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Question about "Another look at reflections" article.
Hi all. I have a doubt about if this text from "Another look at
reflections" reflect the author ideas. "Laboratory and experimental work often requires holding incident voltage constant with variation in loading. A constant-voltage source is usually obtained for this purpose by inserting a pad having 15 to 20 dB attenuation between the generator and the line to absorb the reflected power, preventing it from reaching the generator where it would alter the line coupling and cause the generator output voltage to vary. Because of the absorption of the pad, the generator sees a nearly perfect match for all load conditions and all reflected power is lost - but these are laboratory control conditions required to obtain valid test data." For example this part = "by inserting a pad having 15 to 20 dB attenuation between the generator and the line TO ABSORB the reflected power" ¿Absorb the reflected power or amortiguate the effects of variyng load impedance? As I understand it, even without a conjugated match the reflected power neither returns to the source (Thevenin equivalent source). I give the following example: If we load a generator directly with a resistance of 10 ohms, without any transmission line, there are not traveling waves interfering, therefore there are not stationary waves, there is only a impedance mismatch affecting the source voltage ¿OK?. If then I load it with a line of half wave finished in the same 10 ohms resitance, the generator "will see" a 10 ohms load, and it will behave identically as it made it with the original "direct" load. In both cases there is not reflected power that it can return to the generator... (I admire Walter's work and this consultation should only be read as a question, not as a criticism) Thank you very much in advance. Miguel Ghezzi LU6ETJ |
Question about "Another look at reflections" article.
On 24 mayo, 17:30, Owen Duffy wrote:
lu6etj wrote in news:62702483-6d3c-43e0-aa8b- : Hi all. I have a doubt about if this text from "Another look at reflections" reflect the author ideas. "Laboratory and experimental work often requires holding incident voltage constant with variation in loading. A constant-voltage source is usually obtained for this purpose by inserting a pad having 15 to 20 dB attenuation between the generator and the line to absorb the reflected power, preventing it from reaching the generator where it would alter the line coupling and cause the generator output voltage to vary. Because of the absorption of the pad, the generator sees a nearly perfect match for all load conditions and all reflected power is lost - but these are laboratory control conditions required to obtain valid test data." For example this part = "by inserting a pad having 15 to 20 dB attenuation between the generator and the line TO ABSORB the reflected power" I would not refer to the cascading of a practical source with a 15 to 20dB attenautor at a 'constant-voltage source' as Walt did above. A Thevenin source with Zs=50+j0, with varying load impedances, will always have terminal v/i such that the Vfwd component of a traveling wave decomposition of v is constant. I give the explanation / proof athttp://vk1od.net/blog/?p=1028. Even if a given source has Zs not equal to 50+j0, a sufficiently large cascaded 50 ohm attenuator will result in a combination that is very close to the ideal Zs=50+j0, albeit at lower level. This is a common technique for 'standardising' the equivalent source impedance of a source. Owen- Ocultar texto de la cita - - Mostrar texto de la cita - Thank you for your answer Owen (but I think that it doesn't answer my specific question, possibly because I am not able to express it well due to my translation to English). ...... I agree with what you say in your answer and several things you write in your good article "Is Zs of to HF ham tx typically 50+j0? ". For example that for an idealized Theveninj generator, the incident power is independent of the load impedance (net power Pf-Pr not) (If an impedances adapting device is inserted between the line and generator and we insert directional wattmeter between the matching device and line, of course Pf varies and Pnet also do it). Regarding that output impedance of a rig be quite near to 50+j0 I think that depends on the concrete design; such a condition is not necessary and in my opinion neither optimal. Best regards Miguel Ghezzi LU6ETJ |
Question about "Another look at reflections" article.
On Mon, 24 May 2010 14:23:59 -0700 (PDT), lu6etj
wrote: Regarding that output impedance of a rig be quite near to 50+j0 I think that depends on the concrete design; such a condition is not necessary and in my opinion neither optimal. Hi Miguel, Problems arise from qualitative statements or questions instead of quantitative statements or questions. output impedance of a rig be quite near to 50+j0 is a quantitative fact (depending upon the qualitative "near"). A fact can be measured and compared by in dependant and objective means. in my opinion neither optimal is a qualitative judgment. A judgment is only as authoritative as is the authority making it - and even then it comes at a discount. We get many judgments here by folks with little authority (and such judgments are usually adorned with anti-authority messages). I have measured the characteristic Z of my two transistorized HF rigs by different methods that agree in their results. I have also measured the Z of active loads that employ the same circuitry found in power sources. The values of Z for my HF rigs wander from 35 Ohms to 75 Ohms. The variation is a function of 1. frequency; 2. power; 3. which of the two is being measured. What is notable, for those values that are furthest from 50 Ohms (the design goal), my rigs experience issues (instabilities, poor efficiency, distortion...). The variation by frequency is found at the margins (at the ends of the HF band). The variation by power is found in inabilities to maintain a constant power deliver in every band - which weakly correlates to Z offset from 50. Variation by power also defines Z. Low power operation is not going to give you a 50 Ohm source (the manufacturer designs Z for rated power). The variation by set is due to different capabilities: one set is capable of 150W, and the other 100W; thus the first is more robust. As these variations of Z are not remarkable, trying to turn them into quantitative efficiency results are speculative at best (common designs are not optimal by any stretch when you add in the complication of modes). 73's Richard Clark, KB7QHC |
Question about "Another look at reflections" article.
On 24 mayo, 19:33, Richard Clark wrote:
On Mon, 24 May 2010 14:23:59 -0700 (PDT), lu6etj wrote: Regarding that output impedance of a rig be quite near to 50+j0 I think that depends on the concrete design; such a condition is not necessary and in my opinion neither optimal. Hi Miguel, Problems arise from qualitative statements or questions instead of quantitative statements or questions.output impedance of a rig be quite near to 50+j0 is a quantitative fact (depending upon the qualitative "near"). *A fact can be measured and compared by in dependant and objective means.in my opinion neither optimal is a qualitative judgment. *A judgment is only as authoritative as is the authority making it - and even then it comes at a discount. *We get many judgments here by folks with little authority (and such judgments are usually adorned with anti-authority messages). I have measured the characteristic Z of my two transistorized HF rigs by different methods that agree in their results. *I have also measured the Z of active loads that employ the same circuitry found in power sources. *The values of Z for my HF rigs wander from 35 Ohms to 75 Ohms. *The variation is a function of 1. frequency; 2. power; 3. which of the two is being measured. What is notable, for those values that are furthest from 50 Ohms (the design goal), my rigs experience issues (instabilities, poor efficiency, distortion...). The variation by frequency is found at the margins (at the ends of the HF band). * The variation by power is found in inabilities to maintain a constant power deliver in every band - which weakly correlates to Z offset from 50. *Variation by power also defines Z. *Low power operation is not going to give you a 50 Ohm source (the manufacturer designs Z for rated power). The variation by set is due to different capabilities: *one set is capable of 150W, and the other 100W; thus the first is more robust. As these variations of Z are not remarkable, trying to turn them into quantitative efficiency results are speculative at best (common designs are not optimal by any stretch when you add in the complication of modes). 73's Richard Clark, KB7QHC Hi Richard: As I understand = "output impedance of a rig be near to 50+j0" it is not a cuantitative fact except when it refers to specific study objects, obvously not all study objects (transmitters) will necessarily share those numbers. I think a good general cuantitative assertion would be for example, "all the rig on the market gives near 50 +j0, Zout". I am not deny that experimental fact... When I say "in my opinion" I am just not giving a formal hipothesis or theory and I quite understand technical and logical limitations of such words (for that I use them) Certainly I am definiteley not an authoritative guy on any matter :( (I'm so sorry Richard, although theory of knowledge is a topic of my interest I am not in conditions of discussing about it in a (for me) foreign language :( ) I saw some of this topics discussed time ago in this newsgroup and it was not my intention to return to them because I know they motivated good ponderings and respectfully points of view in its moment. PSE, with the due respect and consideration toward you an the distinguished colleagues and friends, Would you mind return to the original question? (sorry if it is not this the most polite form to ask it) 73 Miguel Ghezzi LU6ETJ |
Question about "Another look at reflections" article.
On May 24, 3:06*pm, lu6etj wrote:
For example this part = "by inserting a pad having 15 to 20 dB I give the following example: If we load a generator directly with a resistance of 10 ohms, without any transmission line, there are not traveling waves interfering, therefore there are not stationary waves, Yes, standing waves are hard to visualize, but there is indeed same- cycle interference involving forward waves and reflected waves. There is a certain delay from the source signal to the load and back that can be calculated if one chooses. The wave reflection model is closer to Maxwell's equations than is the lumped-circuit model where EM waves propagate instantaneously. Maybe the concepts presented in the following paper would help. There is no transmission line for a Tesla coil but reflection effects still exist. The lumped circuit model, to which you allude, incorrectly assumes that signals can travel at faster than the speed of light, an obvious impossibility. http://www.ttr.com/corum/index.htm Incidentally, this is the reason that W8JI measured a 3 ns delay through a foot-long 75m loading coil. There were same-cycle reflections existing in a near-infinite SWR situation. In such a configuration, the phase of the current doesn't change at all yet W8JI assumed the measured phase change was proportional to the delay through the coil. Nothing could be farther from the technical truth. -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On 25/05/2010 07:06, lu6etj wrote:
.... I give the following example: If we load a generator directly with a resistance of 10 ohms, without any transmission line, there are not traveling waves interfering, therefore there are not stationary waves, there is only a impedance mismatch affecting the source voltage ¿OK?. If then I load it with a line of half wave finished in the same 10 ohms resitance, the generator "will see" a 10 ohms load, and it will behave identically as it made it with the original "direct" load. In both cases there is not reflected power that it can return to the generator... I used a similar problem in my discussion entitled "A simple VSWR analysis without mirrors" at http://vk1od.net/blog/?p=1259. The article then goes on to solve a real problem (ie a practical low loss line rather than the lossless line in the first example, and in your example. You should find them relevant. Owen |
Question about "Another look at reflections" article.
On 24 mayo, 21:30, Cecil Moore wrote:
On May 24, 3:06*pm, lu6etj wrote: For example this part = "by inserting a pad having 15 to 20 dB I give the following example: If we load a generator directly with a resistance of 10 ohms, without any transmission line, there are not traveling waves interfering, therefore there are not stationary waves, Yes, standing waves are hard to visualize, but there is indeed same- cycle interference involving forward waves and reflected waves. There is a certain delay from the source signal to the load and back that can be calculated if one chooses. The wave reflection model is closer to Maxwell's equations than is the lumped-circuit model where EM waves propagate instantaneously. Maybe the concepts presented in the following paper would help. There is no transmission line for a Tesla coil but reflection effects still exist. The lumped circuit model, to which you allude, incorrectly assumes that signals can travel at faster than the speed of light, an obvious impossibility. http://www.ttr.com/corum/index.htm Incidentally, this is the reason that W8JI measured a 3 ns delay through a foot-long 75m loading coil. There were same-cycle reflections existing in a near-infinite SWR situation. In such a configuration, the phase of the current doesn't change at all yet W8JI assumed the measured phase change was proportional to the delay through the coil. Nothing could be farther from the technical truth. -- 73, Cecil, w5dxp.com Hi Cecil. thank you very much for your answer. (I am a "fan" of your rationale and very practical multiband open wire antenna feeding and I do not know why it is no so popular as G5RV in my country). ..... I know a model is necessary false by definition because is only a model (an approximation to the hipothetical "real thing"), but I do not imagine what is the tranmission line in the source-load combination of my idealized model circuit example. I am not capable to judge validity of idealized circuit theory in general to this example, neither. Please, could you tell me which would be the transmission line Zo in my resistive divider example? Anyway, my question is about validity of the assertion that reflected wave -in that example- IS ABSORBED by the pad. According to my simple calculations this hipothesis, as I see it, it does not coincide with my early learnings. For example, with a 2 V generator with Zs=1 ohm, Zc=4 ohms and Zo=1 ohm, Pc is 0.64 W. Then Pf =1 W (as Owen says in his article), Pr=0.36 W. Rendering Ef=1 V, and Vf=0.6V. Summing both in phase gives 1,6 V on the line-in point (same as on load point). These are the same power and voltages values that simple resistive divider predicts. The system fulfills the Kirchoff law and all power (as we learn in circuits theory) it flows from source to load. In the example I suggest a half wave transmission line loaded with 4 ohms it is at practical effects indistinguishable from a 4 ohms resistive load directly connected to generator. What would be the reflected power that would be dissipating in Zs (or the pad)? This is not opposed to the conjugate mirror principle explained, neither other propositions given in the cited article. IMHO reflected power never "returns into" generators when they are active (and in steady state); reflected power it is re-reflected in conjugated match, and vectorially composed to render a load impedance to the generator when this is directly connected to the line (when there is not any matching devices inserted). I do not suggesting that reflected power is "virtual" or any similar thing, of course if we insert a circulator to separate both powers, generator now would see 1 ohm load, could develope 1 W incident, 0 W reflected (Pn=1W) on circulator input, 0.36 W would be outputting on the other port to render 0.64 W (Pn) to the load with 1 W Pf and 0,36 W Pr again. If I am in error please give me your explanations. 73 Miguel ghezzi LU6ETJ |
Question about "Another look at reflections" article.
On 24 mayo, 22:09, Owen wrote:
On 25/05/2010 07:06, lu6etj wrote: ... I give the following example: If we load a generator directly with a resistance of 10 ohms, without any transmission line, there are not traveling waves interfering, therefore there are not stationary waves, there is only a impedance mismatch affecting the source voltage ¿OK?. If then I load it with a line of half wave finished in the same 10 ohms resitance, the generator "will see" a 10 ohms load, and it will behave identically as it made it with the original "direct" load. In both cases there is not reflected power that it can return to the generator... I used a similar problem in my discussion entitled "A simple VSWR analysis without mirrors" athttp://vk1od.net/blog/?p=1259. The article then goes on to solve a real problem (ie a practical low loss line rather than the lossless line in the first example, and in your example. You should find them relevant. Owen Well Owen...I have just read your article ¡and you have put me definitively in a problem! :) I was never happened to question the Walter's work, I made effort in learning it being a young student. Now you have put me in a corner and I will must meditate hard about the issue (with your help, certainly). In principle I take a risk to think that somehow it is necessary to solve "what to do with the reflected traveling wave" because the model seems to demand "come off of her" somehow that it doesn't return into the generator, and an explanatory model (and IMHO appropriate) of making it is to postulate its re-reflection to be consistent with it. In conventional theory of circuits we simply apply the second law of Kirchoff because we do not have traveling waves and we accept that the energy can only flow from the generator toward the load (supposing a circuit without reactances), there is not anything to come off (in spanish = deshacerse, desprenderse) :) I am happened to think that your article combine concepts of both models, those of basic circuit theory (without traveling waves) and those of the traveling waves -same as my simple example of course- with the difference that in mine one I only wanted to point that the reflected power is not "absorbed" by the pad, and yours is critic of conjugate mirror model and other postulations. I think the mixture or combination of models -maybe- it would not be "elegant" or consistent although it can arrive to the same numerical results, but I don not dare to advance more than that in my speculations :) 73 Miguel Ghezzi - LU6ETJ |
Question about "Another look at reflections" article.
On Mon, 24 May 2010 17:17:41 -0700 (PDT), lu6etj
wrote: PSE, with the due respect and consideration toward you an the distinguished colleagues and friends, Would you mind return to the original question? (sorry if it is not this the most polite form to ask it) Hi Miguel, I presume by "original question" you mean: On Mon, 24 May 2010 13:06:19 -0700 (PDT), lu6etj wrote: ¿Absorb the reflected power or amortiguate the effects of variyng load impedance? The answer is YES. Now, if you mean by absorb that all absorbtion results in heat, then the answer is NO. If you mean by absorb that all energy is combined in a load, then the answer is YES. The difference between YES and NO is the PHASE differences of the two energies that are combined. 73's Richard Clark, KB7QHC |
Question about "Another look at reflections" article.
On 25/05/2010 16:10, lu6etj wrote:
.... I think the mixture or combination of models -maybe- it would not be "elegant" or consistent although it can arrive to the same numerical results, but I don not dare to advance more than that in my speculations :) Miguel, My explanation uses standard linear circuit theory, and the RLGC model of a transmission line (captured in the Telegrapher's equation). It is a frequency domain model, and it is complete and consistent. One of the things that creates confusion in some peoples minds is that they want one foot in the frequency domain (where you can talk about concepts like reactance, complex impedance, VSWR) and simultaneously, one in the time domain taking about re-re-reflected waves. You can work in either domain, and you can transform between domains, but trying to be in both at the same time creates problems. BTW, if you think the problem is challenging to solve in the frequency domain, don't even think about trying to solve it in the time domain. So, do not worry about re-reflection, it is dealt with as you have discovered by the steady state solution when you load the source with the (steady state) impedance seen looking into the line. The resolution of the wave component voltages and currents with KVL and KCL at the circuit nodes gives the steady state solution. The Telegrapher's equation gives you the amplitude and phase relationship of the wave components for the transmission line, not just for fictitious lossless lines, but for practical lines as the example demonstrates. A steady state frequency domain analysis is quite adequate for most ham problems. You don't see it spelled out as such, but that is how the ham handbooks describe and solve problems. As far as the myths about PAs destroyed by absorbing reflected power, see "Does SWR damage HF ham transmitters?" at http://vk1od.net/blog/?p=1081 . Owen |
Question about "Another look at reflections" article.
On 25 mayo, 03:56, Owen wrote:
On 25/05/2010 16:10, lu6etj wrote: ... I think the mixture or combination of models -maybe- it would not be "elegant" or consistent although it can arrive to the same numerical results, but I don not dare to advance more than that in my speculations :) Miguel, My explanation uses standard linear circuit theory, and the RLGC model of a transmission line (captured in the Telegrapher's equation). It is a frequency domain model, and it is complete and consistent. One of the things that creates confusion in some peoples minds is that they want one foot in the frequency domain (where you can talk about concepts like reactance, complex impedance, VSWR) and simultaneously, one in the time domain taking about re-re-reflected waves. You can work in either domain, and you can transform between domains, but trying to be in both at the same time creates problems. BTW, if you think the problem is challenging to solve in the frequency domain, don't even think about trying to solve it in the time domain. So, do not worry about re-reflection, it is dealt with as you have discovered by the steady state solution when you load the source with the (steady state) impedance seen looking into the line. The resolution of the wave component voltages and currents with KVL and KCL at the circuit nodes gives the steady state solution. The Telegrapher's equation gives you the amplitude and phase relationship of the wave components for the transmission line, not just for fictitious lossless lines, but for practical lines as the example demonstrates. A steady state frequency domain analysis is quite adequate for most ham problems. You don't see it spelled out as such, but that is how the ham handbooks describe and solve problems. As far as the myths about PAs destroyed by absorbing reflected power, see "Does SWR damage HF ham transmitters?" athttp://vk1od.net/blog/?p=1081. Owen Hi Richard and Owen To Richard: What I mean is irrelevant :) relevant is what Walt wanted to say in this sentence: "Because of the absorption of the pad, the generator sees a nearly perfect match for all load conditions and all reflected power is lost " Pllease, tell me what in english means "all reflected power is lost"? I understood (or translate or interpret) that reflected power is dissipated in the pad: Is it a bad translation/interpretation? To Owen: Sincerely thanks for your reasons. You can be sure I will take note about your explanation an take some time tu analize it, but I am not sure about to arrive at at the right conclusion because what I read in this newsgroup is a long-standing discussion here. Miguel |
Question about "Another look at reflections" article.
Another bit of reading that might help shed light on the matter is
http://eznec.com/misc/Food_for_thought.pdf, written eight years ago during one of the many times the subject has come up before on this newsgroup. The chart and discussion in the "Forward and reverse power" section show that the concept of "reflected power" being absorbed in or dissipated by the source is incorrect. I find the concept of traveling waves of average power to be misleading at best, and analyses using this concept lead to impossible conclusions like the supposed absorption of power in the source resistance. Roy Lewallen, W7EL |
Question about "Another look at reflections" article.
On 25/05/2010 18:45, lu6etj wrote:
.... To Owen: Sincerely thanks for your reasons. You can be sure I will take note about your explanation an take some time tu analize it, but I am not sure about to arrive at at the right conclusion because what I read in this newsgroup is a long-standing discussion here. Yes, it is a recurrent discussion item. No doubt someone will be along shortly to add some confusion to the pot. A parting thought, do not confuse the process of establishment of steady state with steady state. The only reason that a steady state solution is valid, is that the system spends most of its time in steady state, or substantially so. If the solution needs to focus mainly on establishment of steady state (in other words, the system never substantially settles), then you should be doing a time domain solution, and VSWR, reactance, complex impedance are not a time domain concepts. If you need to convince yourself, do a hand workup of five of ten transits with a sinusoidal excitation. See that is does converge, and quickly, and that at the load end, the reflected wave relative to the forward wave is a fixed ratio (and hence VSWR) that depends ONLY on Zo and Zl, there is NO influence by the source or any perceived source reflection on the steady state. There are some animations of this on the net, but they seem to confuse people more than enlighten them. Owen |
Question about "Another look at reflections" article.
On May 25, 2:56*am, Roy Lewallen wrote:
The chart and discussion in the "Forward and reverse power" section show that the concept of "reflected power" being absorbed in or dissipated by the source is incorrect. I'm sorry, Roy, but that is a very misleading statement. There are THREE things that can possibly happen to the reflected energy. 1. It can indeed, be absorbed/dissipated by the source but it certainly doesn't have to be. The conditions governing absorption, reflection, and redistribution of EM wave energy are well understood in the field of optics. RF gurus seem to be myopic about interference effects. 2. It can obey the laws of the reflection model for EM waves, e.g. if the source impedance is different from the Z0 of the transmission line, some reflected energy will be re-reflected. (I know that doesn't apply to your "food for thought" examples because the source resistor is equal to the Z0 of the feedline.) 3. The interference phenomenon that you completely ignored in your discussion, presumably through ignorance. This third possibility for the reflected energy redistribution is associated with constructive and destructive interference. It is the same phenomenon that is in operation with the 1/4WL coating on non-reflective glass. Here's a quote from a Florida State University web page with capitals added by me for emphasis: micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/ waveinteractions/index.html "... when two waves of equal amplitude and wavelength that are 180- degrees ... out of phase with each other meet, they are not actually annihilated, ... ALL OF THE PHOTON ENERGY present in these waves must somehow be recovered or REDISTRIBUTED IN A NEW DIRECTION, according to the law of energy conservation ... Instead, upon meeting, the photons are REDISTRIBUTED TO REGIONS THAT PERMIT CONSTRUCTIVE INTERFERENCE, so the effect should be considered as a REDISTRIBUTION OF light (and RF) waves and PHOTON ENERGY rather than the spontaneous construction or destruction of light." This is the concept that you have been missing for eight years. "Redistribution of energy" in a transmission line is simply a reversal of energy flow. Each of your examples have different magnitudes of interference (depending on the phasing between the forward energy and the reflected energy). The phasing is a variable for all of your examples. Why did you completely ignore the interference? Why would you expect a CONSTANT magnitude of reflected energy to be dissipated in examples with VARIABLE phasing and VARIABLE levels of interference??? Here are the possibilities: 1. No Interference at the source resistor - this happens when the forward wave is 90 degrees out of phase with the reflected wave at the source resistor. In this case, the reflected energy is indeed dissipated in the source resistor. I have written a short article on the no interference case at: http://www.w5dxp.com/nointfr.htm 2. Constructive Interference at the source resistor - This happens when the phase angle between the forward wave and reflected wave is less than 90 degrees at the source resistor. The voltage across the source resistor increases and therefore more power is dissipated in the source resistor. In fact, for total constructive interference, all of the forward power and all of the reflected power is dissipated in the source resistor. This happens when the SWR is infinite and the reflected wave arrives at the source resistor in phase with the forward wave from the source. 3. Destructive interference at the source resistor - This happens when the angle between the forward wave and reflected wave is between 90+ degrees and 180 degrees. In this case, reflected energy is redistributed back toward the load and less is dissipated in the source resistor. In fact, for total destructive interference, zero reflected power is dissipated in the source resistor and all of the reflected energy is redistributed back toward the load. This happens when the SWR is infinite and the reflected wave arrives at the source resistor 180 degrees out of phase with the forward wave from the source. I explained all of this in my "WorldRadio" article way back in 2005. Presumably, you have never read it. In order to understand the role that interference plays during the superposition of EM waves, you might want to read it and take a look at the references. The question of where reflected energy goes was answered long ago by optical physicists. Too bad that RF gurus remain so ignorant of that knowledge. Here's that five year old "WorldRadio" article. http://www.w5dxp.com/energy.htm You obviously know how to add voltage phasors but you are obviously ignorant as to what happens to the ExH power in each of those waves. In any optics reference book, you will find the following irradiance equation. Since irradiance uses the same units as the Poynting vector, I have changed I (irradiance) to P (power density) in the irradiance equation. I first saw this equation in Dr. Best's QEX article[1]. Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(theta) When any two EM waves are superposed, whether light waves or RF waves, this is what happens to the component powers. The phase angle, theta, is the relative phase angle between the two electric fields before superposition. The last term in the equation is called the "interference term" and it is easy to see how it modifies the total power in the wave after superposition. Energy must be conserved. If destructive interference occurs between two waves in a transmission line, constructive interference must occur in the opposite direction - and vice versa. By completely ignoring the destructive and constructive interference occurring in your "food for thought" examples, you have ignored the accepted laws of physics (available in any physics optics reference) and arrived at completely false conclusions. Roy, every one of your examples can be explained simply by using the above equation from the field of EM wave optics. Differing types/levels of interference explains every one of your examples perfectly. [1] Best, Steven R., "Wave Mechanics of Transmission Lines, Part 3", QEX, Nov/Dec 2001 -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On May 24, 10:31*pm, lu6etj wrote:
Anyway, my question is about validity of the assertion that reflected wave -in that example- IS ABSORBED by the pad. According to my simple calculations this hipothesis, as I see it, it does not coincide with my early learnings. Miguel, let's switch your example over to an easier to understand example. Assume an ideal signal generator equipped with a resistive circulator load. Let's call such a device an SGCR, a Signal Generator equipped with a Circulator and a Resistor. Assume that 100% of the reflected energy is dissipated in the circulator load resistor (none re-reflected) and none of the reflected energy reaches the source. So here is the block diagram. SGCR--------feedline--------load That model should be easier to discuss than the pad attenuator model. What do you think? -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On Tue, 25 May 2010 00:45:36 -0700 (PDT), lu6etj
wrote: To Richard: What I mean is irrelevant :) relevant is what Walt wanted to say in this sentence: "Because of the absorption of the pad, the generator sees a nearly perfect match for all load conditions and all reflected power is lost " Pllease, tell me what in english means "all reflected power is lost"? I understood (or translate or interpret) that reflected power is dissipated in the pad: Is it a bad translation/interpretation? Hi Miguel, Your translation is fine. However, I have no idea what the pad design looks like, nor do I know the component values. I have calibrated thousands of standard pads at frequencies up to the 12 GHz. They came in either a Pi design, or a T design. Their intended use is in system isolation. That is, they isolate the source from the load OR isolate the load from the source OR isolate both. For certain component values, you can replace the "OR" with "AND." You would isolate the source to keep its frequency and power constant. You would isolate the load to keep line SWR flat. For this line application, it is assumed you are calibrating either a load equal or nearly equal to Zc, or you are measuring RF power. These are the purpose of pads (they also serve the same function in audio circuits). Measuring power in the presence of SWR other than 1:1 requires sophisticated math that is rarely discussed here. Most discussion usually accepts the presumptions of special cases (which are often sufficient) and employ less rigorous formulas (which serve well within the unstated presumptions). In conventional Kirchoff analysis, the resistor that bridges the transmission line opening becomes the source (that is Vs and Rs). Pad design usually makes that one resistor for the Pi pad, or two resistors for the T pad. If you are working in accurate and precise measurement, you then account for the input (source) resistance in parallel/series combinations. This second computation is the numeric analysis of isolation. The higher the ratios of these pad resistors, the higher the isolation. It doesn't normally serve any use to have the pad apparent resistance (what I called Rs above) different from Zc or from Zload, but as this component is a sacrificial one, the designer may choose to put it to use to achieve a desired goal. Pad performance suffers with heat due to energy combinations that come from multiple/single sources. 73's Richard Clark, KB7QHC |
Question about "Another look at reflections" article.
On 25/05/2010 18:56, Roy Lewallen wrote:
Another bit of reading that might help shed light on the matter is http://eznec.com/misc/Food_for_thought.pdf, written eight years ago during one of the many times the subject has come up before on this newsgroup. Roy's notes lead you into thinking about instantaneous power, and the behaviour over a complete AC power cycle. Thinking that through, and the implication about the power/time graphs at various points on the transmission line gives insight into average energy flow, and energy exchange between line sections, line and load and line and source, and the physical bounds of exchange of stored energy during a cycle. The Telegraphers Equation that I mentioned earlier captures the behaviour of the line, and basic AC circuit theory, Joule's law etc takes you the rest of the way. Sure, working these things through takes time... and there isn't much afforded in today's world of instant gratification where appealing explanations are more accepted, irrespective of whether they are correct. Miguel, invest in yourself and your knowledge, don't accept explanations because of a vote count, accept them because they reconcile with things that you truly know as facts (which sometimes means challenging what you think you know). Owen |
Question about "Another look at reflections" article.
On 25 mayo, 12:51, Richard Clark wrote:
On Tue, 25 May 2010 00:45:36 -0700 (PDT), lu6etj wrote: To Richard: What I mean is irrelevant :) *relevant is what Walt wanted to say in this sentence: *"Because of the absorption of the pad, the generator sees a nearly perfect match for all load conditions and all reflected power is lost " Pllease, tell me what in english means "all reflected power is lost"? I understood (or translate or interpret) that reflected power is dissipated in the pad: Is it a bad translation/interpretation? Hi Miguel, Your translation is fine. However, I have no idea what the pad design looks like, nor do I know the component values. *I have calibrated thousands of standard pads at frequencies up to the 12 GHz. *They came in either a Pi design, or a T design. *Their intended use is in system isolation. *That is, they isolate the source from the load OR isolate the load from the source OR isolate both. *For certain component values, you can replace the "OR" with "AND." You would isolate the source to keep its frequency and power constant. You would isolate the load to keep line SWR flat. *For this line application, it is assumed you are calibrating either a load equal or nearly equal to Zc, or you are measuring RF power. *These are the purpose of pads (they also serve the same function in audio circuits). Measuring power in the presence of SWR other than 1:1 requires sophisticated math that is rarely discussed here. *Most discussion usually accepts the presumptions of special cases (which are often sufficient) and employ less rigorous formulas (which serve well within the unstated presumptions). In conventional Kirchoff analysis, the resistor that bridges the transmission line opening becomes the source (that is Vs and Rs). *Pad design usually makes that one resistor for the Pi pad, or two resistors for the T pad. *If you are working in accurate and precise measurement, you then account for the input (source) resistance in parallel/series combinations. *This second computation is the numeric analysis of isolation. *The higher the ratios of these pad resistors, the higher the isolation. It doesn't normally serve any use to have the pad apparent resistance (what I called Rs above) different from Zc or from Zload, but as this component is a sacrificial one, the designer may choose to put it to use to achieve a desired goal. *Pad performance suffers with heat due to energy combinations that come from multiple/single sources. 73's Richard Clark, KB7QHC Dear friends Sincerely it was not my desire to vivify old polemics but to tell the truth, eight years it is a lot of time for not having arrived to a consent!; hey boys this is science non religion! We must have a way of leaving the well! :) Is not possible you are using different models to describe an only one phenomenon?, as looking at the same cat from their muzzle or from his tail believing each one his cat is the true or real "cat" :) I finished reading Cecil's article (http://www.w5dxp.com/nointfr.htm) and I took of his example that of the 12,5 ohm load. I took a Smith's chart and obtained the line input impedance, then I applied basic circuits theory and I obtained the same value of power dissipated in Rs -exactly- As I see, if we use a simple electric model of generators and impedances to solve the problem (maybe like Owen suggests), we can explain the dissipation in Rs without appealing to any reflected power returning into the generator because the interference phenomenon that Cecil describes takes place to form the impedance that generator see. Or alternatingly the dissipation can be described by means of the equations that Cecil shows in its page. In such case we should consider both powers (direct and reflected) operating simultaneously on generator resistance. Same cat, different points of view... :) Perhaps my vision is naive but this situation reminds me an example of Sears and Semansky book "University physics" (third edition) where he explains that energy can be thought as not transported by charges in movement, but for the electromagnetic field associated to them. Last is a little hard to see -Poynting vector et al- :) but it is applicable. Always has been a pleasure for me to read you. I have learneing very much from your enthusiastic discussions. You made me think of things that I never thought without your help. Thank you. Miguel Ghezzi . LU6ETJ PS: Meanwhile I take the Owen advice and I am still studying! |
Question about "Another look at reflections" article.
On Tue, 25 May 2010 20:41:45 -0700 (PDT), lu6etj
wrote: eight years it is a lot of time for not having arrived to a consent! Hi Miguel, You got on this train rather late if all you see is eight years of it. The circular references have entertainment value - so did vaudeville. 73's Richard Clark, KB7QHC |
Question about "Another look at reflections" article.
On May 25, 10:41*pm, lu6etj wrote:
Is not possible you are using different models to describe an only one phenomenon?, as looking at the same cat from their muzzle or from his tail believing each one his cat is the true or real "cat" :) Dr. Corum tells us what the problem is: "Lumped circuit theory fails because it's a *theory* whose presuppositions are inadequate. Every EE in the world was warned of this in their first sophomore circuits course. ... The engineer must either use Maxwell's equations or distributed elements to model reality. ... Distributed theory encompasses lumped circuits and always applies." In particular, *energy flow* is not addressed at all in the lumped circuit model. Some RF gurus are so confused that they imply that there is no Poynting vector power density in reflected waves. Their basic error (for the past 8 years) is that they believe there is no mechanism outside of the reflection model that can redistribute the reflected energy. But what can happen to reflected energy has been known for decades in the field of optical physics. The reflection that one sees in a mirror contains an ExH power density that activates one's human retina. Waves cannot exist without energy. Standing waves cannot exist without forward and reverse traveling waves. That some otherwise knowledgeable and influential RF gurus deny the reality of such is really sad. What they are missing is simple. The FSU web page describes how wave cancellation redistributes the reflected energy back toward the load from what is essentially a Z0-match. The redistribution of reflected energy due to wave cancellation is technically NOT a re-reflection since it involves destructive interference between TWO waves. When the RF gurus broaden their knowledge base to include wave cancellation, they will alleviate their ignorance on how reflected energy is redistributed back toward the load. That knowledge can be obtained from any good optics reference book including "Optics", by Hecht and "Principles of Optics", by Brown and Wolf. Until those gurus admit to themselves that they are not omniscient, the argument will continue. I finished reading Cecil's article (http://www.w5dxp.com/nointfr.htm) Remember that article describes the two special cases where the two superposed waves are 90 degrees apart and therefore do not interfere with each other, i.e. no wave cancellation exists. I have not yet written the other two articles about constructive and destructive interference. -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On May 26, 6:59*am, Cecil Moore wrote:
That knowledge can be obtained from any good optics reference book including "Optics", by Hecht and "Principles of Optics", by Brown and Wolf. Continuing after taking my wife to work: The reason that optical physicists know so much more about energy transfer than RF gurus is that the optical physicists do not have the luxury of measuring the voltage and current associated with an EM wave at light frequencies. They have historically been forced to deal with irradiance, i.e. power density, at every step of their analysis since that is the only thing they could easily quantize through measurements. As a result, they know everything one needs to know about where the energy goes during reflection and wave cancellation. If one wants to catch up on such as it applies to all EM waves, including RF waves, please obtain a copy of "Optics", by Hecht and read the chapters on superposition and interference. It was an eye opener for me and resulted in my WorldRadio energy analysis article. http://www.w5dxp.com/energy.htm Optical physicists usually make power density (irradiance) measurements and then calculate the electric and magnetic fields of the EM wave. RF gurus make voltage and current measurements and ignore energy/power except for net power in and net power out thus losing important details in the process. When they don't understand energy transfer, they dismiss it as unimportant or worse yet, assume that their ignorance somehow proves something as W7EL has done in his "food for thought" article on forward and reflected power. All that he has succeeded in proving is his ignorance of partial or total wave cancellation involving two superposed component waves which can reverse the flow of energy in a transmission line just as easily as can an actual reflection. Here is a diagram of the energy flow at a 50- ohm Z0-match as is common in ham installations. http://www.w5dxp.com/enfig3.gif Pref1 = Zero = P3 + P4 - 2*SQRT(P3*P4) where P3=P4 and the two electric fields are 180 degrees out of phase. This is total destructive interference due to wave cancellation, i.e. out-of-phase superposition. Pfor2 = P1 + P2 + 2*SQRT(P1*P2) This total constructive interference due to in-phase superposition. If W7EL would use the general power density equation on his "food-for- thought" examples Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(theta) he would obtain all the correct answers as to where the reflected energy goes, i.e. the energy analysis would agree exactly with his voltage analysis. That energy analysis would tell us exactly how much reflected power is absorbed in the source resistor and exactly how much is redistributed back toward the load as part of the forward wave. But when W7EL heard these facts of physics from the field of EM wave optics many years ago, he said "Gobbleygook". -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On 26 mayo, 02:26, Richard Clark wrote:
On Tue, 25 May 2010 20:41:45 -0700 (PDT), lu6etj wrote: eight years it is a lot of time for not having arrived to a consent! Hi Miguel, You got on this train rather late if all you see is eight years of it. The circular references have entertainment value - so did vaudeville. 73's Richard Clark, KB7QHC Hi Richard You got on this train rather late if all you see is eight years of it. I am not mean that! I clearly said= "Always has been a pleasure for me to read you. I have learning very much from your enthusiastic discussions. You made me think of things that I never thought without your help. Thank you." Miguel |
Question about "Another look at reflections" article.
On May 26, 2:33*pm, lu6etj wrote:
On 26 mayo, 02:26, Richard Clark wrote: On Tue, 25 May 2010 20:41:45 -0700 (PDT), lu6etj wrote: eight years it is a lot of time for not having arrived to a consent! Hi Miguel, You got on this train rather late if all you see is eight years of it. The circular references have entertainment value - so did vaudeville. 73's Richard Clark, KB7QHC Hi Richard You got on this train rather late if all you see is eight years of it. I am not mean that! I clearly said= "Always has been a pleasure for me to read you. I have learning very much from your enthusiastic discussions. You made me think of things that I never thought without your help. Thank you." Miguel Cecil, have forgotten that because the source resistance of the RF power amp is non-dissipative, none of the reflected power is absorbed therein? Walt |
Question about "Another look at reflections" article.
On May 26, 1:56*pm, walt wrote:
Cecil, have forgotten that because the source resistance of the RF power amp is non-dissipative, none of the reflected power is absorbed therein? Actually, this discussion is based on an example introduced by W7EL which assumed a 50 ohm source resistor in his "food for thought, forward and reflected power" article posted on his web page. The source resistance of an actual RF power amp doesn't matter. W7EL's example specified a 50 ohm resistor as the source impedance. http://eznec.com/misc/Food_for_thought.pdf -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On 26 mayo, 11:09, Cecil Moore wrote:
On May 26, 6:59*am, Cecil Moore wrote: That knowledge can be obtained from any good optics reference book including "Optics", by Hecht and "Principles of Optics", by Brown and Wolf. Continuing after taking my wife to work: The reason that optical physicists know so much more about energy transfer than RF gurus is that the optical physicists do not have the luxury of measuring the voltage and current associated with an EM wave at light frequencies. They have historically been forced to deal with irradiance, i.e. power density, at every step of their analysis since that is the only thing they could easily quantize through measurements. As a result, they know everything one needs to know about where the energy goes during reflection and wave cancellation. If one wants to catch up on such as it applies to all EM waves, including RF waves, please obtain a copy of "Optics", by Hecht and read the chapters on superposition and interference. It was an eye opener for me and resulted in my WorldRadio energy analysis article. http://www.w5dxp.com/energy.htm Optical physicists usually make power density (irradiance) measurements and then calculate the electric and magnetic fields of the EM wave. RF gurus make voltage and current measurements and ignore energy/power except for net power in and net power out thus losing important details in the process. When they don't understand energy transfer, they dismiss it as unimportant or worse yet, assume that their ignorance somehow proves something as W7EL has done in his "food for thought" article on forward and reflected power. All that he has succeeded in proving is his ignorance of partial or total wave cancellation involving two superposed component waves which can reverse the flow of energy in a transmission line just as easily as can an actual reflection. Here is a diagram of the energy flow at a 50- ohm Z0-match as is common in ham installations. http://www.w5dxp.com/enfig3.gif Pref1 = Zero = P3 + P4 - 2*SQRT(P3*P4) where P3=P4 and the two electric fields are 180 degrees out of phase. This is total destructive interference due to wave cancellation, i.e. out-of-phase superposition. Pfor2 = P1 + P2 + 2*SQRT(P1*P2) This total constructive interference due to in-phase superposition. If W7EL would use the general power density equation on his "food-for- thought" examples Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(theta) he would obtain all the correct answers as to where the reflected energy goes, i.e. the energy analysis would agree exactly with his voltage analysis. That energy analysis would tell us exactly how much reflected power is absorbed in the source resistor and exactly how much is redistributed back toward the load as part of the forward wave. But when W7EL heard these facts of physics from the field of EM wave optics many years ago, he said "Gobbleygook". -- 73, Cecil, w5dxp.com Hi Cecil Continuing after taking my wife to work: Yo have a chopper, I have a chopper. Your wife go to work, my wife go to work. Are you my big brother? (older brother? = "hermano mayor", in spanish) :D .... Dr. Corum tells us what the problem is: "Lumped circuit theory fails because it's a *theory* whose presuppositions are inadequate. Well... "theories" are theories, models of "reality". As Einstein said: "free creations of the human mind" = Newton's gravity theory, Einstein's gravity theory... Einstein one of course it is more accurate in certain situations, but we still using Newton's laws to send spacecrafts to Mars. As in the Sears-Zemansky example given, they coexist and it can solve differents problems. Velocity adding fails at very high speeds ("presuppositions are inadequate") and we need relativity, but for common situations it is not necessary use the last one (of course you know it, it is only a note). It is easy for us become tempted to think that Einstein one it is the "true" theory and not "a better aproximation to reality" theory. As some of us agree "all models are false", then, perhaps --only "perhaps"-- (I do not want offend to anyone) some of the others models presented by our distinguished colleages could not be as precise for certain special situations, but still quite adequate to solve problems or explain more simple things, as a Newton laws or "charges in movement"... As in other physics laws, it is not possible to reach a similar consensus here in this regard? Models given, leads to wrong numbers or failed to agree with empiric data? Your examples, Cecil, gve me a light, now I have in my mind three models that (for me) describe "reality" so good. With yours I can think in forward and reflected power flowing simultaneously on Rs -not contradiction- Adding phasors of the forward and reflected travelling waves before the last one reach generator, this one sees different Z line input and not any reflected power there is circulating on it (not contradiction to me) = "adding" in my human mind, of course. Reality only God/Allah/ Yahweh/Manitou/The Force/Zeus, knows- :) . I'm not trying to be syncretic. Same cat, different models... Is it not possible that? 73 Miguel Ghezzi LU6ETJ |
Question about "Another look at reflections" article.
On Wed, 26 May 2010 11:33:08 -0700 (PDT), lu6etj
wrote: "Always has been a pleasure for me to read you. I have learning very much from your enthusiastic discussions. You made me think of things that I never thought without your help. Thank you." Hi Miguel, You are welcome. My comments (beyond your quote above) were in regard to you observing the amount of time Walt's topic has been under discussion. In fact, the agony of source resistance has been painfully with us for as long as newsgroups could support the noise bandwidth. As dangerous as unasked-for advice is, prepare something at your bench to measure all these contentious issues for yourself. Force the issues that are only being discussed rather than measured. Discover the roots of what used to be a "hands on" avocation. Learn the practical reality in relation to the academic meaning. Discover the first principles by making mistakes and having failures that you can correct in front of you, instead of being assisted by an "expert." Compare results with like-minded bench workers who can perform the same examinations you are doing. This is what Walt did - many times. His bench work eclipses ALL discussion of theory. The irony that inhabits this is that his bench work may even eclipse his own explanations. Absolutely no one else has dared to slide up to the bench to demonstrate that, however. The level of "critique" is much like ants scattering at the feet of a giant. There is a lot of math thrown against the wall to prove something. It may or may not be the same thing. What it does prove is: "Models are doomed to succeed." This is demonstrated here at least once a week on average, and is even held up as a hallmark of hazing, initiation, or anti-intellectual snobbery. Math/Models/Simulations/Theories serve many religious wars. 73's Richard Clark, KB7QHC |
Question about "Another look at reflections" article.
On May 26, 4:23*pm, lu6etj wrote:
I'm not trying to be syncretic. Same cat, different models... Is it not possible that? Yes, except that W7EL uses the inadequacy of his model to prove that his model yields contradictory results. I freely admit that his model is inadequate and contradictory which implies that he is using the wrong model. -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On 25 mayo, 11:49, Cecil Moore wrote:
On May 24, 10:31*pm, lu6etj wrote: Anyway, my question is about validity of the assertion that reflected wave -in that example- IS ABSORBED by the pad. According to my simple calculations this hipothesis, as I see it, it does not coincide with my early learnings. Miguel, let's switch your example over to an easier to understand example. Assume an ideal signal generator equipped with a resistive circulator load. Let's call such a device an SGCR, a Signal Generator equipped with a Circulator and a Resistor. Assume that 100% of the reflected energy is dissipated in the circulator load resistor (none re-reflected) and none of the reflected energy reaches the source. So here is the block diagram. SGCR--------feedline--------load That model should be easier to discuss than the pad attenuator model. What do you think? -- 73, Cecil, w5dxp.com Excuse me Cecil: I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. (I hope yours be the only one, I will review all thread tho chek for others). In a early post I wrote = "of course if we insert a circulator to separate both powers, generator now would see 1 ohm load, could develope 1 W incident, 0 W reflected (Pn=1W) on circulator input, 0.36 W would be outputting on the other port to render 0.64 W (Pn) to the load with 1 W Pf and 0,36 W Pr again" Is this result OK for you?. The thread advance toward more deeper issues since :), and now I have been analizing all the matter because it quickly superceed my original doubt. A few minutes ago had started to read your article (http:// www.w5dxp.com/energy.htm) and the Roy's one (http://eznec.com/misc/ Food_for_thought.pdf) and yesterday I have been reloading my old "Transmission lines antennas and wave guides" from King, Mimno & Wing to review the issue from that classical perspective. I am interested in your optic analogy, I can imagine the load as a partially reflecting surface, real part of it as absorbance (transmittance if it was a radiator). line as a unidimensional medium and reflection as the form of "redistribute energy" (is it OK?) and a coherent light source for the voltage source, but I am still trying to visualze the optical equivalent of source resistance and its job to be a good analog, Also I am interested in check other values and conditions in your other article (first part) with 45 degree line. Thank you very much for your helping and inspiration. 73 - Miguel LU6ETJ |
Question about "Another look at reflections" article.
On May 27, 9:50*am, lu6etj wrote:
I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. (I hope yours be the only one, I will review all thread tho chek for others). I am also using Google since ATT dropped Usenet. I liked Thunderbird a lot better than Google's usenet interface but I am adapting. The above information is good to know. Thunderbird has a way to keep up with unread vs read postings but Google doesn't seem to - at least I don't know how to do it on Google. In a early post I wrote = "of course if we insert a circulator to separate both powers, generator now would see 1 ohm load, could develope 1 W incident, 0 W reflected (Pn=1W) on circulator input, 0.36 W would be outputting on the other port to render 0.64 W (Pn) to the load with 1 W Pf and 0,36 W Pr again" Is this result OK for you?. The SGCR source is usually designed for 50 ohms, i.e. the signal generator always "sees" a 50 ohm load because it does not "see" any reflected energy. The ideal circulator is usually designed with 50 ohm line and a 50 ohm load resistor. If we could stick with that particular configuration for the SGCR source, it would aid in my understanding what is the actual system configuration, i.e. not your fault but I am confused by your above posting. I am interested in your optic analogy, I can imagine the load as a partially reflecting surface, real part of it as absorbance (transmittance if it was a radiator). line as a unidimensional medium and reflection as the form of "redistribute energy" (is it OK?) and a coherent light source for the voltage source, but I am still trying to visualze the optical equivalent of source resistance and its job to be a good analog, Also I am interested in check other values and conditions in your other article (first part) with 45 degree line. I don't think a laser source handles reflected energy like an RF amp does. So, to start with, let's avoid reflected energy being incident upon the laser source. Here is a good example to start with, a 1/4WL non-reflective coating on glass. Laser-----air-------|--1/4WL thin-film, r = 1.2222---|---Glass, r = 1.4938---... The 1/4WL thin-film coating on the glass acts exactly like a 1/4WL matching section of transmission line. Reflections at the air to thin- film interface are eliminated by wave cancellation just as the FSU web page says, micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/ waveinteractions/index.html "... when two waves of equal amplitude and wavelength that are 180- degrees ... out of phase with each other meet, they are not actually annihilated, ... All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light." Note that the reflection coefficient, r, is 1.0 for air. Thus the SQRT[(1.0)(1.4938)] = 1.2222 ensures that reflections are eliminated by the r = 1.2222 thin-film coating. The same thing happens at the '+' Z0-match in the following RF system. XMTR---50 ohm coax---+---1/4WL 300 ohm feedline---1800 ohm load Note that SQRT[(50)(1800)] = 300 ensuring that reflections are eliminated. -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On May 27, 9:50*am, lu6etj wrote:
Also I am interested in check other values and conditions in your other article (first part) with 45 degree line. Sorry, I forgot to comment on this. If the line length is fixed at 45 degrees, the reflected wave arrives back at the 50 ohm source resistor 90 degrees out of phase with the source's forward wave. When two waves are 90 degrees out of phase, there is zero interference between them because cos(90) = 0 and the interference term in the following equation disappears. Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(theta) There is no re-reflection of the reflected wave from the 50 ohm source resistor because it matches the coax Z0. There is no redistribution of constructive/destructive interference energy because there is zero interference. Therefore, for the special case where Vfor is 90 degrees out of phase with Vref at the source resistor, all of the reflected power will be dissipated in the source resistor. If the interference at the source resistor is constructive, i.e. less than 90 degrees difference between Vfor and Vref, the power dissipated in the source resistor will be all of the reflected power plus some of the source power. If the interference at the source resistor between Vfor and Vref is destructive, i.e. between 90 degrees and 180 degrees, some of the reflected power will be redistributed (by wave cancellation) back toward the load. Note that the I and Q method of transferring two streams of information on the same carrier relies on this same no interference (at 90 degrees) concept. -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On May 27, 9:50*am, lu6etj wrote:
I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. FYI, It appears that Google has a "Reader" available at the top of the web page, that has the mark-as-read/unread ability with which one can keep track of the read/unread postings. -- 73, Cecil, w5dxp.com |
Question about "Another look at reflections" article.
On 27 mayo, 14:05, Cecil Moore wrote:
On May 27, 9:50*am, lu6etj wrote: I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. FYI, It appears that Google has a "Reader" available at the top of the web page, that has the mark-as-read/unread ability with which one can keep track of the read/unread postings. -- 73, Cecil, w5dxp.com Meanwhile I study carefully your reply and investigate the Google Reader (thanks for tell me) tell you that one ohm examples are only because I often work with normalized impedances. I will move me to 50 ohms neighborhood... :) |
Question about "Another look at reflections" article.
On 27 mayo, 14:05, Cecil Moore wrote:
On May 27, 9:50*am, lu6etj wrote: I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. FYI, It appears that Google has a "Reader" available at the top of the web page, that has the mark-as-read/unread ability with which one can keep track of the read/unread postings. -- 73, Cecil, w5dxp.com Meanwhile I study carefully your reply and investigate the Google Reader (thanks for tell me) tell you that one ohm examples are only because I often work with normalized impedances. I will move me to 50 ohms neighborhood... :) |
Question about "Another look at reflections" article.
On 27 mayo, 14:05, Cecil Moore wrote:
On May 27, 9:50*am, lu6etj wrote: I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. FYI, It appears that Google has a "Reader" available at the top of the web page, that has the mark-as-read/unread ability with which one can keep track of the read/unread postings. -- 73, Cecil, w5dxp.com Meanwhile I study carefully your reply and investigate the Google Reader (thanks for tell me) tell you that one ohm examples are only because I often work with normalized impedances. I will move me to 50 ohms neighborhood... :) |
Question about "Another look at reflections" article.
On 26 mayo, 18:44, Richard Clark wrote:
On Wed, 26 May 2010 11:33:08 -0700 (PDT), lu6etj wrote: "Always has been a pleasure for me to read you. I have learning very much from your enthusiastic discussions. You made me think of things that I never thought without your help. Thank you." Hi Miguel, You are welcome. My comments (beyond your quote above) were in regard to you observing the amount of time Walt's topic has been under discussion. *In fact, the agony of source resistance has been painfully with us for as long as newsgroups could support the noise bandwidth. As dangerous as unasked-for advice is, prepare something at your bench to measure all these contentious issues for yourself. *Force the issues that are only being discussed rather than measured. *Discover the roots of what used to be a "hands on" avocation. *Learn the practical reality in relation to the academic meaning. *Discover the first principles by making mistakes and having failures that you can correct in front of you, instead of being assisted by an "expert." Compare results with like-minded bench workers who can perform the same examinations you are doing. This is what Walt did - many times. *His bench work eclipses ALL discussion of theory. *The irony that inhabits this is that his bench work may even eclipse his own explanations. *Absolutely no one else has dared to slide up to the bench to demonstrate that, however. *The level of "critique" is much like ants scattering at the feet of a giant. There is a lot of math thrown against the wall to prove something. *It may or may not be the same thing. *What it does prove is: * * * * "Models are doomed to succeed." This is demonstrated here at least once a week on average, and is even held up as a hallmark of hazing, initiation, or anti-intellectual snobbery. *Math/Models/Simulations/Theories serve many religious wars. 73's Richard Clark, KB7QHC Hi Richard. thanks for your reply. I recognize so much the Walter's work as I said in my initial post. "Another look at reflections" was one of the my most appreciated readings of my early days as student and Ham. But without wishing to be flattering ("adulador" in spanish), I feel in debt with much others works from you (all) (I do not give more names to not commit injustice omiting anyone). I believe you are a gifted, brilliant, intelligent and supportive Hams sharing your knowledge and experience with us. For that, I am/we are indebted to all of you :) However, I believe for all reasons given above, you will be capable to arrive to a good technical/scientific consensus about the matter. We trust in your ability and capacity to get it. This is very important for us because not all are capable to develop theory from empirical working and we need your agreement to study things that in my experience are very difficult to grasp even for university graduates... 73 Miguel Ghezzi - LU6ETJ |
Question about "Another look at reflections" article.
On Thu, 27 May 2010 17:03:28 -0700 (PDT), lu6etj
wrote: However, I believe for all reasons given above, you will be capable to arrive to a good technical/scientific consensus about the matter. We trust in your ability and capacity to get it. Hi Miguel, A nice sentiment, but even the most silvered authorities disagree. One has only to look at the relativist camp vs. the quantum camp in nuclear physics. This is very important for us because not all are capable to develop theory from empirical working and we need your agreement to study things that in my experience are very difficult to grasp even for university graduates... Then, this will be dissappointing. University is for finding your own way, and that does not come without regrets. There's a maxim that applies he "If you haven't failed, you are not trying hard enough." 73's Richard Clark, KB7QHC |
Question about "Another look at reflections" article.
On 27 mayo, 22:11, Richard Clark wrote:
On Thu, 27 May 2010 17:03:28 -0700 (PDT), lu6etj wrote: However, I believe for all reasons given above, you will be capable to arrive to a good technical/scientific consensus about the matter. We trust in your ability and capacity to get it. Hi Miguel, A nice sentiment, but even the most silvered authorities disagree. One has only to look at the relativist camp vs. the quantum camp in nuclear physics. This is very important for us because not all are capable to develop theory from empirical working and we need your agreement to study things that in my experience are very difficult to grasp even for university graduates... Then, this will be dissappointing. *University is for finding your own way, and that does not come without regrets. *There's a maxim that applies he * * * * "If you haven't failed, you are not trying hard enough." 73's Richard Clark, KB7QHC Yes Richard. I agree, however I said "even for graduated" ("aún para graduados" in spanish), but this a radio Amateur group, isn't it... Hey Richard, do not be hard with us :) your helping is important and valuable for hundred (if not thousands) of present an future Amateurs. Thanks again. Miguel |
Question about "Another look at reflections" article.
On 27 mayo, 12:34, Cecil Moore wrote:
On May 27, 9:50*am, lu6etj wrote: I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. (I hope yours be the only one, I will review all thread tho chek for others). I am also using Google since ATT dropped Usenet. I liked Thunderbird a lot better than Google's usenet interface but I am adapting. The above information is good to know. Thunderbird has a way to keep up with unread vs read postings but Google doesn't seem to - at least I don't know how to do it on Google. In a early post I wrote = "of course if we insert a circulator to separate both powers, generator now would see 1 ohm load, could develope 1 W incident, 0 W reflected (Pn=1W) on circulator input, 0.36 W would be outputting on the other port to render 0.64 W (Pn) to the load with 1 W Pf and 0,36 W Pr again" Is this result OK for you?. The SGCR source is usually designed for 50 ohms, i.e. the signal generator always "sees" a 50 ohm load because it does not "see" any reflected energy. The ideal circulator is usually designed with 50 ohm line and a 50 ohm load resistor. If we could stick with that particular configuration for the SGCR source, it would aid in my understanding what is the actual system configuration, i.e. not your fault but I am confused by your above posting. I am interested in your optic analogy, I can imagine the load as a partially reflecting surface, real part of it as absorbance (transmittance if it was a radiator). line as a unidimensional medium and reflection as the form of "redistribute energy" (is it OK?) and a coherent light source for the voltage source, but I am still trying to visualze the optical equivalent of source resistance and its job to be a good analog, Also I am interested in check other values and conditions in your other article (first part) with 45 degree line. I don't think a laser source handles reflected energy like an RF amp does. So, to start with, let's avoid reflected energy being incident upon the laser source. Here is a good example to start with, a 1/4WL non-reflective coating on glass. Laser-----air-------|--1/4WL thin-film, r = 1.2222---|---Glass, r = 1.4938---... The 1/4WL thin-film coating on the glass acts exactly like a 1/4WL matching section of transmission line. Reflections at the air to thin- film interface are eliminated by wave cancellation just as the FSU web page says, micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/ waveinteractions/index.html "... when two waves of equal amplitude and wavelength that are 180- degrees ... out of phase with each other meet, they are not actually annihilated, ... All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light." Note that the reflection coefficient, r, is 1.0 for air. Thus the SQRT[(1.0)(1.4938)] = 1.2222 ensures that reflections are eliminated by the r = 1.2222 thin-film coating. The same thing happens at the '+' Z0-match in the following RF system. XMTR---50 ohm coax---+---1/4WL 300 ohm feedline---1800 ohm load Note that SQRT[(50)(1800)] = 300 ensuring that reflections are eliminated. -- 73, Cecil, w5dxp.com May we advance in little steps to ensure we share basic assumptions? 1) I did not think of (or is think on?) a laser source, I was one step before, I think only of a "coherent" source to match monofrequency simple AC generator analogy. 2) What would be Rs optical analog? 3) Superposition is a medium phenomenon ¿yes?, for example "eter". Interference an result of it on a other "thing", for example photographic plate or screen. Are we agree? K |
Question about "Another look at reflections" article.
On May 27, 9:10*pm, lu6etj wrote:
On 27 mayo, 12:34, Cecil Moore wrote: On May 27, 9:50*am, lu6etj wrote: I am reading this newsgroup through Google groups web page and I just realized that later replies to previous post are intercalated in the thread, while I expected to see it always at the end of it, for that reason I did not ACK before to it. (I hope yours be the only one, I will review all thread tho chek for others). I am also using Google since ATT dropped Usenet. I liked Thunderbird a lot better than Google's usenet interface but I am adapting. The above information is good to know. Thunderbird has a way to keep up with unread vs read postings but Google doesn't seem to - at least I don't know how to do it on Google. In a early post I wrote = "of course if we insert a circulator to separate both powers, generator now would see 1 ohm load, could develope 1 W incident, 0 W reflected (Pn=1W) on circulator input, 0..36 W would be outputting on the other port to render 0.64 W (Pn) to the load with 1 W Pf and 0,36 W Pr again" Is this result OK for you?. The SGCR source is usually designed for 50 ohms, i.e. the signal generator always "sees" a 50 ohm load because it does not "see" any reflected energy. The ideal circulator is usually designed with 50 ohm line and a 50 ohm load resistor. If we could stick with that particular configuration for the SGCR source, it would aid in my understanding what is the actual system configuration, i.e. not your fault but I am confused by your above posting. I am interested in your optic analogy, I can imagine the load as a partially reflecting surface, real part of it as absorbance (transmittance if it was a radiator). line as a unidimensional medium and reflection as the form of "redistribute energy" (is it OK?) and a coherent light source for the voltage source, but I am still trying to visualze the optical equivalent of source resistance and its job to be a good analog, Also I am interested in check other values and conditions in your other article (first part) with 45 degree line. I don't think a laser source handles reflected energy like an RF amp does. So, to start with, let's avoid reflected energy being incident upon the laser source. Here is a good example to start with, a 1/4WL non-reflective coating on glass. Laser-----air-------|--1/4WL thin-film, r = 1.2222---|---Glass, r = 1.4938---... The 1/4WL thin-film coating on the glass acts exactly like a 1/4WL matching section of transmission line. Reflections at the air to thin- film interface are eliminated by wave cancellation just as the FSU web page says, micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/ waveinteractions/index.html "... when two waves of equal amplitude and wavelength that are 180- degrees ... out of phase with each other meet, they are not actually annihilated, ... All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light." Note that the reflection coefficient, r, is 1.0 for air. Thus the SQRT[(1.0)(1.4938)] = 1.2222 ensures that reflections are eliminated by the r = 1.2222 thin-film coating. The same thing happens at the '+' Z0-match in the following RF system. XMTR---50 ohm coax---+---1/4WL 300 ohm feedline---1800 ohm load Note that SQRT[(50)(1800)] = 300 ensuring that reflections are eliminated. -- 73, Cecil, w5dxp.com May we advance in little steps to ensure we share basic assumptions? 1) I did not think of (or is think on?) a laser source, I was one step before, I think only of a "coherent" source to match monofrequency simple AC generator analogy. 2) What would be Rs optical analog? 3) Superposition is a medium phenomenon ¿yes?, for example "eter". Interference an result of it on a other "thing", for example photographic plate or screen. Are we agree? K |
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