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Revisiting the Power Explanation
Cecil Moore wrote:
Gene Fuller wrote: That quote agrees completely with what I said. Gene, you remind me of an ex-friend of mine who when asked what would happen if he were caught by his wife in bed with his girlfriend, said, "I would just deny it." You said there is no equation for interference. Hecht in "Optics" provided the equation that you said didn't exist. I12 is the symbol for interference between the I1 and I2 waves. Cecil, This is getting good. Are you actually claiming that "I" is the symbol for "interference"? Betcha it ain't. How does Hecht define I1, I2, and I12? I have plenty of optics books, but I don't have a copy of Hecht. I don't plan to buy one. I seriously doubt that he says anything very different from any other author of optics textbooks. 73, Gene W4SZ |
Revisiting the Power Explanation
Gene Fuller wrote:
This is getting good. Are you actually claiming that "I" is the symbol for "interference"? Betcha it ain't. How does Hecht define I1, I2, and I12? I is the symbol for irradiance. I1 is the irradiance in wave 1. I2 is the irradiance in wave 2. I12 is the symbol for the interference between wave 1 and wave 2, you know - the symbol that you said didn't exist. I have plenty of optics books, but I don't have a copy of Hecht. So during all these months of denying what Hecht has written, you have been completely ignorant of what Hecht has written? Why am I not surprised? -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Cecil Moore wrote:
Gene Fuller wrote: This is getting good. Are you actually claiming that "I" is the symbol for "interference"? Betcha it ain't. How does Hecht define I1, I2, and I12? I is the symbol for irradiance. I1 is the irradiance in wave 1. I2 is the irradiance in wave 2. I12 is the symbol for the interference between wave 1 and wave 2, you know - the symbol that you said didn't exist. I have plenty of optics books, but I don't have a copy of Hecht. So during all these months of denying what Hecht has written, you have been completely ignorant of what Hecht has written? Why am I not surprised? Cecil, OK, we are getting somewhere. According to the equation, we know that interference has the same units as irradiance. Interesting. Does that mean that interference doesn't work for fields? Are there multiple definitions for interference? I am not at all ignorant of what Hecht has written, but I will admit to being ignorant of the exact choice of words he uses. 8-) 73, Gene W4SZ |
Revisiting the Power Explanation
Gene Fuller wrote:
OK, we are getting somewhere. According to the equation, we know that interference has the same units as irradiance. Interesting. Does that mean that interference doesn't work for fields? Are there multiple definitions for interference? My IEEE Dictionary is 130 miles away so I cannot look up the technical definition for interference. The units of irradiance are watts per unit area. In a transmission line, we can consider the unit area to be constant and simply drop the units of area. The resultant "irradiance" equation for transmission line power is: Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A) 2*SQRT(P1*P2)cos(A) is the interference term in watts where A is the angle between the two interfering fields. This is all explained at: http://www.w5dxp.com/energy.htm If the above equation is followed, adding powers is easier than superposing voltages and then calculating power. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Cecil Moore wrote in news:DHiPh.3364$YL5.1126
@newssvr29.news.prodigy.net: .... simply drop the units of area. The resultant "irradiance" equation for transmission line power is: Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A) 2*SQRT(P1*P2)cos(A) is the interference term in watts where A is the angle between the two interfering fields. Is this what you mean by 'algebraic sum' as in your words 'resultant disturbance at any point in a medium is the algebraic sum of the separate constituent waves'? Owen |
Revisiting the Power Explanation
On Mar 30, 6:27 pm, Cecil Moore wrote:
Keith Dysart wrote: Well I guess that settles it. You clearly are not aware of the methodologies. Even ones that work on the simplest of examples. Perhaps you could educate me. I am not convinced Please provide an S-Parameter analysis of the math model of the source that you have refused to provide. especially when the student who does not know how to solve the problem attempts to tell the teacher how to do so. There will be opportunity for education when you are prepared to act as a student, listen to what the teacher has to say, think about the information being provided and integrate it into your knowledge base. Until then, you are not ready to be educated. Sorry. Please feel free to come back when you are ready. ....Keith |
Revisiting the Power Explanation
Owen Duffy wrote:
Cecil Moore wrote: simply drop the units of area. The resultant "irradiance" equation for transmission line power is: Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A) 2*SQRT(P1*P2)cos(A) is the interference term in watts where A is the angle between the two interfering fields. Is this what you mean by 'algebraic sum' as in your words 'resultant disturbance at any point in a medium is the algebraic sum of the separate constituent waves'? Of course not. That statement of Hecht's applies to vector (and phasor) fields. Powers are scalars and must be treated accordingly. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Keith Dysart wrote:
There will be opportunity for education when you are prepared to act as a student, listen to what the teacher has to say, think about the information being provided and integrate it into your knowledge base. :-) What are you, Keith, a sophomore in junior college? -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
"Cecil Moore" wrote in message t... Owen Duffy wrote: Cecil Moore wrote: simply drop the units of area. The resultant "irradiance" equation for transmission line power is: Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A) 2*SQRT(P1*P2)cos(A) is the interference term in watts where A is the angle between the two interfering fields. Is this what you mean by 'algebraic sum' as in your words 'resultant disturbance at any point in a medium is the algebraic sum of the separate constituent waves'? Of course not. That statement of Hecht's applies to vector (and phasor) fields. Powers are scalars and must be treated accordingly. and hence my recommendation to abandon power as a useful method of analyzing waves in transmission lines. power is a scalar, you can't add power without resorting to digging out the phase angles that come from the current or voltage waves. so really that p-total equation is what happens when you add 2 current waves and then convert back to powers... you cancel out a whole bunch of Z0 terms but end up having to keep that ugly phase related factor. |
Revisiting the Power Explanation
Dave wrote:
and hence my recommendation to abandon power as a useful method of analyzing waves in transmission lines. Dave, try the "Energy Brain Teaser" problem. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Cecil Moore wrote:
Gene Fuller wrote: OK, we are getting somewhere. According to the equation, we know that interference has the same units as irradiance. Interesting. Does that mean that interference doesn't work for fields? Are there multiple definitions for interference? My IEEE Dictionary is 130 miles away so I cannot look up the technical definition for interference. The units of irradiance are watts per unit area. In a transmission line, we can consider the unit area to be constant and simply drop the units of area. The resultant "irradiance" equation for transmission line power is: Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A) 2*SQRT(P1*P2)cos(A) is the interference term in watts where A is the angle between the two interfering fields. This is all explained at: http://www.w5dxp.com/energy.htm If the above equation is followed, adding powers is easier than superposing voltages and then calculating power. Cecil, This is a perfect example of how the messages get so confused around here. You cannot simply ignore dimensionality, units, vector vs. scalar, etc. That leads to really silly conclusions like expressing interference in the same units as irradiance. I don't know if dimensional analysis is still taught these days, but way back when I was taking science classes it was often stated that the very first thing to do when attempting to solve any problem was to make sure that all of the dimensions and units matched up. The numbers mean nothing if the equations are wrong. I suspect a few others in this newsgroup had similar training. 73, Gene W4SZ |
Revisiting the Power Explanation
Gene Fuller wrote:
This is a perfect example of how the messages get so confused around here. You cannot simply ignore dimensionality, units, vector vs. scalar, etc. That leads to really silly conclusions like expressing interference in the same units as irradiance. Just remember when you are standing at the Pearly Gates outside looking in, it was you and not me who called Eugene Hecht's conclusions "really silly". :-) -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Richard Clark wrote:
Cecil no doubt hesitates to trod this path that leads to the destruction of his other favorite, failed example of anti-glare glass. I found out later that it wasn't anti-glare glass. It was anti-reflective glass. Out of ignorance, I just chose the wrong word. However, everything I said was true about anti-reflective glass. A 1/4WL thin film acts virtually the same way a 1/4WL series matching section acts and reflections are eliminated through wave cancellation. Incidentally, your attempt at superposing powers was why your reflections turned out to be brighter than the surface of the sun. If you had instead used the irradiance equation, your result and mine would have been identical - reflections eliminated. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Richard Clark wrote:
Cecil Moore wrote: reflections eliminated. As I said, you lack of experience in the field of Optics is glaring. I have some anti-reflective glass, Richard. It does a reasonable job of wave cancellation even under less than ideal conditions with ambient light. Our previous example involved an ideal laser and an ideal 1/4WL coating of thin film. The index of refraction of the thin film material was exactly the square root of the index of refraction of the glass. Those are ideal conditions for 100% wave cancellation of reflections as anyone in the field of optics should know. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Sat, 31 Mar 2007 11:11:13 GMT, "Dave" wrote:
and hence my recommendation to abandon power as a useful method of analyzing waves in transmission lines. power is a scalar, you can't add power without resorting to digging out the phase angles that come from the current or voltage waves. Hi Dave, This is not a very compelling reason for method that has a wide application in optics; Cecil has no real experience in this field, and it shows frequently, but the math is quite common. It relates, peripherally, to Keith's advice to investigate lattice diagrams, but Cecil no doubt hesitates to trod this path that leads to the destruction of his other favorite, failed example of anti-glare glass. 73's Richard Clark, KB7QHC |
Revisiting the Power Explanation
On Sat, 31 Mar 2007 12:21:00 -0500, Cecil Moore
wrote: reflections eliminated. As I said, you lack of experience in the field of Optics is glaring. |
Revisiting the Power Explanation
Richard Clark wrote:
Cecil Moore wrote: Your own model showed the falsity of this supposed 100% cancellation. Actually, it showed perfectly the 100% cancellation by two waves each of equal magnitude and opposite phase. It was your superposition of powers that was at fault and you apparently don't even realize your mistake. Do you want to go through it again? Have you tried the brain teaser I posted? It is a lot like non-reflective thin films. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Fri, 30 Mar 2007 11:46:15 -0800, Richard Clark wrote:
On Fri, 30 Mar 2007 18:19:20 GMT, Walter Maxwell wrote: Oh, 'cmon Richard, are you saying that if a load reflected wave Is not the same statement as your earlier one: On Fri, 30 Mar 2007 14:45:56 GMT, Walter Maxwell wrote: The part I feel is contradicted is that when total re-reflection is caused without a total discontinuity such as a physical short or open circuit What is in your "without a total discontinuity" that is now found in your "load reflected wave?" Are we to now parse "total discontinuity" as being wholly different from "partial discontinuity" such that waves suddenly mix from that difference? My example of the classic AT/ATR tube evidences EVERY observation you offer, except it is a necessary load without which those observations would never appear. If I were to replace its "total discontinuity" with a weak tube (it exhibits less than total short); it too would exhibit EVERY observation you offer EXCEPT they would be imperfect or "partial discontinuities" repeated every quarter wave. It is obvious that the effect follows the physical load, not the waves (they haven't changed when the tube went bad). The physical load is the principle in the process of interference. Richard, I don't consider the AT/ATR tube relevant to the discussion of re-reflection of reflected power incident on the output of a pi-network in an RF TX. There is absolutely no example of interference that does not rely on a load to reveal it. If what you just said is true, then how do you explain nulls in a radiation pattern of an antenna having more than one radiator, for example two verticals spaced 1/4 wl and fed in quadrature, thus creating a cardioid pattern with a total null in one direction in azimuth? If the null was not created by interference between the radiations from the two verticals, then how do you explain the formation of the null? 73's Richard Clark, KB7QHC Walt, W2DU |
Revisiting the Power Explanation
On Sat, 31 Mar 2007 17:40:45 GMT, Cecil Moore
wrote: Those are ideal conditions for 100% wave cancellation of reflections as anyone in the field of optics should know. Amusing. You don't really know anyone of them, do you? Certainly none who have to live with the invalidity of this sophomoric treatment. Your own model showed the falsity of this supposed 100% cancellation. As I said, your visiting the topic of Lattice diagrams is not even on the horizon, and behind it, Fresnel's equations are even more remote. (I can well imagine you flipping through the index to find those pages to presume their understanding, your Xeroxing other's work is apparent.) Like I said, your are quite ignorant to the matters of Optics. RF runs a poor second, as your professional life in binary design has corrupted your thinking. |
Revisiting the Power Explanation
Richard Clark wrote:
On Sat, 31 Mar 2007 18:57:02 GMT, Cecil Moore wrote: Do you want to go through it again? Do I need to see you crawl again? Here it is again. Reflections are eliminated at the air to thin-film surface. In the following fixed font diagram, IR is the Index of Refraction. air | 1/4WL thin-film | Glass 1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---... Ifor=1W | Ifor=1.0101W | Ifor=1W Iref=0W | Iref=0.0101W | Iref=0 Note that I is "irradiance", not current. Given: The irradiance reflection coefficient is 0.01 at both interfaces. The irradiance transmission coefficient is 0.99 at both interfaces. Have you tried the brain teaser I posted? It is a lot like non-reflective thin films. proves my point adequately (your fumbling with patch-work proofs like bracing the SQRT with absolutes is funny only once however). Well, Hecht apparently didn't have to deal with nit-pickers as exist on this newsgroup. If such had any importance at all, I'm sure he would have mentioned it. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Sat, 31 Mar 2007 18:57:02 GMT, Cecil Moore
wrote: Do you want to go through it again? Do I need to see you crawl again? No one is really interested in your capacity to hug a second rate explanation; neither am I. The amusement factor may provide comic relief, but your invitation: Have you tried the brain teaser I posted? It is a lot like non-reflective thin films. proves my point adequately (your fumbling with patch-work proofs like bracing the SQRT with absolutes is funny only once however). |
Revisiting the Power Explanation
On Sat, 31 Mar 2007 19:16:23 GMT, Walter Maxwell
wrote: My example of the classic AT/ATR tube evidences EVERY observation you offer, except it is a necessary load without which those observations would never appear. If I were to replace its "total discontinuity" with a weak tube (it exhibits less than total short); it too would exhibit EVERY observation you offer EXCEPT they would be imperfect or "partial discontinuities" repeated every quarter wave. It is obvious that the effect follows the physical load, not the waves (they haven't changed when the tube went bad). The physical load is the principle in the process of interference. Richard, I don't consider the AT/ATR tube relevant to the discussion of re-reflection of reflected power incident on the output of a pi-network in an RF TX. Hi Walt, And yet they are relevant to the larger topic of reflection, are they not? That is their sole purpose after all, they exhibit EVERY observation you've offered, and without them those observations disappear. There is absolutely no example of interference that does not rely on a load to reveal it. If what you just said is true, then how do you explain nulls in a radiation pattern of an antenna having more than one radiator, for example two verticals spaced 1/4 wl and fed in quadrature, thus creating a cardioid pattern with a total null in one direction in azimuth? If the null was not created by interference between the radiations from the two verticals, then how do you explain the formation of the null? I use a load. It is exactly like the internal resistance of a internal resistance where the reflected energy is in phase with the source. The load exhibits no current flow (a null). Shift the phase 180 and the internal resistance exhibits a dramatic current flow (a lobe). The energies are there either way. Remove the internal resistance and nothing conducts EVER. Clearly the internal resistance (the load in this case) is what reveals interference, not the energies. In free space they would pass like ships in the night. 73's Richard Clark, KB7QHC |
Revisiting the Power Explanation
On Sat, 31 Mar 2007 18:01:53 -0800, Richard Clark wrote:
On Sat, 31 Mar 2007 19:16:23 GMT, Walter Maxwell wrote: snip There is absolutely no example of interference that does not rely on a load to reveal it. If what you just said is true, then how do you explain nulls in a radiation pattern of an antenna having more than one radiator, for example two verticals spaced 1/4 wl and fed in quadrature, thus creating a cardioid pattern with a total null in one direction in azimuth? If the null was not created by interference between the radiations from the two verticals, then how do you explain the formation of the null? I use a load. It is exactly like the internal resistance of a internal resistance where the reflected energy is in phase with the source. The load exhibits no current flow (a null). Shift the phase 180 and the internal resistance exhibits a dramatic current flow (a lobe). The energies are there either way. Remove the internal resistance and nothing conducts EVER. Clearly the internal resistance (the load in this case) is what reveals interference, not the energies. In free space they would pass like ships in the night. 73's Richard Clark, KB7QHC Richard, you are dodging the question again! You have not answered my question, 'how do you explain the formation of the null in an antenna pattern' in the example I presented above? Walt |
Revisiting the Power Explanation
On Sun, 01 Apr 2007 01:01:02 GMT, Cecil Moore
wrote: If such had any importance at all, I'm sure he would have mentioned it. So, Hecht is rather limited is he? Absolutely nothing about Fresnel's Equations, hmmm? Now that is shallow - but for who? As you are generally a poor reporter, I'll drop the second shoe for others: Hecht does cover them. |
Revisiting the Power Explanation
Richard Clark wrote:
Cecil Moore wrote: Richard Clark wrote: proves my point adequately (your fumbling with patch-work proofs like bracing the SQRT with absolutes is funny only once however). You deliberately trimmed the above to try to falsify the meaning of my posting through innuendo. Well, Hecht apparently didn't have to deal with nit-pickers as exist on this newsgroup. If such had any importance at all, I'm sure he would have mentioned it. So, Hecht is rather limited is he? Absolutely nothing about Fresnel's Equations, hmmm? Now that is shallow - but for who? As you are generally a poor reporter, I'll drop the second shoe for others: Hecht does cover them. Richard, don't you think falsifying postings is getting a little desparate? As you know, and as indicated by your posting to which I was responding, the subject was which square root to use in the irradiance equations, not Fresnel's equations. I have honestly reproduced the thread above. You have been caught red-handed cutting and editing in an attempt to falsify the meaning of my posting. That's a violation of netnews rules and probably a violation of your ISP's rules. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Sun, 01 Apr 2007 02:44:04 GMT, Walter Maxwell wrote:
On Sat, 31 Mar 2007 18:01:53 -0800, Richard Clark wrote: On Sat, 31 Mar 2007 19:16:23 GMT, Walter Maxwell wrote: snip There is absolutely no example of interference that does not rely on a load to reveal it. If what you just said is true, then how do you explain nulls in a radiation pattern of an antenna having more than one radiator, for example two verticals spaced 1/4 wl and fed in quadrature, thus creating a cardioid pattern with a total null in one direction in azimuth? If the null was not created by interference between the radiations from the two verticals, then how do you explain the formation of the null? I use a load. It is exactly like the internal resistance of a internal resistance where the reflected energy is in phase with the source. The load exhibits no current flow (a null). Shift the phase 180 and the internal resistance exhibits a dramatic current flow (a lobe). The energies are there either way. Remove the internal resistance and nothing conducts EVER. Clearly the internal resistance (the load in this case) is what reveals interference, not the energies. In free space they would pass like ships in the night. 73's Richard Clark, KB7QHC Richard, you are dodging the question again! You have not answered my question, 'how do you explain the formation of the null in an antenna pattern' in the example I presented above? Walt Richard, are you avoiding answering my question because you don't want to admit that the null in an antenna pattern results from interference between two coherent fields? Walt |
Revisiting the Power Explanation
On Sun, 01 Apr 2007 12:41:05 GMT, Cecil Moore
wrote: That's a violation of netnews rules and probably a violation of your ISP's rules. ******* SPOILER FOLLOWS ************ A toothless cowboy outside his moderated day-job! I suppose you itch to snip, delete, or otherwise discard posts like another Texan by the name of Gonzales. Should we call you "Judge" instead of Gabby? |
Revisiting the Power Explanation
On Sun, 01 Apr 2007 16:43:01 GMT, Walter Maxwell wrote:
On Sun, 01 Apr 2007 02:44:04 GMT, Walter Maxwell wrote: On Sat, 31 Mar 2007 18:01:53 -0800, Richard Clark wrote: On Sat, 31 Mar 2007 19:16:23 GMT, Walter Maxwell wrote: snip There is absolutely no example of interference that does not rely on a load to reveal it. If what you just said is true, then how do you explain nulls in a radiation pattern of an antenna having more than one radiator, for example two verticals spaced 1/4 wl and fed in quadrature, thus creating a cardioid pattern with a total null in one direction in azimuth? If the null was not created by interference between the radiations from the two verticals, then how do you explain the formation of the null? I use a load. It is exactly like the internal resistance of a internal resistance where the reflected energy is in phase with the source. The load exhibits no current flow (a null). Shift the phase 180 and the internal resistance exhibits a dramatic current flow (a lobe). The energies are there either way. Remove the internal resistance and nothing conducts EVER. Clearly the internal resistance (the load in this case) is what reveals interference, not the energies. In free space they would pass like ships in the night. 73's Richard Clark, KB7QHC Richard, you are dodging the question again! You have not answered my question, 'how do you explain the formation of the null in an antenna pattern' in the example I presented above? Walt Richard, are you avoiding answering my question because you don't want to admit that the null in an antenna pattern results from interference between two coherent fields? Walt Richard, also consider an open-wire transmission line, equal and opposite currents flowing on each wire, and no common-mode currents. There is zero radiation, because the opposing fields developed by the current flow cancel. Are you denying that the zero radiation results from interferece? Also, consider standing waves on on a line, resulting from the superposition of the forward and reflected waves, where the maximum amplitude results from constructive interference and the minimum amplitude results from destructive interference. Are you denying the existence of interference in this case? Walt |
Revisiting the Power Explanation
Richard Clark wrote:
Cecil Moore wrote: Richard, did you unethically edit my posting to make it appear that I said something different from what I said? I suppose ... OK, please don't do that anymore. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Sun, 01 Apr 2007 23:43:27 -0700, Richard Clark wrote:
On Sun, 01 Apr 2007 16:43:01 GMT, Walter Maxwell wrote: Richard, are you avoiding answering my question because you don't want to admit that the null in an antenna pattern results from interference between two coherent fields? Hi Walt, Of course a null, or a peak, or any point in between results in the interference (the combination of energy phases as power from separate sources, or separate waves) found only in a load. I recall having posted comments specifically to the issue of phase combination several many times in as many days. I will do it several times more he That same point in space without that load has absolutely no effect on any energies that passes through it. Energies do not mix (combine) in linear space; thus one energy can have no effect on other energies. Any single energy is unperturbed by any other energies without that load to offer a point of summation (the point of interference). That unloaded point in space cannot support reflections or force any change on any energy. Interference is not the cessation of energy flow; it is not an impediment to energy flow; it is not an redirection of energy flow; it is not the amplification of energy flow; it is merely the passive observation at a point of the summation in a load of all contributions of energy flow. Further, the load may compound the redistribution of energy flows (AKA directors or reflectors) becoming, as it were, a new and separate source for a yet another remote load to combine new phase relationships into a new null/peak/what-have-you. Remove the load, and those products disappear. Interference follows the load. Interference is caused by source relationships developed at the load. Move ANY of these actors, and the entire map of responsivity changes. All antenna graphical lobe descriptions demonstrate this. 73's Richard Clark, KB7QHC Allright, Richard, then where is the load when there are two vertical radiators spaced 1/4 wl and fed in quadrature, such that their individual fields are omni-directional in azimuth, but when the two fields combine, a maximum greater than either of the individual fields is propagated in one direction, while a null results in the opposite direction, negating the propagation of the individual fields in the direction of the null. What is your explanation of the negation of the propagation of the individual fields when both radiators are radiating equal EM energy? Are you still denying that interference is not the cause of the modification of the resultant fields? So I repeat the question--where is the load in this case? Walt |
Revisiting the Power Explanation
Allright, Richard, then where is the load when there are two vertical
radiators spaced 1/4 wl and fed in quadrature, such that their individual fields are omni-directional in azimuth, but when the two fields combine, a maximum greater than either of the individual fields is propagated in one direction, while a null results in the opposite direction, negating the propagation of the individual fields in the direction of the null. What is your explanation of the negation of the propagation of the individual fields when both radiators are radiating equal EM energy? Are you still denying that interference is not the cause of the modification of the resultant fields? So I repeat the question--where is the load in this case? Walt Shoot Walt, even I can answer that one. The load is called a pair of vertical radiators spaced 1/4 Lambda and fed in quadrature. W4ZCB |
Revisiting the Power Explanation
"Harold E. Johnson" wrote in message news:HBUPh.24578$_c5.3773@attbi_s22... Allright, Richard, then where is the load when there are two vertical radiators spaced 1/4 wl and fed in quadrature, such that their individual fields are omni-directional in azimuth, but when the two fields combine, a maximum greater than either of the individual fields is propagated in one direction, while a null results in the opposite direction, negating the propagation of the individual fields in the direction of the null. What is your explanation of the negation of the propagation of the individual fields when both radiators are radiating equal EM energy? Are you still denying that interference is not the cause of the modification of the resultant fields? So I repeat the question--where is the load in this case? Walt Shoot Walt, even I can answer that one. The load is called a pair of vertical radiators spaced 1/4 Lambda and fed in quadrature. W4ZCB Harold -- I'm going to do a 'Cecil' on you...;) Wrong! The load is the space into which the antennae is radiating! The devil made me do that! Sorry. Dean -- W4IHK |
Revisiting the Power Explanation
Harold -- I'm going to do a 'Cecil' on you...;) Wrong! The load is the space into which the antennae is radiating! The devil made me do that! Sorry. Dean -- W4IHK Hi Dean, these guys have WAY too much time on their hands. If they had a life, I'm sure they wouldn't know what to do with it. Regards W4ZCB |
Revisiting the Power Explanation
On Apr 1, 11:55 pm, "Harold E. Johnson" wrote:
Harold -- I'm going to do a 'Cecil' on you...;) Wrong! The load is the space into which the antennae is radiating! The devil made me do that! Sorry. Dean -- W4IHK Hi Dean, these guys have WAY too much time on their hands. If they had a life, I'm sure they wouldn't know what to do with it. Regards W4ZCB - Hide quoted text - - Show quoted text - Hi Harold, I've found out that it takes a lot of time to spar with Richard C. When he gets ya he won't let go. He and I have somewhat different backgrounds--he's an English major, while mine are math and physics. That's probably why his definition of 'interference' disagrees with mine. As you've probable already determined, I maintain that modification of an antenna pattern from that of a single dipole results from interference between the fields radiated from more than one radiator, and the pattern is determined by the relative phase between the interfering fields. Kraus developed 'interferometers' to analyze the field relationship between the fields radiated from the radiators in his radio telescope. What more does one need to understand the effect of interference? SWR on a transmission line is another example of interference, that between the forward and reflected, is it not? Thanks for permitting me the 'time'. Walt, W2DU |
Revisiting the Power Explanation
On Mar 29, 7:33 pm, Cecil Moore wrote:
Unless it is located at a physical impedance discontinuity, absolutely nothing happens because of the V/I ratio. The last half of the sentence is absolute correct. The V/I ratio is a result, not a cause. 73, ac6xg |
Revisiting the Power Explanation
On Apr 2, 1:51 pm, Richard Clark wrote:
On Sun, 01 Apr 2007 18:30:06 GMT, Cecil Moore wrote: Cecil Moore wrote: Richard, did you unethically edit my posting to make it appear that I said something different from what I said? So, are you still sleeping with Hecht? Richard, it's very uncommon, but on this issue I'm having a difficult time following you. The only reason that I can conclude for my lack of understanding is that our definition of 'interference' must be divergent. So I'll just drop the discussion--OK? Walt |
Revisiting the Power Explanation
On Mar 31, 6:01 pm, Cecil Moore wrote:
In the following fixed font diagram, IR is the Index of Refraction. IR usually means 'infra red' in optics discussions. The letter 'n' is customarily used for the variable representing index of refraction. Sort of obviates the need to define one in every discussion. air | 1/4WL thin-film | Glass 1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---... Ifor=1W | Ifor=1.0101W | Ifor=1W Iref=0W | Iref=0.0101W | Iref=0 Note that I is "irradiance", not current. Also note that 'Watt' isn't a unit of irradiance or current. 73, ac6xg |
Revisiting the Power Explanation
On Mar 28, 7:38 pm, Cecil Moore wrote:
Keith Dysart wrote: Once again the hypothetical equality Preflected = Pdissipated + Pre-reflected 229.6 = 872 + 0 does not hold. I have pointed out your errors and misconceptions 3-4 times and you have refused to correct them. This is the last time I am going to waste my time. The source is a 2A Norton with a shunt 450 ohm resistor. During steady-state, the source sees 75 ohms. Adding 1WL of 75 ohm lossless line doesn't change anything but for the Nth time, points out your errors and misconceptions. source--1WL 75 ohm line--+--1000' 450 ohm line--75 ohm load Pfor1-- Pfor2-- Pload --Pref1 --Pref2 Taking your numbers, Pload = 220.4w, Pfor2 = 450w, and Pref2 = 229.6w Obviously Pfor1 = Pload = 220.4w and Pref1 = ZERO The joules/sec into the impedance discontinuity must equal the joules/sec out of the impedance discontinuity. Let's see if they do. Pfor1 + Pref2 = Pref1 + Pfor2 220.4 + 229.6 = ZERO + 450 450 joules/sec = 450 joules/sec There you have it. You simply made a mistake. There is NO violation of the conservation of energy principle. The same conditions that exist at the impedance discontinuity also exist at the source. Total destructive interference toward the source is accompanied by total constructive interference toward the load. Every sliver of energy is accounted for. You analysis is, once again, simply wrong. -- 73, Cecil http://www.w5dxp.com I didn't catch this first time round, but better late then never. As you say, adding 1 wavelength of 75 Ohm line will not change the steady state. So the condition on the right side of the inserted 75 Ohm line section will be reproduced on the left side of the line section. This means that reflecting back into the generator from the left end of the 75 Ohm line will be the same Pref2 = 229.6 W that exists on the right side of the 75 Ohm line section. If you doubt, do the same energy analysis for the left end of the line that you did for the right end. The generator is putting 450 W forward into the connection to the 75 Ohm line, 229.6 W is being reflected and 220.4 W is entering the 75 Ohm line. All is in balance at the connection at the left end as it is at the right end. But 229.6 W are going back into the generator. Where do these Watts go? Using your analytical approach it may help if you put a one wavelength section of 450 Ohm line between the generator and the 75 Ohm line section previously inserted. This will not change the steady state result but may make it easier to visualize the reflected 229.6 W. ....Keith |
Revisiting the Power Explanation
On Sun, 01 Apr 2007 16:43:01 GMT, Walter Maxwell
wrote: Richard, are you avoiding answering my question because you don't want to admit that the null in an antenna pattern results from interference between two coherent fields? Hi Walt, Of course a null, or a peak, or any point in between results in the interference (the combination of energy phases as power from separate sources, or separate waves) found only in a load. I recall having posted comments specifically to the issue of phase combination several many times in as many days. I will do it several times more he That same point in space without that load has absolutely no effect on any energies that passes through it. Energies do not mix (combine) in linear space; thus one energy can have no effect on other energies. Any single energy is unperturbed by any other energies without that load to offer a point of summation (the point of interference). That unloaded point in space cannot support reflections or force any change on any energy. Interference is not the cessation of energy flow; it is not an impediment to energy flow; it is not an redirection of energy flow; it is not the amplification of energy flow; it is merely the passive observation at a point of the summation in a load of all contributions of energy flow. Further, the load may compound the redistribution of energy flows (AKA directors or reflectors) becoming, as it were, a new and separate source for a yet another remote load to combine new phase relationships into a new null/peak/what-have-you. Remove the load, and those products disappear. Interference follows the load. Interference is caused by source relationships developed at the load. Move ANY of these actors, and the entire map of responsivity changes. All antenna graphical lobe descriptions demonstrate this. 73's Richard Clark, KB7QHC |
Revisiting the Power Explanation
On Sun, 01 Apr 2007 19:04:49 GMT, Walter Maxwell
wrote: Allright, Richard, then where is the load when there are two vertical radiators spaced 1/4 wl and fed in quadrature, such that their individual fields are omni-directional in azimuth, but when the two fields combine, a maximum greater than either of the individual fields is propagated in one direction, while a null results in the opposite direction, negating the propagation of the individual fields in the direction of the null. What is your explanation of the negation of the propagation of the individual fields when both radiators are radiating equal EM energy? Are you still denying that interference is not the cause of the modification of the resultant fields? So I repeat the question--where is the load in this case? Hi Walt, The load is anywhere you place it, obviously. We can even abstract one radiator of these two radiators being a load for the other - and through symmetry, the other way around. They obviously interefere with each other. As I've offered, remove (or simply move) either and the entire picture changes. If you want to add yet another, remote load, that is fine. There is no limit to where it can reside. There is no limit in the number of loads either. All loads reveal interference, however, if you remove them, obviously there is no interference. Fields do not act one upon the other in linear space. The addition of a load may present a new source of radiation. In that case, the new mapping of wave action follows the load. This is demonstrated in every yagi. Remove the load (or yagi element) and the specific mapping changes, following the load. 73's Richard Clark, KB7QHC |
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