees out of phase. I think it may actually make many round trips. There may be multiple reflections. Of course, but like a transmission line, there is only one forward wave and one reflected wave. All the multiple reflections are contained in those two waves. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#4
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Cecil Moore wrote:
Jim Kelley wrote: Seems to me it just implies that current at the end of a dipole isn't really zero. The net current is very close to zero because the forward and reflected currents are very nearly equal and 180 degrees out of phase. Is that what you meant by: "For that general assumption to be true, the reflected current would have to equal the forward current on a standing-wave antenna. But we know it doesn't"? It's not clear whether you're making this point in recognition of the fact that wires are not lossless, or whether you're claiming it's somehow fundamental to the performance of a radiator. I think it may actually make many round trips. There may be multiple reflections. Of course, but like a transmission line, there is only one forward wave and one reflected wave. All the multiple reflections are contained in those two waves. Then apparently you've decided not to completely eschew making at least some steady state assumptions. Seductive indeed. ;-) 73, Jim AC6XG |
#5
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Jim Kelley wrote:
Cecil Moore wrote: The net current is very close to zero because the forward and reflected currents are very nearly equal and 180 degrees out of phase. This is at the tip end of a dipole. Is that what you meant by: "For that general assumption to be true, the reflected current would have to equal the forward current on a standing-wave antenna. But we know it doesn't"? This is apparently not at the tip end of a dipole. It's not clear whether you're making this point in recognition of the fact that wires are not lossless, or whether you're claiming it's somehow fundamental to the performance of a radiator. Both, radiation is a "loss". Of course, but like a transmission line, there is only one forward wave and one reflected wave. All the multiple reflections are contained in those two waves. Then apparently you've decided not to completely eschew making at least some steady state assumptions. Seductive indeed. ;-) There are only two possible directions, forward and reverse, in which energy can flow. Multiple reflections do not create any more directions. I am not opposed to the steady-state solution and use it all the time. I am opposed to people being seduced by the steady-state solution into believing there is not such thing as forward and reflected waves even though standing waves require forward and reflected waves. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#6
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Cecil Moore wrote:
I am opposed to people being seduced by the steady-state solution into believing there is not such thing as forward and reflected waves even though standing waves require forward and reflected waves. I also apologize if this particular dose of reality brought this thread to a screeching halt. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#7
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Cecil Moore wrote: I am opposed to people being seduced by the steady-state solution into believing there is not such thing as forward and reflected waves even though standing waves require forward and reflected waves. Say that in Sacramento, and some knucklehead legislator will pass a law against it. 73, Jim AC6XG |
#8
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Cecil, W5DXP wrote:
"There are only two possible directions, forward and reverse in which energy can flow. Multiple reflections do not create any more directions." True. Further, all the same-frequency, same-direction signals merge. So, as Cecil said, there are only two same-frequency signals on a transmission line, forward and reverse. The interference pattern these signals produce does not represent another signal. Trying to use ordinary circuit analysis on standing-wave antennas is problematic, but it`s been tried in this thread. Here is what R.W.P. King wrote in "Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing, 1945, on page 86: "Inductance and capacitance as used in near-zone circuits with uniform current cannot be defined, and ordinary circuit analysis does not apply." This has not stopped efforts in this thread to analyze LC circuits as if we were dealing with low frequencies. Best regards, Richard Harrison, KB5WZI |
#9
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Richard Harrison wrote: Cecil, W5DXP wrote: "There are only two possible directions, forward and reverse in which energy can flow. Multiple reflections do not create any more directions." True. Further, all the same-frequency, same-direction signals merge. So, as Cecil said, there are only two same-frequency signals on a transmission line, forward and reverse. The interference pattern these signals produce does not represent another signal. So the claim is that the amplitude of the reflection being bandied about is the sum of multiple reflections? I haven't seen any indication of that. It appears to be simply the amplitude of the first reflection. At least, that's the way it appears to me, the uninitiated. 73, Jim AC6XG |
#10
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Richard Harrison wrote:
Trying to use ordinary circuit analysis on standing-wave antennas is problematic, but it`s been tried in this thread. Here is what R.W.P. King wrote in "Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing, 1945, on page 86: "Inductance and capacitance as used in near-zone circuits with uniform current cannot be defined, and ordinary circuit analysis does not apply." This has not stopped efforts in this thread to analyze LC circuits as if we were dealing with low frequencies. Very true. The basic problem, as I see it, is in assuming that the standing- wave current only has one component. For standing-wave antennas, the standing-wave current must necessarily have two components, If and Ir, as explained by Balanis in _Antenna_Theory_, (page 489, 2nd edition). In general, when forward waves and reflected waves exist in the circuit, lumped circuit analysis fails and distributed network analysis is the only method that yields the correct result. So even if one allows that the forward current magnitude is constant through a coil and the reflected current magnitude is constant through a coil, the net current magnitude will not be constant because of the phase differences in the two superposed currents at each end of the coil. I've been told by the gurus that I can ignore the forward and reflected waves and still obtain the correct steady-state solution. A mobile bugcatcher coil on 75m seems to disprove that assertion. But I have been called a "Grasshopper" and thus apparently have a lot yet to learn. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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