|
Reg Edwards wrote:
Regardless of impedances, with a sensibly zero-loss line it's quite obvious ALL the power leaving the generator is dissipated in the load. There's nowhere else for the stuff to go. Over any delta-length of time during steady-state: Forward waves + reflected waves = standing waves + power to the load. The net energy in the standing waves isn't delivered to the load until the source is switched off. -- 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! =----- |
What happens when instead of switching off the generator the load is
disconnected? |
Gene Fuller wrote:
What I find interesting is that there is not one mention of bouncing energy waves or waves that have disappeared but their energy lives on. There's not one mention of "bouncing energy waves" in my postings prior to your introduction of the term so in that area, I appear to agree with Born and Wolf. I also agree with Born and Wolf that a one thin-film layer can be analyzed using the older method of wave analysis and that's exactly what I am doing. I am hoping that once people see how the thin- film works, how an RF match point works will be obvious. As far as energy living on, please complete the following. _____________ is one example of energy being destroyed along with the destruction of the source of the energy. It is my understanding that "energy lives on" no matter what. -- 73, Cecil, W5DXP |
Richard Clark wrote:
wrote: Maxwell's equations yield answers but give no clue as to the detailed physical process involved. Clueless, hmm? Clueless with respect to the underlying processes. Extremely accurate with respect to the net answers. Tomorrow I plan to be in Austin. That statement says nothing about how I plan to get there. Here's a quote from Steve's article Ah, the great satan having been invoked. How'd I peg that so square on the head? Does Steve know that you equate him to "the great satan"? I agree with that statement. But when I ask what happens to the energy in those two cancelled waves, all I get is silence. That's all it merits, Why? It's an honest question. I am absolutely amazed that physicists and engineers don't know what happens to the energy in canceled waves. It cannot be destroyed, it cannot stand still, and it's not incident upon the source. Wonder what happens to it? That's a very simple honest question. So Richard, what happens to the energy in those two cancelled waves? Destroyed? Bleeds off to a parallel universe? Routed through a black hole for constructive interference in the opposite direction? The answer is more than obvious. From those three alternatives drawn from a hat? Three card monte is a more honest game. Those are only three ridiculous possibilities. Another possibility is what the Melles-Griot web page says: The energy apparently "lost" in the destructive interference of two rearward-traveling reflected waves is not lost at all. It appears in (coherently joins) the forward-traveling wave. That's a simple answer to a simple question. Do you have a better one? And here you told us that maxwell's equations were clueless, answers that described nothing and no help at all Maxwell's equations work to obtain the answer as does quantum electrodynamics. But those answers do not contain clues about the process. The process components are what is being discussed here. -- 73, Cecil, W5DXP |
Reg Edwards wrote:
What happens when instead of switching off the generator the load is disconnected? A modern ham radio transmitter will fold back its output power until it can dissipate all of it without damage. The protection circuitry probably dissipates most of the power that had been stored in the transmission line during steady-state full power output. -- 73, Cecil, W5DXP |
Hi Cecil,
In a futile attempt to maintain my rapidly diminishing sanity I try to limit my role of playing Cecil's fool on RRAA to no more than two days a week. I will be off-line for a few days. But first, here is a clue for the answer to your never-ending question. Those reward traveling waves do not merely cancel at some "match point". They cancel everywhere. In other words they do not exist. There is no energy that needs to be explained away. The model was set up to include these wave components, but the solution comes back to say that those components do not really have a non-zero amplitude. No harm done; this sort of thing happens all the time in the solution of science and engineering problems. Oh, one more clue. I have nothing against waves. I make my living dealing with wave phenomena. But as the old song goes, I know when to hold 'em, and I know when to fold 'em. Try hauling your ox back out of the ditch. There's a whole world out there. 73, Gene W4SZ Cecil Moore wrote: Here's a quote from Steve's article: "When the system reaches the steady state, the two rearward-traveling waves at the match point are 180 degrees out of phase with respect to each other and a complete cancellation of both waves occurs." I agree with that statement. But when I ask what happens to the energy in those two cancelled waves, all I get is silence. |
Cecil wrote:
"But when I ask what happens to the energy in these two cancelled waves, all I get is silence." It`s golden! Cecil then wrote: "Maxwell`s equations tell us that all the energy in a Zo-matched system winds up incident upon a load." Would Maxwell lie to you? The question must be hidden between the lines. It may be: What mechanism reverses the wave? Terman may satisfy the questionn, whatever it is. Terman says of the incident wave: "---everywhere on the line Eprime/Iprime=Zo. Terman says of the reflected wave: "---everywhere on the line Edouble prime/ Idouble prime= -Zo. The difference is only the minus sign which indicates the reversed travel direction of the reflected wave. Terman says on a line with an open-circuited load, that at the load, voltages of the incident and reflected waves have the same phase but the current of the reflected wave has the opposite phase from the reflected voltage. For a transmission line with a short-circuited load, behavior is the opposite. Incident and reflected voltages are out of phase but the currents are in phase. But, there is a 180-degree phase difference between volts and amps in the reflected wave as in the open-circuit line case. The point is that at a discontinuity, upon reflection the phase between voltage and current in a wave is reversed. That is, that either the phase of the volts or amps flips upon reflection, not both. My assumption is that were the phase of both volts and amps reversed at the same time and place, you would see the same wave traveling in the same direction but delayed or advanced by 180-degrees. Its travel direction is unchanged. Which is cause and which is effect? Often what is cause and what is effect can be interchanged. Volts across a resistor produce a current. Current in a resistor produces a voltage drop. Take your pick of cause or effect. A reflection may be caused by a phase reversal between voltage and current. What causes a phase reversal? It`s the discontinuity. Stick a mirror that obstructs the path of a light beam in the path and you have a discontinuity. In electrical circuits we should remember Lenz and his immutable law among obstructions. Best regards, Richard Harrison, KB5WZI |
Gene Fuller wrote:
Those reward traveling waves do not merely cancel at some "match point". They cancel everywhere. In other words they do not exist. That's impossible, Gene, since those two cancelled waves originate at opposite ends of the matched antenna system, one at the mismatched load, and one at the physical mismatch at the Z0-match point. Your explanation requires instantaneous action at a distance. Those two waves are coherent but they do not suffer from quantum entanglement. Please feel free to try again. The question is: What reverses the direction and momentum of the energy reflected from a mismatched load? Saying that reflections from a mismatched load don't exist is simply denying reality. We know they must exist in order to cause standing waves which obviously exist. Your argument contains logical contradictions. To the best of my knowledge, mine doesn't. The model was set up to include these wave components, but the solution comes back to say that those components do not really have a non-zero amplitude. No harm done; this sort of thing happens all the time in the solution of science and engineering problems. Again, you are talking about a net solution which is useless in determining what is happening in real time at the component wave level. If net solutions are all you have to offer, you have nothing to offer for this component discussion. Try hauling your ox back out of the ditch. There's a whole world out there. Dripping water wears away a stone, even brains made of stone. I intend to keep this up until someone proves me wrong with a method besides handwaving, lip flapping, and allusions to magic like your quantum entangled waves above. -- 73, Cecil, W5DXP |
On Fri, 12 Mar 2004 09:41:43 -0600, Cecil Moore
wrote: Tomorrow I plan to be in Austin. That statement says nothing about how I plan to get there. This is the sort of consistently gratuitous discussion that renders the entire topic without merit. Here's a quote from Steve's article Ah, the great satan having been invoked. How'd I peg that so square on the head? Does Steve know that you equate him to "the great satan"? This time a gratuitous question. I agree with that statement. But when I ask what happens to the energy in those two cancelled waves, all I get is silence. That's all it merits, Why? It's an honest question. You would have an entertaining time proving it everyone, but that wouldn't mean they were convinced (or entertained, except in the sense of morbid fascination). Again, it has no merit. So Richard, what happens to the energy in those two cancelled waves? Destroyed? Bleeds off to a parallel universe? Routed through a black hole for constructive interference in the opposite direction? The answer is more than obvious. From those three alternatives drawn from a hat? Three card monte is a more honest game. Those are only three ridiculous possibilities. And hence gratuitous, I pegged another one square on the head. And here you told us that maxwell's equations were clueless, answers that described nothing and no help at all Maxwell's equations work to obtain the answer as does quantum electrodynamics. And yet you have never shown a lick of work in that regard, so we must take it on faith? Like I said, all that is missing is your sagging claims of truth, justice and the american way to bolster this supposed knowledge. How many joules in a square cm of sunlight? Such a question will only find specious qualifications and metaphysical implications from you in response - but then, it is only entertainment, cheaper than video on demand and not injurious to the mental couch potato. However, that entertainment value has been drained, we will pause for your public service announcement: |
On Fri, 12 Mar 2004 12:46:03 -0600, Cecil Moore
wrote: Your argument contains logical contradictions. Because his sources acknowledge and perform to the teachings of Einstein and the workers in the fields of Quantum physics which allow such contradictions. To the best of my knowledge, mine doesn't. Still stuck in the antediluvian, Newtonian universe. |
Richard Harrison wrote:
Which is cause and which is effect? Often what is cause and what is effect can be interchanged. Volts across a resistor produce a current. Current in a resistor produces a voltage drop. Take your pick of cause or effect. A resistor's resistance *causes* a certain V/I ratio. If there exists a V/I ratio and no resistor, then the V/I ratio is the *cause* of the resistance (or impedance). In other words, a resistorless resistance cannot be the cause of anything. It is always an effect, often from interference, and is usually lossless. Note that anything that suffers from I^2*R dissipation is a resistor by this definition. That includes a piece of copper wire. A reflection may be caused by a phase reversal between voltage and current. We are past discussing reflections. The present argument is: Given two coherent waves traveling in the same path with the same magnitude and opposite phases, wave cancellation results from destructive interference. The laws of physics tells us that energy cannot be destroyed and if destructive interference exists, an equal magnitude of constructive interference is required to exist. The destructive interference in a Z0-matched system is toward the source. The constructive interference in a Z0-matched system is toward the load. The energy components involved in those two types of interference are traveling in opposite directions. The conclusion is obvious. The reflected energy involved in the wave cancellation process heads back toward the load just as explained on the Melles-Griot web page. That there is such a large well-organized good old boy conspiracy trying to hide these simple facts of physics speaks volumes about the sad state of amateur radio. http://www.mellesgriot.com/products/optics/oc_2_1.htm -- 73, Cecil, W5DXP |
Richard Clark wrote:
wrote: Your argument contains logical contradictions. Because his sources acknowledge and perform to the teachings of Einstein and the workers in the fields of Quantum physics which allow such contradictions. Well, I guess that settles that. A thing is not what it is. Why didn't I think of simply denying reality? I think I get it now. When you lose an argument using a certain math model, declare that math model to be null and void. -- 73, Cecil, W5DXP |
Cecil, W5DXP wrote:
"Given two coherent waves traveling in the same path with the same magnitude and opposite phases, wave cancellation results from destructive interference." The waves don`t cancel out. Anything in their path just receives equal and opposite influences and the effect of the waves is nil. Sound cancellers which generate equal and opposite waves are an example. Either the sound or antisound sources alone might be deafening, but taken together, there is relative quiet. Best regards, Richard Harrison, KB5WZI |
On Fri, 12 Mar 2004 13:08:20 -0600, Cecil Moore
wrote: We are past discussing reflections. The present argument is: Given two coherent waves traveling in the same path with the same magnitude and opposite phases, wave cancellation results from destructive interference. The laws of physics tells us that energy cannot be destroyed and if destructive interference exists, an equal magnitude of constructive interference is required to exist. The destructive interference in a Z0-matched system is toward the source. The constructive interference in a Z0-matched system is toward the load. The energy components involved in those two types of interference are traveling in opposite directions. The conclusion is obvious. The reflected energy involved in the wave cancellation process heads back toward the load just as explained on the Melles-Griot web page. That there is such a large well-organized good old boy conspiracy trying to hide these simple facts of physics speaks volumes about the sad state of amateur radio. http://www.mellesgriot.com/products/optics/oc_2_1.htm It constantly amazes me that the majority of you guys on this thread just can't seem to get (understand) the truth that Cecil's been trying to get through to you. Where does the power in the cancelled reflected waves go? Conservation of energy dictates that it is totally re-reflected when a complete impedance match has been achieved. You want the wave mechanism that accomplishes this feat? I'll get to that shortly. In Steve Best's latest QEX article, Part 3, bottom of Page 43, beginning with the section titled, "The Total Reflection Fallacy" and continuing ONLY on the left column of Page 44, Steve tells it like it is, correctly. Then why does he call it a Fallacy? Please be patient and I'll tell you why. Steve' correct explanation of the matching process in that single column was taken directly from my writings in ARRL journals. Paraphrased, yes, but correctly so. My first publication of this issue appeared in QST, October 1973, entitled, "A View Into the Conjugate Mirror." This article appears in both Eds 1 and 2 of Reflections as Chapter 4, Steve also copied from another of my articles, this one in QEX,, Mar/Apr 1998, entitled "Examining the Mechanics of Wave Interference in Impedance Matching," which also appears in Reflections 2 as Chapter 23. Steve and I have been in contentious controversy on this subject for several years. He continued this controversy by publishing this totally erroneous material in QEX,, erroneous except for the portion in the single column where he presented my material correctly. The remaining portion of his article is simply an unsuccessful attempt to show that my position is incorrect, and therefore calls it a 'fallacy'. In fact, however, the entire portion following the correct portion he copied from me is where the REAL fallacy lies--it proves that he knows very little about the subject of his title, "Wave Mechanics of Transmission Lnes." It also shows he doesn't have a clue concerning the superposition of two rearward traveling waves that are conjugately related at the matching point. In fact, the two waves cancel each other, and establish either a one-way open circuit or a one-way short circuit that totally re-reflects the reflected power, with its voltage and current components traveling in the same phase as t;hose from the source, and therefore adding to the source power. I know that many on this thread believe that no open or short circuit can be established by the superposition of waves. It is true that forward and reflected waves, traveling in OPPOSITE directions establish only the standing wave--no open or short circuits. But it's a different ball game when two waves traveling in the SAME direction are conjugately related. The waves are conjugately related because the canceling wave generated by the matching device is tailored to have the same magnitude but opposite phase as the wave reflected from the mismatched load on the transmission line. Here's why a short or open circuit is established when conjugately related waves join at a matching point. From an analytic viewpoint the voltage appearing at any point on the line can be replaced with a generator delivering the same voltage at the same phase that appears at that point. This generator is called a 'point' generator that delivers an impedance-less EMF. Now consider one generator delivering the voltage appearing in the wave reflected at the mismatched load and a second generator delivering the voltage from the canceling wave reflected by a matching stub, or whatever the matching device, at the same point on the line as the first. The voltage from this second generator has the same magnitude, but opposite phase from that of the first generator. When the voltages delivered by the t wo generators are 180 degrees out of phase we have a short circuit--if they're in phase we have an open circuit. As the result, in either of these two conditions no reflected wave can pass rearward of the matching point. From the simple fact that the impedance at the input of an antenna tuner is 50+ j0 we know that no reflected power is traveling rearward further than the tuner input. Where did the power in the reflected wave go? That energy cannot disappear as if by some sort of magic--it is totally re-reflected by the open or short circuit, and adds to the source power to establish a forward power equal to the sum of the source and reflected power. I hope this helps to end the confusion, and also gives Cecil what he deserves for his attempt to give you guys the straight dope. Walt, W2DU |
On Fri, 12 Mar 2004 13:28:37 -0600, Cecil Moore
wrote: When you lose an argument using a certain math model, declare that math model to be null and void. You were expecting chopped liver? |
Richard Harrison wrote:
Cecil, W5DXP wrote: "Given two coherent waves traveling in the same path with the same magnitude and opposite phases, wave cancellation results from destructive interference." The waves don`t cancel out. Anything in their path just receives equal and opposite influences and the effect of the waves is nil. I assume you are not asserting that canceled waves' effects become nil, i.e. undetectable, until the end of time yet they still possess the same amount of energy as always except now that energy is completely undetectable for the rest of the life of the universe. Consider the ramifications of what you are asserting. Seems to me, canceled waves must *cease to exist* at a point in space-time if they exhibit zero measurable evidence of their existence forever after (plus exactly that same amount of energy is required by the system in the opposite direction). Incidentally, that was Dr. Best's argument. The energy in the canceled waves continues to propagate forever in the opposite direction of the load. Never mind, that exact same amount of energy is required for constructive interference in the opposite direction and cannot just be created out of nothing. That's when he left the newsgroup. For constructive interference to exist, energy from destructive interference MUST be supplied from somewhere real, according to Hecht in _Optics_. The conservation of energy principle agrees with Hecht as it did during the spring of '01 when the arguments were raging between Dr. Best and me. There is still not enough energy only in P1 and P2 to make P1 + P2 + 2*(P1*P2) equal to the forward power. The 2*(P1*P2) is known in the field of optics as the "interference term" and obviously is supplied from destructive interference between two reflected waves. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Clark wrote:
wrote: When you lose an argument using a certain math model, declare that math model to be null and void. You were expecting chopped liver? No, I was expecting a modicum of rationality - silly me. -- 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:
"Seems to me, canceled waves must "cease to exist" at a point in space time if they exhibit zero measurable evidence of their existence forever after---." A union of two waves can make them both disappear if they are exact opposites, but elimination of just one of the two produces the the other. Were the two waves actually annhilated by their coexistence on the path together, they would not be exhibitable on demand by turning-off one or the other constituent. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Cecil, W5DXP wrote: "Seems to me, canceled waves must "cease to exist" at a point in space time if they exhibit zero measurable evidence of their existence forever after---." A union of two waves can make them both disappear if they are exact opposites, but elimination of just one of the two produces the the other. Were the two waves actually annhilated by their coexistence on the path together, they would not be exhibitable on demand by turning-off one or the other constituent. Of course they would, Richard, since the destructive interference ceases when one of them is turned off. What happens while the two waves are engaging in total destructive interference is that their combined energy components flow in the opposite direction as a constructive interference wave. That's why Melles-Griot says the rearward flowing "lost" energy involved in the destructive interference is not lost at all and instead joins the forward wave traveling in the opposite direction. That's why the rearward-flowing reflected wave energy all winds up flowing toward the load in a Z0-matched system. Let's say I am a light year away from two interfering waves at your location. I am measuring zero energy. You switch off one of the signals. How long does it take for me to sense any energy? - a year because the energy pipeline is empty before you switch off one of the signals. There's no energy flowing toward me while both signals are on. -- 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:
"You switch off one of the signals. How long does it take for me to sense any energy?" Yes, if the transit time is one year, that`s how long it takes to sense any change in received signal. I was wrong when I wrote waves would be exhibitable on demand were they annhilated. Cecil was right. It makes no difference whether two equal and opposite signals are received or no signal is received. Same result. What happens at an open circuit on a transmission line? The current is interrupted and must reverse direction as it has nowhere else to go. A changed direction is a reversed polarity so incident and reflected currents add to zero. Energy in the magnetic field is eliminated by the addition to zero of the incident and reflected currents. This canceled energy goes to the only place it can go, into the electric field. This results in doubling the voltage at the open-circuit end of the line. In a short-circuit on a line, the current doubles and the voltage goes to zero. 1/4-wave back from a short, a near open circuit exists. !/4-wave back from an open or a short, conditions are inverted on a low-loss line. Where there is an open-circuit at the end of a line, a near short circuit exists 1/4-wave back. I think that at the open circuit at the end of a line, the excess voltage launches the reflected wave. At the short on a line, excess current launches a reverse wave. Voltage produces current and current produces voltage. Voltage and current are dominnated by the Zo of the line in all movement through the line. Cecil has argued that the high impedance at the open end of a shorted 1/4-wave stub does not inhibit current into the stub. The reflection point is at the short, not at the high impedance point back 1/4-wave from the short. That sounds reasonable to me, but I wonder if it makes any difference. Best regards, Richard Harrison, KB5WZI |
On Fri, 12 Mar 2004 21:12:23 GMT, Walter Maxwell wrote:
My first publication of this issue appeared in QST, October 1973, entitled, "A View Into the Conjugate Mirror." This article appears in both Eds 1 and 2 of Reflections as Chapter 4, Steve also copied from another of my articles, this one in QEX,, Mar/Apr 1998, entitled "Examining the Mechanics of Wave Interference in Impedance Matching," which also appears in Reflections 2 as Chapter 23. Steve and I have been in contentious controversy on this subject for several years. He continued this controversy by publishing this totally erroneous material in QEX,, erroneous except for the portion in the single column where he presented my material correctly. The remaining portion of his article is simply an unsuccessful attempt to show that my position is incorrect, and therefore calls it a 'fallacy'. In fact, however, the entire portion following the correct portion he copied from me is where the REAL fallacy lies--it proves that he knows very little about the subject of his title, "Wave Mechanics of Transmission Lnes." It also shows he doesn't have a clue concerning the superposition of two rearward traveling waves that are conjugately related at the matching point. In fact, the two waves cancel each other, and establish either a one-way open circuit or a one-way short circuit that totally re-reflects the reflected power, with its voltage and current components traveling in the same phase as t;hose from the source, and therefore adding to the source power. I know that many on this thread believe that no open or short circuit can be established by the superposition of waves. It is true that forward and reflected waves, traveling in OPPOSITE directions establish only the standing wave--no open or short circuits. But it's a different ball game when two waves traveling in the SAME direction are conjugately related. The waves are conjugately related because the canceling wave generated by the matching device is tailored to have the same magnitude but opposite phase as the wave reflected from the mismatched load on the transmission line. Here's why a short or open circuit is established when conjugately related waves join at a matching point. From an analytic viewpoint the voltage appearing at any point on the line can be replaced with a generator delivering the same voltage at the same phase that appears at that point. This generator is called a 'point' generator that delivers an impedance-less EMF. Now consider one generator delivering the voltage appearing in the wave reflected at the mismatched load and a second generator delivering the voltage from the canceling wave reflected by a matching stub, or whatever the matching device, at the same point on the line as the first. The voltage from this second generator has the same magnitude, but opposite phase from that of the first generator. When the voltages delivered by the t wo generators are 180 degrees out of phase we have a short circuit--if they're in phase we have an open circuit. As the result, in either of these two conditions no reflected wave can pass rearward of the matching point. From the simple fact that the impedance at the input of an antenna tuner is 50+ j0 we know that no reflected power is traveling rearward further than the tuner input. Where did the power in the reflected wave go? That energy cannot disappear as if by some sort of magic--it is totally re-reflected by the open or short circuit, and adds to the source power to establish a forward power equal to the sum of the source and reflected power. I hope this helps to end the confusion, and also gives Cecil what he deserves for his attempt to give you guys the straight dope. Walt, W2DU This is a post script to go along with the above discussion. Sorry, Guys, I forgot to mention that for those of you who don't have access to my QEX article, or Chapter 23 in Reflections, Chapter 23 appears in PDF form for downloading from my web page at http://home.iag.net/~w2du. I would also like to add that from the first QST article I published on this subject in 1973, and the QEX article in 1998, I have received more than one hundred responses from RF engineers saying that my explanation of impedance matching via wave mechanics gave them the first real understanding of how the impedance matching process works. No one, other than Steve Best, has disputed my explantion. If the more than one hundred responses seems exaggerated please take a look at the hit count on my web page. At this time the hit count reads 19,927, and I've had NO responses disputing any of the material presented there. So who do you want to believe? Steve Best or me? Walt, W2DU |
Richard Harrison wrote:
Cecil has argued that the high impedance at the open end of a shorted 1/4-wave stub does not inhibit current into the stub. The reflection point is at the short, not at the high impedance point back 1/4-wave from the short. That sounds reasonable to me, but I wonder if it makes any difference. Only to those who think the stub is a physical open-circuit at the mouth of the stub and therefore, no current flows in the stub. :-) -- 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! =----- |
On Sat, 13 Mar 2004 15:53:34 -0600, Cecil Moore
wrote: Richard Harrison wrote: Cecil has argued that the high impedance at the open end of a shorted 1/4-wave stub does not inhibit current into the stub. The reflection point is at the short, not at the high impedance point back 1/4-wave from the short. That sounds reasonable to me, but I wonder if it makes any difference. Cecil wrote: Only to those who think the stub is a physical open-circuit at the mouth of the stub and therefore, no current flows in the stub. :-) In a stub constructed of lossless material the only current that flows in the stub is that required to bring it up to the steady statecondition. In a practical stub with attenuation current flows into the stub continually, but only that sufficient to compensate for the loss due to attenuation to retain it's steady state condition. Walt, W2DU |
Richard Harrison wrote:
I was wrong when I wrote waves would be exhibitable on demand were they annhilated. Cecil was right. It makes no difference whether two equal and opposite signals are received or no signal is received. Same result. Which means that if two coherent waves flowing in the same direction are completely cancelled, the E-fields and the H-fields in that direction are also cancelled and therefore, zero energy propagates in that direction. So the question remains: What happens to the energy in those two rearward- traveling cancelled waves? Melles-Griot says the energy is not lost and instead appears as enhanced intensity in the forward-traveling wave. From rearward-traveling energy to forward-traveling energy certainly sounds like a direction reversal to me. "Enhanced Intensity" is another name for constructive interference and in a matched system, it is "total constructive interference". "Complete Wave Cancellation" is another name for "total destructive interference". Bottom line: In a Z0-matched system, the reflected energy flows rearward to the Z0-match point, changes direction at the Z0-match point (because of interference), and joins the forward energy wave flowing toward the load. Anyone got a better word than "re-reflected" to describe that event? From looking at a diagram of laser light encountering a perfect 1/4WL thin-film coating on glass, it is apparent that all the rearward-traveling reflected irradiance (power) encounters a conjugate "mirror" and joins the forward-traveling energy wave. -- 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! =----- |
Walter Maxwell wrote:
In a stub constructed of lossless material the only current that flows in the stub is that required to bring it up to the steady statecondition. In a practical stub with attenuation current flows into the stub continually, but only that sufficient to compensate for the loss due to attenuation to retain it's steady state condition. All that is true for *NET* current, Walt. But a full magnitude of forward and reflected current is flowing in and out of the mouth of the stub. All of the reflections in a stub occur at the shorted end. What happens at the mouth of the stub is superposition and interference, not reflection. The forward and reflected voltages superpose (in phase) to a high net value and the forward and reflected currents superpose (out of phase) to a low net value. (|Vfwd|+|Vref|)/(|Ifwd|-|Iref|) = very high V/I ratio = very high impedance, but that high impedance is an effect and not the cause of anything (except arguments on r.r.a.a :-) Inside the stub, Vfwd/Ifwd = Z0 and Vref/Iref = Z0 Anyone can prove this to himself. Install a wattmeter 1/8WL down into the 1/4WL shorted stub. One will read high forward power and high reflected power. From those powers, one can actually calculate the forward and reflected voltages and currents at that halfway point within the stub given that halfway point is equal to 45 degrees. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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