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Jim Kelley wrote:
I don't think we should expect to see any momentum "calculations" from Cecil. He can't even explain how the energy magically reverses direction and heads back toward the load. It's not magic and is explained on the Melles-Groit web page at: http://www.mellesgriot.com/products/optics/oc_2_1.htm Here's a quote: "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then REFLECTED WAVEFRONTS INTERFERE DESTRUCTIVELY, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be ZERO." Destructive interference stops the energy flow and momentum of the two reflected waves toward the source (at the Z0-match point 'x' below. Continuing: "In the absence of absorption or scatter, the principle of conservation of energy indicates all "lost" reflected intensity will appear as ENHANCED INTENSITY (constructive interference) in the transmitted beam" (traveling toward the load). The energy involved in the destructive interference is not lost and appears as constructive interference in the opposite direction. That energy has changed direction and had its momentum reversed. The fact is, in a matched system, energy obviously isn't reflected from the load. No reflections from the load? Neglecting losses, I calculate 178 watts reflected from the load and 278 watts of forward power on the 450 ohm feedline in the following matched system. 100w XMTR---50 ohm feedline---x---1/2WL 450 ohm feedline---50 ohm load -- 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! =----- |
Steve Nosko wrote:
"I am missing just what it is that gets to the "Loss-less resistance" conclusion." If part of the source resistance were not lossless, efficiency would be limited to 50%. Fact is, Class-C amplification frequently is 2/3 or 66.6% efficient. Twice the power is delivered to the load as is lost in the source. If we deliver 1000 watts into 50 ohms, we have 224 volts and 4.47 amps as a load. Our source is identical,, 224 volts and 4.47 amps. Its volts to amps is the same ratio, but its loss is not the same power as the power delivered to the load because part of the source resistance is lossless because it is the product of interrupted energy delivery, not energy conversion into heat. With 2/3 efficiency, when we have 1000 watts into the load, we have 500 watts lost in the source. 500 watts lost means only 1/2 the dissipative resistance as we have resistance in the load, and thus the source resistance consists of 25 ohms of dissipative resistance and 25 ohms of non-dissipative resistance. The total source resistance is 50 ohms which matches the load resistance. A match allows maximum power transfer. The less than 360 degrees of energy supplied to the load makes 25 ohms of dissipationless resistance as a part of our 50-ohm source in our example. Keep working on the idea and eventually you may get the model. Best regards, Richard Harrison, KB5WZI |
Steve Nosko wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load Assuming yes... I think you are saying that we have a 450 ohm line and a 50 ohm load and therefore there is a reflection. Is that where you are? Yes, there are reflections on the 450 ohm ladder line and none on the 50 ohm coax. There is no reflected energy on the 50 ohm coax and therefore no momentum in the reflected waves on the 50 ohm coax. I think you contradicted yourself here. Are you saying that there IS or IS NOT reflected energy/waves on the 50 ohm section (to the left of the "--x--"?? No contradiction. There's no reflected energy on the 50 ohm coax because those two reflections are cancelled at the Z0-match point 'x'. They are equal in magnitude and opposite in phase. So what changes the direction and momentum of the energy wave reflected from the load? You lost me here. Perhaps you are asking; "If there is a wave reflected at the end with the 50 ohm, back toward the left, how does this power then get absorbed IN that load?" Is this the question? No, there is 178 joules/sec rejected by the load on the 450 ohm line. That energy possesses direction and momentum toward the source. What reverses the direction and momentum of that reflected energy? The wave, in the 450 section, reflected from the load is NOT the power being delivered TO the load, so it does not have to be "changed' to be delivered there. There is 100 joules/sec traveling to the right and none to the left on the 50 ohm coax. There is 278 joules/sec traveling to the right and 178 joules/sec to the left on the 450 ohm ladder-line. How does the rearward traveling 178 joules/sec, rejected by the load, get turned around and join the forward traveling 100 source joules/sec in order to add up to 278 joules/sec of forward power on the ladder line that is incident upon the load? It's a simple question: How does energy rejected by the load become energy incident upon the load? -- 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! =----- |
Cecil, W5DXP wrote:
"It`s not magic and is explained on the Melles-Groit web page---. (how the energy magically reverses direction and heads back toward the load.) I agree. It isn`t magic. Optical examples are good because we can see reflections. I think we have impedances which are inherent from the physical and electrical characteristics that describe the path the electrical energy takes. The path enforces the voltage to current ratio called impedance. First example is Zo. Second example is 377 ohms of free-space. Third example is resistor type resistance. Resistance is a property based on configuration, dimensions, material, and temperature which make a certain voltage to current ratio, which is measured in ohms. One ohm is the resistance at zero degrees C of a uniform column of mercury 106.300 cm long and weighing 14.451 grams. One volt across a resistor of one ohm causes a current of one ampere. A wave travels down a transmission line and it consists of a traveling electric field and an associated traveling magnetic field. Relative strengths of the electric and magnetic fields on a practical uniform transmission line are forced into a ratio Zo which is the square root of the ratio of the inductance per unit length divided by the capacitance per unit length. Zo is measured in ohms. If the line is terminated in an impedance other than Zo, one of the two fields is in limited supply as compared with its travel partner. As the termination only accepts energy in its fixed ohms ratio of voltage to current, surplus energy in the ample field is all rejected and reflected as it has nowhere else to go other than to reverse its course. The reflected wave must conform to Zo the same as the incident must. Seems to me there`s no magic. The waves just do what they must. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Seems to me there`s no magic. The waves just do what they must. Reckon why some people believe that wave energy is allowed to change directions at the load but not allowed to change directions at the match point? One of the problems in the field of RF (that the optics people don't have) is the effective reflection coefficient Vs the physical reflection coefficient. I have never heard an optics engineer say, "Since reflections are eliminated at the thin-film surface, the effective index of refraction of the thin-film is 1.0." The optics reflection coefficient doesn't change with the magnitude of reflected energy. It is always (n2-n1)/(n2+n1) where 'n' is the index of refraction. Yet RF engineers will say, "Since reflections are eliminated at the 50 ohm to 450 ohm impedance discontinuity, the reflection coefficient is zero." Why isn't the reflection coefficient always (Z2-Z1)/(Z2+Z1) as it is in the field of optics? Note: [(Z2-Z1)/(Z2+Z1)]^2 = [(n2-n1)/(n2+n1)]^2 The term on the left side of the equation is the RF power reflection coefficient. The term on the right side of the equation is the Reflectance (the irradiance optical reflection coefficient). Irradiance is energy per unit time per unit area and is equal to power per unit area. Thus the irradiance of a confined laser beam is equivalent to power in a confined transmission line. -- 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! =----- |
Interesting question and actually BEFORE I read R. Clark's post I decided to say that I will begg off at this point. I may be overcome with curiosity on this question in the near future (like I did for the QST WATTMETER article calculations), but for now I prefer not to take the challenge. Joules are much more than I want to consider. Job, Wife, Mother, Mother-in-law, sister-in-law all needing care and a job that is full of crap. I use this for enjoyable discussions that I can contribute to and this a way more brain power that I want to devote. I suspect you, Cecil, can explain it. I know that a 1/2 wave repeats the load Z, smith Chart and all that and can work with that. you win. 73 -- Steve N, K,9;d, c. i My email has no u's. "Cecil Moore" wrote in message ... Steve Nosko wrote: 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load Assuming yes... I think you are saying that we have a 450 ohm line and a 50 ohm load and therefore there is a reflection. Is that where you are? Yes, there are reflections on the 450 ohm ladder line and none on the 50 ohm coax. There is no reflected energy on the 50 ohm coax and therefore no momentum in the reflected waves on the 50 ohm coax. I think you contradicted yourself here. Are you saying that there IS or IS NOT reflected energy/waves on the 50 ohm section (to the left of the "--x--"?? No contradiction. There's no reflected energy on the 50 ohm coax because those two reflections are cancelled at the Z0-match point 'x'. They are equal in magnitude and opposite in phase. So what changes the direction and momentum of the energy wave reflected from the load? You lost me here. Perhaps you are asking; "If there is a wave reflected at the end with the 50 ohm, back toward the left, how does this power then get absorbed IN that load?" Is this the question? No, there is 178 joules/sec rejected by the load on the 450 ohm line. That energy possesses direction and momentum toward the source. What reverses the direction and momentum of that reflected energy? The wave, in the 450 section, reflected from the load is NOT the power being delivered TO the load, so it does not have to be "changed' to be delivered there. There is 100 joules/sec traveling to the right and none to the left on the 50 ohm coax. There is 278 joules/sec traveling to the right and 178 joules/sec to the left on the 450 ohm ladder-line. How does the rearward traveling 178 joules/sec, rejected by the load, get turned around and join the forward traveling 100 source joules/sec in order to add up to 278 joules/sec of forward power on the ladder line that is incident upon the load? It's a simple question: How does energy rejected by the load become energy incident upon the load? -- 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! =----- |
High Wind alert!
"Richard Harrison" wrote in message ... Steve Nosko wrote: "I am missing just what it is that gets to the "Loss-less resistance" conclusion." If part of the source resistance were not lossless, efficiency would be limited to 50%. I disagree. DC or RF: Real source = XX volts Real source resistance = 10 ohms Real load 50 ohms Efficiency is 50% DONE. (and I have no "loss-less resistance" anywhere) I can't figure out where you are going nor what the hole is in the technology that needs this extra stuff... are there formulas for it? Fact is, Class-C amplification frequently is 2/3 or 66.6% efficient. Twice the power is delivered to the load as is lost in the source. Yea. We both know all this. I'm trying to get to the WHY part. The issue is WHERE or WHAT is this loss-less resistance? Where is it? Why is is even needed? If we deliver 1000 watts into 50 ohms, we have 224 volts and 4.47 amps as a load. Our source is identical,, 224 volts and 4.47 amps. Its volts to amps is the same ratio, but its loss is not the same power as the power delivered to the load because part of the source resistance is lossless I stop right here and say: It is because the source resistanace is *less than* the load resistance. Simple as that! I can not come to any other conclusion. because it is the product of interrupted energy delivery, not energy conversion into heat. My eyes glaze over here. "interrupted energy delivery" -- can't get a grip on this. With 2/3 efficiency, when we have 1000 watts into the load, we have 500 watts lost in the source. Yep. I believe the true issue is WHY does this happen? What is the cause of this effect? "Loss-less resistance" or Rs RL? I say the latter, I think you say the former. 500 watts lost means only 1/2 the dissipative resistance as we have resistance in the load, and thus the source resistance consists of 25 ohms of dissipative resistance and 25 ohms of non-dissipative Why is this needed? I think I see. You are saying that because we have a "match" We are at the "conjugate match" condition and therefore *MUST* be at Rs=RL. Here's where I disagree. I believe this is absolutely NOT the case as I say above. It's as simple as the fact that the amplifire IS NOT internal resistance limited, but either dissipation limited or perhaps breakdown voltage limited or cathode emission limited. NOW I THINK I GET IT (your tack)!. You believe that the source resistance MUST equal the load resistance.. Well I say Nope! DC or RF: Real source = XX volts Real source resistance = 10 ohms Real load 50 ohms Efficiency is 50% DONE. What was that guy's name Ocam? of the Ocam's razor fame:"Take the simple answer", or something to that effect. resistance. The total source resistance is 50 ohms which matches the load resistance. I beileve this is where your error of reasoning is. I can not accept that there is some other resistance which is not quantifiable/observable/measureable. The source resistance must be less than the load R for Eff 50%. I believe you are making this much more complicated that is really is. A match allows maximum power transfer. I believe this is what is leading you astray. I believe you are assuming that the maximum power of the maximum power theorem, and the maximum power out of the real transmitter, are the same maximum power. This is where I believe the error in reasoning is that requires the loss-less resistance. They are not the same maximum power. The typical DC power supply is operated with Rs RL. The power output is limited by factors other than the maximum power transfer theorem (its output resistance) suggest. The power supply is not limited by its internal resistanse, but usually its ability to dissipate what little amount of heat (relative to the load power) that is can. I'll stick my neck out (because I believe I have a firm understanding of what physical laws can not be violated) and say that my conclusion is that the tube amplifier we commonly think of, when "tuned up" absolutely can not be operating in an Rs=RL configuration. This is for the very reasons you have stated. Because the power dissipated in the tube is less than in the load, Rs does not = RL. This is is, of course, looking at the load R presented to the tube by its output tank circuit. I believe that this (the RL.-RS stuff) is the natural law which can not be violated. There is indeed SOME load which the tube likes to see in order to get whatever power it can provide to come out. And it AIN'T Rs. The less than 360 degrees of energy supplied to the load makes 25 ohms of dissipationless resistance as a part of our 50-ohm source in our example. Keep working on the idea and eventually you may get the model. Best regards, Richard Harrison, KB5WZI I 'spose it's briefly is an interesting idea for discussion, but sorry Richard, I find it hard to beieve it actually has any merit. My (I think generally accepted) model works fine, it can explain all the phenomona(sp) so far observed and... I work in the field and have never met anyone with this idea and I have yet to see anything in the (uh, oh, here it comes. nose in the air Engineer snob talk) profession which suggests this lossless stuff is there, nor is there anything that remains unexplained which needs some other effect. I guess we close disagreeing. If it works for you... Interesting journey into some serious examination of principles, however. -- Steve N, K,9;d, c. i My email has no u's. |
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Steve Nosko wrote:
you win. 73 It's not a contest, Steve, it is a search for the truth. I thought maybe you could offer something compelling. The following two questions are very similar: 1. How does RF power (joules/sec) rejected by the load wind up being incident upon that same load some time later in a Z0-matched system? 2. How does irradiance (joules/sec) rejected by a pane of glass wind up being incident upon that same glass pane some time later in a non- glare air/thin-film/glass system? Question #2 was answered decades ago by optics physicists. Question #1 still remains unanswered by RF physicists and engineers even though it has virtually the same answer as Question #2. -- 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! =----- |
Cecil,
With reference to question #2: Who says such a silly thing? Not Melles-Griot, who appear to be your favorite optical reference source. Not Born and Wolf in "Principles of Optics", which is the ultimate optics reference book. Not any professional optics experts I have ever encountered. If you go ahead and solve the antireflective glass problem using standard Maxwell's equations (sorry, Reg and Peter) with standard boundary conditions for E and H fields you will find there is not the slightest bit of confusion. This analysis is shown in many optics and E&M textbooks. In the perfectly antireflective case all of the waves keep moving in the same direction, from air to thin film to glass. If one postulates the existence of a wave in the reverse direction it will be discovered that the amplitude of that wave is zero, meaning it does not exist. There is no "bouncing back and forth" of confused energy not knowing where and how to turn around. The interference model is useful and intuitive for what it is meant to explain. However, don't expect this sort of simple handwaving model to be extendable to all sorts of silliness about energy and momentum transfer. 73, Gene W4SZ Cecil Moore wrote: Steve Nosko wrote: you win. 73 It's not a contest, Steve, it is a search for the truth. I thought maybe you could offer something compelling. The following two questions are very similar: 1. How does RF power (joules/sec) rejected by the load wind up being incident upon that same load some time later in a Z0-matched system? 2. How does irradiance (joules/sec) rejected by a pane of glass wind up being incident upon that same glass pane some time later in a non- glare air/thin-film/glass system? Question #2 was answered decades ago by optics physicists. Question #1 still remains unanswered by RF physicists and engineers even though it has virtually the same answer as Question #2. |
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