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Cecil's Math a Blunder
W5DXP wrote: Jim Kelley wrote: Waves that propagate to a dissipative load transfer energy to that dissipative load. How does the wave about to be dissipated differ from the wave that is about to be canceled? Hint: they are identical. The waves themselves are indeed identical, i.e. you can't measure a difference between them on a transmission line. But one might indeed be transferring energy while the other is not. For example, take a length of 50 ohm transmission line with a short at one end. Think about this circuit with and without a circulator at the source. With the circulator in circit, energy is transferred from the source to the circulator load. Without the circulator, the source transfers no energy to a load. If there are no re-reflections from the source, is there a measureable difference? How does the wave know ahead of time whether it is going to encounter a dissipative load or not? It obviously cannot know ahead of time so all waves with the same V and I in phase carry the same amount of energy. This is a misconception on your part. I don't claim anything has to know anything. That is your claim, and I don't agree. I think it's silly. A wave about to encounter a dissipative load or the same wave about to be canceled carry the same amount of energy. They have the same potential to transfer energy. The do not necessarily transfer that energy. That depends on boundary conditions. The wave gives up that energy in both cases. In the first case it gives up its energy as heat. In the second case, it gives its energy to the constructive interference. I understand that is your theory. I even like the sound of your theory. But you need to understand that it is only your theory. It is an unproven, and unsupported theory. I've tried to come up with support for it, but find none. Yes, the energy which would have been reflected does appear in the transmitted direction. We know this from conservation of energy - energy incident equals energy transmitted plus energy reflected. What does not appear to happen is all the bouncing around you describe. It can be shown that at the first boundary, two reflected waves destructively interfere, producing zero reflected energy. It can also be seen at that boundary that the two waves traveling in the forward direction yield all of the forward energy due to their constructive interference. I understand that if you take the interference term and change its sign you get the same number, and its not just a coincidence. Yes, you can say the amount cancelled in the reflected direction equals the amount of enhancement in the forward direction. But that doesn't mean energy had to turn around in order to accomplish that. That would only be true if energy were indeed traveling in the reverse direction to begin with. When we realize it's not, there's nothing left to account for Cecil. There's no conservation of energy problem to solve. It's all there, just as it says in the physics books, and enegineering books as well. If you want to believe that energy bounces all over the place like, so be it. Just don't expect to be able to sell paraphysical phenomena as science. But as I've tried to tell you many times before, when you can show that Poynting vector solution changing direction solely as a result of destructive interference, you'll have a case - and my vote. Until then, it looks like a non-starter. 73, ac6xg |
Jim Kelley wrote:
The waves themselves are indeed identical, i.e. you can't measure a difference between them on a transmission line. But one might indeed be transferring energy while the other is not. How can two waves that are identical possibly be doing different things. It doesn't make any sense at all. Waves cannot exist without intrinsic energy!!! If any wave is destroyed, it gives up its energy. There is simply no other place for the energy to go. How does the wave know ahead of time whether it is going to encounter a dissipative load or not? It obviously cannot know ahead of time so all waves with the same V and I in phase carry the same amount of energy. This is a misconception on your part. I don't claim anything has to know anything. That is your claim, and I don't agree. I think it's silly. That the wave has to be smart falls out from your assertions. Your waves appear to exist without intrinsic energy so they have to know ahead of time whether to exist without energy or not. Only a wave existing without energy could be destroyed and not give up any energy in the process. But you need to understand that it is only your theory. It is an unproven, and unsupported theory. I've tried to come up with support for it, but find none. Please reference the cross-posting from sci.physics.electromag. Somebody predicted that those guys would tear me apart. On the contrary, they tend to agree with me. It can be shown that at the first boundary, two reflected waves destructively interfere, producing zero reflected energy. It can also be seen at that boundary that the two waves traveling in the forward direction yield all of the forward energy due to their constructive interference. Those reflected waves not only destructively interfere, they are destroyed so they MUST give up any intrinsic energy in the waves. Nothing else is possible. Yes, you can say the amount cancelled in the reflected direction equals the amount of enhancement in the forward direction. But that doesn't mean energy had to turn around in order to accomplish that. Huh??? "Reflected" to "forward" isn't a turn around? Can I have a hit off whatever you are smokin'? :-) That would only be true if energy were indeed traveling in the reverse direction to begin with. When we realize it's not, there's nothing left to account for Cecil. If reflected energy from a mismatched load is not traveling in the reverse direction, pray tell what direction is it traveling in? I can see you haven't referenced the reflected power flow vector. There's no conservation of energy problem to solve. There certainly is in your scenario. The reflected energy doesn't ever change directions but it somehow joins the forward wave anyway. Can you spell B-A-F-F-L-E-G-A-B? We know reflected energy is traveling rearward from a mismatched load. Your mantras cannot change that fact of physics. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
W5DXP wrote: Jim Kelley wrote: The waves themselves are indeed identical, i.e. you can't measure a difference between them on a transmission line. But one might indeed be transferring energy while the other is not. How can two waves that are identical possibly be doing different things. It doesn't make any sense at all. Waves cannot exist without intrinsic energy!!! If any wave is destroyed, it gives up its energy. There is simply no other place for the energy to go. Here's a quote from Born and Wolfe, pg. 48, talking about waves encountering the interface between dense and less dense media: "Although there is a field in the second medium, it is easy to see that no energy flows across the boundary. More precisely it will be shown that, although the components of the Poynting vector in the direction normal to the boundary is in general finite, its time average vanishes; this implies that the energy flows to and fro, but that there is no lasting flow into the second medium." Please reference the cross-posting from sci.physics.electromag. Somebody predicted that those guys would tear me apart. On the contrary, they tend to agree with me. I tended to agree with what you said there too. You can be very agreeable when you want to be. :-) I notice you didn't mention anything about photons dematerializing and rematerializing over there. Give that one a try. :-) Those reflected waves not only destructively interfere, they are destroyed so they MUST give up any intrinsic energy in the waves. Nothing else is possible. If you mean they give up their energy to the two waves traveling in the other direction, then I agree. Can I have a hit off whatever you are smokin'? :-) I've got a pack of Camel lights out in the car. You can have 'em. That would only be true if energy were indeed traveling in the reverse direction to begin with. When we realize it's not, there's nothing left to account for Cecil. If reflected energy from a mismatched load is not traveling in the reverse direction, pray tell what direction is it traveling in? P2 is energy that had been traveling in the reflected direction and was re-reflected. That's it. Good thing too, because any more than P2 and P1 and you'd have a perpetual motion machine! :-) I can see you haven't referenced the reflected power flow vector. I've seen you reference it, but I have seen you define one - or write it out mathematically. There's no conservation of energy problem to solve. There certainly is in your scenario. The reflected energy doesn't ever change directions but it somehow joins the forward wave anyway. Can you spell B-A-F-F-L-E-G-A-B? You wrote the bafflegab - whatever that is. I certainly never said reflected energy joins the forward energy. That's your invention. 73, ac6xg |
W5DXP wrote:
Jim Kelley wrote: Here's a quote from Born and Wolfe, pg. 48, talking about waves encountering the interface between dense and less dense media: "Although there is a field in the second medium, it is easy to see that no energy flows across the boundary. This is just another one of your *NET* energy statements. Actually it's just a quote from an optics book describing, among other things, how a field can cross a boundary without energy necessarily flowing across it. It happens to be the same thing I've been trying to explain to you. I notice you didn't mention anything about photons dematerializing and rematerializing over there. Give that one a try. :-) I did, and no one objected. But we don't need quantum physics to explain HF RF waves. I searched the entire thread. Not one mention of photons dematerializing or rematerializing. Upon what technical point do we still disagree? If the rearward-traveling waves give up their energy to the two waves traveling in the forward direction, the energy has changed direction, by definition. Actually you have it backwards. If energy had been reflected, the forward waves would have given up their energy to the reflected waves. The reflected waves only give up their energy in a sense by not taking it in the first place. There is not enough energy in P1 and P2 to support Pfwd2. Constructive interference energy must be added into Pfwd2. P1 and P2 are the only terms in the equation, Cecil. Obviously they have enough energy to produce the correct answer. But I know what you are getting at, Cecil, and it is interesting. But the example Joseph Legris gave you (the same one I described previously) should illustrate for you the problem with what you are saying. It's apparent from that example that energy from the lasers goes directly to the constructive interference side without having to come from somewhere else (other than the source). And it's also true that that only occurs when destructive interference is also occurring elsewhere. According to Hecht, that constructive interference energy can only originate from destructive interference. What he's saying is that you can't have one without the other. 73, Jim AC6XG |
Jim Kelley wrote:
Actually it's just a quote from an optics book describing, among other things, how a field can cross a boundary without energy necessarily flowing across it. It happens to be the same thing I've been trying to explain to you. But nobody cares about your *NET* energy quotations. This thread is about the energy components underlying the steady-state solution. What is it about that statement that you don't understand? If you want to discuss net energy, please start another thread. I searched the entire thread. Not one mention of photons dematerializing or rematerializing. It was in another thread, Jim. Actually you have it backwards. If energy had been reflected, the forward waves would have given up their energy to the reflected waves. The forward wave *DOES* give up (Pfwd*|rho|^2) energy to the reflected wave at a mismatched load. I'm sorry to pull an argumentum ad populum on you but everybody on this newsgroup knows that except you. The reflected waves only give up their energy in a sense by not taking it in the first place. More Bafflegab. Please research the reflected power Poynting vector from the previously provided reference of Ramo and Whinnery and get back to us. P1 and P2 are the only terms in the equation, Cecil. Which leads you to believe that P1+P2 P1+P2? What is it about destructive interference energy feeding energy to the constructive interference event that you do not understand? Assume Wave(1) is associated with P1 and Wave(2) is associated with P2. P1 + P2 + 2*Sqrt(P1*P2) is the proper equation when Wave(1) and Wave(2) are in zero phase. On the other side of the impedance discontinuity we have a similar equation: P3 + P4 - 2*Sqrt(P3*P4). Agree so far? Pref1 = P3 + P4 - 2*Sqrt(P3*P4) = zero reflected power toward the source. So P3 + P4 = 2*Sqrt(P3*P4) = 100% destructive interference during wave cancellation, i.e. Wave(3) and Wave(4) are destroyed. The destructive interference equals the total of P3+P4 *because* those waves were destroyed. Wave(3) and Wave(4) give up their combined energy components as destructive interference energy. Hecht tells us that the magnitude of the destructive interference must equal the magnitude of the constructive interference. Guess what? He is right! |2*Sqrt(P3*P4)| does indeed equal |2*Sqrt(P1*P2)| and we can simply say that according to Hecht, |2*Sqrt(P1*P2)| comes from |2*Sqrt(P3*P4)|. The destructive interference event supplies the power for the constructive interference event. I'm getting tired of typing the following quote from _Optics_, by Hecht: "The principle of Conservation of Energy makes it clear (to everyone except Jim) that if there is constructive interference at one point, the 'extra' energy at that location must have come from ... destructive interference somewhere else." I don't know how Hecht could have said it any clearer. The constructive interference energy is supplied from the destructive interference event. So the valid equation becomes P1+P2+2*Sqrt(P3*P4) = P1+P2+P3+P4, i.e. all the power in the entire Z0-matched system winds up flowing toward the load but we already knew that. How you can argue with that fact is simply unbelievable. What he's saying is that you can't have one without the other. Yes, and the destructive interference event *supplies* energy to the constructive interference event. It certainly cannot be vice versa as you are attempting to imply. Hecht was very careful to say that constructive interference requires a source of energy from a destructive interference event. See above quote. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Roy Lewallen wrote:
So where do you draw the required surface over which to integrate the Poynting vector and thus obtain the "reflected power"? As you know, a one-dimensional transmission line is a lot easier to work with than the 3D space involved with radiation from antennas. In fact, all we need for a transmission line Poynting vector is a plane. Since the field area of the coax is constant and known, we can treat it as a constant normalized value. It allows us to come up with a simplified version of what you posted. Quoting from, "Fields and Waves in Communications Electronics", by Ramo, Whinnery, and Van Duzer: "... we are often most interested in the ratio of power in the reflected wave to that in the incident wave, and this ratio is given by ... Pz-/Pz+ = |rho|^2" where Pz- is the reflected power Poynting vector, Pz+ is the forward power Poynting vector, and rho is the reflection coefficient. Pretty simple stuff that I learned 45 years ago. Again from the same book: Pz+ = (E+)x(H+) and Pz- = -(E-)x(H-) -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Jim Kelley wrote:
W5DXP wrote: But nobody cares about your *NET* energy quotations. Nobody said anything about *NET* energy in the quotation. But it is obvious that they are referring to *NET* energy. Please stop trying to sneak *NET* energy stuff into the argument. The forward wave *DOES* give up (Pfwd*|rho|^2) energy to the reflected wave at a mismatched load. I'm sorry to pull an argumentum ad populum on you but everybody on this newsgroup knows that except you. At a single boundary, yes. If it exists at a single boundary, then it exists. Which leads you to believe that P1+P2 P1+P2? I don't believe that. If you believe that all the power comes from P1 and P2 in the following equation, then yes, you do believe that. P1 + P2 + 2*Sqrt(P1*P2) = Pfwd2 Hint: the 2*Sqrt(P1*P2) power does NOT come from P1 and P2. If it did you would be violating the conservation of energy principle and creating free energy. That term is the *INTERFERENCE TERM* which is supplied from the destructive interference going on on the other side of the impedance discontinuity. Why do you have such a problem with the concept of conservation of energy? The destructive interference equals the total of P3+P4 *because* those waves were destroyed. Wave(3) and Wave(4) give up their combined energy components as destructive interference energy. There exists no mechanism for them to "give up" their energy in the way you describe. The mechanism is conservation of energy involving destructive interference feeding energy to the constructive interference event. The guys over on sci.physics.electromag can explain it to you. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Cecil, W5DXP wrote:
"They", above, are reflected waves. The two waves traveling in the other direction are forward waves." Forward waves? Being, I suppose, a voltage wave and a current wave, as it is impossible to have two voltage waves or two current waves of identical frequency traveling in the same direction on the same line. Two identical frequency waves, voltage or current, form a resultant and merge. They can`t maintain a separate existence and become inseparable in a single waveform. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Cecil, W5DXP wrote: "They", above, are reflected waves. The two waves traveling in the other direction are forward waves." Forward waves? Being, I suppose, a voltage wave and a current wave, as it is impossible to have two voltage waves or two current waves of identical frequency traveling in the same direction on the same line. Two identical frequency waves, voltage or current, form a resultant and merge. They can`t maintain a separate existence and become inseparable in a single waveform. Of course they superpose. In the s-parameter equation, b1=s11*a1+s12*a2, those two reflected waves superpose to b1. In the s-parameter equation, b2=s21*a1+s22*a2, those two forward waves superpose to b2. They superpose at the point where they both appear together for the first time. But the point is that s11*a1 and s12*a2 are flowing in the direction of the source when they cancel (interfere destructively). s21*a1 and s22*a2 are flowing in the direction of the load when they interfere constructively. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Jim Kelley wrote:
I've never disputed the solutions produce the same number. I disputed the validity of the approach you took to arrive at the equation. I guess I owe the group an apology. The guys over on sci.physics.electromag are inferring things that I never meant to imply. I thought the meaning of my words was clear but I guess that's not the case. I apologize to anyone I accused of deliberately misinterpreting what I said. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
W5DXP wrote:
Jim Kelley wrote: I've never disputed the solutions produce the same number. I disputed the validity of the approach you took to arrive at the equation. Here's a question for you, Jim. How much power is transferred from the mismatched load rearward by Pref2? First I have to know where it is being transferred. I don't see two ends of a path. Is there a load resistor hiding in a circulator somewhere? Is the 33.33 watts extra power that I could drive another antenna with, but is otherwise going unused? Or is it instead, simply a field that is excited by the boundary conditions at the load that is being measured. source---50 ohm feedline---+---150 ohm feedline---load Pfwd1=200W -- Pfwd2=133.33W -- -- Pref1=100W -- Pref2=33.33W If, as you have asserted, Pref2 can transfer zero watts when it measures 33.33W, can it also transfer 100W when it measures 33.33W? The meter measures voltage, assumes a transfer of energy, and displays power based on that assumption. 73, ac6xg |
Jim Kelley wrote:
First I have to know where it is being transferred. This is the basic flaw in your argument. The power goes where it goes. That you need to know something indicates that you believe what you know dictates reality. Sorry, Jim, you are NOT that powerful. Mother Nature doesn't care what you know or don't know. Reality is exactly the same either way and will be exactly the same way after you are six feet under. Why do you think you are so important that what you know affects reality? -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
W5DXP wrote: Jim Kelley wrote: First I have to know where it is being transferred. This is the basic flaw in your argument. The power goes where it goes. So, where does the 33.33 watts in your question go? 73, ac6xg |
Jim Kelley wrote:
But where does that 33.33 watts actually go? If you have 100 watts coming from the source, and 100 watts traversing the first and second boundaries, where does the 33.33 watts go? Through a two step process, it merges in phase with the forward wave at the Z0-match point. A directional wattmeter reads 133.33W forward power. 100W + 33.33W = 133.33W. 100W/|1-rho|^2 = 133.33W Forward power - reflected power = power delivered to the load. The standing waves confirm that there is indeed 133.33W forward and 33.33W reflected. Let's look at the following system with two sources. Each source is a signal generator equipped with a circulator and load. SGCL1 is equipped with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm circulator and load. The two signal generators are phase locked but can be turned on and off independently. rho=0.5 100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2 Pfwd1-- Pfwd2-- --Pref1 --Pref2 Using the principle of superposition: With SGCL1 on and SGCL2 off, Pfwd1=100W, Pref1=25W, Pfwd2=75W, and Pref2=0W. SGCL2 dissipates 75W (Pfwd1)(|rho|^2) and SGCL1 dissipates 25W (Pfwd1)(1-|rho|^2). This is how the s11 and s21 s-parameters are measured. With SGCL2 on and SGCL1 off, Pfwd1=0W, Pref1=25W, Pfwd2=8.33W, Pref2=33.33W. SGCL2 dissipates 8.33W (Pref2)(|rho|^2) and SGCL1 dissipates 25W (Pref2)( 1-|rho|^2). This is how the s12 and s22 s-parameters are measured. Note that SGCL1 dissipates 25W in both of the above cases. With SGCL1 and SGCL2 both on, Pfwd1=100W, Pref1=0W, Pfwd2=133.33W, and Pref2=33.33W. This is similar to the earlier single source example. Note that the two 25W waves obviously engage in wave cancellation and their combined 50W of destructive interference joins the forward wave as constructive interference. Assume there is a set of feedline lengths that will accomplish the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm feedline will do that if the voltages from the two sources are max+ at the same time. b1 = s11*a1 + s12*a2 = 0V s11*a1 and s12*a2 engage in destructive interference. Pref1 = Pfwd1(|rho|^2) + Pref2(1-|rho|^2) - destructive interference 0W = 25W + 25W - 50W Pfwd2 = Pfwd1(1-|rho|^2) + Pref2(|rho|^2) + constructive interference 133.33W = 75W + 8.33W = 50W -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
Cecil, W5DXP wrote:
"Let`s look at the following system with two sources," Is this tempest about creation of an oppositely phased wave to cancel a reflection? A way to avoid reflected waves within a waveguide is mentioned by Terman on page 148 of his 1955 edition: "Create a reflected wave near the load that is equal in magnitude but opposite in phase from the wave reflected by the load; in this way the two reflected waves cancel each other." "Some of these (matching arrangements) are analogous to the impedance matching arrangements employed in transmission lines (described in Sec. 4-11) while others are unique to waveguides." Best regards, Richard Harrison, KB5WZI |
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W5DXP wrote: A directional wattmeter reads 133.33W forward power. 100W + 33.33W = 133.33W. 100W/|1-rho|^2 = 133.33W Forward power - reflected power = power delivered to the load. The standing waves confirm that there is indeed 133.33W forward and 33.33W reflected. So where does the 33.33 watts actually go? The 100 watt number seems to account for all of the transfer of energy. Let's look at the following system with two sources. Each source is a signal generator equipped with a circulator and load. SGCL1 is equipped with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm circulator and load. The two signal generators are phase locked but can be turned on and off independently. rho=0.5 100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2 Pfwd1-- Pfwd2-- --Pref1 --Pref2 Using the principle of superposition: With SGCL1 on and SGCL2 off, Pfwd1=100W, Pref1=25W, Pfwd2=75W, and Pref2=0W. SGCL2 dissipates 75W (Pfwd1)(|rho|^2) and SGCL1 dissipates 25W (Pfwd1)(1-|rho|^2). With SGCL2 on and SGCL1 off, Pfwd1=0W, Pref1=25W, Pfwd2=8.33W, Pref2=33.33W. SGCL2 dissipates 8.33W (Pref2)(|rho|^2) and SGCL1 dissipates 25W (Pref2)( 1-|rho|^2). This is how the s12 and s22 s-parameters are measured. Note that SGCL1 dissipates 25W in both of the above cases. With SGCL1 and SGCL2 both on, Pfwd1=100W, Pref1=0W, Pfwd2=133.33W, and Pref2=33.33W. This is similar to the earlier single source example. Except for the amount of power input. Although you didn't mention it explicitly, I assume SGCL2 inputs 33.33 watts. Obviously there's that much more power being input to the network in this example that the others we've worked on. Note that the two 25W waves obviously engage in wave cancellation and their combined 50W of destructive interference joins the forward wave as constructive interference. I note that interference takes place. Assume there is a set of feedline lengths that will accomplish the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm feedline will do that if the voltages from the two sources are max+ at the same time. Well, you won't get a phase reversal from reflection at the 150/50 ohm boundary as you would with the 150/450 ohm boundary in our other problem, so I think the 150 ohm feedline would need to be an odd number of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first boundary. But since we're really interested in the relative phase between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm boundary, you would want to maintain the phase of Vref2 by using an even multiple of 1/4 wave lengths of the 150 ohm line. b1 = s11*a1 + s12*a2 = 0V s11*a1 and s12*a2 engage in destructive interference. Pref1 = Pfwd1(|rho|^2) + Pref2(1-|rho|^2) - destructive interference 0W = 25W + 25W - 50W Pfwd2 = Pfwd1(1-|rho|^2) + Pref2(|rho|^2) + constructive interference 133.33W = 75W + 8.33W = 50W I'll trade you a plus sign for your equal sign. ;-) We also know that 133.33W = 100W + 33.33W. So 100W = 100W + 33.33W - 33.33W. Which is equally revealing. ;-) 73, Jim AC6XG |
Jim Kelley wrote:
So where does the 33.33 watts actually go? The 100 watt number seems to account for all of the transfer of energy. I told you, the 33.33W joins the 100W forward wave at the impedance discontinuity and heads toward the load where a new 33.33W is rejected by the mismatched load. Maybe that is the point of your confusion. The 33.33W that makes a round trip to the impedance discontinuity and back to the load is not the same 33.33W that is reflected by the load. TV ghosting proves that to be true. Let's look at the following system with two sources. Each source is a signal generator equipped with a circulator and load. SGCL1 is equipped with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm circulator and load. The two signal generators are phase locked but can be turned on and off independently. rho=0.5 100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2 Pfwd1-- Pfwd2-- --Pref1 --Pref2 Except for the amount of power input. Although you didn't mention it explicitly, I assume SGCL2 inputs 33.33 watts. "33.33W SGCL2" seems pretty explicit to me. Obviously there's that much more power being input to the network in this example that the others we've worked on. We are also taking out more power than we previously were. I note that interference takes place. Please note that the 25W reflected when SGCL1 only is on is 35.36V at zero deg, and 0.707A at 180 deg. When SGCL2 only is on, the 25W not re-reflected is 35.36V at 180 deg, and 0.707A at zero deg. That's why they cancel when both are on. Assume there is a set of feedline lengths that will accomplish the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm feedline will do that if the voltages from the two sources are max+ at the same time. Well, you won't get a phase reversal from reflection at the 150/50 ohm boundary as you would with the 150/450 ohm boundary in our other problem, so I think the 150 ohm feedline would need to be an odd number of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first boundary. But since we're really interested in the relative phase between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm boundary, you would want to maintain the phase of Vref2 by using an even multiple of 1/4 wave lengths of the 150 ohm line. Nope, you are wrong about that but I'll let you figure our your own error. The only voltage phase reversal in the given example is in the re-reflected wave associated with the 8.33W. Vfwd1 travels 1WL while Vref2 travels 1/2WL. That is enough to put them 180 degrees out of phase. Vfwd1(|rho|^2) undergoes zero phase shift. Vref2(1-|rho|^2) undergoes zero phase shift. -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
Richard Harrison wrote:
Terman said about the same thing about cancellation of the reflected wave in a waveguide. Terman offered several methods to generate the offsetting wave in a waveguide, and there are several ways to generate a wave which produces cancellation on a transmission line, too. Trouble is, Terman didn't say anything about what happens to the intrinsic energy that existed in the waves before they were canceled. I say the energy in rearward-traveling canceled reflected waves is reflected back toward the load. Jim (and others) disagree even though we know that all the energy winds up incident upon the load. -- 73, Cecil, W5DXP |
Jim Kelley wrote:
W5DXP wrote: The 33.33W that makes a round trip to the impedance discontinuity and back to the load is not the same 33.33W that is reflected by the load. TV ghosting proves that to be true. TV ghosting is primarily the result of multipathing, which is external to the antenna system. :-) I'm talking about TV ghosting due to reflections on the transmission line between a TV test generator and a TV receiver. The only "multipathing" involved is the different paths taken by the reflected waves inside the transmission line. The only voltage phase reversal in the given example is in the re-reflected wave associated with the 8.33W. How are you determining your phase reversals? Don't forget that phase reversal on reflection occurs when a wave encounters a more dense medium (or a higher impedance). Nope, a voltage phase reversal occurs when a wave encounters a *lower* impedance. A voltage wave encountering a higher impedance does not reverse phase. In the earlier example, the only lower impedance encountered is by the rearward-traveling wave flowing back into the impedance discontinuity. I repeat, Vref2(-rho) is the only voltage phase reversal. -- 73, Cecil, W5DXP |
Richard Harrison wrote:
"Create a reflected wave near the load that is equal in magnitude but opposite in phase from the wave reflected by the load; in this way the two reflected waves cancel each other." Unfortunately, he didn't say what happens to the intrinsic energy necessary for those two canceled waves to have existed in the first place. The question is: At the moment that two waves are canceled, do they give up their intrinsic energy components? Or, as in the sour grapes tale, were they never associated with any intrinsic energy to begin with? -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
they're called "evanescent waves"
get a book "Richard Harrison" wrote in message news:4061-3F2059E0-171@storefull-yada It is in there. The waves go away. Terman said: blah blah |
Richard Harrison wrote:
Terman says the reflected energies cancel. Does he define, "cancel"? Energy components cannot cease to exist. I suspect he means they engage in destructive interference which requires a constructive interference elsewhere according to Hecht. Equal and opposite reflected energies at the junction are a short. A short takes the voltage to zero and doubles the current. Are you sure that is what happens during wave cancellation? The energy can`t disappear. It does not go back toward the load. Half has already been there. The other half has alredy been reflected by Zstub. Only one route is left to the reflected energy, but Terman says there`s no reflection on the generator side of the junction. Your Bird wattmeter won`t see any reflected power. What this says is that the stub produces an impedance match and the incident energy becomes the same as the generator output and load energies. Would you please publish an example? -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
H. Adam Stevens wrote:
they're called "evanescent waves" get a book In one-dimensional transmission lines? -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Cecil, W5DXP wrote:
"Would you publish an example?" I am error prone as anyone reading many of my postings may attest. The 1955 edition of Terman`s "Electronic and Radio Engineering" is the culmination of about 25 or more years of Terman`s work which was checked, rechecked, then checked again. It`s stood the test of time ever since its publication too. In Sect. 4-11 are answers to questions of impedance matching to transmission lines. I think Terman should be consulted straight away for a simple logical cause of no reflections. It is called impedance matching. The match does away with reflections. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Cecil, W5DXP wrote: "Would you publish an example?" I am error prone as anyone reading many of my postings may attest. The 1955 edition of Terman`s "Electronic and Radio Engineering" is the culmination of about 25 or more years of Terman`s work which was checked, rechecked, then checked again. It`s stood the test of time ever since its publication too. In Sect. 4-11 are answers to questions of impedance matching to transmission lines. I'll look up the book next time I am at Texas A&M. I think Terman should be consulted straight away for a simple logical cause of no reflections. It is called impedance matching. The match does away with reflections. Not directed at Richard, but I've seen the following logic used often with EM waves. What causes the match? The cancellation of reflections. What causes the cancellation of reflections? The match. Does anyone else detect the circular logic problem in the above? Another one: 'rho' is zero because there are no reflections. There are no reflections because 'rho' is zero. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Cecil, W5DXP wrote:
"What causes the match? The cancellation of reflections." This is like Abbot and Costello. You got it! Who`s on first! You have a source which only delivers energy at a certain volts to amps ratio, period! You have a load which only accepts energy at a certain volts to amps ratio, period! When the above ratios are in fact identical there is no problem. When the above ratios are different something has to give and it isn`t either ratio, Zo or ZL. Instead, the load takes what the source can deliver and rejects the surplus volts or amps whichever is the case, as are created by ZL, and the limited deliverability created by the Zo. The surplus generates a reflected wave as Cunningham recounted with his description of missing inductance and capacitance in the missinng continuation of the transmission line (A short on the line vitiates the capacitance etc.). The explanation is logical and I recommend his broadcast antenna book. What`s not to understand about matching? When the load is not matched, you add the incremental receptivity required to take the surplus current or voltage at the load, so that the load, and the stub in Terman`s example, are a perfect match and there is no longer any surplus to be reflected. The impedance of the load has been adjusted, by the addition of a stub in Terman`s example, to Zo. Terman says the wave from the stub cancels the wave from the mismatched load. No doubt a shorted stub makes a reflection. The reflected volts from the stub are exactly equal and opposite to the reflected surplus volts from the load at the junction, and Terman says there is no reflection toward the generator. I never won an argument with Terman, though I tried. I believe there is no reflection past the stub back toward the generator. If so, and it certainly is so, I`ve put those stubs out at the antenna myself and they create a match when adjusted properly, the energy on a lossless line will be the same at the generator and anywhere on lhe line. It`s flat. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
What`s not to understand about matching? The energy and momentum in EM waves, particularly, the energy and momentum in the wave reflected from a mismatched load. What happens to that energy and momentum? We know it joins the forward wave. How does that energy and momentum get reversed? A virtual impedance is a result and not the cause of anything. The reflected volts from the stub are exactly equal and opposite to the reflected surplus volts from the load at the junction, and Terman says there is no reflection toward the generator. I'm not arguing that point. What happens to the energy in the two waves that have their voltage and current amplitudes equal in amplitude and opposite in phase? There is no doubt those waves cancel and there is no reflection toward the generator. But what happens to the intrinsic energy that pre-existed the cancellation event? I never won an argument with Terman, though I tried. I believe there is no reflection past the stub back toward the generator. No argument about that. The question is where did the energy and momentum go when those two waves were canceled thus eliminating reflections back toward the generator. Waves can be canceled. The energy in those waves cannot be destroyed. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
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Cecil wrote:
But that is not what happens when two waves cancel. For an idea of what happens, please reference Reflections II, page 23-8,9. If the two voltages are 180 degrees out of phase and the two currents are 180 degrees out of phase, neither one of them can double. Calling that a virtual short or virtual open is incorrect, IMO. Cecil, this is the error on Reflections that I told you about last week when you cited the above reference in Reflections. I inadvertantly called the effect a 'short circuit', while it is actually an open circuit. If a 3rd edition of Reflections is ever printed this error will be corrected. Now when two waves interfere with their voltages 180 degrees out of phase, but with their currents in phase, we have a short circuit. Either of these conditions occur at the matching stub point, depending on the resistance component of the load impedance and the distance from the load to the stub point. Cecil also wrote: "What happens to the energy in the two waves that have their voltage and current amplitudes equal in amplitude and opposite in phase?" These two waves continue to circulate. I forgot to say they are traveling in the same direction and that they are coherent. The two waves disappear from existence. What happens to the energy they contained? I thought it was common knowledge that the reflected energy is totally re-reflected in the forward direction when encountering the open or short circuit at the matching point. Walt, W2DU |
Walter Maxwell wrote:
I thought it was common knowledge that the reflected energy is totally re-reflected in the forward direction when encountering the open or short circuit at the matching point. The voltages and currents are listed at the Z0-match point in the following example: rho=0.5 200W source---50 ohm feedline--+--1/2WL 150 ohm feedline---50 ohm load Vfwd1=100V-- Vfwd2=200V-- Ifwd1=2A-- Ifwd2=1.333A-- --Vref1=0V --Vref2=100V --Iref1=0A --Iref2=0.667A Vref1 = Vfwd1(rho) + Vref2(tau) (this is the same as b1=s11*a1+s12*a2) 0V = 50V at zero deg + 50V at 180 deg i.e. the voltages cancel Iref1 = Ifwd1(rho) + Iref2(tau) 0A = 1.0A at 180 deg + 1.0A at zero deg i.e. the currents cancel Through wave cancellation, both the voltage and current go to zero in the direction of the source. That's like a short for voltage and an open for current. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Richard Clark wrote:
Your failure to confirm or respond to Walt's second correction offers compelling further evidence that you are an unreliable correspondent. Actually, it confirms that I am a polite correspondent. Walt and I are discussing this in private email. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
On Sun, 27 Jul 2003 13:07:13 -0500, W5DXP
wrote: Richard Clark wrote: Your failure to confirm or respond to Walt's second correction offers compelling further evidence that you are an unreliable correspondent. Actually, it confirms that I am a polite correspondent. Walt and I are discussing this in private email. So this makes the third occurence. |
Richard Clark wrote:
W5DXP wrote: Richard Clark wrote: Your failure to confirm or respond to Walt's second correction offers compelling further evidence that you are an unreliable correspondent. Actually, it confirms that I am a polite correspondent. Walt and I are discussing this in private email. So this makes the third occurence. There is no pleasing you guys. Many say, "Take it to private email!", and you say, "Don't take it to private email." What is a poster to do? -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
On Sun, 27 Jul 2003 14:28:00 -0500, W5DXP
wrote: Richard Clark wrote: W5DXP wrote: Richard Clark wrote: Your failure to confirm or respond to Walt's second correction offers compelling further evidence that you are an unreliable correspondent. Actually, it confirms that I am a polite correspondent. Walt and I are discussing this in private email. So this makes the third occurence. There is no pleasing you guys. Many say, "Take it to private email!", and you say, "Don't take it to private email." What is a poster to do? Admit your repeated public errors On Sun, 27 Jul 2003 00:33:47 GMT, "Walter Maxwell" wrote: Cecil, this is the error on Reflections that I told you about last week that have already been solved in private correspondence. Why do you make Walt come to this public forum to correct you for an error he has already corrected in "private" emails? You continue to support the evidence of not having read your correspondence from Walt, or ignoring it to offer a knowingly erroneous premise. This is unreliable correspondence at two levels. 73's Richard Clark, KB7QHC |
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