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Revisiting the Power Explanation
Dr. Honeydew wrote:
Clearly the 40 ohm load dissipates more power than the 50 ohm load, so we don't see how your answer and Mr. Harrison's posting can both be correct. It appears that the internal resistance of the Thevenin equivalent circuit was chosen to be ~1 ohm. Maximum power transfer of ~13 watts should occur when the load is ~1 ohm and the current is ~3.6 amps with ~3.6 volts across the 1 ohm load and across the ~1 ohm generator impedance. How one gets 13 watts out of a source rated for one watt is a non-linear physical design problem, not a linear theory problem. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Mar 23, 11:05 am, Dan Bloomquist wrote:
Your example assumes that the reflected power will see the 50 ohms of the generator. And I had shown you a condition where it will see a short no mater what the 'output' impedance of the generator. Truly, we have found the root of the disconnect. Kindly compute the reflection coefficient at the connection where a 50 Ohm line is driving a 75 Ohm line. For your convenience, recall that RC = (Z2-Z1)/(Z2+Z1). Now redo the same, except that the 75 Ohm line is one-half wavelength long terminated in a short. If you get the same answer, then you will see why the reflected voltage in the example does not encounter a discontinuity at the entrance to the generator and is therefore not re-reflected. If you get a different answer, then some study of reflection coefficient is in order. ....Keith |
Revisiting the Power Explanation
Keith Dysart wrote:
Truly, we have found the root of the disconnect. Truly, we have but it is not what you think. A source doesn't obey the passive reflection rules. The V/I ratio encountered within a source is *active*, not passive. Active V/I ratios can and do cause reflections. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Cecil Moore wrote:
Reflected power doing damage is an overly simplistic way of viewing things. The reflected wave is capable of causing damage without giving up any of its energy. It is not the energy in the reflected wave, per se, that causes the damage. It is the interference pattern established by superposition of the forward wave and the reflected wave that causes the damage. If the reflected voltage arrives back at the source in phase with the forward voltage, that constructive voltage interference can cause an over-voltage condition that blows the finals. This over-voltage condition actually reduces the dissipation in the finals because it is accompanied by a destructive interference, reduced-current condition. The finals are actually cooler than normal when the over-voltage begins to fry them. Cecil, Ya know, it is possible to simply add and subtract voltages. It is not required to determine an "interference pattern". Solid state electronics are damaged by high voltage or high power dissipation (or both). No "interference patterns" required. Perhaps the manufacturers could add exactly the correct length of transmission line inside the transceivers so that the wrong kind of interference could never occur at the finals. 8-) 73, Gene W4SZ |
Revisiting the Power Explanation
Cecil Moore wrote:
[snip] Since the generator is not delivering any power and there is a forward power and a reflected power, the reflected power is supplying the forward power, i.e. being 100% re-reflected. The re-reflection is associated with total destructive interference toward the generator and total constructive interference toward the load. Anything else violates the conservation of energy principle. Cecil, You have highlighted a really useful mathematical aid. Namely, include the desired answer as a basic condition for setting up the problem, and the proof becomes easy. 8-) 73, Gene W4SZ |
Revisiting the Power Explanation
Gene Fuller wrote:
Ya know, it is possible to simply add and subtract voltages. It is not required to determine an "interference pattern". It is absolutely necessary to understand the interference patterns if one wants to track the energy through the system. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Gene Fuller wrote:
Namely, include the desired answer as a basic condition for setting up the problem, and the proof becomes easy. Yep, no need to make up new laws of physics to explain things that obviously follow the old laws of physics. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Cecil Moore wrote:
Keith Dysart wrote: Truly, we have found the root of the disconnect. Truly, we have but it is not what you think. A source doesn't obey the passive reflection rules. The V/I ratio encountered within a source is *active*, not passive. Active V/I ratios can and do cause reflections. Well said. Thanks, Dan. |
Revisiting the Power Explanation
On Thu, 22 Mar 2007 12:59:20 -0800, Richard Clark wrote:
On Thu, 22 Mar 2007 15:55:40 GMT, Walter Maxwell wrote: On Wed, 21 Mar 2007 08:18:14 -0500, "Richard Fry" wrote: "Walter Maxwell" wrote (RF): And if so, would that also mean that such a tx would not be prone to producing r-f intermodulation components when external signals are fed back into the tx from co-sited r-f systems? This issue is irrelevant, because the signals arriving from a co-sited system would not be coherent with the local source signals, while load- reflected signals are coherent. The destructive and constructive interference that occurs at the output of a correctly loaded and tuned PA requires coherence of the source and reflected waves to achieve the total re-reflection of the reflected waves back into the direction toward the load. Hi Walt, It is not irrelevant, merely illustrative of the concept of reflection that is consistent with a coherent source. Your points of phase are the sine non quo to the discussion, but all too often arguers only take the half of the 360 degrees available to argue a total solution. Even more often, they take only one or two degrees of the 360. Richard, it's been my observation that many of those who argue are clueless concerning the phase relationships required to obtain the destructive and constructive interference that achieves the re-reflection of the reflected waves. A reflection resulting from a discontinuity in the path of a signal delivered by a souce is guaranteed to be coherent with the source wave. If there is no coherence between the reflected wave and the source wave there may be an interference, but none of the type that results in total destructive and constructive interference relevant to impedance matching. I don't understand what you mean by 'taking only one of two degrees of the 360.' But even for coherent reflections, if the PA tank circuit has very low loss for incident power (which it does), why does it not have ~ equally low loss for load reflections of that power? Such would mean that load reflections would pass through the tank to appear at the output element of the PA, where they can add to its normal power dissipation. The paragraph above seems to me to imply that RF doesn't understand the destructive and constructive interference phenomena involved with re-reflection. This is the symmetry of the illustration of external signals. You used external signals yourself as part of your case study; hence the relevance has been made by you. Whoa, Richard! You'll have to point out where I've discussed external signals in any case study involving phase relationships between forward and reflected waves. I've never done so knowingly. Also, does not the result of combining the incident and reflected waves in the tx depend in large part on the r-f phase of the reflection there relative to the r-f phase of the incident wave? And the r-f phase of the reflection is governed mostly by the number of electrical wavelengths of transmission line between the load reflection and the plane of interest/concern -- which is independent of how the tx has been tuned/loaded. And we return to the sine non quo for the discussion: phase. That's true, but although RF apparently realizes that the phase relationship is relevant, he doesn't seem to understand the details of the phase requirements that achieve the necessary interferences that accomplish the impedance matching. If the ham transmitter designs that your paper applies to produce a total re-reflection of reverse power seen at their output tank circuits, then there would be no particular need for "VSWR foldback" circuits to protect them. Yet I believe these circuits are fairly common in ham transmitters, aren't they? They certainly are universal in modern AM/FM/TV broadcast transmitters, and are the result of early field experience where PA tubes, tx output networks, and the transmission line between the tx and the antenna could arc over and/or melt when reflected power was sufficiently high. RF Richard, your statement above begs the question, "Are you aware of the phase relationships between forward and reflected voltages and between forward and reflected currrents that accomplish the impedance-matching effect at matching points such as with stub matching and also with antenna tuners? It seems he is on the face of it, doesn't it? Afterall, he is quite explicit to this in the statement you are challenging. No Richard, I don't believe he is. I don't see the 'explicitness' you seem to find. It's the complete lack of the explicitness that makes me believe he doesn't quite get it. When the matching is accomplished the phase relationship between the foward and reflected voltages can become either 0° or 180°, resulting in a total re-reflection of the voltage. If the resultant voltage is 0°, then the resultant current is 180°, thus voltage sees a virtual open circuit and the current sees a virtual short circuit. The result is that the reflected voltage and current are totally re-reflected IN PHASE with the source voltage and current. This is the reason the forward power in the line is greater than the source power when the line is mismatched at the load, but where the matching device has re-reflected the reflected waves. Nothing here contradicts anything either of you have to say. True, but RF just hasn't said it all, because, as I said above, I don't believe he understands the details of the phase requirements to achieve the match. This phenomenon occurs in all tube transmitters in the ham world when the tank circuit is adjusted for delivering all available power at a given drive level. This introduces the two concepts of the "need for match" and the "match obtained." They are related only through an action that spans from one condition to the other. They do not describe the same condition, otherwise no one would ever need to perform the match: I don't comprehend your statements in the paragraph above. When this condition occurs the adjustment of the pi-network has caused the relationship between the forward and reflected voltages to be either 0° or 180° and vice versa for currents, as explained above. When this condition occurs, destructive interference between the forward and reflected voltages, as well as between the forward and reflected currents, causes the reflected voltage and current to cancel. However, due to the conservation of energy, the reflected voltage and current cannot just disappear, so the resulting constructive interference following immediately, causes the reflected voltage and current to be reversed in direction, now going in the foward direction along with and in phase with the forward voltage and current. If a tree were to fall onto the antenna, a new mismatch would occur. Would the transmitter faithfully meet the expectations of the Ham unaware of the accident? No, reflected (0-179 degrees) energy would undoubtedly offer a 50% chance of excitement in the shack. The consequences of dissipation would be quite evident on that occasion. For the other 180 (180-359) degrees of benign combination; then perhaps not. If a tree were to fall onto the antenna the new mismatch would surely detune the transmitter, causing unwanted dissipation, of course, but only a lid would fail to retune the transmitter before removing the tree. In transmitters with tubes and a pi-network output coupling circuit there is no 'fold back' circuitry to protect the amp, because none is needed, due to the total re-reflection of the reflected power. That would more probably be due to cost averse buying habits of the Amateur community, and the explicit assumption of risk by them to react appropriately in the face of mismatch. Tubes were far more resilient to these incidents than transistors of yore. It is only in solid-state transmitters that have no circuitry to achieve destructive and constructive interference that requires fold back to protect the output transistors. They too have access to the services of a transmatch that is probably more flexible than the tubes' final. If they didn't use a tuner, then the foldback would render many opportunistic antennas as useless. Again, as a cost item, this solution (fold-back) is dirt cheap and was driven by the market economies of a more onerous and costly repair through a lengthy bench time to replace the transistor (which has an exceedingly high probability of a quicker failure for a poor job). IMHO, Richard, the mfgrs of solid-state rigs with no means of matching the output to a load other than 50 ohms short changed the ham, thus requiring him to be satisfied with the power fold back, or buy an antenna tuner. Walt, W2DU |
Revisiting the Power Explanation
On Mar 21, 11:48 pm, Cecil Moore wrote:
Keith wrote: For some years now, you have been arguing the reality of 'reverse power'. Nope, for the last few years I have been arguing the reality of a reverse or reflected EM energy wave. Energy is what moves and is the essence of an EM wave moving at the speed of light. All I am arguing is the validity of the distributed network reflection model, something that has stood the test of time for a century or so. But there are some challenges to the premise of 'reverse power': - where does the 'reverse power' go? - why does the change in dissipation of a generator when 'reverse power' changes depend more on the design of the generator than on the magnitude of the 'reverse power'? Reflected energy waves obey the principles of conservation of energy and superposition some of which is discussed in my WorldRadio energy article at: http://www.w5dxp.com/energy.htm I would suggest that you try trodding this path. Make a list of phenomena that you think are explained by 'reverse power'. Actually, "reflected energy" rather than "reverse power". Here is very close to an experiment we did at Texas A&M during the 50's. We observed the ghosting and the professor explained reflected energy waves to us. TVSG-----1000 feet 450 ohm ladder-line---75 ohm TV RCVR If the TV Signal Generator is not equipped with a circulator to get rid of the reflected energy wave, ghosts will appear on the TV RCVR. The ghosts are exactly where they should be if reflected wave energy exists. How would you explain the ghosting? Ghosting is explained by reflected voltages and currents. I notice that you frequently use this example, unfortunately never completing the analysis. Since you leave it unspecified, let me choose the output impedance of the TVSG to be 450 Ohms. Now there is 'reverse power' in the ladder line but no ghosts. Where did the 'reverse power' go such that it does not produce ghosts? If it is re-reflected, then it is a delayed copy of the original, i.e. a ghost. So where does it go? Seems to me there is only two ways and the way that does not produce a ghost is into the generator, so we need an accounting of this power within the generator, iff it is real power. Of course there is no issue once you let go of the notion of 'reverse power'. The reverse voltage and current pass into the generator since the generator is matched to the line and nothing is reflected. And since multiplying reverse voltage by reverse current does not, in general, produce a real power, there is no power to account for. But, again, to argue that 'reverse power' is real, it becomes necessary to explain where it went. You could start by providing a list of phenomena for which you think the reality of 'reverse power' is the only viable explanation and offer a willingness to learn about alternative explanations. Please see above. And please abandon the words, "reverse power" in favor of reverse or reflected EM energy wave. Love to, once it is no longer used as an explanation for these phenomena. Simply changing the name to something else won't help. ....Keith |
Revisiting the Power Explanation
On Fri, 23 Mar 2007 19:30:56 GMT, Walter Maxwell
wrote: I don't understand what you mean by 'taking only one of two degrees of the 360.' Hi Walt, I offered: they take only one or two degrees of the 360. Arguments that are confined only at 0 or 180 (the one OR two degrees) and are submitted as proofs as though they boxed the compass. All too often I've seen one condition (at one phase angle) offered as a negation of internal heating to prove the source lacks its own internal resistance. When I've taken exactly the same circuit and explored the 180th degree alternative, I've demonstrated melt-down clear and simple. The dissipation of energies does not always lead to this consequence, but if we were to average the analysis over a complete 360 degrees, we can only arrive at the obvious evidence of resistance and calories expended. But even for coherent reflections, if the PA tank circuit has very low loss for incident power (which it does), why does it not have ~ equally low loss for load reflections of that power? Such would mean that load reflections would pass through the tank to appear at the output element of the PA, where they can add to its normal power dissipation. The paragraph above seems to me to imply that RF doesn't understand the destructive and constructive interference phenomena involved with re-reflection. Then asking a question to clarify would be in order. To me, it reads quite ordinarily as a statement of symmetry. In my own words, it would say that if a tank circuit can pass energy from source to load in one direction, it can certainly perform the same transformation in the opposite direction. After all, that is the function of transformation and a passive circuit composed of L and C is strictly linear. Circuit analysis allows us to treat a load as a source in the complete circuit description. This is the symmetry of the illustration of external signals. You used external signals yourself as part of your case study; hence the relevance has been made by you. Whoa, Richard! You'll have to point out where I've discussed external signals in any case study involving phase relationships between forward and reflected waves. I've never done so knowingly. It seems to me that in your initial post in the original thread (that was largely ignored for comment) you made mention of injecting a signal from an external source into the mouth of the dragon for the purposes of measuring the source Z. Am I wrong? And we return to the sine non quo for the discussion: phase. That's true, but although RF apparently realizes that the phase relationship is relevant, he doesn't seem to understand the details of the phase requirements that achieve the necessary interferences that accomplish the impedance matching. That is not what I read. It seems he is on the face of it, doesn't it? Afterall, he is quite explicit to this in the statement you are challenging. No Richard, I don't believe he is. I don't see the 'explicitness' you seem to find. It's the complete lack of the explicitness that makes me believe he doesn't quite get it. That has not been my impression of the complete post. Nothing here contradicts anything either of you have to say. True, but RF just hasn't said it all, because, as I said above, I don't believe he understands the details of the phase requirements to achieve the match. That has not been my impression of the complete post. This phenomenon occurs in all tube transmitters in the ham world when the tank circuit is adjusted for delivering all available power at a given drive level. This introduces the two concepts of the "need for match" and the "match obtained." They are related only through an action that spans from one condition to the other. They do not describe the same condition, otherwise no one would ever need to perform the match: I don't comprehend your statements in the paragraph above. The initial mismatch and its correction do not describe the same condition. There is a first state, and then the operator imposes a second state in reaction. If a tree were to fall onto the antenna, a new mismatch would occur. Would the transmitter faithfully meet the expectations of the Ham unaware of the accident? No, reflected (0-179 degrees) energy would undoubtedly offer a 50% chance of excitement in the shack. The consequences of dissipation would be quite evident on that occasion. For the other 180 (180-359) degrees of benign combination; then perhaps not. If a tree were to fall onto the antenna the new mismatch would surely detune the transmitter, causing unwanted dissipation, of course, but only a lid would fail to retune the transmitter before removing the tree. Of course, but then this is a single instance: a mismatch causing possible increased dissipation within the source. "Possible" is in proportion to phase relationships; dissipation is always. If you retune to correct the mismatch's potential to destruction (or to provide full power through-put); then you have moved to another state or configuration. That new state admits the potential negative consequence of the first state. IMHO, Richard, the mfgrs of solid-state rigs with no means of matching the output to a load other than 50 ohms short changed the ham, thus requiring him to be satisfied with the power fold back, or buy an antenna tuner. A tube rig requires the same means of matching. It's a wash. 73's Richard Clark, KB7QHC |
Revisiting the Power Explanation
Keith Dysart wrote:
Ghosting is explained by reflected voltages and currents. Z0 has a large resistive component and a small reactive component. Vref*Iref = Vref^2/Z0 = Iref^2*Z0 = real watts You cannot have reflected voltages and currents that contain zero energy. In fact, a reflected TEM wave must have its E-field and H-field 90 degrees apart. The watts contained in the reflected EM wave are ExB, i.e. real watts. In a flat-matched system the forward traveling wave is the only one that exists. Would you also argue that the forward traveling wave doesn't contain any energy? All traveling waves contain energy including reflected traveling waves. I notice that you frequently use this example, unfortunately never completing the analysis. Since you leave it unspecified, let me choose the output impedance of the TVSG to be 450 Ohms. Now there is 'reverse power' in the ladder line but no ghosts. Where did the 'reverse power' go such that it does not produce ghosts? Into heating the 450 ohm circulator load resistor that you would have to install to get the generator impedance to exhibit 450 ohms to the reflected waves. Otherwise, the generator will exhibit a mismatch to the 450 ohm line. It is *impossible* for the generator to see 450 ohms when looking into a Z0=450 ohm line with reflections. So it is impossible for the reflected wave to encounter a V/I ratio of 450 ohms in the active generator (unless a 450 ohm circulator resistor is installed). And since multiplying reverse voltage by reverse current does not, in general, produce a real power, there is no power to account for. Suggest you review transmission line theory. If reverse voltage and reverse current exists, the transmission line forces their ratio to be Z0, i.e. 450 ohms, i.e. in phase. V*I*cos(0) = watts. Since reflections are causing ghosting, it is prima facie evidence that "real power" is being extracted from the reflected wave. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On 23 Mar 2007 09:21:07 -0700, "Dr. Honeydew"
wrote: Thank you, Mr. Clark. But I'm having trouble reconciling what you wrote with what Mr. Harrison posted about the output impedance being equal to the conjugate of the load which dissipates the most power. Clearly the 40 ohm load dissipates more power than the 50 ohm load, so we don't see how your answer and Mr. Harrison's posting can both be correct. It is a problem of apples and oranges. Your source is not representative of either a commercial transmitter, nor a amateur transmitter. It is a precision source. It has no need to exhibit impedance matching, and is thus not designed to perform that purpose. It is happy to run quite inefficiently as efficiency is a discardable quality. Rather, it is incumbent upon you to enforce matching by careful load selection. Otherwise, its utility is wasted and makes it a rather expensive choice for a general purpose signal generator. 73's Richard Clark, KB7QHC |
Revisiting the Power Explanation
On Mar 23, 7:18 pm, Cecil Moore wrote:
Keith Dysart wrote: I notice that you frequently use this example, unfortunately never completing the analysis. Since you leave it unspecified, let me choose the output impedance of the TVSG to be 450 Ohms. Now there is 'reverse power' in the ladder line but no ghosts. Where did the 'reverse power' go such that it does not produce ghosts? Into heating the 450 ohm circulator load resistor that you would have to install to get the generator impedance to exhibit 450 ohms to the reflected waves. Otherwise, the generator will exhibit a mismatch to the 450 ohm line. It is *impossible* for the generator to see 450 ohms when looking into a Z0=450 ohm line with reflections. So it is impossible for the reflected wave to encounter a V/I ratio of 450 ohms in the active generator (unless a 450 ohm circulator resistor is installed). You are trying too hard. Even your problem statement said "no circulator". So I offer a simple example of such a generator: a voltage source connected by a 450 Ohm resistor to the output terminal. Voila: a generator with 450 Ohm output impedance. Not too efficient perhaps, but the design is used when quality source matching is desired or when reflections need to be suppressed but it is impractical to terminate the load in the characteristic impedance of the line. Now back to the problem... and the questions... Actually, "reflected energy" rather than "reverse power". Here is very close to an experiment we did at Texas A&M during the 50's. We observed the ghosting and the professor explained reflected energy waves to us. TVSG-----1000 feet 450 ohm ladder-line---75 ohm TV RCVR If the TV Signal Generator is not equipped with a circulator to get rid of the reflected energy wave, ghosts will appear on the TV RCVR. The ghosts are exactly where they should be if reflected wave energy exists. How would you explain the ghosting? I notice that you frequently use this example, unfortunately never completing the analysis. Since you leave it unspecified, let me choose the output impedance of the TVSG to be 450 Ohms. Now there is 'reverse power' in the ladder line but no ghosts. Where did the 'reverse power' go such that it does not produce ghosts? If it is re-reflected, then it is a delayed copy of the original, i.e. a ghost. So where does it go? Seems to me there is only two ways and the way that does not produce a ghost is into the generator, so we need an accounting of this power within the generator, iff it is real power. Of course there is no issue once you let go of the notion of 'reverse power'. The reverse voltage and current pass into the generator since the generator is matched to the line and nothing is reflected. And since multiplying reverse voltage by reverse current does not, in general, produce a real power, there is no power to account for. Suggest you review transmission line theory. If reverse voltage and reverse current exists, the transmission line forces their ratio to be Z0, i.e. 450 ohms, i.e. in phase. V*I*cos(0) = watts. Since reflections are causing ghosting, it is prima facie evidence that "real power" is being extracted from the reflected wave. Oh I can do the math, and the units are definitely Watts but that doesn't necessarily mean they offer any reality. The toaster in my kitchen has 120 Volts on it and the lamp is drawing 1 Amp (and the two are in phase), but that does not mean that 120 toaster-lamp Watts has any meaning. ....Keith |
Revisiting the Power Explanation
Keith Dysart wrote:
You are trying too hard. Even your problem statement said "no circulator". So I offer a simple example of such a generator: a voltage source connected by a 450 Ohm resistor to the output terminal. Voila: a generator with 450 Ohm output impedance. The generator is an *active dynamic source*. It does not have a 450 ohm output impedance just because you install a 450 ohm resistor. Just as soon as reflected waves are allowed to reach the source, the source impedance dynamically shifts away from the 450 ohm value. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Mar 23, 9:21 pm, Cecil Moore wrote:
Keith Dysart wrote: You are trying too hard. Even your problem statement said "no circulator". So I offer a simple example of such a generator: a voltage source connected by a 450 Ohm resistor to the output terminal. Voila: a generator with 450 Ohm output impedance. The generator is an *active dynamic source*. It does not have a 450 ohm output impedance just because you install a 450 ohm resistor. Just as soon as reflected waves are allowed to reach the source, the source impedance dynamically shifts away from the 450 ohm value. I notice that you've skipped the important question posed in the example. I'll try again. Use my generator: voltage source, 450 Ohm resistor to the output. No circulator. Drive your system: 450 Ohm line terminated with a 75 Ohm TV receiver. TVSG-----1000 feet 450 ohm ladder-line---75 ohm TV RCVR Ghosts or no? If no? Where did the 'reverse power' go? If ghosts? Well, think very carefully before choosing this option because there aren't. ....Keith |
Revisiting the Power Explanation
Keith Dysart wrote:
If ghosts? Well, think very carefully before choosing this option because there aren't. There will be ghosting but because of the attenuation in the 450 ohm series resistor, the ghosting will be below the level detectable on a TV screen by the human eye. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Mar 23, 11:06 pm, Cecil Moore wrote:
Keith Dysart wrote: If ghosts? Well, think very carefully before choosing this option because there aren't. There will be ghosting but because of the attenuation in the 450 ohm series resistor, the ghosting will be below the level detectable on a TV screen by the human eye. It would be much more convincing if you were to compute the magnitude of the ghost rather than handwaving an answer. Though when you do so, you will find that it is exactly zero, so be prepared. ....Keith |
Revisiting the Power Explanation
"Keith Dysart" wrote in message oups.com... On Mar 23, 11:06 pm, Cecil Moore wrote: Keith Dysart wrote: If ghosts? Well, think very carefully before choosing this option because there aren't. There will be ghosting but because of the attenuation in the 450 ohm series resistor, the ghosting will be below the level detectable on a TV screen by the human eye. It would be much more convincing if you were to compute the magnitude of the ghost rather than handwaving an answer. Though when you do so, you will find that it is exactly zero, so be prepared. ...Keith its all ghosts and mirrors! forget power, forget energy, they are products of other calculations. all that is needed is impedance and voltage OR current. once you know voltage OR current and the impedance you can calculate all the rest. |
Revisiting the Power Explanation
Keith Dysart wrote:
It would be much more convincing if you were to compute the magnitude of the ghost rather than handwaving an answer. I made the rough computation of -25 dB before I responded. It is below the ability of the human eye to detect TV ghosting. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Dave wrote:
forget power, forget energy, they are products of other calculations. all that is needed is impedance and voltage OR current. once you know voltage OR current and the impedance you can calculate all the rest. What would you do if you were dealing with EM light waves instead of EM RF waves? Optical physicists don't have the luxury of dealing with "impedance and voltage OR current". They are forced to deal with the essence of EM waves which is photonic energy. They have done it for over a century and have learned a lot about the nature of EM waves. They did not "forget power, forget energy". -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Mar 24, 8:45 am, Cecil Moore wrote:
Keith Dysart wrote: It would be much more convincing if you were to compute the magnitude of the ghost rather than handwaving an answer. I made the rough computation of -25 dB before I responded. It is below the ability of the human eye to detect TV ghosting. Would you care to reveal the details of the calculations so that we can locate the error(s) that lead to a non-zero answer? ....Keith |
Revisiting the Power Explanation
Keith Dysart wrote:
On Mar 24, 8:45 am, Cecil Moore wrote: I made the rough computation of -25 dB before I responded. It is below the ability of the human eye to detect TV ghosting. Would you care to reveal the details of the calculations so that we can locate the error(s) that lead to a non-zero answer? As I said, it was a rough computation based on your 450 ohm series padding resistor and a lot of simplified assumptions about the configuration of the system. In order to avoid ambiguity, would you please give us better specifications for the source? Ramo and Whinnery warn us not to draw any conclusions about what happens inside a Thevenin equivalent source so your specification must be an achievable well-defined real-world source. How about a class-A linear amplifier with no filters, no impedance transformation, and no protection circuitry? Vrms/Irms = 75 ohms Vrms*Irms*cos(0) = 1 watt The amount of interference cannot be calculated without magnitude and phase. Shall we assume that the 450 ohm line is an integer number of wavelengths and lossless? i.e. Z0-matched to 75 ohms? If you require detailed calculations, more specified information is needed. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Mar 24, 10:02 am, Cecil Moore wrote:
Keith Dysart wrote: On Mar 24, 8:45 am, Cecil Moore wrote: I made the rough computation of -25 dB before I responded. It is below the ability of the human eye to detect TV ghosting. Would you care to reveal the details of the calculations so that we can locate the error(s) that lead to a non-zero answer? As I said, it was a rough computation based on your 450 ohm series padding resistor and a lot of simplified assumptions about the configuration of the system. In order to avoid ambiguity, would you please give us better specifications for the source? Ramo and Whinnery warn us not to draw any conclusions about what happens inside a Thevenin equivalent source so your specification must be an achievable well-defined real-world source. How about a class-A linear amplifier with no filters, no impedance transformation, and no protection circuitry? Vrms/Irms = 75 ohms Vrms*Irms*cos(0) = 1 watt The amount of interference cannot be calculated without magnitude and phase. Shall we assume that the 450 ohm line is an integer number of wavelengths and lossless? i.e. Z0-matched to 75 ohms? If you require detailed calculations, more specified information is needed. Standard analysis can tell us quite a bit about the situation without the need for further information. The information we have: - Generator with 450 Ohm output impedance - connected directly to a 1000 foot line with 450 Ohm characteristic impedance - connected to a load with 75 Ohm input impedance. Remembering that reflection coefficient is RC = (Z2-Z1)/(Z2+Z1) let use compute the RCs at all the connection points: RCgen-line = (450-450)/(450+450) - 0 RCline-load = (75-450)/(75+450) - -0.714 RCline-gen = (450-450)/(450+450) - 0 Let the output of the generator into a 450 Ohm load be Vout. Forward voltage supplied by the generator into the line: Vfwd = Vout This is also the Vfwd that reaches the load (for the purposes of example we are dealing with lossless lines). The reflected voltage at the load: Vref = Vfwd * RCline-load = Vfwd * -0.714 = Vout * -0.714 The voltage applied to the load: Vload = Vfwd + Vref = Vout * 0.286 The voltage reflected from the load that arrives back at the generator: Vref = Vout * -0.714 The amount of Vref reaching the generator that is reflected back to the load: Vrefref = RCline-gen * Vref = 0 * Vref = 0 There for no ghosts since there is no re-reflection. A few other things of possible interest... Power delivered to the load: Pload = Vload * Vload / 75 = Vout * Vout * 0.0816 / 75 = Vout * Vout * .0011 This is the actual power that flows in the line. Forwarwd power as indicated by a directional wattmeter in the line: Pfwd = Vfwd * Vfwd / 450 Reverse power as indicated by a directional wattmeter in the line: Prev = Vref * Vref / 450 Net power flowing in the line: Pfwd - Prev = (Vfwd*Vfwd)/450 - (Vref*Vref)/450 = Vout*Vout/450 - Vout*RC*Vout*RC/450 = Vout*Vout*(1-(RC*RC))/450 = Vout * Vout * 0.0011 Which is the same as the load power; this being as expected since directional wattmeters are useful for obtaining the net power. It is also useful to know what we can not compute from the provided information.... The voltage at the generator output and its phase with respect to the current since this is dependant on the velocity factor of the line and the signal frequency. Though if the output is DC (e.g. a step function) than this can be easily computing for the situation after one round trip delay. Generator dissipation since we would need to know the internal construction of the generator and the actual voltage and current phase at the output of the generator. ----- Still, using conventional techniques we have derived quite a bit about the system without knowing the internals of the generator. Cecil, are you saying that your techniques can not produce any derivations without knowing the internal construction of the generator and the length of the line (in wavelengths)? ....Keith |
Revisiting the Power Explanation
"Cecil Moore" wrote in message ... Dave wrote: forget power, forget energy, they are products of other calculations. all that is needed is impedance and voltage OR current. once you know voltage OR current and the impedance you can calculate all the rest. What would you do if you were dealing with EM light waves instead of EM RF waves? Optical physicists don't have the luxury of dealing with "impedance and voltage OR current". They are forced to deal with the essence of EM waves which is photonic energy. They have done it for over a century and have learned a lot about the nature of EM waves. They did not "forget power, forget energy". -- yes they do. but photons are propagating em waves, they aren't currents and voltages, they are E and B fields. you can do the exact same calculations using either E or B and the refractive index or epsilon/mu. again, all you need is 2 of them, and you can compute the third, and then you know it all. |
Revisiting the Power Explanation
Keith Dysart wrote:
Standard analysis can tell us quite a bit about the situation without the need for further information. The information we have: - Generator with 450 Ohm output impedance There are a number of techniques to ensure that the reflected waves encounter an impedance of 450 ohms but none of those techniques are implemented in amateur radio transmitters. Your generator is obviously equipped with a circulator load of 450 ohms, 450 ohm attenuation pads, or some active feedback. Given that, your analysis is correct but moot. Your point is already known and accepted by every initiated person including me. I freely admit that if the proper circulator load exists or if 450 ohm attenuator pads exist, they will dissipate the reflected power by heating up the circulator load resistor or the pads. But it's tough to argue that there is no energy in the reflected waves while they are being used to boil water. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Dave wrote:
yes they do. but photons are propagating em waves, they aren't currents and voltages, they are E and B fields. you can do the exact same calculations using either E or B and the refractive index or epsilon/mu. again, all you need is 2 of them, and you can compute the third, and then you know it all. True, but how do they measure two of them? Ever tried to measure the E-field of visible photons? Hint: It is an irradiance (power density) time- averaged measurement proportion to E^2. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Richard Clark wrote:
Cecil Moore wrote: your analysis is correct but moot Is he stealing your style? "Moot" is an interesting word, Richard. From Webster's - "moot - 1. a: debatable, b: disputed" -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
On Sat, 24 Mar 2007 11:40:37 -0500, Cecil Moore
wrote: your analysis is correct but moot Is he stealing your style? |
Revisiting the Power Explanation
On Mar 24, 12:40 pm, Cecil Moore wrote:
Keith Dysart wrote: Standard analysis can tell us quite a bit about the situation without the need for further information. The information we have: - Generator with 450 Ohm output impedance There are a number of techniques to ensure that the reflected waves encounter an impedance of 450 ohms but none of those techniques are implemented in amateur radio transmitters. Your generator is obviously equipped with a circulator load of 450 ohms, 450 ohm attenuation pads, or some active feedback. While it might be, none of the above are necessary nor included in the example I have analyzed. A simple voltage source in series with 450 Ohms resistor or a current source in parallel with one will produce exactly the same results. Look carefully at the analysis and you will find no evidence of any of the items you have mentioned above, unless you consider the 450 Ohm resistor to be an attenuation pad. (Though it is this resistor which matches the line and ensures there is no reflection). Given that, your analysis is correct but moot. Your point is already known and accepted by every initiated person including me. Not you, I suspect, since you seem to find it necessary to add some components which were not in the description of the setup. I freely admit that if the proper circulator load exists or if 450 ohm attenuator pads exist, they will dissipate the reflected power by heating up the circulator load resistor or the pads. Yes, indeed they would. The more challenging case for you is the example I presented without these components. How do you explain the absence of ghosts? And the unaccountability of the 'reflected power'. While an example with a circulator makes it easy to locate and account for the 'reflected power', there is more to be learned by studying the example without. If you can not demonstrate a flaw in the analysis then you may wish to explore why the example without a circulator (or other additional components) causes you difficulty. But it's tough to argue that there is no energy in the reflected waves while they are being used to boil water. While you can construct examples where the 'reflected power' will boil water, there are alternate examples where the total system dissipation drops in the presence of 'reflected power'. So sometimes 'reflected power' boils water and sometimes it does not. Not a very reliable source of energy it seems. But I do suggest going back to the example. If there is no flaw in the analysis, please explain where the 'reflected power' goes. ....Keith |
Revisiting the Power Explanation
Cecil Moore wrote in news:9wcNh.95$Q23.73
@newssvr17.news.prodigy.net: I freely admit that if the proper circulator load exists or if 450 ohm attenuator pads exist, they will dissipate the reflected power by heating up the circulator load resistor or the pads. Considering only the attenuator case, you suggest in such a general statement that reflected power always increases the dissipation in a real source that includes an attenuator for the purpose of constraining the equivalent source impedance. Such a suggestion is incorrect in the steady state, some loads may increase the generator's internal dissipation, some may decrease it depending on the attenuator circuit. Owen |
Revisiting the Power Explanation
In message , Cecil Moore
writes Richard Clark wrote: Cecil Moore wrote: your analysis is correct but moot Is he stealing your style? "Moot" is an interesting word, Richard. From Webster's - "moot - 1. a: debatable, b: disputed" Have a look here. http://en.wikipedia.org/wiki/Moot_hall I remember 'Moot Hall' from my days at primary school (some 60 years ago), learning about the Anglo-Saxons. I guess the word may possibly be associated with 'meet', ie a meeting hall where things were debated. However, my Anglo-Saxon is a bit rusty (not much call for it these days). Ian. -- |
Revisiting the Power Explanation
On Sat, 24 Mar 2007 17:08:48 GMT, Cecil Moore
wrote: Richard Clark wrote: Cecil Moore wrote: your analysis is correct but moot Is he stealing your style? "Moot" is an interesting word, Richard. From Webster's - "moot - 1. a: debatable, b: disputed" Correct but "disputed." Hmmm. Correct but "debatable." Hmmm. Sounds like the definition of masturbation without orgasm. Yes, I recognize your style there. |
Revisiting the Power Explanation
Ian Jackson wrote:
In message , Cecil Moore writes Richard Clark wrote: Cecil Moore wrote: your analysis is correct but moot Is he stealing your style? "Moot" is an interesting word, Richard. From Webster's - "moot - 1. a: debatable, b: disputed" Have a look here. http://en.wikipedia.org/wiki/Moot_hall I remember 'Moot Hall' from my days at primary school (some 60 years ago), learning about the Anglo-Saxons. I guess the word may possibly be associated with 'meet', ie a meeting hall where things were debated. However, my Anglo-Saxon is a bit rusty (not much call for it these days). Cecil was using "moot" in its legal sense: that a point had become irrelevant, or no longer needed to be decided because of a change in circumstances. Or at least, Cecil tried to claim that a point made by Keith had become moot. But Keith disputed that... and so it rumbles on. -- 73 from Ian GM3SEK |
Revisiting the Power Explanation
On Fri, 23 Mar 2007 15:10:40 -0800, Richard Clark wrote:
On Fri, 23 Mar 2007 19:30:56 GMT, Walter Maxwell wrote: I don't understand what you mean by 'taking only one of two degrees of the 360.' Hi Walt, I offered: they take only one or two degrees of the 360. Arguments that are confined only at 0 or 180 (the one OR two degrees) and are submitted as proofs as though they boxed the compass. All too often I've seen one condition (at one phase angle) offered as a negation of internal heating to prove the source lacks its own internal resistance. When I've taken exactly the same circuit and explored the 180th degree alternative, I've demonstrated melt-down clear and simple. The dissipation of energies does not always lead to this consequence, but if we were to average the analysis over a complete 360 degrees, we can only arrive at the obvious evidence of resistance and calories expended. But even for coherent reflections, if the PA tank circuit has very low loss for incident power (which it does), why does it not have ~ equally low loss for load reflections of that power? Such would mean that load reflections would pass through the tank to appear at the output element of the PA, where they can add to its normal power dissipation. The paragraph above seems to me to imply that RF doesn't understand the destructive and constructive interference phenomena involved with re-reflection. Then asking a question to clarify would be in order. To me, it reads quite ordinarily as a statement of symmetry. In my own words, it would say that if a tank circuit can pass energy from source to load in one direction, it can certainly perform the same transformation in the opposite direction. After all, that is the function of transformation and a passive circuit composed of L and C is strictly linear. Circuit analysis allows us to treat a load as a source in the complete circuit description. Richard, assume a mismatched load has produced both voltage and current refleftions on the line that result in a particular reactive impedance at the line input. The line input is connected to the output of the transceiver that has a pi-network output coupling circuit. When the network has been adjusted to deliver all the available power into the line the output source impedance is the complex conjugate of the line input impedance. In this condition the reflected voltage and current waves are totally re-reflected back into the line, while adding in phase to the voltage and current waves from the source, respectively. Consequently, the reflected waves do not pass rearward through the network to be incident on the plate. Only if the network is mistuned, such as being connected to the reactive input of the line without being retuned to resonance, in which case the excessive plate current due to being mistuned will result in an inordinate amount of heating of the plate. This is the symmetry of the illustration of external signals. You used external signals yourself as part of your case study; hence the relevance has been made by you. Whoa, Richard! You'll have to point out where I've discussed external signals in any case study involving phase relationships between forward and reflected waves. I've never done so knowingly. It seems to me that in your initial post in the original thread (that was largely ignored for comment) you made mention of injecting a signal from an external source into the mouth of the dragon for the purposes of measuring the source Z. Am I wrong? Yes, you are wrong here, because I made no mention of the 'mouth of a dragon'. That comment must have come from another poster, twarn't I. And we return to the sine non quo for the discussion: phase. That's true, but although RF apparently realizes that the phase relationship is relevant, he doesn't seem to understand the details of the phase requirements that achieve the necessary interferences that accomplish the impedance matching. That is not what I read. It seems he is on the face of it, doesn't it? Afterall, he is quite explicit to this in the statement you are challenging. No Richard, I don't believe he is. I don't see the 'explicitness' you seem to find. It's the complete lack of the explicitness that makes me believe he doesn't quite get it. That has not been my impression of the complete post. Nothing here contradicts anything either of you have to say. True, but RF just hasn't said it all, because, as I said above, I don't believe he understands the details of the phase requirements to achieve the match. That has not been my impression of the complete post. Richard, try this on for size and then determine whether you believe RF understands the function of the phasing in impedance matching: Assume a 150-ohm pure resistance terminates a 50-ohm line, producing a voltage reflection coefficient rhoV =0.5 at 0°, a current reflection coefficient rhoi, yielding a 3:1 mismatch. We want to place a series stub at the appropriate position on the line to yield a match at that point. The appropriate position on the line is where the real portion of the line impedance is 50 ohms, with a residual reactance, which in this case is -j57.7 ohms, determined by the 3:1 mismatch. A series stub having a terminal impedance of +j57.7 ohms cancels the residual reactance, achieving a match at the stub point. Now to the phase relationships that occur here that I believe RF has not considered. First, the line rhoV at the stub is 0.5 at -60°, and the stub rhoV is 0.5 at +60°, with the resultant rho = 0°. Second, the line rhoi at the stub is 0.5 at +120°, and the stub rhoi is 0.5 at -120°, with the resultant rho = 180°. Third, with resultant voltage and current at 0° and 180°, respectively, we have achieved a virtual open circuit to waves reflected from the 150-ohm mismatch, causing total re-reflection of the reflected waves at the stub point. Proof that total re-reflection has occurred is by observing that there is no evidence of any reflected waves rearward from the stub point to the source. Now, when adjusting the output network of a tube-type transceiver to deliver all the available power into a line having reflections, the adjustment of the network accomplishes the same function as the stub on the line in the above discussion. Consequently, this is the reason why the reflected power is totally re-reflected at the output terminals of the network, and is never seen at the plate of the amp to cause heating. This is the concept I believe Richard Fry is not appreciating. If I'm wrong on this I hope he'll straighten me out. Walt |
Revisiting the Power Explanation
Keith Dysart wrote:
While an example with a circulator makes it easy to locate and account for the 'reflected power', there is more to be learned by studying the example without. You are not allowed to deny the existence of the 450 ohm real world load while asserting that it still exists. If it exists, it dissipates the reflected energy. If it doesn't exist, it reflects the reflected energy. Please choose one or the other - obviously, you cannot have both at the same time. While you can construct examples where the 'reflected power' will boil water, there are alternate examples where the total system dissipation drops in the presence of 'reflected power'. So sometimes 'reflected power' boils water and sometimes it does not. Not a very reliable source of energy it seems. If reflected energy is not dissipated, it undergoes destructive interference and is redirected back toward the load as constructive interference instead of being incident upon the source. Why are you having difficulty with that concept from page 388 of "Optics", by Hecht, 4th edition? But I do suggest going back to the example. If there is no flaw in the analysis, please explain where the 'reflected power' goes. Again, reflected power doesn't flow so it doesn't go anywhere. Until you understand that simple fact of physics, further discussion is unlikely to yield valid results. "Reflected power that goes" is only one of your many conceptual flaws. So your first logical step would be proving that reflected power actually flows. After you do that, we can continue to your other conceptual flaws. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Owen Duffy wrote:
Considering only the attenuator case, you suggest in such a general statement that reflected power always increases the dissipation in a real source that includes an attenuator for the purpose of constraining the equivalent source impedance. Owen, why do you feel compelled to lie about what I said? If the reflected energy is dissipated within the source, it increases the dissipation. If the reflected energy is not dissipated within the source, it does not increase the dissipation. Why is that so hard for you to understand? -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Richard Clark wrote:
Correct but "disputed." Hmmm. i.e. 92% correct. Correct but "debatable." Hmmm. i.e. 85% correct. -- 73, Cecil http://www.w5dxp.com |
Revisiting the Power Explanation
Cecil Moore wrote:
*** 1 *** If reflected energy is not dissipated, it undergoes destructive interference and is redirected back toward the load as constructive interference instead of being incident upon the source. Why are you having difficulty with that concept from page 388 of "Optics", by Hecht, 4th edition? *** 2 *** Again, reflected power doesn't flow so it doesn't go anywhere. Until you understand that simple fact of physics, further discussion is unlikely to yield valid results. "Reflected power that goes" is only one of your many conceptual flaws. So your first logical step would be proving that reflected power actually flows. After you do that, we can continue to your other conceptual flaws. Cecil, Paragraphs 1 and 2 appear to declare exactly the opposite behavior for energy (power). Is there some subtle re-definition going on to allow "redirected back toward the load" and "it doesn't go anywhere" in the same message? What is it that you are trying to say? 73, Gene W4SZ |
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