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Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
Could you better describe how you determine that the source has a Z0 equal to the line Z0? I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. This will do it. We have a black box. We use a 12vdc battery and a current meter to measure the impedance of the black box. The current meter reads zero when we connect the 12vdc battery to the black box terminals. What is the impedance inside the black box since test V/I = infinity? What is actually in the black box is a very low source impedance battery in series with a 50 ohm resistor. This is what you are up against when reflections arrive at your source. The ideal voltage source does *NOT* exhibit zero ohms. Gary Coffman once likened it to spitting down a fire hose. If the reflections are in phase with the source voltage, the source exhibits infinite ohms to the reflections and of course, 100% re-reflection results. The reflection coefficient encountered by the reflected waves is variable and depends upon the relative phase between the source and the reflections. The series resistor has very little effect on the reflection coefficient. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Roy Lewallen wrote:
A perfect voltage source has a zero impedance, so if it's connected to a transmission line with no series resistance, it presents a reflection coefficient of -1. But that is only when the source voltage is zero. What gives you the right to assert such a thing when the source is turned on? Please quote references and be specific. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Roy Lewallen wrote:
Roger wrote: Could you better describe how you determine that the source has a Z0 equal to the line Z0? I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. Probably the simplest way is to put the entire source circuitry into a black box. Measure the terminal voltage with the box terminals open circuited, and the current with the terminals short circuited. The ratio of these is the source impedance. If you replace the box with a Thevenin or Norton equivalent, this will be the value of the equivalent circuit's impedance component (a resistor for most of our examples). If the driving circuitry consists of a perfect voltage source in series with a resistance, the source Z will be the resistance; if it consists of a perfect current source in parallel with a resistance, the source Z will be the resistance. You can readily see that the open circuit V divided by the short circuit I of these two simple circuits equals the value of the resistance. The power output of the Thévenin equivalent circuit follows the load. Sorry, I don't understand this. Can you express it as an equation? There seems to be some confusion as to the terms "Thévenin equivalent circuit", "ideal voltage source", and how impedance follows these sources. Two sources we all have access to are these links: Voltage source: http://en.wikipedia.org/wiki/Voltage_source Thévenin equivalent circuit: http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem Three important observations about the ideal voltage source: 1. The voltage is maintained, no matter what current flows through the source. Presumably, this would also mean that the voltage would be maintained if a negative current flowed through the source. Thus, if we set the voltage to 1 volt, the voltage would remain 1v even if we supplied an infinite amount of power to the source. 2. The idea voltage source can ADSORB an infinite amount of power while maintaining voltage. In this capacity, it is like a variable resistor, capable of changing resistance to maintain a design voltage, no matter what current is supplied to it. 3. The idea voltage source has an infinite impedance at the design voltage, not a zero impedance as many have suggested. Zero internal resistance is assumed to reasonably allow the ideal voltage source to supply or adsorb current without changing voltage internally. It is not zero impedance with the result that voltage drops to zero if external power FLOWS INTO the ideal source. Current flows into the ideal voltage source when the applied voltage exceeds the design voltage. At the point the current reverses, we have voltage/zero current, which is infinite impedance. Two things to notice about the Thévenin equivalent circuit: 1. It contains an ideal voltage source "IN SERIES" with a resistor. This has important implications when externally supplied voltage exceeds the design voltage. Any returning power would not only reverse the current flow in the ideal voltage source, it would develop voltage across the internal series resistor. The output Thevenin voltage would be the design voltage plus the voltage developed in the resistor. 2. The impedance of the Thevenin equivalent source would be infinite at the design voltage because a voltage will existed from the ideal source but current does not flow. This is true no matter what the design impedance is for the Thevenin source. Please notice in the link about the Thevenin circuit, a reference to a "Thévenin-equivalent resistance". This resistance uses the ideal voltage source set to zero. This appears to be the circuit that entered the discussion at some point, justifying a negative 1 reflection factor. Therefore, when the load delivers power, the Thévenin equivalent circuit adsorbs power. Right? Certainly, any energy leaving the transmission line must enter the circuitry to which it's connected. Is that what you mean? Roy Lewallen, W7EL If the rules found in the links are acceptable, I could agree that energy be allowed to enter the connected circuitry. I think we will find that the returning power from a reflected wave is either completely reflected with no change in energy, or adds to the voltage, thus increasing the current flow and power contained in the system. 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
Roger wrote:
If the rules found in the links are acceptable, I could agree that energy be allowed to enter the connected circuitry. I think we will find that the returning power from a reflected wave is either completely reflected with no change in energy, or adds to the voltage, thus increasing the current flow and power contained in the system. I think you are on the right track, Roger. Another way of saying it is: If the principles of superposition are used, the superposition must necessarily be implemented. It appears to me that the principles of superposition are being used but the results of the ensuing necessary superposition are, for some ulterior motive, being completely ignored. When the superposition interference patterns between the source waves and the reflected waves are taken into account, everything becomes perfectly clear (including the deliberate obfuscations). -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
On Jan 1, 1:20 pm, Roger wrote: Discussing forward and reflecting waves, when is stability reached. Roy Lewallen wrote: If "stability" means steady state, a transmission line with any resistance at either end or both ends is less complicated to analyze than the particularly difficult lossless case I used for my analysis which never reaches a true steady state. The presence of resistance allows the system to settle to steady state, and that process can easily and quantifiably be shown. And in two special cases, the process from turn-on to steady state is trivially simple -- If the line is terminated with Z0 (technically, its conjugate, but the two are the same for a lossless line since Z0 is purely resistive), steady state is reached just as soon as the initial forward wave arrives at the far end of the line. No reflections at all are present or needed for the analysis. The second simple case is when the source impedance equals Z0, resulting in a source reflection coefficient of zero. In that case, there is a single reflection from the far end (assuming it's not also terminated with Z0), but no re-reflection from the source, and steady state is reached as soon as the first reflected wave arrives at the source. Roy Lewallen, W7EL Could you better describe how you determine that the source has a Z0 equal to the line Z0? I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. This will do it. As will a Norton with the parallel resistor set to Z0. The power output of the Thévenin equivalent circuit follows the load. Therefore, when the load delivers power, the Thévenin equivalent circuit adsorbs power. Right? This apparently simple question has a very complicated answer that depends on what precisely is meant by "load delivers power" and "circuit absorbs power". If by "load delivers power", you mean the reflected wave, then this may or may not (depending on the phase), mean that energy is transfered into the generator. If you mean that the time averaged product of the actual voltage and current at the generator terminals show a transfer of energy into the generator, then energy is indeed flowing into the generator. If by "circuit absorbs power", you mean that there is an increase in the energy dissipated in the generator, this can not be ascertained without detailed knowledge of the internal arrangement of the generator and also depends the meaning of "load delivers power", discussed above. ...Keith I think we will find it simpler than that. I posted some links to descriptions of ideal voltage sources. After reading them, I am taking another look at how we define impedance from an ideal voltage source in a separate posting. You may wish to read the links, and hopefully, my posting. Here are the links: Voltage source: http://en.wikipedia.org/wiki/Voltage_source Thévenin equivalent circuit: http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
Roger wrote:
Roy Lewallen wrote: Roger wrote: The power output of the Thévenin equivalent circuit follows the load. Sorry, I don't understand this. Can you express it as an equation? There seems to be some confusion as to the terms "Thévenin equivalent circuit", "ideal voltage source", and how impedance follows these sources. Two sources we all have access to are these links: Voltage source: http://en.wikipedia.org/wiki/Voltage_source Thévenin equivalent circuit: http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem As far as I can see, I've used the terms entirely consistently with the definitions and descriptions in the linked articles. If you can find any instance in which I haven't, please point it out so I can correct it. Three important observations about the ideal voltage source: 1. The voltage is maintained, no matter what current flows through the source. Presumably, this would also mean that the voltage would be maintained if a negative current flowed through the source. Actually, the wikipedia article correctly states this explicitly. Thus, if we set the voltage to 1 volt, the voltage would remain 1v even if we supplied an infinite amount of power to the source. You have to be very careful with using infinite amounts of anything because the underlying mathematics runs into problems. For example, look up "impulses" in any circuits text, and you'll find situations where a pulse has zero width and infinite height yet finite area. So let's just say the voltage stays at 1 volt if you apply any finite amount of current. 2. The idea voltage source can ADSORB an infinite amount of power while maintaining voltage. Yes, that's correct. In this capacity, it is like a variable resistor, capable of changing resistance to maintain a design voltage, no matter what current is supplied to it. No, that's not correct, unless you're willing to work with negative resistances. V/I is a negative number when you supply current to a voltage source. So it doesn't in any way act like a resistance, changing or not. 3. The idea voltage source has an infinite impedance at the design voltage, not a zero impedance as many have suggested. No, that's not correct. Zero internal resistance is assumed to reasonably allow the ideal voltage source to supply or adsorb current without changing voltage internally. Yes, that's correct. It is not zero impedance with the result that voltage drops to zero if external power FLOWS INTO the ideal source. Sorry, I don't understand that statement. The voltage never changes, regardless of the current flowing in or out. Review the wikipedia article or any circuits text. Current flows into the ideal voltage source when the applied voltage exceeds the design voltage. You can't "apply" a voltage to an ideal voltage source, except through an impedance. If you did, for example by connecting it to another voltage source of different value, the conditions defined for the voltage sources are contradictory and can't exist at the same time. If you do apply a voltage through an impedance, the current is simply the voltage across the impedance divided by the impedance. At the point the current reverses, we have voltage/zero current, which is infinite impedance. It sounds like you're trying to calculate some sort of "instantaneous impedance". This isn't a generally accepted concept, and you'll have to develop a fairly complete set of rules and mathematics to cover it. Two things to notice about the Thévenin equivalent circuit: 1. It contains an ideal voltage source "IN SERIES" with a resistor. This has important implications when externally supplied voltage exceeds the design voltage. Not really. The current simply equals the voltage across the resistor divided by the resistance, in accordance with Ohm's law. Any returning power would not only reverse the current flow in the ideal voltage source, it would develop voltage across the internal series resistor. The output Thevenin voltage would be the design voltage plus the voltage developed in the resistor. Your problem here, and I suspect elsewhere, is that you've embraced the idea that there are waves of flowing and reflecting power. Your otherwise reasoned questions and arguments are a good illustration into the traps this concept leads a person into. You're seeing that in order to support this mistaken concept, you have to reject some very fundamental principles of electrical circuit operation. So you end up with only two choices: 1. Re-write a great deal of fundamental circuit theory and reject the foundation which has consistently given demonstrably correct results for over a century, or 2. Realize that the flowing-power concept is flawed and abandon it. The choice is yours. 2. The impedance of the Thevenin equivalent source would be infinite at the design voltage because a voltage will existed from the ideal source but current does not flow. This is true no matter what the design impedance is for the Thevenin source. You're trying to define the impedance of a source as the ratio of its voltage to the current being drawn from it. That's not the impedance of the source, it's the impedance of the load. The two are not the same. The impedance of the source is its open circuit voltage divided by its short circuit current. Another way to determine the impedance of the source is to apply various loads and see how much the voltage drops. Any test you run will confirm that the ideal voltage source has a zero resistance, as my postings, the wikipedia article, or any circuits text tell you. Please notice in the link about the Thevenin circuit, a reference to a "Thévenin-equivalent resistance". This is the resistance component of the Thevenin equivalent circuit. It's equal to the resistance of the circuitry for which the Thevenin equivalent is being substituted. This resistance uses the ideal voltage source set to zero. Can you explain this? There is no requirement that the source be set to zero in determining or using the Thevenin equivalent resistance. This appears to be the circuit that entered the discussion at some point, justifying a negative 1 reflection factor. My analysis did not contain a Thevenin equivalent to any circuit. It contained only an ideal voltage source. Therefore, when the load delivers power, the Thévenin equivalent circuit adsorbs power. Right? Certainly, any energy leaving the transmission line must enter the circuitry to which it's connected. Is that what you mean? Roy Lewallen, W7EL If the rules found in the links are acceptable, I could agree that energy be allowed to enter the connected circuitry. I think we will find that the returning power from a reflected wave is either completely reflected with no change in energy, or adds to the voltage, thus increasing the current flow and power contained in the system. You're building a logical argument from a flawed premise. I'm afraid you've only just begun to see the problems you'll be having as you pursue this path. Roy Lewallen, W7EL |
Standing-Wave Current vs Traveling-Wave Current
"Cecil Moore" wrote in message .net... Roger wrote: If the rules found in the links are acceptable, I could agree that energy be allowed to enter the connected circuitry. I think we will find that the returning power from a reflected wave is either completely reflected with no change in energy, or adds to the voltage, thus increasing the current flow and power contained in the system. I think you are on the right track, Roger. Another way of saying it is: If the principles of superposition are used, the superposition must necessarily be implemented. It appears to me that the principles of superposition are being used but the results of the ensuing necessary superposition are, for some ulterior motive, being completely ignored. When the superposition interference patterns between the source waves and the reflected waves are taken into account, everything becomes perfectly clear (including the deliberate obfuscations). -- 73, Cecil http://www.w5dxp.com I have been down wrong tracks before! I sure hope this is the right track. Thanks. The principles of superposition are mathematically usable, not too hard, and I think very revealing. Yes, if we use part of the model, we must use it all the way. To do otherwise would be error, or worse. You have been a supporter of this theory for a long time. 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
Roy Lewallen wrote:
Your problem here, and I suspect elsewhere, is that you've embraced the idea that there are waves of flowing and reflecting power. Roy, your problem is that you have embraced the idea that reflected waves contain zero energy such that zero reflected energy can be measured as zero reflected power passing a measurement point. However, zero energy waves are impossible!!! 2. Realize that the flowing-power concept is flawed and abandon it. Why don't you realize that reflected waves of zero energy are impossible? Can you explain this? There is no requirement that the source be set to zero in determining or using the Thevenin equivalent resistance. It is a requirement of the superposition principle that the source voltage be set to zero for determining the voltage and current due only to the reflected wave. I'm surprised you are trying to sweep that fact under the rug. My analysis did not contain a Thevenin equivalent to any circuit. It contained only an ideal voltage source. An ideal voltage source plus a series resistor *IS* a Thevenin equivalent circuit. Nice try at obfuscation. You're building a logical argument from a flawed premise. Nobody's premise is more flawed than yours. You are only performing 1/2 of the pre-superposition phase. Everything can be explained if you actually go through with the superposition of which you are so afraid. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Roger wrote:
The principles of superposition are mathematically usable, not too hard, and I think very revealing. Yes, if we use part of the model, we must use it all the way. To do otherwise would be error, or worse. Roy and Keith don't seem to realize that the zero source impedance for the ideal voltage source is only when the source is turned off for purposes of superposition. They conveniently avoid turning the source voltage on to complete the other half of the superposition process. When the source signal and the reflected wave are superposed at the series source resistor, where the energy goes becomes obvious. Total destructive interference in the source results in total constructive interference toward the load. See below. You have been a supporter of this theory for a long time. Yes, I have. I am a supporter of the principles and laws of physics. Others believe they can violate the principle of conservation of energy anytime they choose because the principle of conservation of energy cannot be violated - go figure. Roger, I have explained all of this before. If you are capable of understanding it, I take my hat off to you. Here's what happens in that ideal voltage source using the rules of superposition. 1. With the reflected voltage set to zero, the source current is calculated which results in power in the source resistor. Psource = Isource^2*Rsource 2. With the source voltage set to zero, the reflected current is calculated which results in power in the source resistor. Pref = Iref^2*Rsource Now superpose the two events. The resultant power equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) where 'A' is the angle between the source current and the reflected current. Note this is NOT power superposition. It is the common irradiance (power density) equation from the field of optics which has been in use in optical physics for a couple of centuries so it has withstood the test of time. For the voltage source looking into a 1/2WL open circuit stub, angle 'A' is 180 degrees. Therefore the total power in the source resistor is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0 Ptot in the source resistor is zero watts but we already knew that. What is important is that the last term in that equation above is known as the "interference term". When it is negative, it indicates destructive interference. When Ptot = 0, we have "total destructive interference" as defined by Hecht in "Optics", 4th edition, page 388. Whatever the magnitude of the destructive interference term, an equal magnitude of constructive interference must occur in a different direction in order to satisfy the conservation of energy principle. Thus we can say with certainty that the energy in the reflected wave has been 100% re-reflected by the source. The actual reflection coefficient of the source in the example is |1.0| even when the source resistor equals the Z0 of the transmission line. I have been explaining all of this for about 5 years now. Instead of attempting to understand these relatively simple laws of physics from the field of optics, Roy ploinked me. Hopefully, you will understand. All of this is explained in my three year old Worldradio energy article on my web page. http://www.w5dxp.com/energy.htm -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
On Jan 1, 3:59*pm, Richard Clark wrote:
On Tue, 1 Jan 2008 06:12:48 -0800 (PST), Keith Dysart wrote: To illustrate some of these weaknesses, consider an example where a step function from a Z0 matched generator is applied to a transmission line open at the far end. Hi Keith, It would seem we have either a Thevenin or a Norton source (again, the ignored elephant in the living room of specifications). *This would have us step back to a Z0 in series with 2V or a Z0 in parallel with V - it seems this would be a significant detail in the migration of what follows: I do not think so. Regardless of the output impedance of the generator, the step impresses some voltage V which begins to propagate down the line. Since the experiment ends before the reflection returns to the generator, the impedance of the generator is irrelevant. The step function eventually reaches the open end where the current can no longer flow. The inductance insists that the current continue until the capacitance at the end of the line is charged to the voltage which will stop the flow. This voltage is double the voltage of the step function applied to the line (i.e 2*V). Fine (with omissions of the fine grain set-up) However, what follows is so over edited as to be insensible:Once the infinitesimal capacitance at the end of the line is charged, energy has reached the "end of the line" so to speak; and yet:the current now has to stop just a bit earlier TIME is backing up? * Ah. The challenge of written precisely. Consider the generator to be on the left end of the line and the open to be at the right end. When the capacitance at the right end charges to 2 * V, the current now has to stop a little bit more to the left. Are we at the edge of an event horizon?and this charges the inifinitesimal capacitance a bit further from the end. BEYOND the end of the line? *Just how long can this keep up? Again, the poor writing meant a bit further to left from the end. Very strange stuff whose exclusion wouldn't impact the remainder:So a step in the voltage propagates back along the line towards the source. In front of this step, current is still flowing. Behind the step, the behind the reflected step, rather?current is zero and voltage is 2*V. Want to explain how you double the stored voltage in the distributed capacitance of the line without current? * The energy formerly present in the inductance of the line has been transferred to the capacitance. The definition of capacitance is explicitly found in the number of electrons (charge or energy) on a surface; which in this case has not changed.The charge that is continuing to flow from the source is being used to charge the distributed capacitance of the line. It would appear now that charge is flowing again, but that there is a confusion as to where the flow comes from. * Using the new reference scheme, charge to the left of the leftward propagating step continues to flow, while charge to the right has stopped flowing. Why would the source at less voltage provide current to flow into a cap that is rising in potential above it? * The beauty of inductance. You get extra voltage when you try to stop the flow. Rolling electrons uphill would seem to be remarkable. Well yes, but they don't roll for long. And you don't get more out than you put in. Pity. Or I could be quite rich. Returning to uncontroversial stuff:The voltage that is propagating backwards along the line has the value 2*V, but this can also be viewed as a step of voltage V added to the already present voltage V. The latter view is the one that aligns with the "no interaction" model; the total voltage on the line is the sum of the forward voltage V and the reverse voltage V or 2*V. If this is the "latter view" then the former one (heavily edited above?) is troubling to say the least. Challenges with the referent. The former view is that a voltage of 2*V is propagating back along the line. The latter view is that it is a step of V above the already present V. In this model, the step function has propagated to the end, been reflected and is now propagating backwards. Implicit in this description is that the step continues to flow to the end of the line and be reflected as the leading edge travels back to the source. This is a difficult read. *You have two sentences. *Is the second merely restating what was in the first, or describing a new condition (the reflection)? Agreed. What is a good word to describe the constant voltage that follows the actual step change in voltage? The "tread" perhaps? "The tread continues to flow to the end of the line and be reflected as the leading edge travels back to the source." I am not sure that that is any clearer. And this is the major weakness in the model. Which model? * The "no interaction" model. The latter? or the former? It claims the step function is still flowing in the portion of the line that has a voltage of 2*V and *zero* current. Does a step function flow? * Perhaps the "tread"? But then should the step change be called the "riser"? As for "zero" current, that never made sense in context here. Somewhat clarified, I hope. But for clarity, the current to the right of the leftward propagating step is zero. Now without a doubt, when the voltages and currents of the forward and reverse step function are summed, the resulting totals are correct. * In this thread, that would be unique. Or a miracle? But it seems to me that this is just applying the techniques of superposition. And when we do superposition on a basic circuit, we get the correct totals for the voltages and currents of the elements but we do not assign any particular meaning to the partial results. Amen. Unfortunately, more confusion:A trivial example is connecting to 10 volt batteries in parallel through *a .001 ohm resistor. Parallel has two outcomes, which one? *"Through" a resistor to WHERE? In series? *In parallel? * Much to ambiguous. I know. Trying to conserve words leads to confusion. Try: negative to negative, positives connected using the resistor. The partial results show 10000 amps flowing in each direction in the resistor with a total of 0. This would suggest in parallel to the parallel batteries, but does not resolve the bucking parallel or aiding parallel battery connection possibilities. *The 0 assignment does not follow from the description, mere as one of two possible solutions. But I do not think that anyone assigns significance to the 10000 amp intermediate result. Everyone does agree that the actual current in the resistor is zero. Actually, no. *Bucking would have 0 Amperes. *Aiding would have 20,000 Amperes. However, by this forced march through the math, it appears there are two batteries in parallel; (series) bucking; with a parallel resistor. So in the end, successful communication of the schematic. The "no interaction" model, Is this the "latter" or former model?while just being superposition, seems to lend itself to having great significance applied to the intermediate results. Partially this may be due to poor definitions. Certainly as I read it.If the wave is defined as just being a voltage wave, then all is well. Still ambiguous. And then deeper:But, for example, when looking at a solitary pulse, it is easy (and accurate) to view the wave as having more than just voltage. One can compute the charge, the current, the power, and the energy. It would seem if you knew the charge, you already know the energy; but the power? Just energy per unit time. We know the energy distribution on the line, so we know the power at any point and time. But when two waves are simultaneously present, it is only legal to superpose the voltage and the current. And illegal if only one is present? * No. Legal to also compute the power. Odd distinction. *Is there some other method like superposition that demands to be used for this instance?But it is obvious that a solitary wave has voltage, current, power, etc. But when two waves are present it is not legal to.... etc., etc. The "no interaction" model does not seem to resolve this conflict well, and some are lead astray. I was lost on a turn several miles back. Perhaps try again, with the clarifications. And it was this conflict that lead me to look for other ways of thinking about the system. I can only hope for clarity from this point on. But given the history, disappointment will be no surprise. Eh? Earlier you asked for an experiment. How about this one.... Take two step function generators, one at each end of a transmission line. Start a step from each end at the same time. When the steps collide in the middle, the steps can be viewed as passing each other without interaction, or reversing and propagating back to their respective sources. Why just that particular view? Those seem to be the common alternatives. If there are more, please share. We can measure the current at the middle of the line and observe that it is always 0. Is it? *When? Always. If, for some infinitesimal line section, there is no current through it, then there is no potential difference across it. Or did you mean along it? Hence, the when is some infinitesimal time before the waves of equal potential meet - and no current flow forever after. Therefore the charge that is filling the capacitance and causing the voltage step which is propagating back towards each generator How did that happen? *No potential difference across an infinitesimal line section, both sides at full potential (capacitors fully charged, or charging at identical rates). *Potentials on either side of the infinitesimal line section are equal to each other and to the sources, hence no potential differences anywhere, *No potential differences, no current flow, no charge change, no reflection, no more wave. The last bit of induction went to filling the last capacitance element with the last charge of current. *Last gasp. *No more gas. *Nothing left. Finis. must be coming from the generator to which the step is propagatig because no charge is crossing the middle of the line. Do you like it? Not particularly. *What does it demonstrate? That they bounce rather than pass silently. ...Keith 73's Richard Clark, KB7QHC ...Keith |
Standing-Wave Current vs Traveling-Wave Current
On Jan 1, 4:16*pm, Cecil Moore wrote:
Keith Dysart wrote: The Norton or Thévenin equivalent circuits seem *capable of positive reflection coefficients. * Either can be positive, negative, or zero depending on the value of the output impedance compared to Z0. Would you please quote a reference that addresses the subject of reflection coefficients from Thevenin or Norton equivalent sources? To the best of my knowledge, there is absolutely no requirement that a Thevenin or Norton equivalent circuit exhibit the same reflection coefficient at the circuit it replaces. It happens without effort because the output impedance of the Norton/Thevenin equivalent circuit is the same as the circuit it replaces, otherwise it is not an equivalent. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
Cecil Moore wrote:
Gene Fuller wrote: Cecil Moore wrote: As you well know, the convention is to apply a negative sign to positive energy flowing in the opposite direction from the "forward" energy which is arbitrarily assigned a plus sign. Let's see even one reference that mentions explicitly the concept of applying a negative sign to positive energy. Not power, not voltage, not current, not waves, but energy. If you keep feigning ignorance like that Gene, you are going to lose all respect. If the Poynting vector has a negative sign, as used by Ramo & Whinnery, that sign is an indication of the *direction of energy flow*, see quote below. From Ramo & Whinnery: The Poynting vector is "the vector giving *direction* and magnitude of *energy flow*". When Ramo & Whinnery hang a sign on a Poynting vector in a transmission line, it is an indication of the direction of energy flow. For pure standing waves, "The average [NET] value of Poynting vector is zero at every cross-sectional plane; this emphasizes the fact that on the average as much energy is carried away by the reflected wave as is brought by the incident wave." What? Reflected waves "carrying" energy? Shame on Ramo & Whinnery for contradicting the rraa gurus. It is impossible to satisfy you, Gene. When I quote reference after reference about reflected power, you say power doesn't reflect. When I change it to reflected energy, you ask for a reference. Cecil, Still up to your tricks? I ask for reference on a scalar quantity, and you respond with some stuff about vectors. If you want to continue to misinterpret the experts and believe that power, energy, or whatever flows in opposite directions at a single point at the same time, go right ahead. I suppose such beliefs expounded on RRAA are quite harmless in the grand scheme of world affairs. For future reference, however, just remember: Fields first, then power or energy. That's the way superposition really works. (By the way, I am not the one who made the point about power vs. energy. That must have been someone else.) 73, Gene W4SZ |
Standing-Wave Current vs Traveling-Wave Current
On Jan 1, 4:57*pm, Cecil Moore wrote:
Keith Dysart wrote: Could you better describe how you determine that the source has a Z0 equal to the line Z0? *I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. This will do it. We have a black box. We use a 12vdc battery and a current meter to measure the impedance of the black box. The current meter reads zero when we connect the 12vdc battery to the black box terminals. What is the impedance inside the black box since test V/I = infinity? Recall that impedance is the slope and to obtain the slope you need to do two measurements. Only if the entity is completely passive can you make only one measurement because in that case the line passes through the origin which is the implicit other measurement. With two measurements you will obtain the correct impedance. ....Keith |
Standing-Wave Current vs Traveling-Wave Current
On Jan 1, 6:00*pm, Roger wrote:
Roy Lewallen wrote: Roger wrote: Could you better describe how you determine that the source has a Z0 equal to the line Z0? *I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. Probably the simplest way is to put the entire source circuitry into a black box. Measure the terminal voltage with the box terminals open circuited, and the current with the terminals short circuited. The ratio of these is the source impedance. If you replace the box with a Thevenin or Norton equivalent, this will be the value of the equivalent circuit's impedance component (a resistor for most of our examples). If the driving circuitry consists of a perfect voltage source in series with a resistance, the source Z will be the resistance; if it consists of a perfect current source in parallel with a resistance, the source Z will be the resistance. You can readily see that the open circuit V divided by the short circuit I of these two simple circuits equals the value of the resistance. The power output of the Thévenin equivalent circuit follows the load. Sorry, I don't understand this. Can you express it as an equation? There seems to be some confusion as to the terms "Thévenin equivalent * circuit", "ideal voltage source", and how impedance follows these sources. Two sources we all have access to are these links: Voltage source: *http://en.wikipedia.org/wiki/Voltage_source Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem I don't disagree with anything I read there. But you may not quite have the concept of impedance correct. The impedance of the Thevenin/Norton equivalent source is not V/I but rather the slope of the line representing the relationship of the voltage to the current. When there is a source present, this line does not pass through the origin. Only for passive components does this line pass through the origin in which case it becomes V/I. Because the voltage to current plot for an ideal voltage source is horizontal, the slope is 0 and hence so is the impedance. For an ideal current source the slope is vertical and the impedance is infinite. Hoping this helps clarify.... ...Keith |
Standing-Wave Current vs Traveling-Wave Current
On Jan 1, 9:03*pm, Cecil Moore wrote:
Roger wrote: The principles of superposition are mathematically usable, not too hard, *and I think very revealing. *Yes, if we use part of the model, we must use it all the way. *To do otherwise would be error, or worse. Roy and Keith don't seem to realize that the zero source impedance for the ideal voltage source is only when the source is turned off for purposes of superposition. They conveniently avoid turning the source voltage on to complete the other half of the superposition process. When the source signal and the reflected wave are superposed at the series source resistor, where the energy goes becomes obvious. Total destructive interference in the source results in total constructive interference toward the load. See below. You have been a supporter of this theory for a long time. Yes, I have. I am a supporter of the principles and laws of physics. Others believe they can violate the principle of conservation of energy anytime they choose because the principle of conservation of energy cannot be violated - go figure. Roger, I have explained all of this before. If you are capable of understanding it, I take my hat off to you. Here's what happens in that ideal voltage source using the rules of superposition. 1. With the reflected voltage set to zero, the source current is calculated which results in power in the source resistor. Psource = Isource^2*Rsource 2. With the source voltage set to zero, the reflected current is calculated which results in power in the source resistor. Pref = Iref^2*Rsource Now superpose the two events. The resultant power equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) where 'A' is the angle between the source current and the reflected current. Note this is NOT power superposition. It is the common irradiance (power density) equation from the field of optics which has been in use in optical physics for a couple of centuries so it has withstood the test of time. For the voltage source looking into a 1/2WL open circuit stub, angle 'A' is 180 degrees. Therefore the total power in the source resistor is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) * or Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0 Ptot in the source resistor is zero watts but we already knew that. What is important is that the last term in that equation above is known as the "interference term". When it is negative, it indicates destructive interference. When Ptot = 0, we have "total destructive interference" as defined by Hecht in "Optics", 4th edition, page 388. Whatever the magnitude of the destructive interference term, an equal magnitude of constructive interference must occur in a different direction in order to satisfy the conservation of energy principle. Thus we can say with certainty that the energy in the reflected wave has been 100% re-reflected by the source. The actual reflection coefficient of the source in the example is |1.0| even when the source resistor equals the Z0 of the transmission line. I have been explaining all of this for about 5 years now. Instead of attempting to understand these relatively simple laws of physics from the field of optics, Roy ploinked me. Hopefully, you will understand. All of this is explained in my three year old Worldradio energy article on my web page. http://www.w5dxp.com/energy.htm Can you make this all work for a pulse, or a step function? How do you compute Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) for a pulse or a step? Or is your approach limited to sinusoids? ...Keith |
Standing-Wave Current vs Traveling-Wave Current
"Cecil Moore" wrote in message ... Roger wrote: The principles of superposition are mathematically usable, not too hard, and I think very revealing. Yes, if we use part of the model, we must use it all the way. To do otherwise would be error, or worse. clip some Roger, I have explained all of this before. If you are capable of understanding it, I take my hat off to you. Here's what happens in that ideal voltage source using the rules of superposition. Well, I would like to think I could understand it, but maybe something is wrong like Roy suggests. I think it is all falling into place, but our readers are not all in agreement. Could I ask a couple of questions to make sure I am understanding your preconditions? Is this a Thevenin source? If so, what is the internal resistor set to in terms of Z0? 1. With the reflected voltage set to zero, the source current is calculated which results in power in the source resistor. Psource = Isource^2*Rsource For condition number 2 below, is this a Thevenin equivalent resistance? 2. With the source voltage set to zero, the reflected current is calculated which results in power in the source resistor. Pref = Iref^2*Rsource Now superpose the two events. The resultant power equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) where 'A' is the angle between the source current and the reflected current. Note this is NOT power superposition. It is the common irradiance (power density) equation from the field of optics which has been in use in optical physics for a couple of centuries so it has withstood the test of time. OK. A few reactions. 1. To add the total power of each wave and then also add the power reduction of the phase relationship, would be creating power unless the cos(A) is negative. Something seems wrong here, probably my understanding. 2. The two voltages should be equal, therfore the power delivered by each to the source resister would be equal. 3. The power delivered to the source resistor will arrive a different times due to the phase difference between the two waves. 4. If the reflected power returns at the same time as the delivered power (to the source resistor), no power will flow. This because the resistor in a Thevenin is a series resistor. Equal voltages will be applied to each side of the resitor. The voltage difference will be zero. 5. If the reflected power returns 180 degrees out of phase with the applied voltage (to the source resistor), the voltage across the resistor will double with each cycle, resulting in an ever increasing current.(and power) into the reflected wave. For the voltage source looking into a 1/2WL open circuit stub, angle 'A' is 180 degrees. Therefore the total power in the source resistor is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0 This is correct for this condition. The problem with the equation comes when cos(90) which can easily happen. I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is the angle between positive peaks of the two waves. This angle will rotate twice as fast as the signal frequency due to the relative velocity between the waves.. Ptot in the source resistor is zero watts but we already knew that. What is important is that the last term in that equation above is known as the "interference term". When it is negative, it indicates destructive interference. When Ptot = 0, we have "total destructive interference" as defined by Hecht in "Optics", 4th edition, page 388. Whatever the magnitude of the destructive interference term, an equal magnitude of constructive interference must occur in a different direction in order to satisfy the conservation of energy principle. Thus we can say with certainty that the energy in the reflected wave has been 100% re-reflected by the source. The actual reflection coefficient of the source in the example is |1.0| even when the source resistor equals the Z0 of the transmission line. I have been explaining all of this for about 5 years now. Instead of attempting to understand these relatively simple laws of physics from the field of optics, Roy ploinked me. Hopefully, you will understand. All of this is explained in my three year old Worldradio energy article on my web page. http://www.w5dxp.com/energy.htm -- 73, Cecil http://www.w5dxp.com Light, especially solar energy, is a collection of waves of all magnitudes and frequencies. We should see only about 1/2 of the power that actually arrives due to superposition. The reflection from a surface should create standing waves in LOCAL space that will contain double the power of open space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be correct for a collection of waves, but incorrect for the example. I tried to read your article a couple of weeks ago, but I found myself not understanding. I have learned a lot since then thanks to you, Roy, and others. I will try to read it again soon. So what do you think Cecil? 73, Roger, W7WkB |
Standing-Wave Current vs Traveling-Wave Current
Correction. Please note the change below. I apologize for the error.
Roger Sparks wrote: "Cecil Moore" wrote in message ... Roger wrote: The principles of superposition are mathematically usable, not too hard, and I think very revealing. Yes, if we use part of the model, we must use it all the way. To do otherwise would be error, or worse. clip some Roger, I have explained all of this before. If you are capable of understanding it, I take my hat off to you. Here's what happens in that ideal voltage source using the rules of superposition. Well, I would like to think I could understand it, but maybe something is wrong like Roy suggests. I think it is all falling into place, but our readers are not all in agreement. Could I ask a couple of questions to make sure I am understanding your preconditions? Is this a Thevenin source? If so, what is the internal resistor set to in terms of Z0? 1. With the reflected voltage set to zero, the source current is calculated which results in power in the source resistor. Psource = Isource^2*Rsource For condition number 2 below, is this a Thevenin equivalent resistance? 2. With the source voltage set to zero, the reflected current is calculated which results in power in the source resistor. Pref = Iref^2*Rsource Now superpose the two events. The resultant power equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) where 'A' is the angle between the source current and the reflected current. Note this is NOT power superposition. It is the common irradiance (power density) equation from the field of optics which has been in use in optical physics for a couple of centuries so it has withstood the test of time. OK. A few reactions. 1. To add the total power of each wave and then also add the power reduction of the phase relationship, would be creating power unless the cos(A) is negative. Something seems wrong here, probably my understanding. 2. The two voltages should be equal, therfore the power delivered by each to the source resister would be equal. 3. The power delivered to the source resistor will arrive a different times due to the phase difference between the two waves. 4. If the reflected power returns at the same time as the delivered power (to the source resistor), no power will flow. This because the resistor in a Thevenin is a series resistor. Equal voltages will be applied to each side of the resitor. The voltage difference will be zero. 5. If the reflected power returns 180 degrees out of phase with the applied voltage (to the source resistor), the voltage across the resistor will double with each cycle, resulting in an ever increasing current.(and power) into the reflected wave. For the voltage source looking into a 1/2WL open circuit stub, angle 'A' is 180 degrees. Therefore the total power in the source resistor is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0 This is correct for this condition. The problem with the equation comes when cos(90) which can easily happen. I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is the angle between positive peaks of the two waves. This angle will rotate twice as fast as the signal frequency due to the relative velocity between the waves.. No, this angle is fixed by system design. If the system changes, this angle will rotate twice as fast as the angle change of a single wave. The equation should be OK. Ptot in the source resistor is zero watts but we already knew that. What is important is that the last term in that equation above is known as the "interference term". When it is negative, it indicates destructive interference. When Ptot = 0, we have "total destructive interference" as defined by Hecht in "Optics", 4th edition, page 388. Whatever the magnitude of the destructive interference term, an equal magnitude of constructive interference must occur in a different direction in order to satisfy the conservation of energy principle. Thus we can say with certainty that the energy in the reflected wave has been 100% re-reflected by the source. The actual reflection coefficient of the source in the example is |1.0| even when the source resistor equals the Z0 of the transmission line. I have been explaining all of this for about 5 years now. Instead of attempting to understand these relatively simple laws of physics from the field of optics, Roy ploinked me. Hopefully, you will understand. All of this is explained in my three year old Worldradio energy article on my web page. http://www.w5dxp.com/energy.htm -- 73, Cecil http://www.w5dxp.com Light, especially solar energy, is a collection of waves of all magnitudes and frequencies. We should see only about 1/2 of the power that actually arrives due to superposition. The reflection from a surface should create standing waves in LOCAL space that will contain double the power of open space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be correct for a collection of waves, but incorrect for the example. I tried to read your article a couple of weeks ago, but I found myself not understanding. I have learned a lot since then thanks to you, Roy, and others. I will try to read it again soon. So what do you think Cecil? 73, Roger, W7WkB |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
On Jan 1, 6:00 pm, Roger wrote: Roy Lewallen wrote: Roger wrote: Could you better describe how you determine that the source has a Z0 equal to the line Z0? I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. Probably the simplest way is to put the entire source circuitry into a black box. Measure the terminal voltage with the box terminals open circuited, and the current with the terminals short circuited. The ratio of these is the source impedance. If you replace the box with a Thevenin or Norton equivalent, this will be the value of the equivalent circuit's impedance component (a resistor for most of our examples). If the driving circuitry consists of a perfect voltage source in series with a resistance, the source Z will be the resistance; if it consists of a perfect current source in parallel with a resistance, the source Z will be the resistance. You can readily see that the open circuit V divided by the short circuit I of these two simple circuits equals the value of the resistance. The power output of the Thévenin equivalent circuit follows the load. Sorry, I don't understand this. Can you express it as an equation? There seems to be some confusion as to the terms "Thévenin equivalent circuit", "ideal voltage source", and how impedance follows these sources. Two sources we all have access to are these links: Voltage source: http://en.wikipedia.org/wiki/Voltage_source Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem I don't disagree with anything I read there. But you may not quite have the concept of impedance correct. The impedance of the Thevenin/Norton equivalent source is not V/I but rather the slope of the line representing the relationship of the voltage to the current. When there is a source present, this line does not pass through the origin. Only for passive components does this line pass through the origin in which case it becomes V/I. Because the voltage to current plot for an ideal voltage source is horizontal, the slope is 0 and hence so is the impedance. For an ideal current source the slope is vertical and the impedance is infinite. Hoping this helps clarify.... ...Keith Yes, I can understand that there is no change in voltage no matter what the current load is, so there can be no resistance or reactive component in the source. The ideal voltage link also said that the ideal source could maintain voltage no matter what current was applied. Presumably a negative current through the source would result in the same voltage as a positive current, which is logical if the source has zero impedance. During the transition from negative current passing through the source to positive current passing through the source, the current at some time must be zero. How is the impedance of the perfect source defined at this zero current point? How is the impedance of the attached system defined? The resistor in the Thevenin/Norton equivalent source is selected with some criteria in mind. What I would like to do is to design a Thevenin source to provide 1v across a 50 ohm transmission line INFINATELY long, ignoring ohmic resistance. I would like the source resistor to absorb as much power as the line, so that if power ever returns under reflected wave conditions, it can all be absorbed by the resistor. I think the size of such a resistor will be 50 ohms. Thanks for the clarification. 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
Second analysis: +0.5 input reflection coefficient
I got enough email responses to my request to make me believe it might be worthwhile to put together another analysis. The analysis I did previously is of limited use because of the total lack of any loss in the system. It has also caused conceptual difficulties because of the perfect source's -1 reflection coefficient. So I'll do another analysis with a more realistic source. The following analysis will illustrate the startup and progress to steady state of a lossless, one wavelength, open ended 50 ohm transmission line as before. But connected to the input end of the line will be a perfect voltage source in series with a 150 ohm resistance. This is not a Thevenin equivalent circuit, because it isn't meant to be an equivalent of anything; it's simply a perfect source and a resistance, both of which are common idealized linear circuit models. The only difference between this setup and the one I analyzed earlier is the addition of the source resistance. I'm going to make a slight change in notation and call the initial forward and reverse waves vf1 and vr1 respectively, instead of just vf and vr as before. I apologize if this causes any confusion. When we initially connect or turn on the voltage source, the source/resistance combination sees Z0 looking into the line. So the voltage at the line input is vs(t) * Z0 / (Z0 + Rs), where vs(t) is the voltage of the perfect voltage source and Rs is the 150 ohm source resistance, until the reflection of the original forward wave returns. So I'm going to specify that vs(t) = 4 * sin(wt) so that the initial voltage at the line input is simply sin(wt). At any point that the forward wave has reached, then, vf1(t, x) = sin(wt - x) The open far end of the line has a reflection coefficient of +1, so as before the returning wave is: vr1(t, x) = sin(wt + x) (The change from -x to +x is due to the reversal of direction of propagation. You'll see this with every reflection.) At any time between when vf1 reflects from the output end (t = 2*pi/w) and when it reaches the input end of the line (t = 4*pi/w), the total voltage at any point the returning wave has reached is vtot = vf1(t, x) + vr1(t, x) = sin(wt - x) + sin(wt + x) [Eq. 1] When the returning wave vr1 reaches the input end of the line, it re-reflects to form a new forward wave vf2. But because of the added 150 ohm resistance, the source reflection coefficient is +0.5 instead of the -1 of the previous analysis, so vf2(t, x) = .5 * sin(wt - x) Just after the reflection, the voltage at the line input will be vf1(t, 0) + vr1(t, 0) + vf2(t, 0) = sin(wt) + sin(wt) + .5 * sin(wt) = 2.5 * sin(wt) So we'll see a 2.5 peak volt sine wave at the input from the time the first returning wave reaches the source until the next one does, or from t = 4*pi/w to t = 8*pi/w. And anywhere on the line which vf2 has reached, we'll see vtot(t, x) = 1.5 * sin(wt - x) + sin(wt + x) When vf2 hits the far end and reflects, it creates the second reflected wave vr2(t, x) = .5 * sin(wt + x) So now at any point where vr2 has reached we'll have vtot(t, x) = 1.5 * sin(wt - x) + 1.5 * sin(wt + x) [Eq. 2] There's something worth pointing out here. Look at the similarity between Eq. 1, which was the total voltage with only one forward and one reflected wave, and Eq. 2, which is the total with two forward and two reflected waves. They're exactly the same except in amplitude -- Eq. 2 vtot is 1.5 times Eq. 1 vtot. Also notice that the equation for vf2 is exactly the same as for vf1 except a constant, and likewise vr2 and vr1. Every time a new forward wave is created by reflection from the input end of the line, a new reflected wave is created by the reflection from the far end. The reflection coefficient at the far end doesn't change, so the relationship between each forward wave and the corresponding reflected wave is the same as for any other forward wave and its reflection. So the ratio of the sum of all forward waves to the sum of all reflected waves is the same as for any single forward wave and its reflection. The relationship between forward and reverse waves determines the SWR, so this is why the source reflection coefficient doesn't play any role in determining the line SWR. No matter what it is, it creates a new pair of waves having the same ratio as every other pair. Let's look at just one more pair of waves. vf3 = .5 * .5 * sin(wt - x) = 0.25 * sin(wt - x) vr3 = .5 * .5 * sin(wt + x) = 0.25 * sin(wt + x) From the time vf3 is created until vr3 returns to the input, or from t = 8*pi/w to t = 10*pi/w, at the input end of the line we'll see v(t, 0) = vf1(t, 0) + vr1(t, 0) + vf2(t, 0) + vr2(t, 0) + vf3(t, 0) = 3.25 * sin(wt) For the previous round trip period the sine wave amplitude was 2.5 volts and for this round trip period it's 3.25 volts. Where is this going to end after an infinite number of reflections? Well, it had better end at 4 volts, the voltage of the perfect source! At steady state, the impedance looking into the line is infinite, so the current from the source is zero. (A comparable analysis of the current waves would show convergence to itot = 0 at the input.) So there's no drop across the 150 ohm resistor and the line input voltage equals the voltage of the source. Let's see if we can show this convergence mathematically. The trick is to take advantage of a simple formula for the sum of an infinite geometric series. If F = a + ra + r^2*a + r^3*a, . . . an infinite number of terms then F = a / (1 - r) if |r| 1. It's a very useful formula. I learned it in high school algebra, but see that it's in at least one my calculus and analytic geometry texts. Try it out, if you want, with a pocket calculator or spreadsheet. Going back and looking at the analysis, we had vtot(t, x) = vf1(t, x) + vr1(t, x) for the first set of waves. The next set was exactly 1/2 the value of the first set, due to the +0.5 source reflection coefficient. The next set was 1/2 that value, and so forth. So the first term of the series (a) is sin(wt - x) + sin(wt - x) and the ratio of terms is 0.5, so vtot(t, x)(steady state) = (sin(wt - x) + sin(wt + x)) / (1 - 0.5) = 2 * sin(wt - x) + sin(wt + x) Using the same trig identity as in the earlier analysis, vtot(t, x)(steady state) = 4 * sin(wt) * cos(x) This is the correct steady state solution. You can see that it includes the standing wave envelope cos(x), as it must. Because superposition applies, we can add up the infinite number of forward and reverse waves any way we want, and the total will always be the same. So let's separately add the forward waves and reverse waves. vf3/vf2 = vf2/vf1 = 0.5, as it is for the ratio of any two successive forward waves vr3/vr2 = vr2/vr1 = 0.5, as it is for the ratio of any two successive reverse waves So vf(t, x)(total) = sin(wt - x) / (1 - 0.5) = 2 * sin(wt - x) and vr(t, x)(total) = sin(wt + x) / (1 - 0.5) = 2 * sin(wt + x) These can be used for steady state analysis, as though there is only one forward and one reverse wave, and the result will be exactly the same as if the system was analyzed for each of the infinite number of waves individually. This is almost universally done; the run-up from the initial state is usually only of academic interest. If we add the total forward and reverse waves to get the total voltage, the result is exactly the same as it was when we summed the pairs of waves to get the total. Again I ran a SPICE simulation of the circuit, using a 5 wavelength line for clarity. http://eznec.com/images/TL2_input.gif is the voltage at the line input. As predicted, it starts at 1 volt peak, remains there until the first reflected wave returns to the input end (at t = 10 sec.), then jumps to 2.5 volts. After another round trip, it goes to 3.25. http://eznec.com/images/TL2_1_sec.gif shows the voltage one wavelength (1 second) from the source. Here you can see that the voltage is zero until the initial forward wave arrives at t = 1 sec. From then until the reflected wave arrives at t = 9 sec., it's one volt peak. From then (t = 9) until the reflected wave re-reflects from the source and returns (t = 11), we have vf1 + vr1 = sin(wt - x) + sin(wt + x). x is 2*pi radians or 360 degrees, so the voltage is simply 2 * sin(wt). And that's what the plot shows - a sine wave of 2 volts amplitude. Then vf2 gets added in at t = 11 sec, for a total of sin(wt - x) + sin(wt + x) + 0.5 * sin(wt - x), or at the observation x point, 2.5 * sin(wt). At t = 19 sec., vr2 arrives and adds another 0.5 * sin(wt) to the total, raising the amplitude to 3 volts peak. At t = 21 sec., vf3 arrives, adding another 0.25 * sin(wt) for a total amplitude of 3.25 volts. And so forth. As with the previous analysis, SPICE shows exactly what the analysis predicts. http://eznec.com/images/TL2_5_sec.gif is the voltage at the open end of the line. For the first 5 seconds, the voltage is zero because the initial wave hasn't arrived. At t = 5 sec., the voltage at the end becomes vf1 + vr1 = 2 * sin(wt). At 15 sec., it becomes vf1 + vr1 + vf2 + vr2 = 3 * sin(wt). And so forth, just as predicted by the analysis. Although I've used a line of the convenient length of one wavelength (or 5 wavelengths for the SPICE run), an open circuited end, and a resistive source, none of these are required. Exactly the same kind of analysis can be done with complex loads of any values at the line input and output, and with any line length. But in the general case, returning waves don't add directly in phase or out of phase with forward waves, and a wave undergoes some phase shift other than 0 or 180 degrees upon reflection. So phase angles have to be included in the descriptions of all voltages. In the general case it's much easier to revert to phasor notation, but otherwise the analytical process is identical and the results just as good. So far I haven't seen any analysis using alternative theories, ideas of how sources work, or using power waves, which also correctly predict the voltage at all times and in steady state. Because there's so much interest in power, I'll calculate the power and energy at the line input. But I'll put it in a separate posting. Roy Lewallen, W7EL |
Standing-Wave Current vs Traveling-Wave Current
Here's an analysis of the input power and energy of the system similar
to the one described in my "Second analysis: +0.5 input reflection coefficient" Please refer to that posting when reading this. I'm going to make just one change to the system in that analysis: I'll use the 5 wavelength line of the SPICE analysis. This has absolutely no effect on any of the values shown in the analysis except where I gave periods of time when various equations were valid (e.g. t = 2*pi/w) -- those will be five times greater. The effect of this change is to allow use of the SPICE output as a visual aid. When speaking of the SPICE plot, I'll be referring to http://eznec.com/images/TL2_input.gif, which is the voltage at the input end of the line. For this analysis I'm going to let f = 1 Hz, so w = 2*pi Hz. From the voltage analysis and the SPICE plot, the initial voltage at the input of the line is sin(wt). So the voltage across the input resistor is 3 * sin(wt) (+ toward the source), and the current flowing into the line is (3 * sin(wt)) / 150 = 20 * sin(wt) mA. The average power being delivered to the line is Vin(rms) * Iin(rms) (since the voltage and current are in phase) = (0.7071 v. * 14.14 mA) = 10 mW. Since the line initially presents an impedance of Z0, this should also be Vin(rms)^2 / Z0 or Iin(rms)^2 / Z0. Let's see: Vin(rms) = 0.7071, so Vin(rms)^2 / Z0 = 0.5 / 50 = 10 mW; Iin(rms) = 0.01414 A, so Iin(rms)^2 * Z0 = 0.0002 * 50 W = 10 mW. Ok. The input voltage remains unchanged until the first reflected wave returns at t = 10 sec (remember, we're using a 5 wavelength line and frequency of 1 Hz). So during that time we put 10 mW * 10 sec = 0.1 joule of energy into the line. Note that I could have calculated the instantaneous power for each instant, then integrated it over the first 10 seconds to get the total energy. The result would be identical. During that first 10 seconds, the resistor is dissipating an average of Vr(rms) * Ir(rms) = 2.121 * 14.14 = 30 mW. The source is producing Vs(rms) * Is(rms) = 2.828 * 14.14 = 40 mW. So here's the story so far: For the first 10 seconds (one round trip): The source produces 40 mW, for a total of 400 mj (millijoules) The resistance dissipates 30 mW, for a total of 300 mj The line gets 10 mW, for a total of 100 mj All the power and energy is totally accounted for, so far. The sum of power and energy delivered to the resistor and line equals the power and energy produced by the source. As you can see from the SPICE output or the voltage analysis, the input voltage jumps to 2.5 volts at t = 10 seconds and stays there for the next 10 seconds. During that time, the voltage across the resistor is (4 - 2.5) * sin(wt), so the current drops to 1.5 sin(wt) / 150 = 10 sin(wt) mA which has an RMS value of about 7.071 mA. Using the same methods as before, For the next 10 seconds: The source produces 20 mW, for a total of 200 mj The resistance dissipates 7.50 mW, for a total of 75 mj The line gets 12.5 mW, for a total of 125 mj We've again accounted for all the power and energy, with no need to invoke any kind of power waves. Next the input voltage goes to 3.25 volts, so the current drops to 5 * sin(wt) mA and For the next 10 seconds, The source produces 10 mW, for a total of 100 mj The resistance dissipates 1.875 mW, for a total of 18.75 mj The line gets 8.125 mW, for a total of 81.25 mj There's something interesting about this energy flow. Notice that the amount of power produced by the source decreases by a factor of two each cycle, and the amount of power dissipated by the resistor decreases by a factor of four each cycle. These are to be expected, since the line input voltage is increasing each time. But look at the power supplied to the line -- it actually increases during the second cycle relative to the first, then drops back. What's happening? Well, the line has an apparent impedance of Vin/Iin at any given time. Let's see what it is (the peak values have the same ratio as the RMS values, so I'll use those): For t = 0 to 10, Vin/Iin = 1/0.020 = 50 ohms For t = 10 to 20, Vin/Iin = 2.5/0.010 = 250 ohms For t = 20 to 30, Vin/Iin = 3.75/0.005 = 750 ohms The input impedance will, of course, approach an infinite value as time goes on, Vin approaches Vs, and Iin approaches zero. But during the time interval of 10 to 20 seconds, the apparent line impedance provided a better "impedance match" for the 150 ohm source than at other times, which increased the line power input during that time. Only because we have no energy leaving the line can be calculate a total energy delivered by the source and dissipated by the resistance over an arbitrarily long time period. That is, we can find the total energy delivered if the line were connected and the source left on forever. We can use the same formula for summing an infinite series as used in the voltage analysis to find that: The source produces a total of 400 / (1 - 0.5) = 800 mj The resistor dissipates a total of 300 / (1 - 0.25) = 400 mj from which we see that a total of 400 mj has been supplied to the line. If we were to quickly replace the source and resistor with a 50 ohm resistor across the line and no source, what should happen? Well, we have forward and reverse waves of 2 volts peak traveling on the line, so the end voltage should immediately drop to 2 * sin(wt) as the reverse traveling wave exits with no further reflections from the input end of the line, and the forward wave moves toward the far end. So the 50 ohm resistor would have an RMS voltage of 1.414 volts across it, and a dissipation of (1.414)^2 / 50 = 40 mW. This constant dissipation should continue until the line is completely discharged, which will take one round trip time or 10 sec. The total energy removed from the line and dissipated during this time is 40 mW * 10 sec = 400 mj, which is exactly the amount we put into the line during the charging process. All the energy produced by the source is accounted for. I think I can set this maneuver up with SPICE, and I'll do so if anyone who might be skeptical would be convinced by the SPICE output. A caution is in order. This analysis was greatly simplified by the lack of any energy storing devices other than the transmission line. That is, there was no reactance at either end of the line. An equally accurate analysis could be done with reactances present, but calculation of energy from power would have to account for the temporary energy storage in the reactive components. Also, if a resistance-containing load is connected to the output, it will also dissipate power and energy, so that would also have to be accounted for. The bottom line is that care should be taken in applying this method to more complex circuits without suitable modification. I've completely accounted for the power and energy leaving the voltage source, being dissipated in the resistor, and entering the line, at all times from startup to steady state, and done it quantitatively with numerical results. And I did this without any mention of propagating waves of power or energy. It was done very simply using Ohm's law and elementary power relationships, with the only variable being the line input voltage calculated from the voltage wave analysis. I offer a challenge to those who embrace the notion of traveling waves of average power or other alternative theories to do the same using mathematics and premises consistent with the alternative theory. Until such an analysis is produced, I remain unconvinced that any such theory is valid. *Any* analysis has to produce results consistent with the law of the conservation of energy, as this one has. Roy Lewallen, W7EL |
Standing-Wave Current vs Traveling-Wave Current
On Tue, 1 Jan 2008 18:44:17 -0800 (PST), Keith Dysart
wrote: Since the experiment ends before the reflection returns to the generator, Is it a step, or is it a pulse? Makes a huge difference in the analysis. the impedance of the generator is irrelevant. The impedance perhaps, but not the voltage. I specifically offered a difference of sources (Norton vs. Thevenin). The impedance is the same either way, the voltage is not. As you introduce power later in this response, power at the source is even more contentious. As such, we can drop this altogether. Ah. The challenge of written precisely. Consider the generator to be on the left end of the line and the open to be at the right end. When the capacitance at the right end charges to 2 * V, the current now has to stop a little bit more to the left. Still doesn't make sense, and I wouldn't attribute it to precision seeing that you've written the same thing twice. Again, the poor writing meant a bit further to left from the end. Yes, if you are in a left-right progression, further is more to the right. There is the ambiguity of "end effect." Want to explain how you double the stored voltage in the distributed capacitance of the line without current? * The energy formerly present in the inductance of the line has been transferred to the capacitance. That is always so. You haven't offered anything differentiable from always, where previously (and later in your response, here) you suggested current stopped flowing even when voltage was doubling. The definition of capacitance is explicitly found in the number of electrons (charge or energy) on a surface; which in this case has not changed.The charge that is continuing to flow from the source is being used to charge the distributed capacitance of the line. It would appear now that charge is flowing again, but that there is a confusion as to where the flow comes from. * Using the new reference scheme, What new reference scheme? charge to the left of the leftward propagating step continues to flow, while charge to the right has stopped flowing. If precision was ever called for, now is the time. Why would the source at less voltage provide current to flow into a cap that is rising in potential above it? * The beauty of inductance. You get extra voltage when you try to stop the flow. You don't explain where the extra voltage is. There is a double voltage to be sure, but you have not exactly drawn a cause-and-effect relationship. Challenges with the referent. The former view is that a voltage of 2*V is propagating back along the line. The latter view is that it is a step of V above the already present V. Sounds more like a problem for the challenged (i.e. there is no difference between 2*V and V+V). In this model, the step function has propagated to the end, been reflected and is now propagating backwards. Implicit in this description is that the step continues to flow to the end of the line and be reflected as the leading edge travels back to the source. This is a difficult read. *You have two sentences. *Is the second merely restating what was in the first, or describing a new condition (the reflection)? Agreed. What is a good word to describe the constant voltage that follows the actual step change in voltage? The "tread" perhaps? Does it matter? It is another step. That should be obvious for the sake of superposition. "The tread continues to flow to the end of the line and be reflected as the leading edge travels back to the source." I am not sure that that is any clearer. Not particularly, since we now have three things moving: 1. Step; 2. Tread; 3. Leading Edge. It would seem to me that both 1 and 2 have a 3; so what are you trying to say with the novel introduction of two more terms? And this is the major weakness in the model. Which model? * The "no interaction" model. And which model is the "no interaction" model? The former, or the latter? The latter? or the former? It claims the step function is still flowing in the portion of the line that has a voltage of 2*V and *zero* current. Does a step function flow? * Perhaps the "tread"? But then should the step change be called the "riser"? So now we have four things moving: 1. Step; 2. Tread; 3. Leading Edge. 4. Riser It would seem to me that both 1,2 and 4 have a 3; so what are you trying to say with the novel introduction of three more terms? As for "zero" current, that never made sense in context here. Somewhat clarified, I hope. But for clarity, the current to the right of the leftward propagating step is zero. So now the leftward propagating step is 1. Step; 2. Tread; 3. Leading Edge; 4. Riser? If I simply discard the last three invented terms; went out on the thin limb of interpretation; then, in my mind's eye I would see that, yes, no current is flowing to the physical right of the transient (the only thing that is moving - as evidenced through voltage). A trivial example is connecting to 10 volt batteries in parallel through *a .001 ohm resistor. Parallel has two outcomes, which one? *"Through" a resistor to WHERE? In series? *In parallel? * Much to ambiguous. I know. Trying to conserve words leads to confusion. Try: negative to negative, positives connected using the resistor. Makes quite a difference. Perhaps not in the math, but certainly in the concept. However, by this forced march through the math, it appears there are two batteries in parallel; (series) bucking; with a parallel resistor. So in the end, successful communication of the schematic. Actually no. You describe a resistor in a series loop; I describe a resistor in parallel. Consider the repetition of your last statement above: positives connected using the resistor. Positives connected through the resistor (the sense of "using" the resistor as connector)? Or positives connected, then using the resistor (where it is also connected, but unstated as such) to the negatives? It would seem if you knew the charge, you already know the energy; but the power? Just energy per unit time. We know the energy distribution on the line, so we know the power at any point and time. Time has not been quantified, whereas voltage, hence charge, hence energy has. Why introduce a new topic without enumerating it, when its inclusion adds nothing anyway? You don't develop anything that explains the step in terms of power. You don't use power anywhere. In fact we then step into the philosophy of does a line actually move power, or energy? That debate would cloud any issue of a step's migration, reflection, or any of a spectrum of characteristics that power has no sway over. But when two waves are simultaneously present, it is only legal to superpose the voltage and the current. And illegal if only one is present? * No. Legal to also compute the power. Please note you start your sentence with a "But." Sentences (much less paragraphs) are not started with coordinating conjunctions. When you use "but," the logical implication is that you are coordinating: two-waves with an equal ranking and unexpressed: one-wave. One may also use "but" at the beginning of a sentence as a logical connector between equal ideas (a transitional adverb); however, you don't have anything in the preceding, original paragraph other than a single wave. I do not see anything distinctive about two waves over one wave until your recent injection of power (not at all in the original), and to no effect except to raise the objection (rather circular and unnecessary inclusions to abandon the discussion of reflection in rhetorical limbo). Apparently your transition spans many postings to a festering point by Cecil (this is, of course, another one of my interpretations). Earlier you asked for an experiment. How about this one.... Take two step function generators, one at each end of a transmission line. Start a step from each end at the same time. When the steps collide in the middle, the steps can be viewed as passing each other without interaction, or reversing and propagating back to their respective sources. Why just that particular view? Those seem to be the common alternatives. If there are more, please share. Neither view works. We can measure the current at the middle of the line and observe that it is always 0. Is it? *When? Always. You seem to be in self-contradiction when you describe voltage doubles without current flow. If, for some infinitesimal line section, there is no current through it, then there is no potential difference across it. Or did you mean along it? A point well made in the scheme of precise language. Yet and all, taken singly (one/either conductor) or doubly (a transmission line section); then the identical statement remains true with both interpretations for the step condition. Of course, a lot may be riding on whether you are speaking of a step function, or a pulse. Seeing the original was explicit to step, and never introduces pulse; then there is no current flow as I responded: Hence, the when is some infinitesimal time before the waves of equal potential meet - and no current flow forever after. Therefore the charge that is filling the capacitance and causing the voltage step which is propagating back towards each generator How did that happen? *No potential difference across an infinitesimal line section, both sides at full potential (capacitors fully charged, or charging at identical rates). *Potentials on either side of the infinitesimal line section are equal to each other and to the sources, hence no potential differences anywhere, *No potential differences, no current flow, no charge change, no reflection, no more wave. The last bit of induction went to filling the last capacitance element with the last charge of current. *Last gasp. *No more gas. *Nothing left. Finis. must be coming from the generator to which the step is propagatig because no charge is crossing the middle of the line. Do you like it? Not particularly. *What does it demonstrate? That they bounce rather than pass silently. How do you introduce recoil or maintain momentum without energy? Odd. Please remove my need to perform interpretation and give me a more succinct accounting. Further, limit this to one scenario (this last failed example is certainly dead in the water as far as collisions go). 73's Richard Clark, KB7QHC |
Standing-Wave Current vs Traveling-Wave Current
On Tue, 01 Jan 2008 15:00:54 -0800, Roger wrote:
There seems to be some confusion as to the terms .... Two things to notice about the Thévenin equivalent circuit: 1. It contains an ideal voltage source "IN SERIES" with a resistor. Hi Roger, One confusion would seem to originate in your reliance in resistors. A resistor (R) is NOT the series (or for Norton the parallel) passive device - it is an Impedance (Z). This misapplication (R) was perpetuated by Edison when he battled Westinghouse for funding of power generation projects. He would invariably craft the resistor (R) into the equation for the bankers to prove DC was more efficient that AC systems, when in fact the engineering (Z) proves quite the contrary. I suppose the bankers bought it (R) at the outset, but market economics (Z) hammered Edison into the ground when the bookkeeping of AC made their investors rich at the expense of DC. This also raises issues of the confusion over conjugate matching and Z0 matching where many correspondents here freely intermix the two's characteristics as though they belong to one or the other (or both, or neither). 73's Richard Clark, KB7QHC |
Standing-Wave Current vs Traveling-Wave Current
Correction:
Roy Lewallen wrote: . . . Reflection coefficients are complex numbers, so they can't properly be described as "positive" or "negative" except in the special cases of +1 and -1. In all other cases, the can only be described by their magnitude and angle, or real and imaginary component. . . It's also reasonable to talk about positive or negative reflection coefficients when you're restricting the possibilities to ones having an angle of zero or 180 degrees. This would be the case, for example, when dealing with lossless lines (purely real Z0) connected to other lossless lines or to a purely resistive source or load. Roy Lewallen, W7EL |
Standing-Wave Current vs Traveling-Wave Current
On Jan 2, 1:41*am, Roger wrote:
Keith Dysart wrote: On Jan 1, 6:00 pm, Roger wrote: Roy Lewallen wrote: Roger wrote: Could you better describe how you determine that the source has a Z0 equal to the line Z0? *I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. Probably the simplest way is to put the entire source circuitry into a black box. Measure the terminal voltage with the box terminals open circuited, and the current with the terminals short circuited. The ratio of these is the source impedance. If you replace the box with a Thevenin or Norton equivalent, this will be the value of the equivalent circuit's impedance component (a resistor for most of our examples). If the driving circuitry consists of a perfect voltage source in series with a resistance, the source Z will be the resistance; if it consists of a perfect current source in parallel with a resistance, the source Z will be the resistance. You can readily see that the open circuit V divided by the short circuit I of these two simple circuits equals the value of the resistance. The power output of the Thévenin equivalent circuit follows the load. Sorry, I don't understand this. Can you express it as an equation? There seems to be some confusion as to the terms "Thévenin equivalent * circuit", "ideal voltage source", and how impedance follows these sources. Two sources we all have access to are these links: Voltage source: *http://en.wikipedia.org/wiki/Voltage_source Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem I don't disagree with anything I read there. But you may not quite have the concept of impedance correct. The impedance of the Thevenin/Norton equivalent source is not V/I but rather the slope of the line representing the relationship of the voltage to the current. When there is a source present, this line does not pass through the origin. Only for passive components does this line pass through the origin in which case it becomes V/I. Because the voltage to current plot for an ideal voltage source is horizontal, the slope is 0 and hence so is the impedance. For an ideal current source the slope is vertical and the impedance is infinite. Hoping this helps clarify.... ...Keith Yes, I can understand that there is no change in voltage no matter what the current load is, so there can be no resistance or reactive component in the source. *The ideal voltage link also said that the ideal source could maintain voltage no matter what current was applied. *Presumably a negative current through the source would result in the same voltage as a positive current, which is logical if the source has zero impedance. This is true. During the transition from negative current passing through the source to positive current passing through the source, the current at some time must be zero. * This would occur, for example, when the output terminals are open circuited. Or connected to something that had the same open-circuit voltage as your source. How is the impedance of the perfect source defined at this zero current point? * This is not a different question than: How is the resistance of a resistor defined when it has no current flowing in it? It continues to satisfy the relation V = I * R, though one can not compute R from the measured V and I. How is the impedance of the attached system defined? It is unknown, but has a voltage equal to the voltage of the source. The resistor in the Thevenin/Norton equivalent source is selected with some criteria in mind. *What I would like to do is to design a Thevenin * source to provide 1v across a 50 ohm transmission line INFINATELY long, ignoring ohmic resistance. I would like the source resistor to absorb as much power as the line, so that if power ever returns under reflected wave conditions, it can all be absorbed by the resistor. *I think the size of such a resistor will be 50 ohms. That is the correct value to not have any reflections at the source. But do not expect the power dissipated in the resistor to increase by the same amount as the "reflected power". In general, it will not. This is what calls into question whether the reflected wave actually contains energy. Do some simple examples with step functions. The math is simpler than with sinusoids and the results do not depend on the phase of the returning wave, but simply on when the reflected step arrives bach at the source. Examine the system with the following terminations on the line: open, shorted, impedance greater than Z0, and impedance less than Z0. Because excitation with a step function settles to the DC values, the final steady state condition is easy to compute. Just ignore the transmission line and assume the termination is connected directly to the Thevenin generator. When the line is present, it takes longer to settle, but the final state will be the same with the line having a constant voltage equal to the voltage output of the generator which will be the same as the voltage applied to the load. Then do the same again, but use a Norton source. You will find that conditions which increase the dissipation in the resistor of the Thevenin equivalent circuit reduce the dissipation in the resistor of the Norton equivalent circuit and vice versa. This again calls into question the concept of power in a reflected wave, since there is no accounting for where that "power" goes. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
On Jan 2, 2:59*am, Richard Clark wrote:
On Tue, 1 Jan 2008 18:44:17 -0800 (PST), Keith Dysart wrote: Since the experiment ends before the reflection returns to the generator, Is it a step, or is it a pulse? As stated, a step. Makes a huge difference in the analysis. the impedance of the generator is irrelevant. The impedance perhaps, but not the voltage. *I specifically offered a difference of sources (Norton vs. Thevenin). *The impedance is the same either way, the voltage is not. *As you introduce power later in this response, power at the source is even more contentious. *As such, we can drop this altogether. The generality of the specified voltage (V) would seem to adequately cover both the Thevenin and Norton generators. Ah. The challenge of written precisely. Consider the generator to be on the left end of the line and the open to be at the right end. When the capacitance at the right end charges to 2 * V, the current now has to stop a little bit more to the left. Still doesn't make sense, and I wouldn't attribute it to precision seeing that you've written the same thing twice. Given your previous writings, I suspect that you have a solid understanding of the behaviour of an open-circuited transmission line excited with a step function. Perhaps you could make an attempt at writing a clear description of the behaviour of such a system in terms of charge flow and storage. Since "wave" is a word overloaded with meanings, it would be good not to use it in the description. Once a clear description exists, I can extend it using the same clear terminology to illustrate the points of interest. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
It happens without effort because the output impedance of the Norton/Thevenin equivalent circuit is the same as the circuit it replaces, otherwise it is not an equivalent. False. There can be a large difference in the output impedance of an amplifier designed to drive a 50 ohm load and a 50 ohm Thevenin equivalent circuit. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
On Jan 1, 9:03*pm, Cecil Moore wrote:
Roger wrote: The principles of superposition are mathematically usable, not too hard, *and I think very revealing. *Yes, if we use part of the model, we must use it all the way. *To do otherwise would be error, or worse. Roy and Keith don't seem to realize that the zero source impedance for the ideal voltage source is only when the source is turned off for purposes of superposition. I am not sure you have the methodology quite correct. The source is not turned off; its output is set to 0. It does what every ideal voltage source will do when set to a voltage; maintain that voltage. Through all of this, the impedance of the ideal source remains 0. Now it turns out that an ideal voltage source set to zero volts can be replaced by a short which also has an impedance of 0 and produces no volts. But this does not alter that the ideal source always has an impedance of 0. Analogously, an ideal current source always has an infinite impedance. When set to 0 amps, it behaves exactly like an open circuit. They conveniently avoid turning the source voltage on to complete the other half of the superposition process. When the source signal and the reflected wave are superposed at the series source resistor, where the energy goes becomes obvious. Total destructive interference in the source results in total constructive interference toward the load. See below. You have been a supporter of this theory for a long time. Yes, I have. I am a supporter of the principles and laws of physics. Others believe they can violate the principle of conservation of energy anytime they choose because the principle of conservation of energy cannot be violated - go figure. You should really stop repeating this to yourself. No one is attempting to violate the principle of conservation of energy. By continually repeating this mantra, you convince yourself that you do not need to examine the claims of those who disagree with you. So you do not examine and understand their claims. This seriously limits your capability to learn. If you truly wish to demolish the claims, you should study them in great detail, then write an even better and more persuasive description of the claim than did the original author. Then identify and point out the flaws. As it stands, you do not examine the claims, but immediately coat them with the tar of "violates conservation of energy" or some other mantra and walk away. It does not lead to learning. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
Gene Fuller wrote:
For future reference, however, just remember: Fields first, then power or energy. That's the way superposition really works. Way back before optical physicists could measure light wave fields, they were dealing with reflectance, transmittance, and irradiance - all involving power or energy. They are still using those concepts today proven valid over the past centuries. Optical physicists calculate the fields *AFTER* measuring the power density and they get correct consistent answers. Use whatever method works for you but don't try to change or replace the body of the laws of physics that was in place before your grandfather was born. Your rejection of those laws of physics from the past centuries is why you are so confused today by your steady-state short cuts. It's why Keith doesn't recognize a 1.0 reflection coefficient when it is staring him in the face. It's why Roy rejects energy in reflected waves. Optical physicists have known for centuries where the energy goes. That RF engineers are incapable of performing an energy analysis is sad. Irradiance (intensity) is a power density. Many problems in physics can be solved without even knowing or caring about the strength of the fields. Here is one such lossless line problem for you. 100w--50 ohm line--+--1/2WL 300 ohm line--50 ohm load Pfor1--|--Pfor2 Pref1--|--Pref2 Without using fields, voltages, or currents: Calculate the magnitudes of the four P terms above. Using the RF power reflection-transmission coefficients, please explain the magnitude of Pref1. If you cannot do that, you really need to broaden your horizons and alleviate your ignorance. Quoting HP AN 95-1: "The previous four equations show that s-parameters are simply related to power gain and mismatch loss, quantities which are often of more interest than the corresponding voltage functions." -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
On Jan 1, 4:16*pm, Cecil Moore wrote:
Keith Dysart wrote: The Norton or Thévenin equivalent circuits seem *capable of positive reflection coefficients. * Either can be positive, negative, or zero depending on the value of the output impedance compared to Z0. Would you please quote a reference that addresses the subject of reflection coefficients from Thevenin or Norton equivalent sources? Seshardi, "Fundamentals of transmission lines and electoromagnetic fields", page 19-21, explains reflection diagrams and the generator is styled after Thevenin. Page 21 says RHOg = (Rg-R0)/(Rg+R0) RHOg is the reflection coefficient at the generator and Rg is the value of the output resistor. The exposition has not yet generalized to Z so is still in terms of R. But do not feel limited to Seshardi. Any decent book on transmission lines will cover the subject. Or even easier, google ''"lattice diagram" reflection', "reflection diagram" or "bounce diagram". You will find many examples using Thevenin sources. But this same information has been repeatedly provided and ignored. Will this time be different? ...Keith |
Standing-Wave Current vs Traveling-Wave Current
On Jan 2, 8:39*am, Cecil Moore wrote:
Keith Dysart wrote: It happens without effort because the output impedance of the Norton/Thevenin equivalent circuit is the same as the circuit it replaces, otherwise it is not an equivalent. False. There can be a large difference in the output impedance of an amplifier designed to drive a 50 ohm load and a 50 ohm Thevenin equivalent circuit. Then your Thevenin circuit is not an equivalent for the amplifier, is it? Please study "equivalent circuit" and report back. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
With two measurements you will obtain the correct impedance. With two measurements you will obtain an impedance. It will not be the impedance needed to calculate the reflection coefficient seen by the reflected waves. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
On Jan 2, 9:24*am, Cecil Moore wrote:
Keith Dysart wrote: With two measurements you will obtain the correct impedance. With two measurements you will obtain an impedance. It will not be the impedance needed to calculate the reflection coefficient seen by the reflected waves. You should really spend some time looking for a reference to support your assertion that "It will not be the impedance needed to calculate the reflection coefficient seen by the reflected waves." You will not find one. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
The impedance of the Thevenin/Norton equivalent source is not V/I but rather the slope of the line representing the relationship of the voltage to the current. However, after the interference patterns have been established, the reflected waves do not encounter that source impedance. That is why the reflection coefficient seen by the reflected waves is relatively unrelated to the value of resistance in a Thevenin equivalent circuit. You need to complete step 3 of the superposition process to realize exactly what is happening. Reference the irradiance equation from the field of EM wave optics to ascertain the interference levels. What do you have to lose by alleviating your ignorance? Keith, until you take time to understand destructive and constructive interference, you will never understand what is happening inside a source and will be forever confused by your blinders-on-come-hell-or-high-water method of thinking. Optical physicists figured out a couple of centuries ago exactly what you are wrestling with now. Your present problem was already solved before your grandfather was born. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
Can you make this all work for a pulse, or a step function? Please reference a good book on optical EM waves for a complete answer. It is *not me* making it work. It is a body of physics knowledge that has existed since long before you were born. It should have been covered in your Physics 201 class. That you are apparently unaware of such is a display of basic ignorance of the science of EM waves. The basic theory applies specifically to coherent waves (which are the only EM waves capable of truly interfering). CW RF waves are close enough to ideal coherency that the theory works well. It would no doubt work for a coherent Fourier series as well but I don't want to spend the time necessary to prove that assertion. How do you compute Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) for a pulse or a step? The above equation applies to coherent signals. It is known not to work when the signals are not coherent because the angle 'A' never reaches a fixed steady-state value. Or is your approach limited to sinusoids? Again, it is not *my* approach and is described in any textbook on "Optics" including Hecht and Born & Wolf. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
On Jan 2, 9:38*am, Cecil Moore wrote:
Keith Dysart wrote: The impedance of the Thevenin/Norton equivalent source is not V/I but rather the slope of the line representing the relationship of the voltage to the current. However, after the interference patterns have been established, the reflected waves do not encounter that source impedance. That is why the reflection coefficient seen by the reflected waves is relatively unrelated to the value of resistance in a Thevenin equivalent circuit. I assume that you have not provided a reference to support this assertion because you have not been able to find one. You need to complete step 3 of the superposition process to realize exactly what is happening. Reference the irradiance equation from the field of EM wave optics to ascertain the interference levels. What do you have to lose by alleviating your ignorance? Unfortunately optics do not do well at explaining transmission lines since they do not extend down to DC. Keith, until you take time to understand destructive and constructive interference, you will never understand what is happening inside a source and will be forever confused by your blinders-on-come-hell-or-high-water method of thinking. Optical physicists figured out a couple of centuries ago exactly what you are wrestling with now. Your present problem was already solved before your grandfather was born. I have yet to find anything about transmission lines that needs constructive and destructive interference for explanation. Volts, amps and superposition seem to be able to do it all, and have the added benefit of explaining the behaviour for step functions and pulses. With the volts, amps and superposition, sinusoids are just a special case of the general one. I am unsure why some are content to constrain themselves to solution techniques and explanations that only work on the special case of sinusoids. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
On Jan 2, 9:59*am, Cecil Moore wrote:
Keith Dysart wrote: Can you make this all work for a pulse, or a step function? I accept your "NO" and agree that EM waves are incapable of providing solutions for pulse or step excitation. But why don't you just say so clearly. Please reference a good book on optical EM waves for a complete answer. Given that optical EM waves are only capable of solving a subset of the uses of transmission lines, it is not obvious why I should study them when I can invest in learning approaches that will do the whole job. It is *not me* making it work. True. And as you have said, it does not work for pulses or steps. It is a body of physics knowledge that has existed since long before you were born. It should have been covered in your Physics 201 class. That you are apparently unaware of such is a display of basic ignorance of the science of EM waves. Some who claim to have studied them thoroughly seem to be constrained by their limitations. Is that better? The basic theory applies specifically to coherent waves (which are the only EM waves capable of truly interfering). CW RF waves are close enough to ideal coherency that the theory works well. It would no doubt work for a coherent Fourier series as well but I don't want to spend the time necessary to prove that assertion. How do you compute * Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) for a pulse or a step? The above equation applies to coherent signals. It is known not to work when the signals are not coherent because the angle 'A' never reaches a fixed steady-state value. Or is your approach limited to sinusoids? Again, it is not *my* approach and is described in any textbook on "Optics" including Hecht and Born & Wolf. Well, others more knowledgeable than I in optics have disputed whether *your* approach accurately represents those described in the textbooks. In any case, being applicable only to sinusoids limits the general applicability to transmission lines which happily work at DC. ...Keith |
Standing-Wave Current vs Traveling-Wave Current
Roger Sparks wrote:
Well, I would like to think I could understand it, but maybe something is wrong like Roy suggests. I think it is all falling into place, but our readers are not all in agreement. Anyone who will take the time to understand and is capable of understanding, will understand. What I am reporting are centuries-old proven scientific concepts. Readers need only reference an optics textbook for irradiance, reflectance, and transmittance. "Optics", by Hecht is easy reading. Is this a Thevenin source? If so, what is the internal resistor set to in terms of Z0? The example was a Thevenin source driving a 1/2WL open line through a source resistor equal to Z0, e.g. 50 ohms. For condition number 2 below, is this a Thevenin equivalent resistance? Yes, all the rules of superposition and Thevenin equivalency are being followed. 1. To add the total power of each wave and then also add the power reduction of the phase relationship, would be creating power unless the cos(A) is negative. Something seems wrong here, probably my understanding. Yep, I haven't given the other corresponding equation. The pair of equations that satisfy the conservation of energy requirements are (in Ramo&Whinnery Poynting vector notation): |Pz+| = P1 + P2 + 2*SQRT(P1*P2)cos(A) eq.1 |Pz-| = P3 + P4 + 2*SQRT(P3*P4)cos(-A) eq.2 In a transmission line, the Poynting vector for Pz+ and the Poynting vector for Pz- are pointed in opposite directions. The destructive interference, as you observed, carries a negative sign. This negative sign does NOT create energy. It means that some excess amount of destructive interference energy must go somewhere. It naturally goes into constructive interference somewhere else. Please see: http://www.mellesgriot.com/products/optics/oc_2_1.htm http://micro.magnet.fsu.edu/primer/j...ons/index.html 2. The two voltages should be equal, therfore the power delivered by each to the source resister would be equal. Under each of the two superposition steps, yes. However, when the signals are combined, interference results, and the power in the source resistor is reduced to zero. That "excess" energy must go somewhere and it appears as constructive interference redistributed toward the load. See the web pages above. 3. The power delivered to the source resistor will arrive a different times due to the phase difference between the two waves. We are talking average power. Please don't get bogged down in instantaneous power which would be the same when integrated. 4. If the reflected power returns at the same time as the delivered power (to the source resistor), no power will flow. This because the resistor in a Thevenin is a series resistor. Equal voltages will be applied to each side of the resitor. The voltage difference will be zero. Yes, the result is destructive interference. 5. If the reflected power returns 180 degrees out of phase with the applied voltage (to the source resistor), the voltage across the resistor will double with each cycle, resulting in an ever increasing current.(and power) into the reflected wave. Good time to switch over to a Norton equivalent. This is correct for this condition. The problem with the equation comes when cos(90) which can easily happen. That's no problem at all. It is just constructive interference which implies destructive interference somewhere else. Destructive interference must always equal constructive interference to avoid violation of the conservation of energy principle. Light, especially solar energy, is a collection of waves of all magnitudes and frequencies. We should see only about 1/2 of the power that actually arrives due to superposition. The reflection from a surface should create standing waves in LOCAL space that will contain double the power of open space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be correct for a collection of waves, but incorrect for the example. A collection of waves is not likely to be coherent so the equation would not work. That equation works for any coherent EM wave. CW RF waves are coherent. I tried to read your article a couple of weeks ago, but I found myself not understanding. I have learned a lot since then thanks to you, Roy, and others. I will try to read it again soon. You might want to pick up a copy of "Optics", by Hecht available from http://www.abebooks.com Please pay close attention to the chapters on interference and superposition. So what do you think Cecil? Also try HP's AN 95-1, an s-parameter app-note available on the web. Pay close attention to what they say about power and what happens when you square the s-parameter equations. Surprise! You get the power interference equation above. -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Roger wrote:
The resistor in the Thevenin/Norton equivalent source is selected with some criteria in mind. What I would like to do is to design a Thevenin source to provide 1v across a 50 ohm transmission line INFINATELY long, ignoring ohmic resistance. I would like the source resistor to absorb as much power as the line, so that if power ever returns under reflected wave conditions, it can all be absorbed by the resistor. I think the size of such a resistor will be 50 ohms. If you allow reflected energy to flow through the source resistor, destructive interference is often the result and the resistor therefore will not dissipate the reflected power. The only way to get the source resistor to dissipate all of the reflected power is to cause total constructive interference within the source. That requires all of the energy from the load side of the system. When the source rejects reflected energy due to destructive interference, that energy has no choice except to reverse direction and flow back toward the load as constructive interference energy. Maybe you should reconsider the circulator+load-resistor? -- 73, Cecil http://www.w5dxp.com |
Standing-Wave Current vs Traveling-Wave Current
Follow up and more corrections.
Roger wrote: Correction. Please note the change below. I apologize for the error. Roger Sparks wrote: "Cecil Moore" wrote in message ... Roger wrote: The principles of superposition are mathematically usable, not too hard, and I think very revealing. Yes, if we use part of the model, we must use it all the way. To do otherwise would be error, or worse. clip some Roger, I have explained all of this before. If you are capable of understanding it, I take my hat off to you. Here's what happens in that ideal voltage source using the rules of superposition. Well, I would like to think I could understand it, but maybe something is wrong like Roy suggests. I think it is all falling into place, but our readers are not all in agreement. Could I ask a couple of questions to make sure I am understanding your preconditions? Is this a Thevenin source? If so, what is the internal resistor set to in terms of Z0? 1. With the reflected voltage set to zero, the source current is calculated which results in power in the source resistor. Psource = Isource^2*Rsource For condition number 2 below, is this a Thevenin equivalent resistance? 2. With the source voltage set to zero, the reflected current is calculated which results in power in the source resistor. Pref = Iref^2*Rsource Now superpose the two events. The resultant power equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) where 'A' is the angle between the source current and the reflected current. Note this is NOT power superposition. It is the common irradiance (power density) equation from the field of optics which has been in use in optical physics for a couple of centuries so it has withstood the test of time. OK. A few reactions. 1. To add the total power of each wave and then also add the power reduction of the phase relationship, would be creating power unless the cos(A) is negative. Something seems wrong here, probably my understanding. 2. The two voltages should be equal, therfore the power delivered by each to the source resister would be equal. 3. The power delivered to the source resistor will arrive a different times due to the phase difference between the two waves. 4. If the reflected power returns at the same time as the delivered power (to the source resistor), no power will flow. This because the resistor in a Thevenin is a series resistor. Equal voltages will be applied to each side of the resitor. The voltage difference will be zero. 5. If the reflected power returns 180 degrees out of phase with the applied voltage (to the source resistor), the voltage across the resistor will double with each cycle, resulting in an ever increasing current.(and power) into the reflected wave. For the voltage source looking into a 1/2WL open circuit stub, angle 'A' is 180 degrees. Therefore the total power in the source resistor is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0 This is correct for this condition. The problem with the equation comes when cos(90) which can easily happen. I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is the angle between positive peaks of the two waves. This angle will rotate twice as fast as the signal frequency due to the relative velocity between the waves.. No, this angle is fixed by system design. If the system changes, this angle will rotate twice as fast as the angle change of a single wave. The equation should be OK. The equation can NOT be OK for anything except a single interaction point. It can not be used to plot the interaction between waves because the wave location information is not present. Ptot in the source resistor is zero watts but we already knew that. What is important is that the last term in that equation above is known as the "interference term". When it is negative, it indicates destructive interference. When Ptot = 0, we have "total destructive interference" as defined by Hecht in "Optics", 4th edition, page 388. Whatever the magnitude of the destructive interference term, an equal magnitude of constructive interference must occur in a different direction in order to satisfy the conservation of energy principle. Thus we can say with certainty that the energy in the reflected wave has been 100% re-reflected by the source. The actual reflection coefficient of the source in the example is |1.0| even when the source resistor equals the Z0 of the transmission line. I have been explaining all of this for about 5 years now. Instead of attempting to understand these relatively simple laws of physics from the field of optics, Roy ploinked me. Hopefully, you will understand. All of this is explained in my three year old Worldradio energy article on my web page. http://www.w5dxp.com/energy.htm -- 73, Cecil http://www.w5dxp.com Light, especially solar energy, is a collection of waves of all magnitudes and frequencies. We should see only about 1/2 of the power that actually arrives due to superposition. The reflection from a surface should create standing waves in LOCAL space that will contain double the power of open space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be correct for a collection of waves, but incorrect for the example. I tried to read your article a couple of weeks ago, but I found myself not understanding. I have learned a lot since then thanks to you, Roy, and others. I will try to read it again soon. So what do you think Cecil? 73, Roger, W7WkB In the last 24 hours, Roy posted a revised analysis that contains results useful here. He presented a voltage example that resulted in a steady state with steady state voltages 4 time initial value. Under superposition, this would equate to 4 times the initial power residing on the transmission line under the conditions presented. This concurs with other authors who predict power on the transmission line may exceed the delivered power due to reflected waves. Your equation "Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)" is taken from illumination and radiation theory to describe power existing at a point in space near a reflecting surface. If we consider space to be a transmission media, and the reflecting surface to be a discontinuity in the transmission media, then we have a situation very similar to an electrical transmission line near a line discontinuity. It is entirely reasonable to consider that the reflection ratio between the space transmission media and the reflective surface would result in an storage factor equaling 4 times the peak power of the initial forward wave. By storage factor, I simply mean the ratio of forward power to total power on the transmission media under standing wave conditions. Under open circuit conditions, a half wavelength transmission line will have a storage factor of 2. Roy presented an example where the storage factor was 4. Perhaps the space transmission media also has a storage factor of 4 under some conditions, as described by "Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)". 73, Roger, W7WKB |
Standing-Wave Current vs Traveling-Wave Current
Keith Dysart wrote:
On Jan 2, 1:41 am, Roger wrote: Keith Dysart wrote: On Jan 1, 6:00 pm, Roger wrote: Roy Lewallen wrote: Roger wrote: Could you better describe how you determine that the source has a Z0 equal to the line Z0? I can guess that you use a Thévenin equivalent circuit and set the series resistor to Z0. Probably the simplest way is to put the entire source circuitry into a black box. Measure the terminal voltage with the box terminals open circuited, and the current with the terminals short circuited. The ratio of these is the source impedance. If you replace the box with a Thevenin or Norton equivalent, this will be the value of the equivalent circuit's impedance component (a resistor for most of our examples). If the driving circuitry consists of a perfect voltage source in series with a resistance, the source Z will be the resistance; if it consists of a perfect current source in parallel with a resistance, the source Z will be the resistance. You can readily see that the open circuit V divided by the short circuit I of these two simple circuits equals the value of the resistance. The power output of the Thévenin equivalent circuit follows the load. Sorry, I don't understand this. Can you express it as an equation? There seems to be some confusion as to the terms "Thévenin equivalent circuit", "ideal voltage source", and how impedance follows these sources. Two sources we all have access to are these links: Voltage source: http://en.wikipedia.org/wiki/Voltage_source Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem I don't disagree with anything I read there. But you may not quite have the concept of impedance correct. The impedance of the Thevenin/Norton equivalent source is not V/I but rather the slope of the line representing the relationship of the voltage to the current. When there is a source present, this line does not pass through the origin. Only for passive components does this line pass through the origin in which case it becomes V/I. Because the voltage to current plot for an ideal voltage source is horizontal, the slope is 0 and hence so is the impedance. For an ideal current source the slope is vertical and the impedance is infinite. Hoping this helps clarify.... ...Keith Yes, I can understand that there is no change in voltage no matter what the current load is, so there can be no resistance or reactive component in the source. The ideal voltage link also said that the ideal source could maintain voltage no matter what current was applied. Presumably a negative current through the source would result in the same voltage as a positive current, which is logical if the source has zero impedance. This is true. During the transition from negative current passing through the source to positive current passing through the source, the current at some time must be zero. This would occur, for example, when the output terminals are open circuited. Or connected to something that had the same open-circuit voltage as your source. How is the impedance of the perfect source defined at this zero current point? This is not a different question than: How is the resistance of a resistor defined when it has no current flowing in it? It continues to satisfy the relation V = I * R, though one can not compute R from the measured V and I. How is the impedance of the attached system defined? It is unknown, but has a voltage equal to the voltage of the source. The resistor in the Thevenin/Norton equivalent source is selected with some criteria in mind. What I would like to do is to design a Thevenin source to provide 1v across a 50 ohm transmission line INFINATELY long, ignoring ohmic resistance. I would like the source resistor to absorb as much power as the line, so that if power ever returns under reflected wave conditions, it can all be absorbed by the resistor. I think the size of such a resistor will be 50 ohms. That is the correct value to not have any reflections at the source. But do not expect the power dissipated in the resistor to increase by the same amount as the "reflected power". In general, it will not. This is what calls into question whether the reflected wave actually contains energy. Do some simple examples with step functions. The math is simpler than with sinusoids and the results do not depend on the phase of the returning wave, but simply on when the reflected step arrives bach at the source. Examine the system with the following terminations on the line: open, shorted, impedance greater than Z0, and impedance less than Z0. Because excitation with a step function settles to the DC values, the final steady state condition is easy to compute. Just ignore the transmission line and assume the termination is connected directly to the Thevenin generator. When the line is present, it takes longer to settle, but the final state will be the same with the line having a constant voltage equal to the voltage output of the generator which will be the same as the voltage applied to the load. Then do the same again, but use a Norton source. You will find that conditions which increase the dissipation in the resistor of the Thevenin equivalent circuit reduce the dissipation in the resistor of the Norton equivalent circuit and vice versa. This again calls into question the concept of power in a reflected wave, since there is no accounting for where that "power" goes. ...Keith Very interesting! It makes sense to me. I must be gone most of the day today so will be QRT for a while. I am gaining an appreciation for the time boundaries here. To properly account for the power, we would need to integrate the source power over the entire time from switch on to switch off plus the time required for the reflected wave to completely return and die out. If we consider the extended time period, within that period the power delivered by the source should equal the power dissipated by the resistor. The forward and reflected waves should only serve to be storage for the power during portions of the extended time period. 73, Roger, W7WKB |
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