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Re-Normalizing the Smith Chart (Changing the SWR into the
Keith wrote:
"2) At quarter wave points along this line, voltages and currents which are always 0 can be observed -- the standing wave." This is proof positive in Keith`s own words that the first 0-volt point encountered by either forward or reflected wave is completely ineffective in halting or abating progress of the waves. Keith also wrote: "6) From 2) and 5), the power (rate of energy flow) at quarter wave points will be 0." The contradiction is obvious. Were energy flow or power reduced by SWR zeros, you would not have multiple zeros along the line. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Keith wrote: "2) At quarter wave points along this line, voltages and currents which are always 0 can be observed -- the standing wave." This is proof positive in Keith`s own words that the first 0-volt point encountered by either forward or reflected wave is completely ineffective in halting or abating progress of the waves. Keith also wrote: "6) From 2) and 5), the power (rate of energy flow) at quarter wave points will be 0." The contradiction is obvious. Were energy flow or power reduced by SWR zeros, you would not have multiple zeros along the line. Your analysis overlooks that we are discussing an ideal line that has reached steady state. Before the line reaches steady state, energy does cross the quarter wave points, but, of course, while this is happening the voltages and currents at the quarter wave points are not yet zero. As an aside, were we to discuss a line that was not open or shorted but rather terminated in other than its characteristic impedance, we would still find voltage and current minimas. At these minima, rather than the energy flow being 0 as it is for a shorted or open line, the energy flow (power) is exactly the energy being delivered down the line. Something similar occurs while an opern or shorted line is approaching steady state. ....Keith |
Keith wrote:
"Is it step 7)?" "7) From 6), the energy crossing quarter wave points is zero" False, as are several other of Keith`s speculations. Standing waves throughout a long transmission line with a hard short or open-circuit at its end are proof enough that SWR nulls other than the one at the actual discontinuity don`t bring the energy crossing 1/4-wave points to zero. SWR nulls other than the one at the actual open or short have no effect on traveling waves, absent an additional actual discontinuity. SWR can`t exist without energy flow in both directions. Energy flow is continuous. It doesn`t start and stop at SWR nulls. This continuity is proved by measurements at the nulls, taken in one direction at a time. No speculation or even math is needed to observe this behavior on an actual line. It is usless to try to conform the observable to some theory. It is far more productive to conform theory to the observable. Best regards, Richard Harrison, KB5WZI |
Keith wrote:
"At these minima, rather than the energy flow being 0 as it is for an open or shorted line, the energy flow (power) is exactly the energy being delivered down the line." The power is not zero at an SWR zero, provided that the line has power flow. The power flow at the specified null is equal in forward and reflected wave directions as shown by the complete zero. Power delivered by the source and to the load is the forward power minus the reflected power. In the case of a complete reflection, the difference between forward and reflected power is zero, so the load is rejecting all the forward power, which is turned around and becomes the reflected power. In a lossless line, forward and reflected powers flow unabated from end to end of the line and are thus have their same amplitudes wherever they are measured in the line, including SWR null points. Best regards, Richard Harrison, KB5WZI |
wrote:
Your analysis overlooks that we are discussing an ideal line that has reached steady state. Before the line reaches steady state, energy does cross the quarter wave points, but, of course, while this is happening the voltages and currents at the quarter wave points are not yet zero. Please stop discussing NET energy and start discussing component energies. -- 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 Harrison wrote:
Keith wrote: "Why are you convinced the reflection point had to move?" This is the case when a line is extended to greater length, and the short or open also is moved to the new end of the line. Ever had a short or open somewhere in the middle of a line? Actually, no. I have, and can testify that the signal doesn`t get past a real hard short or open in the line. As energy can`t be destroyed it had to be reflected by the hard short or open. Or just stopped and stored, perhaps. This is certainly what happens when you excite the line with a step function rather than a sinusoid. When excited by a sinusoid, no energy moves at the quarter wave points where the voltage or current is always 0. As you move away from the quarter wave points, more and more energy moves on each cycle until a maximum amount of energy is moved at the point half way between the quarter wave points. And then it decreases back towards the next quarter wave point. All of this is easily visualized by observing the amplitude of the p(t) function at various points along the line. In the case where the line short is moved to a point nearer the source, or where the short or open is moved to a place more distant on the line from the source, we know the energy travels all the way to the actual short or open where it is reflected because nulls occur at several points along a long transmission line. "we know" is rather strong. I would strongly suggest that no energy crosses those points in the line where the voltage and current are always zero since p(t) is always zero at these points. If the energy were turned around before it reached the end of the line, nulls more distant from the source than the turnaround point would not exist. Not so, the line has reached steady-state. Now the nulls exist. They did not exist before the line reached steady-state. There would be no energy at the actual short or open at the end of the line were the energy turned around before it reached the end of the line. Not necessarily. Only once the line has been charged, does the energy move back and forth between the quarter wave points, while not crossing them. Try visualizing how a step function charges the line. How the voltage step propagates down the line. How the voltage step is reflected at the open end and starts travelling back towards the start. How between the start of the line and the returning voltage step, energy is flowing to charge the line, but between the returning voltage step and the open end of the line, the energy previously delivered is statically stored in the capacitance of the line. How once the line is completely charged, no current flows, there is no power, and the energy delivered during charging is stored in the capacitance. Sinusoidal excitation is more difficult to visualize, but the prinicipal is the same. ....Keith |
Richard Harrison wrote:
Keith wrote: "Is it step 7)?" "7) From 6), the energy crossing quarter wave points is zero" False, as are several other of Keith`s speculations. I do not see any flaw in step 7). Assuming that step 6) "6) From 2) and 5), the power (rate of energy flowing) at quarter wave points will be 0" is correct, then step 7) "7) From 6), the energy crossing quarter wave points is 0" must also be true, since, If the rate of energy flowing is zero, then there is no energy flowing so there can not be any energy crossing the point. I do not find that step 7) is in error. If there is an error in the final conclusion then I do not think that it is an error in step 7) which causes the final error, but, rather, the error must be in one of the earlier steps. ....Keith |
wrote:
I do not see any flaw in step 7). Try opening your eyes. If the rate of energy flowing is zero, then there is no energy flowing so there can not be any energy crossing the point. There is no NET energy flow. There is plenty of component energy flow as defined by Ramo & Whinnery's forward Poynting vector and reflected Poynting vector. Why do you continue to ignore those vectors? Why do you continue to ignore the wealth of knowledge contained in light interference patterns? -- 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! =----- |
W5DXP wrote:
wrote: When excited by a sinusoid, no energy moves at the quarter wave points where the voltage or current is always 0. You keep saying that and it keeps being a false statement. There is absolutely nothing magic about sinusoids. Seems to me that the sinusoidal standing wave with minima and maxima at the quarter wave points can only arise with single frequency sinusoidal excitation of the line. Are there other signals which will produce this result? "we know" is rather strong. I would strongly suggest that no energy crosses those points in the line where the voltage and current are always zero since p(t) is always zero at these points. What about Ramo and Whinnery's forward Poynting vector and reflected Poynting vector? Why do you choose to ignore them? I haven't used them because I don't need them to arrive at an answer. Basic electricity, a dash of circuit theory, a bit of knowledge of trigonometry, some basic calculus and the ability to think is all that is required. Why make the solution more complex than necessary? Just to scare off the neophyte? Not necessarily. Only once the line has been charged, does the energy move back and forth between the quarter wave points, while not crossing them. That has been shown to be a false assertion regarding component waves. Perhaps. Or maybe component waves are not the answer. There is no impedance discontinuity to cause any reflections. Therefore, the waves do not move back and forth. Do you really believe that the energy in a bright interference ring is trapped inside the ring? Get serious! In this context, we are discussing transmission lines. I make NO assertions about light, how rings happen, or don't, or whether the theory and practice of optics is in way analogous to what happens on a transmission line. Transmission lines and their understanding can stand on their own without the help of optics. ....Keith |
wrote:
Seems to me that the sinusoidal standing wave with minima and maxima at the quarter wave points can only arise with single frequency sinusoidal excitation of the line. Are there other signals which will produce this result? Probably not, but that makes sinusoids unique, not magic. What about Ramo and Whinnery's forward Poynting vector and reflected Poynting vector? Why do you choose to ignore them? I haven't used them because I don't need them to arrive at an answer. You need them to keep from making the same mistakes over and over. Why make the solution more complex than necessary? Just to scare off the neophyte? Nope, your solution is simple-minded. Perhaps. Or maybe component waves are not the answer. Then go argue with Ramo & Whinnery. In this context, we are discussing transmission lines. I make NO assertions about light, how rings happen, or don't, or whether the theory and practice of optics is in way analogous to what happens on a transmission line. You will never understand the present topic unless you understand that EM wave interference doesn't affect the flow of energy in the individual waves. That is true for light and RF. -- 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! =----- |
W5DXP wrote:
You will never understand the present topic unless you understand that EM wave interference doesn't affect the flow of energy in the individual waves. That is true for light and RF. I should have added: "in the absence of a physical impedance discontinuity." If a physical impedance discontinuity exists, then of course, the flow of energy in the individual waves is affected by reflections. -- 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! =----- |
I wrote:
"As energy can`t be destroyed it had to be reflected by a hard short or open." Keith replied: "Or just stopped and stored." Wave energy is energy in motion. No motion, no waves. Keith also wrote: "All of this is easily visualized by observing the amplitude of the P(t) function at various points along the line." Power as a function of time has the same amplitude in the forward wave everywhere along the line. Same is true of the reflected wave. Interference as demonstrated by standing waves has no effect on this. Keith wrote: "I would strongly suggest that no energy crosses these points in the line where the voltage and current are always zero since p(t) is always zero." See my comment above on power as a function of time. Keith erred in saying "points in the line where the voltage and current are always zero", as where SWR volts are zero, amps are max, and vice versa. I wrote: "If energy were turned around before it reached the end of the line, nulls more distant from the source than the turnaround point would not exist." Keith wrote: "Not so,---." There is no argument that can make wave interference where there are no waves. In a lossless line, pre-existing waves could circulate forever. But, our discussion relates to effects on actual lines. Keith wrote: "Try visualizing how a step function charges the line." Totally irrelevant. SWR nulls are the result of phase opposition at specific points produced by alternating waves. A step function changes only when it starts or stops. Only during the changes does Zo apply unless a line is terminated in its Zo. In this case there`s no reflected wave to make a null. As Reg once said, "If your line were infinite in length, you could measure Zo with your ohmmeter. Best regards, Richard Harrison, KB5WZI |
Keith wrote:
"I do not find that step 7) is in error." Step 7) declares power is zero at quarter wave points where volts or amps are zero in the standing wave pattern. Power flow does not stop at an actual short or open in a line. It merely changes direction. At zero volt or amp points, where there is no actual short or open, no reversal of direction occurs. Power flow is affected in no way by standing wave nulls or maxima where no impedance discontinuity exists. Energy exists in every SWR null. It is in the two waves which produce the null as these experience no change in energy due to standing waves. Best regards, Richard Harrison, KB5WZI |
What is missing here, IMO, is that the physics of energy flow has not
been adequately explained. Cecil provided a reference to it but did not elaborate for the general readership. The energy in forward and reflected waves has been well documented for many years. From Kraus, Electromagnetics, McGraw Hill, 1953, Chapter 9, Section 9-13, "Energy Relations in a Standing Wave": EQ 9-145 We = 2eEo^2[cos^2wt*sin^2Bx] EQ 9-147 Wm = 2uHo^2[sin^2wt*cos^2Bx] Note: The energy in the E field [We] is a function of the sin^2(Bx). The energy in the H field [Wm] is a function of the cos^2(Bx). You will remember from trigonometry the the maxima [or minima] of a sin and cos are displaced by 90 degrees. Conclusion: when the E field is zero the H field is maximum; when the H field is zero the E field is maximum. Ergo! Energy is conserved and propagates through the zero E field as an H field; also, when the H field is zero the energy is in the E field. This is what Cecil is referring to when he refers to the Poynting vector. It is analogous to a parallel tuned circuit. When the instantaneous voltage across the capacitor is zero we don't claim there is no energy in the circuit. We know that the energy is stored in the inductor. Conversely, when the instantaneous current in the inductor is zero we don't claim there is no energy in the circuit. We know the energy is stored in the capacitor. In a TEM wave the energy cycles between the E field and the H field and the energy components are 90 degrees out of phase. Deacon Dave, W1MCE |
wrote:
What about Ramo and Whinnery's forward Poynting vector and reflected Poynting vector? Why do you choose to ignore them? I haven't used them because I don't need them to arrive at an answer. Yes, but you need them to arrive at the correct answer. :-) Basic electricity, a dash of circuit theory, a bit of knowledge of trigonometry, some basic calculus and the ability to think is all that is required. Apparently, that is not all that is required. Here's a neat web page that will allow you to visualize what is happening. Note the forward and reflected waves do not change energy or momentum at a zero voltage point. http://www.gmi.edu/~drussell/Demos/s....html#standing -- 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! =----- |
Keith wrote:
"Basic electricity, a dash of circuit theory, a bit of knowledge of trigonometry, some basic calculus, and the ability to think is all that is required." An electrical education limited to d-c familiarity leads to mistaken assumptions when dealing with a-c in Keith`s case it seems. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Keith wrote: "I do not find that step 7) is in error." Step 7) declares power is zero at quarter wave points where volts or amps are zero in the standing wave pattern. Unfortunately, probably due to my poor description, you have missed the point I was trying to make. You are, I think, disagreeing with the RESULT of step 7. Even though the result may be wrong, step 7) can be correct. Step 7) again: "7) From 6), the energy crossing quarter wave points is 0" This is saying that if the result of step 6) is correct, then the result of step 7) is correct. The transformation from 6) to 7) merely integrated power to get energy; a common and correct thing to do. So if you disagree with the result of step 7), it is in step 6) which you must search for the error since step 7) is correct. If step 6) is correct, then search in step 2) and step 5) for the error. I suspect that like Cecil, you will end up at step 2) as the source of what you perceive to be an error. Then, when we resolve whether 2) is in error, we will know whether the result of 7) is in error. ....Keith |
W5DXP wrote:
wrote: Seems to me that the sinusoidal standing wave with minima and maxima at the quarter wave points can only arise with single frequency sinusoidal excitation of the line. Are there other signals which will produce this result? Probably not, but that makes sinusoids unique, not magic. Magic was your moniker not mine. But I am glad that you agree. What about Ramo and Whinnery's forward Poynting vector and reflected Poynting vector? Why do you choose to ignore them? I haven't used them because I don't need them to arrive at an answer. You need them to keep from making the same mistakes over and over. I've being using p(t) = v(t) * i(t) and simply arguing that when v(t) or i(t) is zero for all t, then there is no power. Poynting won't change a thing. There is no P when E or H is zero. So the debate can be had with the simpler p(t) = v(t) * i(t). There is no need to complexify. ....Keith |
Richard Harrison wrote:
Keith wrote: "Are there other signals which would produce this result?" A short-circuit produces a voltage inversion. The plus volts and the minus volts make a zero in a short. Imagine a couple of identical positive pulses separated by some distance and traveling in the same direction along a long transmission line which has a short approached by the pulses. When the first pulse hits the short it is inverted and reflected to travel back toward its source as a minus voltage pulse. When the first pulse encounters the second pulse they disappear at the instant of coincidence, but otherwise, travel on their merry ways. What happens at coincidence is transfer of the energy associated with volts to the energy wave associated with amps as energy can`t be obliterated while the volts disappear. I do like pulses. Thinking about pulses on the line certainly helped clarify my understanding of how lines worked. So what happens when two pulses collide? I don't think you will like this answer either, but here goes. First, remember that power and energy are not the same. Power can become zero while the energy remains. It simply means that the energy is now stored in the capacitance or inductance (or equally, the E field or H field) but is not moving. So no energy is lost or destroyed when the power goes to zero. So what happens to these two pulses? They bounce off of each other and return whence they came. This must happen since the voltage on the region of the line where the two pulses collide remains zero and since p(t) = v(t) * i(t) there is no power in this region so there must be no energy flowing. You would observe exactly the same result were you to short the line at the point of collision just before the collision occurred. Two pulses colliding is just like a reflection. As a simple analogue, consider two identical elastic balls rolling towards each other. They bounce back after the collision. On a line this is easier to visualize with two negative pulses colliding. Each pulse consists of a clump of charge. When these clumps of charge collide, remembering that like charge repels, they bounce back. What actually happens during the collision? With balls, the energy is stored in the deformation of the ball and then released as the balls separate. With pulses, the power stops (current is zero), and the energy is stored in the capacitance of the line in the region of the collision. After all the energy has stopped flowing (no power, zero current, all energy in the E field), the energy is released into the reflected pulses and once again, there is energy moving (power) on the line. ....Keith |
Richard Harrison wrote:
I wrote: "As energy can`t be destroyed it had to be reflected by a hard short or open." Keith replied: "Or just stopped and stored." Wave energy is energy in motion. No motion, no waves. There is no doubt that energy moves. The point of disagreement is on how far it moves. Keith wrote: "I would strongly suggest that no energy crosses these points in the line where the voltage and current are always zero since p(t) is always zero." See my comment above on power as a function of time. Keith erred in saying "points in the line where the voltage and current are always zero", as where SWR volts are zero, amps are max, and vice versa. This last is true, but p(t) = v(t) * i(t); volts and amps must be present simultaneously for there to be power. I wrote: "If energy were turned around before it reached the end of the line, nulls more distant from the source than the turnaround point would not exist." Keith wrote: "Not so,---." There is no argument that can make wave interference where there are no waves. In a lossless line, pre-existing waves could circulate forever. But, our discussion relates to effects on actual lines. There are many assumptions in this discussion which means it only applies to ideal lines. The extension to real lines, retains the fundamentals but the details need tuning. As a simple example, on a real line, the nulls are never 0. But including this in the discussion would just make it more difficult to locate the points of disagreement. Keith wrote: "Try visualizing how a step function charges the line." Totally irrelevant. Understanding a step will help with understanding line behaviour. This knowledge can then assist in understanding sinusoidal steady state. ....Keith |
wrote:
I suspect that like Cecil, you will end up at step 2) as the source of what you perceive to be an error. Which step the error is in depends upon whether you are talking about NET energy or the forward and reflected component energies. -- 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! =----- |
wrote:
Poynting won't change a thing. There is no P when E or H is zero. Of course, there are the two component Poynting vectors which each contain power. How else could (Pz-/Pz+) = |rho|^2 You are continuing to confuse NET power with component power. The NET Poynting vector is zero. The component Poynting vectors are Pz- and Pz+ and NOT zero as explained in Ramo & Whinnery. The sum of the two Poynting vectors is zero at certain points because they are 180 degrees out of phase at those points. 1/4WL away, they are in phase. -- 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! =----- |
wrote:
So what happens to these two pulses? They bounce off of each other and return whence they came. This has been disproved many, many times. Do a web search for "superposition" and learn how those waves flow unaffected right through each other. -- 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! =----- |
wrote:
Understanding a step will help with understanding line behaviour. This knowledge can then assist in understanding sinusoidal steady state. Looks like you are never going to understand the principles of superposition and interference until you read and understand those chapters in _Optics_ or similar reference. You continue to make the same mistakes over and over in spite of the obvious mental violations of the principles of physics which have been explained 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 100,000 Newsgroups - 19 Different Servers! =----- |
Keith wrote:
"The last is true, but p(t) = v(t)*i(t); volts and amps must be present simultaneously for there to be power." By the same token, a-c flow is discontinuous at all zero crossings! I don`t think so. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Proof is that a directional sensor finds the same power flow, forward or reflected, at a standing wave zero point as it does at a standing wave maximum, or at any point in between. Richard's statement about what the directional sensor "finds" is perfectly correct - but unfortunately it cannot be used as proof. The reason is that so-called "directional wattmeters" don't physically sense directional power flow. All they sense from the transmission line are the current and the voltage, as two separate samples. Then they add or subtract these samples to give the sensor its directional properties. All the meter reads is a detected RF *voltage*, which changes to a different value when the sensor is reversed. If you want to know what those meter readings mean, you need transmission-line theory in order to understand them. You can then calibrate the meter to read forward and reverse power - but you cannot do that without using transmission-line theory to do it, and that theory is the subject of this entire discussion. Therefore the readings of a "directional wattmeter" cannot be used as evidence for either side, because that argument would be circular - you cannot use any theory to prove itself! However, this attempt to use inadmissible evidence doesn't necessarily affect Richard's wider argument about power flow. If that argument is correct, there definitely *will* be other physical evidence to prove it. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
Ian White, G3SEK wrote:
However, this attempt to use inadmissible evidence doesn't necessarily affect Richard's wider argument about power flow. If that argument is correct, there definitely *will* be other physical evidence to prove it. The biggest clue that I have noticed is that nobody has been able to generate standing waves in a single source, single feedline, single load system without the existence of a reflected wave. -- 73, Cecil, W5DXP |
Richard Harrison wrote:
Keith wrote: "The last is true, but p(t) = v(t)*i(t); volts and amps must be present simultaneously for there to be power." By the same token, a-c flow is discontinuous at all zero crossings! I don`t think so. There is certainly no power at the zero crossings. This variation in the rate of energy flow is why the power dudes really prefer 3 phase; energy flow is constant. ....Keith |
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W5DXP wrote: wrote: There is certainly no power at the zero crossings. There is no NET power at the zero crossings. I think you mean there's no instantaneous power at the zero crossings. 73, ac6xg |
Jim Kelley wrote:
W5DXP wrote: There is no NET power at the zero crossings. I think you mean there's no instantaneous power at the zero crossings. Well, since the NET voltage is always zero at a voltage node when the forward power and reflected power are equal, the instantaneous voltage is always zero, i.e. the steady-state voltage is always zero. If we have equal power flow vectors in opposite directions, the NET power is zero at all points up and down the line. -- 73, Cecil, W5DXP |
W5DXP wrote: Jim Kelley wrote: W5DXP wrote: There is no NET power at the zero crossings. I think you mean there's no instantaneous power at the zero crossings. Well, since the NET voltage is always zero at a voltage node when the forward power and reflected power are equal, the instantaneous voltage is always zero, i.e. the steady-state voltage is always zero. I've never actually seen the voltage at a node in a standing wave pattern referred to as an instantaneous voltage - especially considering that it doesn't vary with time. Instantaneous usually means the solution to a function f(t) at time t (not f(x) and position x.) Nodes and zero crossings aren't necessarily the same thing. 73, Jim AC6XG |
Jim Kelley wrote:
Nodes and zero crossings aren't necessarily the same thing. They are for standing waves on lossless unterminated lines, by definition. -- 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! =----- |
W5DXP wrote:
Jim Kelley wrote: Nodes and zero crossings aren't necessarily the same thing. They are for standing waves on lossless unterminated lines, by definition. I think not. Standing waves are spatial. At certain points on the line the (NET) voltage is always zero: nodes. At other points on the line, the (NET) voltage is sinusoidal and has 2 zero crossings per cycle. The amplitude of these sinusoids varies spatially along the line resulting in the standing wave. In an ideal line terminated by Zo, no matter where you attach your oscillograph to the line, you will observe a sinusoid of the same amplitude. This sinusoid will have zero crossings and power at the time of these zero crossings will be zero; no energy will be flowing at the time of the zero crossing. And going back to the comment that started this sub-thread, it is this cyclical variation in energy flow which prompted the power dudes to invent three phase lines in which the energy flow does not vary cyclically; power is constant. ....Keith |
W5DXP wrote:
wrote: So are you saying that like charge does not repel? No, I'm not saying that at all. In fact, there may not be any charges crossing the NET voltage = zero point and there doesn't have to be for energy to be flowing in both directions. This does seem to be the sticking point, doesn't it. Starting with p = v * i, i = charge_moved/time, if no charge moves, there can be no power, and from p = work/time, if there is no power, no work is being done, and if no work is being done, no energy has moved. Ergo, no energy crosses the boundary. If you find an error in the above derivation, I will happily allow you that energy flows in both directions. Two waves flowing in opposite directions in a constant Z0 environment superpose but have no effect on each other. You are continuing to be confused by the NET values. Actually, I am not confused by NET at all. I contend that NET is the only thing of importance when determining if energy flows. Did you take a look at that Java-driven web page that I posted a couple of days ago. If so, I don't see how you could still be confused. Are you referring to http://www.mellesgriot.com/products/optics/oc_2_1.htm? These dudes do seem to have it right. They sum amplitudes to get the resultant (NET) amplitude and then compute the intensity (power) from the resultant (NET) amplitude. They do NOT state that energy flows across a point with zero resultant amplitude nor do they sum the power (intensity) to determine interference patterns. As an analogy, consider two equal magnitude Tsunami waves flowing in opposite directions in the ocean. According to you, these waves will reflect off of each other. But it is known that those two waves will simply flow through each other and continue their original paths unabated. How can you tell whether it reflected, or flowed through? The look and feel will be the same. Remember the collision of two identical balls. Superficial examination of the collision might cause one to conclude that the energy was transferred between the balls, especially if you viewed this collision after viewing the collision of a moving ball with a stationary one where energy is indeed transferred. Only by looking at the details of the collision, in particular at the interface where the collision occurs, and realizing that f * d is always 0 do you learn that no energy was transferred. Similarly for transmission lines at points where v(t) * i(t) is always 0 and, from the web page, apparently for light as well. ....Keith |
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W5DXP wrote: Jim Kelley wrote: Nodes and zero crossings aren't necessarily the same thing. They are for standing waves on lossless unterminated lines, by definition. I think not. Standing waves are spatial. At certain points on the line the (NET) voltage is always zero: nodes. For the infinite number of times a snapshot of the voltage is not zero, the node *IS* a zero-crossing point. That is more than obvious from the JAVA applets on this web page. http://www.gmi.edu/~drussell/Demos/s....html#standing In an ideal line terminated by Zo, ... That configuration is not covered by my statement above which applies only to standing waves on lossless unterminated 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 100,000 Newsgroups - 19 Different Servers! =----- |
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W5DXP wrote: wrote: So are you saying that like charge does not repel? No, I'm not saying that at all. In fact, there may not be any charges crossing the NET voltage = zero point and there doesn't have to be for energy to be flowing in both directions. This does seem to be the sticking point, doesn't it. There is no sticking point. It is all explained in _Fields_and_Waves_... by Ramo, Whinnery, and Van Duzer. It is also explained in _Optics_, by Hecht. If you find an error in the above derivation, I will happily allow you that energy flows in both directions. That a forward Poynting vector and a reflected Poynting vector exists is not enough for you? Here's my argument analogous to yours: My GMC pickup is white. Until you can prove my GMC pickup is not white, your argument is invalid. Your above argument is just as irrelevant as the color of my pickup. Actually, I am not confused by NET at all. I contend that NET is the only thing of importance when determining if energy flows. What about the forward Poynting vector and the reflected Poynting vector? Net is the thing to consider if all you are worried about is net energy. Net doesn't hack it when you take a look at the forward and reflected Poynting vectors. How can you tell whether it reflected, or flowed through? The look and feel will be the same. Not exactly. Each Tsunami has its own modulation signature. Many, many experiments over hundreds of years have proved that waves flow right through each other. Similarly for transmission lines at points where v(t) * i(t) is always 0 and, from the web page, apparently for light as well. You really believe that the energy in a bright interference ring is trapped between the dark rings by some magical force that exists in your mind? You really need to read the superposition and interference chapters in _Optics_. -- 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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We have a choice of two rho for this situation: Correction: We have a choice of two reflection coefficients each with its own unique definition. black box - 0, computed from the surge impedance of the line and the steady state impedance of the load Actually, Sqrt(Pref/Pfwd), the definition of rho. open box - 0.5, computed from the surge impedance of the line and the surge impedance of the load Actually, (150-50)/(150+50), the definition of s11. In any case, what we have in this experiment is a case where there IS an impedance discontinuity and yet there is no reflection (if you use the "black box" rho, as is often done). This is technically not true. The NET reflections are zero. There are two non-zero component reflections as seen from the s-parameter equation: b1 = s11*a1 + s12*a2 These three terms are all reflections. b1 is the NET reflections toward the source. Since b1 = zero, s11*a1 = -s12*a2, i.e. the two component reflections are of equal magnitude and opposite phase and therefore cancel. This is explained in the last three paragraphs on the Melles-Griot web page. So, if we are allowed to say in the first experiment that rho is 0 despite an impedance discontinuity, we are equally allowed to say for the second that rho is -1 despite the absence of a discontinuity. There is NOT an absence of a discontinuity. The discontinuity is as large as it can possibly be, an unterminated transmission line. There is an infinite SWR on the 1/4WL line. The apparent zero impedance at the input is simply a V/I ratio where Vnet=Vfwd+Vref=0 The forward wave is carrying Vfwd^2*Z0 watts associated with the forward Poynting vector and the reflected wave is carrying Vref^2*Z0 watts associated with the reflected Poynting vector. A directional wattmeter will indicate the same thing. There is NO physical discontinuity at the black box. The only physical discontinuity is the open end of the stub. That's where 100% of the reflection action takes place. -- 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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