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The Rest of the Story
Keith Dysart wrote:
The simulator at that web site does seem to have its issues. Ask it to simulate 700 nm + 680 nm at the same amplitude and see if the result represents reality. The duration of each calculation appears to be about one second and then a reboot. -- 73, Cecil http://www.w5dxp.com |
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On Mar 27, 2:06 am, Roger Sparks wrote:
Cecil Moore wrote: Roger Sparks wrote: You need to take a look at the spreadsheets. Roger, in a nutshell, what is the bottom line? The bottom line in a nutshell? I'll try. First, I added a note to both spreadsheets indicating that zero degrees is CURRENT zero degrees. This because the source turns out to be reactive, with current peak 45 degrees from voltage peak. http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf I was not able to discern the derivation of the various equations though the data in the columns looked somewhat reasonable. Were the equation really based on the opening paragraph statement of 100 Vrms, or is it scaled to a different source voltage. The sin functions have an amplituted of 100, which suggests a source of 200 volts or Vrms of 141.4 volts. The spreadsheet addresses the following issues: Does the traveling wave carry power? Yes. The spreadsheet was built assuming that power is carried by traveling waves. Because the resulting wave form and powers seem correct, the underlaying assumption seems correct. It was not obvious which columns were used to draw this correlation. However, even if this experiment is consistent with the hypothesis it only takes one experiment which is not to disprove the hypothesis. Is power conserved on the transmission line, meaning, can the energy contained in power be conserved and located over time on the transmission line? Yes, the spreadsheet was built assuming that power could be conserved and traced over time so the underlaying assumption seems correct. Does interference occur in this example? The spreadsheet was built assuming that voltage and currents from superpose in a manner consistent with constructive and destructive interference, so the underlaying assumption seems correct. Is power stored in the reactive component for release in later in the cycle or during the next half cycle? Yes, power is stored on the transmission line during the time it takes for power to enter the line, travel to the end and return. The time of wave travel on the transmission line is related to the value of the reactive component. Does the direction of wave travel affect the measurement of voltage and the application of power to a device? Yes. A wave loses energy (and therefore voltage) as it travels through a resistance. As a result, power from the prime source is ALWAYS applied across the sum of the resistance from the resistor AND transmission line. I am not sure that I would describe this as the wave losing energy, but rather as the voltage dividing between the two impedances. If the source resistance was replaced by another transmission, which could easily be set to provide a 50 ohm impedance, would you still describe it as the wave losing energy? The spreadsheet was built using this assumption and seems correct. (At times during the cycle, the forward and reflected waves oppose, resulting in very little current through the resistor. During those times, the power applied to the transmission line is much HIGHER because the reflected wave reflects from the load and source, and merges/adds to the forward wave from the source.) I am not convinced. When there is very little current through the resistor, there is also very little current into the transmission line. This suggests to me that the power applied to the transmission line is low. ....Keith |
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On Mar 27, 11:10 am, Cecil Moore wrote:
Keith Dysart wrote: So you are having difficulty doing the math to justify your hypothesis. Actually no, the math is not difficult. I'm pre- occupied with something else and think it's just time to agree with Hecht that instantaneous power is "of limited utility". Have you taken a look at Roger's spreadsheets? Yes. Conservation of energy requires that the total quantity of energy in the system not change. :-) Isn't the whole purpose of a transmitting antenna to radiate energy away from the antenna system? And that radiation continues to be "lost" from the system space for some time after the source power is removed? Wouldn't you have to define the "system" as the entire universe for your statement to be true? Sort of, but there is an easy work-around. Since we are not particularly interested in what happens to the energy going to the source resistor, for example, we define the boundary of the system to pass through the resistor and simply account for the energy as a flow out of the system. And we do the same for the source since we are not particularly interested in where the energy for the source comes from; we just account for it as a flow into the system. And lastly, we don't really care where the energy on the line goes to or comes from, so we just account for it as a flow out of the system. You may have noticed that we have removed all the components from the system, it is a null system, and all we are doing is accounting for the flows into and out of the system. Hence Ps(t) = Prs(t) + Pg(t) The powers must sum to 0 to satisfy conservation of energy. That may be true, but there's still no conservation of instantaneous power principle. To conserve energy however, i.e. to have no energy accumulate in the null system described above, the flows must balance at all times, that is, instantaneously. A hot resistor continues to radiate heat long after any power source is removed. If you want to be that complicated you can; energy is delivered instantaneously to the resistor according to Vr*Ir (or equivalently, V**2/R), stored in the resistor as heat and dissipated to the environment later. But for our purposes, we can stop that analysis at the point where the energy enters the resistor and use Vr(t) * Ir(t) to compute the instantaneous flow into the resistor. You could simply do the derivation for an example that demonstrates your hypothesis. Already done on my web page. My only actual hypothesis concerns average power. I've wasted too much time bantering about something that Hecht says is "of limited utility". Unfortunately for your hypothesis, average power is insufficient to account for energy which might be in the reflected wave. An average power analysis agrees with your hypothesis, while a more detailed instaneous analysis disproves it. the idea that Pfor and Pref describe actual energy flows is very dubious. Again, look yourself in the mirror and tell yourself that what you are seeing contains no energy. The theory that some EM waves contain energy and some do not is not new to you. Dr. Best was the first to theorize that canceled waves continue to propagate forever devoid of energy. Someone else asserted that canceled waves never contained any energy to start with. I strongly suspect that what you are seeing in the mirror are the waves that didn't cancel and that do contain energy. :-) If it was just a 'suspicion' you could probably let go of the idea long enough to learn what is really happening and why it is not inconsistent with the idea that reflected wave energy is a dubious concept. So the energy is not being stored in the reactive component of the line input impedance. Assuming you have not made an error, You *have* found it hard to the do the math; otherwise, you could detect an error, if there was one. so what? Energy stored in the reactance is only one of the possibilities that I listed earlier. As I said in an earlier posting which you declared a non-sequitor (sic), one or more of the following is true: 1. The source adjusts to the energy requirements. 2. The reactance stores and delivers energy. 3. Wave energy is redistributed during superposition. 4. Something I haven't thought of. Your explanation is not complete until you can identify the element that stores and returns the energy and its energy transfer function. The ExH reflected wave energy exists and cannot be destroyed. It goes somewhere and its average value is dissipated in the source resistor in my special case example. Or not, as has been shown with the detailed analysis. You are attempting to destroy the reflected wave energy using words and math presumably knowing all along that reflected wave energy cannot be destroyed. Energy can not be destroyed. This leads inexorably to the conclusion that the reflected wave does not necessarily contain energy. ....Keith |
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On Mar 27, 11:37 am, Cecil Moore wrote:
Keith Dysart wrote: I had not realized that you had these alternate sources for the interference energies, not having seen that in your papers. I only posted it three times here and you chose to ignore all of those postings. I have published only one paper with three more to go. The special case Part 1 contains zero average interference so there is no alternate source for average interference and indeed, none is needed for Part 1. But it is one way to sidestep the issue; different rules for the expectations of superposition and interference in different scenarios. That's why I have four parts only one of which has been published. The rules are not different but the conditions within the examples are different. Part 2 will be an example with the condition of average destructive interference existing at the source resistor. Although there are no ordinary reflections because the reflection coefficient is 0.0, there will exist something that looks a lot like a reflection caused by superposition/interference. The FSU web page calls it a "redistribution", not a "reflection". I am satisfied with FSU's word "redistribution" for the results of coherent wave interaction. I am surprised then, for the example of Fig 1-1 with 12.5 ohms, that you don't just say "There is a source nearby, that *must* be where the unaccounted energy comes from", and leave it at that. Since my special case example contains zero average interference, the average power output of the source is constant and unaffected by zero interference. There is zero average energy unaccounted for. Part 2 will illustrate the source adjusting its power output to compensate for destructive interference. Perhaps you should complete part one so that it fully accounts for the energy flows before progressing to writing part two. Average is not a full accounting. ....Keith |
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On Sat, 29 Mar 2008 12:45:48 -0700 (PDT)
Keith Dysart wrote: On Mar 27, 2:06 am, Roger Sparks wrote: Cecil Moore wrote: Roger Sparks wrote: You need to take a look at the spreadsheets. Roger, in a nutshell, what is the bottom line? The bottom line in a nutshell? I'll try. First, I added a note to both spreadsheets indicating that zero degrees is CURRENT zero degrees. This because the source turns out to be reactive, with current peak 45 degrees from voltage peak. http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf I was not able to discern the derivation of the various equations though the data in the columns looked somewhat reasonable. Were the equation really based on the opening paragraph statement of 100 Vrms, or is it scaled to a different source voltage. The sin functions have an amplituted of 100, which suggests a source of 200 volts or Vrms of 141.4 volts. I simplified and changed the phase for the equation in column C from '100sin(wt-90) + 100sin(wt-180)' to '100sin(wt+90) + 100sin(-wt)' so that the total voltage in column D better matches with the voltages from my formula in column G. The equation for column B is just the sine wave for the base wave, showing the voltage resulting to the source resistor from the peak current. The equation in column C is the same (from current) but recognizes that current from the reflected wave will cancel the current from the forward wave coming through Rs. The two waves are traveling in opposite directions so one angle must be labeled with a negative sign so that it will rotate in the opposite direction. In column C, the first term is the reflected wave and the second term is the base wave. My formula, displayed in column G, uses the applied voltage of 141.4 volts. I thought it would be easier to see how the spreadsheet was built using current to find the voltages. You can see how the results of column D match well with those from column G. The spreadsheet addresses the following issues: Does the traveling wave carry power? Yes. The spreadsheet was built assuming that power is carried by traveling waves. Because the resulting wave form and powers seem correct, the underlaying assumption seems correct. It was not obvious which columns were used to draw this correlation. Column E and column F display power. Column F recognizes that if 100 watts is applied continueously (on the average) over an entire 360 degree cycle, the final power applied over time would be 360 * 100 = 36000 watt-degrees. Part of the power comes from the reflection, part comes directly. Obviously, interference is very much at work in this example. However, even if this experiment is consistent with the hypothesis it only takes one experiment which is not to disprove the hypothesis. True! Is power conserved on the transmission line, meaning, can the energy contained in power be conserved and located over time on the transmission line? Yes, the spreadsheet was built assuming that power could be conserved and traced over time so the underlaying assumption seems correct. Does interference occur in this example? The spreadsheet was built assuming that voltage and currents from superpose in a manner consistent with constructive and destructive interference, so the underlaying assumption seems correct. Is power stored in the reactive component for release in later in the cycle or during the next half cycle? Yes, power is stored on the transmission line during the time it takes for power to enter the line, travel to the end and return. The time of wave travel on the transmission line is related to the value of the reactive component. Does the direction of wave travel affect the measurement of voltage and the application of power to a device? Yes. A wave loses energy (and therefore voltage) as it travels through a resistance. As a result, power from the prime source is ALWAYS applied across the sum of the resistance from the resistor AND transmission line. I am not sure that I would describe this as the wave losing energy, but rather as the voltage dividing between the two impedances. If the source resistance was replaced by another transmission, which could easily be set to provide a 50 ohm impedance, would you still describe it as the wave losing energy? It should be OK to think of the voltage dividing between two impedances. The important thing is to consider how the reflected voltage sums with the forward voltage where it is measured. Because the two waves are traveling in opposite directions, the measured voltage is not the voltage applied to either Rs or the transmission line. This is why columns B and C must be added to find the total voltage across Rs. The spreadsheet was built using this assumption and seems correct. (At times during the cycle, the forward and reflected waves oppose, resulting in very little current through the resistor. During those times, the power applied to the transmission line is much HIGHER because the reflected wave reflects from the load and source, and merges/adds to the forward wave from the source.) I am not convinced. When there is very little current through the resistor, there is also very little current into the transmission line. This suggests to me that the power applied to the transmission line is low. I don't follow you here. Right, power into the transmission line is low when the current in is low, and it is high when the current is high. It is clear that peak current and peak voltage do not occur at the same time except when measured across the resistor in column D. ...Keith This is just one example, but it seems like the power is accounted for here. We need another example where power can NOT be accounted for. -- 73, Roger, W7WKB |
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On Mar 29, 7:18 pm, Roger Sparks wrote:
On Sat, 29 Mar 2008 12:45:48 -0700 (PDT) Keith Dysart wrote: On Mar 27, 2:06 am, Roger Sparks wrote: Cecil Moore wrote: Roger Sparks wrote: You need to take a look at the spreadsheets. Roger, in a nutshell, what is the bottom line? The bottom line in a nutshell? I'll try. First, I added a note to both spreadsheets indicating that zero degrees is CURRENT zero degrees. This because the source turns out to be reactive, with current peak 45 degrees from voltage peak. http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf I was not able to discern the derivation of the various equations though the data in the columns looked somewhat reasonable. Were the equation really based on the opening paragraph statement of 100 Vrms, or is it scaled to a different source voltage. The sin functions have an amplituted of 100, which suggests a source of 200 volts or Vrms of 141.4 volts. I simplified and changed the phase for the equation in column C from '100sin(wt-90) + 100sin(wt-180)' to '100sin(wt+90) + 100sin(-wt)' so that the total voltage in column D better matches with the voltages from my formula in column G. The equation for column B is just the sine wave for the base wave, showing the voltage resulting to the source resistor from the peak current. The equation in column C is the same (from current) but recognizes that current from the reflected wave will cancel the current from the forward wave coming through Rs. The two waves are traveling in opposite directions so one angle must be labeled with a negative sign so that it will rotate in the opposite direction. In column C, the first term is the reflected wave and the second term is the base wave. My formula, displayed in column G, uses the applied voltage of 141.4 volts. I thought it would be easier to see how the spreadsheet was built using current to find the voltages. You can see how the results of column D match well with those from column G. I am still having difficulty matching these equations with mine. Just to make sure we are discussing the same problem.... The circuit is a voltage source Vs(t) = 141.4 sin(wt-45) driving a source resistor of 50 ohms and 45 degress of 50 ohm transmission line that is shorted at the end. My calculations suggest that Vrs.total(t) = 100 sin(wt-90) Irs.total(t) = 2 sin(wt-90) which agrees with you column D but not the introduction which says that the zero current is at 0 degrees. With -100 volts across the source resistor at 0 degrees, the current should be -2 amps through the source resistor at 0 degrees; in other words, a current maximum. Using superposition, I compute the the contribution of the source to be Vrs.source(t) = 70.7 sin(wt-45) and the contribution from the reflected wave to be Vrs.reflected(t) = -70.7 sin(wt+45) which sum to Vrs.total(t) = 100 sin(wt-90) as expected. So while I can construct an expression for Column D, the contributing values do not agree with those you have provided in Columns B and C. Column E follows from Column D and my calculations agree. And also with Column F. The spreadsheet addresses the following issues: Does the traveling wave carry power? Yes. The spreadsheet was built assuming that power is carried by traveling waves. Because the resulting wave form and powers seem correct, the underlaying assumption seems correct. It was not obvious which columns were used to draw this correlation. Column E and column F display power. Column F recognizes that if 100 watts is applied continueously (on the average) over an entire 360 degree cycle, the final power applied over time would be 360 * 100 = 36000 watt-degrees. Part of the power comes from the reflection, part comes directly. Obviously, interference is very much at work in this example. Column E is the power dissipated in the resistor, and Column F is the integral of Column and represents the total energy which has flowed in to the resistor over the cycle. It is also the average energy per cycle. If you were to extend your analysis to compute the energy in each degree of the reflected wave and add it to the energy in each degree of Vrs.source(t) and sum these, you would find that the instantaneous energy from Vrs.source and Vrs.reflected does not agree with the instantaneous energy dissipated in the source resistor. It is this disagreement that is the root of my argument that the power in the reflected wave is a dubious concept. Using averages, the computed powers support the hypothesis, but when examined with finer granularity, they do not. However, even if this experiment is consistent with the hypothesis it only takes one experiment which is not to disprove the hypothesis. True! Is power conserved on the transmission line, meaning, can the energy contained in power be conserved and located over time on the transmission line? Yes, the spreadsheet was built assuming that power could be conserved and traced over time so the underlaying assumption seems correct. Does interference occur in this example? The spreadsheet was built assuming that voltage and currents from superpose in a manner consistent with constructive and destructive interference, so the underlaying assumption seems correct. Is power stored in the reactive component for release in later in the cycle or during the next half cycle? Yes, power is stored on the transmission line during the time it takes for power to enter the line, travel to the end and return. The time of wave travel on the transmission line is related to the value of the reactive component. Does the direction of wave travel affect the measurement of voltage and the application of power to a device? Yes. A wave loses energy (and therefore voltage) as it travels through a resistance. As a result, power from the prime source is ALWAYS applied across the sum of the resistance from the resistor AND transmission line. I am not sure that I would describe this as the wave losing energy, but rather as the voltage dividing between the two impedances. If the source resistance was replaced by another transmission, which could easily be set to provide a 50 ohm impedance, would you still describe it as the wave losing energy? It should be OK to think of the voltage dividing between two impedances. The important thing is to consider how the reflected voltage sums with the forward voltage where it is measured. Because the two waves are traveling in opposite directions, the measured voltage is not the voltage applied to either Rs or the transmission line. This is why columns B and C must be added to find the total voltage across Rs. Whether one needs to add or subtract is more a matter of the convention being used for the signs of the values. When Vf and Vr are derived using Vtot = Vf + Vr; Itot = If + Ir one would expect to have to add the negative of Vr to the contribution from Vs to arrive at the total voltage across the source resistor. The spreadsheet was built using this assumption and seems correct. (At times during the cycle, the forward and reflected waves oppose, resulting in very little current through the resistor. During those times, the power applied to the transmission line is much HIGHER because the reflected wave reflects from the load and source, and merges/adds to the forward wave from the source.) I am not convinced. When there is very little current through the resistor, there is also very little current into the transmission line. This suggests to me that the power applied to the transmission line is low. I don't follow you here. Right, power into the transmission line is low when the current in is low, and it is high when the current is high. It is clear that peak current and peak voltage do not occur at the same time except when measured across the resistor in column D. Power into the transmission line is low when either the voltage or the current is low; when either is zero, the power is zero. Since the highest voltage occurs with zero current and the highest current occurs with zero voltage, maximum power into the transmission line occurs when the voltage and current are both medium; more precisely, when they are both at .707 of their maximum values. This is just one example, but it seems like the power is accounted for here. We need another example where power can NOT be accounted for. I suggest it is the same example, but the granularity of the analysis needs to be increased. ....Keith |
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Keith Dysart wrote:
Unfortunately for your hypothesis, average power is insufficient to account for energy which might be in the reflected wave. An average power analysis agrees with your hypothesis, while a more detailed instaneous analysis disproves it. One more time: My hypothesis doesn't apply to instantaneous powers at all so there is nothing to disprove. Please leave me out of any discussion of instantaneous powers. If it was just a 'suspicion' you could probably let go of the idea long enough to learn what is really happening and why it is not inconsistent with the idea that reflected wave energy is a dubious concept. One look in a mirror should convince you otherwise. You *have* found it hard to the do the math; Nope, I just think that instantaneous power math is a waste of my time. Seems to also be a waste of your time. Your explanation is not complete until you can identify the element that stores and returns the energy and its energy transfer function. I simply don't have anything at all to say about instantaneous powers. In my opinion, such is a waste of my time. You might as well be demanding that I produce the math for how many angels can dance on the head of a pin. Energy can not be destroyed. Yet, you are trying your best to destroy the energy in the reflected wave. Since you cannot destroy reflected energy at the average power level, you are trying to destroy it at the instantaneous level. -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
Perhaps you should complete part one so that it fully accounts for the energy flows before progressing to writing part two. Average is not a full accounting. Average is the only accounting that I consider to be important and the only accounting that I am going to do. I have added a disclaimer about instantaneous power to my Part 1 article. I personally don't think that anyone else cares about instantaneous powers. If you need an instantaneous power article written, please feel free to write it yourself. I wish you luck but I personally consider it to be a waste of time. -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
Using averages, the computed powers support the hypothesis, but when examined with finer granularity, they do not. Well Keith, yours also falls apart at finer granularity where you are required to determine the position and momentum of the individual charge carriers. -- 73, Cecil http://www.w5dxp.com |
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On Mar 30, 8:28*pm, Cecil Moore wrote:
Keith Dysart wrote: Unfortunately for your hypothesis, average power is insufficient to account for energy which might be in the reflected wave. An average power analysis agrees with your hypothesis, while a more detailed instaneous analysis disproves it. One more time: My hypothesis doesn't apply to instantaneous powers at all so there is nothing to disprove. Please leave me out of any discussion of instantaneous powers. You state that your hypothesis is that for this specific circuit, "the energy in the reflected wave is dissipated in the source resistor". This claim is amenable to analysis using instantaneous energy flows. When so analyzed, the hypothesis fails. I can see why you want to limit the kinds of analysis applied to your hypothesis, but that is not how science works. If you wish to narrow your hypothesis to "the average energy in the reflected wave is simply numerically equal to the increase in the average dissipation in the source resistor" I will not object since that hypothesis would be completely accurate and not misleading. As long as you wish to claim that there is actual energy in the reflected wave, and that this energy is dissipated in the source resistor, you must be prepared to offer a full accounting. If it was just a 'suspicion' you could probably let go of the idea long enough to learn what is really happening and why it is not inconsistent with the idea that reflected wave energy is a dubious concept. One look in a mirror should convince you otherwise. You *have* found it hard to the do the math; Nope, I just think that instantaneous power math is a waste of my time. Seems to also be a waste of your time. Not at all. I've learned a new (at least to me) technique for analyzing where the energy goes. I've increased my skills with complex numbers in Excel. I have a firmer grasp about power computations. I've explored the tautology of Ptotal = Pforward - Preflected. I've learned a bit about macros in Excel. I've tried Calc and found it wanting. I've looked at energy flows in circulators. I've explored active and passive circulators. And I've proved your hypothesis to be false. Not a waste at all. Your explanation is not complete until you can identify the element that stores and returns the energy and its energy transfer function. I simply don't have anything at all to say about instantaneous powers. In my opinion, such is a waste of my time. A higly convenient opinion since it allows you to continue to believe your hypothesis. You might as well be demanding that I produce the math for how many angels can dance on the head of a pin. No. Instantaneous flows *can* be computed. Angels on a pin are a little more problematic. Energy can not be destroyed. Yet, you are trying your best to destroy the energy in the reflected wave. Since you start with an unshakeable belief in the existance of energy in the reflected wave, this would be your natural conclusion. Those of us whose beliefs are not so fixed, evaluate the proofs and come to the conclusion that energy in the reflected wave is a dubious concept. Since you cannot destroy reflected energy at the average power level, you are trying to destroy it at the instantaneous level. "Average" is just a data reduction from "instantaneous". An agreement of averages is not a proof of anything except that the "averages are numerically equal". ...Keith |
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