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On Mon, 14 Apr 2008 09:10:20 -0500
Cecil Moore wrote: Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Your equation above completely ignores reflections which is the subject of the thread. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Even my ten year old grandson can solve that problem and he's no future rocket scientist. -- 73, Cecil http://www.w5dxp.com This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. My view is that any source must both absorb and deliver power at some none zero impedance. As justification for this view, I offer that current always flows from high voltage to lower voltage, so a real voltage source would have to absorb energy if the external voltage exceeded the voltage of the voltage source. While it can be agrued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers power into the source. Charging a battery with zero internal resistance is a good example. Another example is the observation that a generator becomes a motor when the externally suppied voltage exceeds the voltage supplied by the generator. Yes, we can make the assumption that the voltage source can not absorb power at any time, but the assumption takes us into an unreal world and gives answers that are impossible to duplicate with measurements. Some would call that a world of science fiction. -- 73, Roger, W7WKB |
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On Apr 14, 10:10*am, Cecil Moore wrote:
Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: * *Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Ahhh, but there is then agreement that energy flows (power) must be in balance to satisfy conservation of energy. Your equation above completely ignores reflections which is the subject of the thread. That it does. We get to that later. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. They are quite different. I am quite will to explore (and am doing so) the concept of imputed energy flow in the reflected wave. It is just that it comes up short since there is no explanation of where the energy goes. Were an adequate explanation to be offered, I would quite accept it. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. You should tread back through the posts, the question was answered. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. Three! The source can also take energy from the system. Since you have overlooked the source, the rest of your post is quite flawed in its conclusions. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Well that is the question, isn't it? It could be in the source. Or, if it can not be determined where the energy in Pr goes, then the only other answer is that Pr does not represent an energy flow. Think Sherlock: "when the impossible has been eliminated the residuum, however improbable, must contain the truth." Even my ten year old grandson can solve that problem and he's no future rocket scientist. Ah yes, but was he presented with ALL the options? ...Keith |
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Roger Sparks wrote:
While it can be argued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers power into the source. I thought I had already addressed that topic when I added the one-wavelength of transmission line to the example in between the source and source resistance. But here's an example that may allow better tracking of the energy flow. Let's modify my Part 1, Fig. 1-1 to add a 50 ohm circulator and load to the ground leg of the source. Everything else remains the same. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL \ / 3 | 50 ohms | GND How much power is dissipated in the circulator resistor? How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? Does this example answer your questions? -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
Ahhh, but there is then agreement that energy flows (power) must be in balance to satisfy conservation of energy. You can probably answer your own question by figuring out how the light energy in an interference pattern gets from the dark ring to the bright ring some distance away. Do you think it happens faster than the speed of light? It is just that it comes up short since there is no explanation of where the energy goes. Were an adequate explanation to be offered, I would quite accept it. I am satisfied with the destructive/constructive interference explanation. That you have come up short in tracking those component energies is not unexpected given your prejudices. If you told me your car disappeared from existence because you cannot find it, I wouldn't believe you either. You should tread back through the posts, the question was answered. "It depends" was no answer - that was just mealy-mouthing. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. Three! The source can also take energy from the system. The Vs source has zero resistance rendering dissipation impossible. I repeat: There are only two sources of power *DISSIPATION* in the entire system, the load resistor and the source resistor. The source can certainly throttle back its output when there is destructive interference in the source resistor and increase its output when there is constructive interference in the source resistor, but it CANNOT dissipate any power. Since you have overlooked the source, the rest of your post is quite flawed in its conclusions. Nope, you are confused. The source can adjust its output but the source cannot *DISSIPATE* power. As I said, the only sources of power *DISSIPATION* are the two resistors. No amount of obfuscation is going to change that. Perhaps adding a circulator to my Part 1, Fig. 1-1 will allow you to see things in a clearer light. Of course, using light would be even better. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL \ / 3 | 50 ohms | GND How much power is dissipated in the circulator resistor? How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? -- 73, Cecil http://www.w5dxp.com |
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On Apr 14, 12:06*pm, Roger Sparks wrote:
On Mon, 14 Apr 2008 09:10:20 -0500 Cecil Moore wrote: Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: * *Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Your equation above completely ignores reflections which is the subject of the thread. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Even my ten year old grandson can solve that problem and he's no future rocket scientist. -- 73, Cecil *http://www.w5dxp.com This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. *My view is that any source must both absorb and deliver power at some none zero impedance. * I am not sure why you desire a non-zero impedance. The usual definition of an ideal voltage source is that it provides or sinks what ever current is needed to hold the desired output voltage. When it is sourcing current then it is providing energy. No statement is made about where this energy comes from. When it is sinking current, it is absorbing energy. No statement is made about where this energy is going. A non-zero impedance is not required to make any of the above behaviour work. If you include a non-zero impedance, then you have a more real world source which can often be modeled using the Thevenin equivalent circuit; an ideal voltage source (zero impedance) in series with a resistor representing the impedance of the real world source. As justification for this view, I offer that current always flows from high voltage to lower voltage, so a real voltage source would have to absorb energy if the external voltage exceeded the voltage of the voltage source. This is true. While it can be agrued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers *power into the source. *Charging a battery with zero internal resistance is a good example. *Another example is the observation that a generator becomes a motor when the externally suppied voltage exceeds the voltage supplied by the generator. Yes, we can make the assumption that the voltage source can not absorb power at any time, but the assumption takes us into an unreal world and gives answers that are impossible to duplicate with measurements. *Some would call that a world of science fiction. ...Keith |
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On Apr 14, 1:36*pm, Cecil Moore wrote:
Keith Dysart wrote: Ahhh, but there is then agreement that energy flows (power) must be in balance to satisfy conservation of energy. You can probably answer your own question by figuring out how the light energy in an interference pattern gets from the dark ring to the bright ring some distance away. Do you think it happens faster than the speed of light? What is your explanation for this phenomenon? It is just that it comes up short since there is no explanation of where the energy goes. Were an adequate explanation to be offered, I would quite accept it. I am satisfied with the destructive/constructive interference explanation. That you have come up short in tracking those component energies is not unexpected given your prejudices. I suppose. If you are happy with energy equations that don't balance. If you told me your car disappeared from existence because you cannot find it, I wouldn't believe you either. You should tread back through the posts, the question was answered. "It depends" was no answer - that was just mealy-mouthing. Except, that it does depend. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. Three! The source can also take energy from the system. The Vs source has zero resistance rendering dissipation impossible. I repeat: There are only two sources of power *DISSIPATION* in the entire system, the load resistor and the source resistor. You have to read more carefully. I did not use the word dissipation. This is because there are ways other than dissipation to remove energy from the system. Just as the source provides energy and we do not care where it comes from, it can remove energy and we do not care where it goes. Examine Ps(t). You will find that for some of the time energy is being absorbed by the source. This occurs when the sign of Ps(t) is negative. The source can certainly throttle back its output when there is destructive interference in the source resistor and increase its output when there is constructive interference in the source resistor, but it CANNOT dissipate any power. Since we don't know the internals of the source, we do not know if it is dissipating or not. But we do know that when the sign of Ps(t) is negative, the source is absorbing energy from the system, exactly analagous to it providing energy when the sign of Ps(t) is positive. Since you have overlooked the source, the rest of your post is quite flawed in its conclusions. Nope, you are confused. The source can adjust its output but the source cannot *DISSIPATE* power. As I said, the only sources of power *DISSIPATION* are the two resistors. No amount of obfuscation is going to change that. You really should rethink this a bit. When current flows into a voltage source, the voltage source is absorbing energy. Perhaps adding a circulator to my Part 1, Fig. 1-1 will allow you to see things in a clearer light. Of course, using light would be even better. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL * * * *\ / * * * * 3 * * * * | * * *50 ohms * * * * | * * * *GND How much power is dissipated in the circulator resistor? This circulator, while often used, does not in any way add clarity. Changing the circuit changes the results, especially when you add a circulator which alters rather dramatically the energy flows. How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? This depends on the design of the generator and the length of the line. With a shorted line that is 90 degrees long, and a generator constructed using the circuit of Fig 1-1, the source supplies no energy to maintain an imputed forward power of 50 watts on the line. ...Keith |
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Keith Dysart wrote:
When it is sourcing current then it is providing energy. No statement is made about where this energy comes from. The question is: Is that energy being created or dissipated as needed according to your omnipotent whims? -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
What is your explanation for this phenomenon? You first! :-) I suppose. If you are happy with energy equations that don't balance. My energy equations balance perfectly. Yours are the energy equations that don't balance in violation of the conservation of energy principle. "It depends" was no answer - that was just mealy-mouthing. Except, that it does depend. Sounds more like religion than anything else. You have to read more carefully. I did not use the word dissipation. You have to read more carefully. I used the word "dissipation" and you disagreed with me. Please read it again to verify that fact. Since we don't know the internals of the source, we do not know if it is dissipating or not. Sorry, the source is, by definition, lossless. All of the source dissipation is lumped in the source resistance drawn separately on the diagram as Rs. You really should rethink this a bit. When current flows into a voltage source, the voltage source is absorbing energy. But the source is NOT *DISSIPATING* energy because Rs is not inside the source. Rs is clearly drawn outside the lossless source so the source generates *ZERO* heat. I am amazed at the lengths to which you will go to try to obfuscate the discussion. Changing the circuit changes the results, ... When you get more serious about the technical facts of physics than you are about saving face, please get back to us, but please, not before. Everyone is tired of your delusions of grandeur where the laws of physics obey your every whim. -- 73, Cecil http://www.w5dxp.com |
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On Mon, 14 Apr 2008 16:54:47 GMT
Cecil Moore wrote: Roger Sparks wrote: While it can be argued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers power into the source. I thought I had already addressed that topic when I added the one-wavelength of transmission line to the example in between the source and source resistance. I thought the addition of a one wavelength transmission line did not address the issue, and only added more reflections. We still need a reason to assume that a voltage source should *not* absorb power. But here's an example that may allow better tracking of the energy flow. Let's modify my Part 1, Fig. 1-1 to add a 50 ohm circulator and load to the ground leg of the source. Everything else remains the same. Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL \ / 3 | 50 ohms | GND How much power is dissipated in the circulator resistor? How much power does the source have to supply to maintain 50 watts of forward power on the transmission line? Does this example answer your questions? -- 73, Cecil http://www.w5dxp.com No, I'm sorry but no. I offered the examples of two real sources that will absorb power when the returning voltage exceeds the output voltage (a battery and a generator turned into a motor). I think that we must allow our voltage source to have that same real property. I do understand that when we allow the source to receive power, then we need to address source impedance. If we assign a single impedance, then we expect reflections from the source. The simple solution that I propose is to add a source property of absorbing all reflections. This can be accomplished in the real world by making the transmission so long that reflections never return from the source over any reasonable time, or by making the tranmission line sufficiently lossy to absorb reflections. Your example uses the first method. Does the idea of source receiving power run counter to what you were planning to write in Parts 2 and 3? I am trying to understand why you have such great reluctance to accept that the source could receive power for part of a cycle, especially when it could easily bring the instantaneous power and energy calculations into balance. -- 73, Roger, W7WKB |
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On Mon, 14 Apr 2008 13:40:20 -0700 (PDT)
Keith Dysart wrote: On Apr 14, 12:06*pm, Roger Sparks wrote: On Mon, 14 Apr 2008 09:10:20 -0500 Cecil Moore wrote: Keith Dysart wrote: All the elements of the system are completely specified in Fig 1-1 and we used circuit theory to compute the energy flows. Not surprisingly, they completely balanced: * *Ps(t) = Prs(t) + Pg(t) Yes, but that is only *NET* energy flow and says nothing about component energy flow. Everything is already known about net energy flow and there are no arguments about it so you are wasting your time. Your equation above completely ignores reflections which is the subject of the thread. You object to me being satisfied with average energy flow while you satisfy yourself with net energy flow. I don't see one iota of conceptual difference between our two positions. After hundreds of postings, all you have proved is that Eugene Hecht was right when he said instantaneous powers are "of limited utility", such that you cannot even tell me how many joules there are in 100 watts of instantaneous power when it is the quantity of those very joules that are required to be conserved and not the 100 watts. The limit in your quest for tracking instantaneous energy is knowing the position and momentum of each individual electron. Good luck on that one. I am going to summarize the results of my Part 1 article and be done with it. In the special case presented in Part 1, there are only two sources of power dissipation in the entire system, the load resistor and the source resistor. None of the reflected energy is dissipated in the load resistor because the chosen special conditions prohibit reflections from the source resistor. Therefore, all of the energy not dissipated in the load resistor is dissipated in the source resistor because there is no other source of dissipation in the entire system. Only RL and Rs exist. Pr is not dissipated in RL. Where is Pr dissipated? Even my ten year old grandson can solve that problem and he's no future rocket scientist. -- 73, Cecil *http://www.w5dxp.com This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. *My view is that any source must both absorb and deliver power at some none zero impedance. * I am not sure why you desire a non-zero impedance. The usual definition of an ideal voltage source is that it provides or sinks what ever current is needed to hold the desired output voltage. When it is sourcing current then it is providing energy. No statement is made about where this energy comes from. When it is sinking current, it is absorbing energy. No statement is made about where this energy is going. A non-zero impedance is not required to make any of the above behaviour work. My thought needed more developement. When the source delivers power, we readily accept that the impedance of delivery will be the impedance of the attached circuit. We make the same assumption when a reflection is returned to the source. If we make the assumption that the source has the same impedance as the refection, then no reflection from the source is expected. So yes, I agree with your observation. If you include a non-zero impedance, then you have a more real world source which can often be modeled using the Thevenin equivalent circuit; an ideal voltage source (zero impedance) in series with a resistor representing the impedance of the real world source. -- 73, Roger, W7WKB |
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Roger Sparks wrote:
I offered the examples of two real sources that will absorb power when the returning voltage exceeds the output voltage (a battery and a generator turned into a motor). I think that we must allow our voltage source to have that same real property. A battery converts electrical energy to chemical energy, i.e. it transforms the electrical energy. A motor converts electrical energy into physical work, i.e. it transforms the electrical energy. An ideal source does not dissipate power and there is no mechanism for storing energy. It seems what you are objecting to is the artificial separation of Vs and Rs. I do understand that when we allow the source to receive power, then we need to address source impedance. The series source impedance is zero. It acts like a short circuit to reflections, i.e. there are no reflections. However, there seem to be 100% reflection from the GND on the other side of the source. Does the idea of source receiving power run counter to what you were planning to write in Parts 2 and 3? The source will be shown to adjust its output until an energy balance is achieved. It will throttle back when destructive interference occurs at the source resistor and will gear up when constructive interference requires more energy. I am trying to understand why you have such great reluctance to accept that the source could receive power for part of a cycle, especially when it could easily bring the instantaneous power and energy calculations into balance. There is no known mechanism that would allow an ideal source to dissipate or store energy. Consider that the energy you see flowing back into the source is reflected back through the source by the ground on the other side and becomes part of the forward wave out of the source. That would satisfy the distributed network model and explain why interference exists in the source. -- 73, Cecil http://www.w5dxp.com |
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On Apr 14, 5:28*pm, Cecil Moore wrote:
Keith Dysart wrote: What is your explanation for this phenomenon? You first! :-) I suppose. If you are happy with energy equations that don't balance. My energy equations balance perfectly. Yours are the energy equations that don't balance in violation of the conservation of energy principle. "It depends" was no answer - that was just mealy-mouthing. Except, that it does depend. Sounds more like religion than anything else. You have to read more carefully. I did not use the word dissipation. You have to read more carefully. I used the word "dissipation" and you disagreed with me. Please read it again to verify that fact. Since we don't know the internals of the source, we do not know if it is dissipating or not. Sorry, the source is, by definition, lossless. All of the source dissipation is lumped in the source resistance drawn separately on the diagram as Rs. You really should rethink this a bit. When current flows into a voltage source, the voltage source is absorbing energy. But the source is NOT *DISSIPATING* energy because Rs is not inside the source. Rs is clearly drawn outside the lossless source so the source generates *ZERO* heat. I am amazed at the lengths to which you will go to try to obfuscate the discussion. You do need to back up a bit and review voltage sources. When conventional current is flowing out of the positive terminal of a voltage source, it is usually agreed that the voltage source is providing energy to the circuit. This comes from P(t) = V(t) * I(t) when V and I are the same sign, P is positive representing a flow of energy from the source to the other elements of the circuit. What do you explain is happening when conventional current is flowing into the positive terminal of the source? Is the source still providing energy to the circuit now that P is negative? Or is it accepting energy from the circuit, the negative P representing an energy flow into the source? ...Keith |
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Cecil Moore wrote:
A battery converts electrical energy to chemical energy, i.e. it transforms the electrical energy. Gee, I'm just a silly old engineer, but I thought batteries converted chemical energy to electrical energy. Unless you are perhaps speaking of charging it. tom K0TAR |
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On Tue, 15 Apr 2008 01:22:11 GMT
Cecil Moore wrote: Roger Sparks wrote: I offered the examples of two real sources that will absorb power when the returning voltage exceeds the output voltage (a battery and a generator turned into a motor). I think that we must allow our voltage source to have that same real property. A battery converts electrical energy to chemical energy, i.e. it transforms the electrical energy. A motor converts electrical energy into physical work, i.e. it transforms the electrical energy. An ideal source does not dissipate power and there is no mechanism for storing energy. It seems what you are objecting to is the artificial separation of Vs and Rs. No, the separation of Vs and Rs was made to better understand why no interference would occur in Figure 1-1. found at http://www.w5dxp.com/nointfr.htm. Here is a quote from Part 1. "4. Since the transmission line is 1/8 wavelength (45 degrees) long and the load is purely resistive, the reflected wave incident upon the source resistor will be 2(45) = 90 degrees out of phase with the forward wave at the source resistor. This is the necessary and sufficient condition to produce zero interference at the source resistor." The problem is that the source and reflected waves behave as two power sources out of time by 90 degrees. As a result, the current flows as the result of two sine waves, and can be described by only one sine wave. The one sine wave description necessarily shows that power *does* flow into the source during part of the cycle. Interference techniques are used to combine the two sine waves into one wave so it would appear that statement 4 is incorrect. I do understand that when we allow the source to receive power, then we need to address source impedance. The series source impedance is zero. It acts like a short circuit to reflections, i.e. there are no reflections. However, there seem to be 100% reflection from the GND on the other side of the source. It is not the reflections from the source that is the root of the problem. The root is the way two sine waves combine into one wave that runs at a third phase compared to either of the source waves. Does the idea of source receiving power run counter to what you were planning to write in Parts 2 and 3? The source will be shown to adjust its output until an energy balance is achieved. It will throttle back when destructive interference occurs at the source resistor and will gear up when constructive interference requires more energy. I am trying to understand why you have such great reluctance to accept that the source could receive power for part of a cycle, especially when it could easily bring the instantaneous power and energy calculations into balance. There is no known mechanism that would allow an ideal source to dissipate or store energy. Consider that the energy you see flowing back into the source is reflected back through the source by the ground on the other side and becomes part of the forward wave out of the source. That would satisfy the distributed network model and explain why interference exists in the source. I can understand a voltage source that throttles up and down but I can't understand why the throttle all has to be on the plus side. What logic prevents the power from returning to the ideal source from whence it just left? Our real limit is that only one current can flow for only one voltage for each instant at any place in the circuit. This is how we justify a "one sine wave" description. It is why whenever we have a reflection, we also have interference. It is also the reason that we must have power flowing back into the source for part of the cycle. -- 73, Roger, W7WKB |
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Tom Ring wrote:
Cecil Moore wrote: A battery converts electrical energy to chemical energy, i.e. it transforms the electrical energy. Gee, I'm just a silly old engineer, but I thought batteries converted chemical energy to electrical energy. Unless you are perhaps speaking of charging it. Here was the preceding comment in context which you trimmed: "I offered the examples of two real sources that will absorb power when the returning voltage exceeds the output voltage (a battery and a generator turned into a motor)." The context was a battery absorbing power when the charging voltage exceeds the battery's output voltage. A battery converts electrical energy to chemical energy during that charging process. -- 73, Cecil http://www.w5dxp.com |
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Roger Sparks wrote:
On Tue, 15 Apr 2008 01:22:11 GMT Cecil Moore wrote: An ideal source does not dissipate power and there is no mechanism for storing energy. It seems what you are objecting to is the artificial separation of Vs and Rs. No, the separation of Vs and Rs was made to better understand why no interference would occur in Figure 1-1. I wasn't talking about my article. I was talking about Vs & Rs models in general. In the real world, Vs is not separated from Rs. That only occurs in the ideal model. In the ideal model, all dissipation is confined to Rs and there is none in Vs. The problem is that the source and reflected waves behave as two power sources out of time by 90 degrees. Not quite correct. The problem is that the forward waves and reflected waves flowing through the source behave as two power sources out of time by 90 degrees. The source wave is the net superposition of the forward wave and reflected wave. An ideal 50 ohm directional wattmeter in the circuit will not read the source power. It will read a forward power which is a different magnitude than the source power. In any case, only Rs and RL dissipate power in the system. I can understand a voltage source that throttles up and down but I can't understand why the throttle all has to be on the plus side. It is not all on the plus side. Whatever energy flows, flows. Sometimes the flow is forward and sometimes it is backwards. That's the way AC works. If destructive interference is present, the source reduces its output power. If constructive interference is present, the source increases its output power. But the ideal source does not dissipate power, i.e. doesn't heat up. All of the heat generated in the entire system comes from Rs and RL. Our real limit is that only one current can flow for only one voltage for each instant at any place in the circuit. You are, of course, talking about the *net* voltage and the *net* current after superposition of all the components. But this discussion is not about net voltage and net current. This is how we justify a "one sine wave" description. It is why whenever we have a reflection, we also have interference. It is also the reason that we must have power flowing back into the source for part of the cycle. I don't know where you got the idea that energy doesn't flow back into the source for part of the cycle. Since it is AC, it does flow forward and backward but none is dissipated, i.e. none is turned into heat in an ideal source. An equal amount of destructive and constructive interference occurs during each complete cycle. -- 73, Cecil http://www.w5dxp.com |
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Keith Dysart wrote:
Or is it accepting energy from the circuit, the negative P representing an energy flow into the source? Of course there can be an energy flow into the source and through the source. The point is that the ideal source doesn't dissipate that energy, i.e. it doesn't heat up. All of the heating (power dissipation) in the entire example occurs in Rs and RL because they are the only resistive components in the entire system. Any additional heating in the ideal source would violate the conservation of energy principle. -- 73, Cecil http://www.w5dxp.com |
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On Apr 15, 6:16*am, Cecil Moore wrote:
Keith Dysart wrote: Or is it accepting energy from the circuit, the negative P representing an energy flow into the source? Of course there can be an energy flow into the source Good. and through the source. It is not clear what you mean by "through the source". The source provides or absorbs energy. It does not have a "through", since it only has a single port. The point is that the ideal source doesn't dissipate that energy, i.e. it doesn't heat up. It is not obvious why you want to draw a distinction between elements that remove energy from a circuit by heating and those that do so in some other manner. Could you expand? Is there some reason why you think that it is only necessary to account for the energy removed from the circuit by heating? And that you can ignore energy being removed in other ways? And how do you know the ideal source does not dispose of the energy it receives by getting warm? Nowhere do I find in the specification of an ideal source any hint of how it disposes of its excess energy. It could be by heat, could it not? And the resistor, could it not also radiate some of the energy it receives? Perhaps even as visible light? Would that make it less of a resistor because it was not 'dissipating' the energy? All of the heating (power dissipation) in the entire example occurs in Rs and RL because they are the only resistive components in the entire system. Any additional heating in the ideal source would violate the conservation of energy principle. This is quite a leap. The energy flows into the source. We have accounted for that energy. We don't know where it goes from there. How would it violate conservation of energy if it was dissipated rather than going somewhere else? In your model, what things could be done with the energy that would not violate conservation of energy? What other things (besides heating) would violate conservation of energy? ...Keith |
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Keith Dysart wrote:
It is not clear what you mean by "through the source". The source provides or absorbs energy. It does not have a "through", since it only has a single port. Good grief! The source is a two terminal device with one terminal tied to ground. Since it has a zero resistance, an EM reverse wave can flow right through it, encounter the ground, and be 100% re-reflected by that ground. I have already explained that. Did you bother to read it? If you put an ideal 50 ohm directional wattmeter between the source and its ground, what will it read? dir GND---watt----Vs--Rs-----------------------RL meter Did you bother to analyze this configuration? Rs=50 ----50-ohm----/\/\/\/\----50-ohm---- 125w-- 100w 50w-- --25w --50w This explains everything at the average power level. I suspect it also works at the instantaneous level. It is not obvious why you want to draw a distinction between elements that remove energy from a circuit by heating and those that do so in some other manner. Could you expand? We are dealing with an ideal closed system. The ultimate destination for 100% of the ideal source power is heat dissipation in the two ideal resistors, Rs and RL. What happens between the power being sourced and the power being dissipated as heat is "of limited utility". Is there some reason why you think that it is only necessary to account for the energy removed from the circuit by heating? Heating is the only way that the energy can be removed from the ideal closed system. And that you can ignore energy being removed in other ways? No energy is removed in other ways. There is no other way for energy to leave the ideal closed system. And how do you know the ideal source does not dispose of the energy it receives by getting warm? An ideal source has zero source impedance, by definition. All of the source impedance is contained in Rs, the ideal source resistor. And the resistor, could it not also radiate some of the energy it receives? Perhaps even as visible light? Would that make it less of a resistor because it was not 'dissipating' the energy? God could also suck up the energy or the universe could end. Why muddy the waters with irrelevant obfuscations? Please deal with the ideal boundary conditions as presented. This is quite a leap. Nope, it is simple physics through which you must have been asleep. How would it violate conservation of energy if it was dissipated rather than going somewhere else? It can only be dissipated in the resistances, by definition. Nothing other than the resistances in an ideal closed system dissipates power. EE102. In your model, what things could be done with the energy that would not violate conservation of energy? FIVE TMES, I listed three things in previous postings. Since you avoided reading it FIVE TIMES already, I'm just going to point you to the "Optics" chapters on "Interference" and "Superposition". -- 73, Cecil http://www.w5dxp.com |
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On Apr 15, 12:47*pm, Cecil Moore wrote:
Keith Dysart wrote: Is there some reason why you think that it is only necessary to account for the energy removed from the circuit by heating? Heating is the only way that the energy can be removed from the ideal closed system. And that you can ignore energy being removed in other ways? No energy is removed in other ways. There is no other way for energy to leave the ideal closed system. And how do you know the ideal source does not dispose of the energy it receives by getting warm? An ideal source has zero source impedance, by definition. All of the source impedance is contained in Rs, the ideal source resistor. I see we need to back up a bit further. Consider a 10 VDC ideal voltage source. When 2 amps are flowing out of the positive terminal, the ideal voltage source is delivering 20 joules per second to the circuit. Q1. Where does this energy come from? When 1.5 amps is flowing into the positive terminal, the ideal voltage source is absorbing 15 joules per second from the circuit. Q2. Where does this energy go? Answer to both. We do not know and we do not care. An ideal voltage source can deliver energy to a circuit and it can remove energy from a circuit; that is part of the definition of an ideal voltage source. It does not matter how it does it. But just as easily as it can supply energy, it can remove it. Without understanding these basics of the ideal voltage source, it will be impossible to correctly analyze circuits that include them. Perhaps this has been the root cause of the misunderstandings. ...Keith |
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On Tue, 15 Apr 2008 14:40:28 -0700 (PDT)
Keith Dysart wrote: On Apr 15, 12:47*pm, Cecil Moore wrote: Keith Dysart wrote: Is there some reason why you think that it is only necessary to account for the energy removed from the circuit by heating? Heating is the only way that the energy can be removed from the ideal closed system. And that you can ignore energy being removed in other ways? No energy is removed in other ways. There is no other way for energy to leave the ideal closed system. And how do you know the ideal source does not dispose of the energy it receives by getting warm? An ideal source has zero source impedance, by definition. All of the source impedance is contained in Rs, the ideal source resistor. I see we need to back up a bit further. Consider a 10 VDC ideal voltage source. When 2 amps are flowing out of the positive terminal, the ideal voltage source is delivering 20 joules per second to the circuit. Q1. Where does this energy come from? When 1.5 amps is flowing into the positive terminal, the ideal voltage source is absorbing 15 joules per second from the circuit. Q2. Where does this energy go? Answer to both. We do not know and we do not care. An ideal voltage source can deliver energy to a circuit and it can remove energy from a circuit; that is part of the definition of an ideal voltage source. It does not matter how it does it. But just as easily as it can supply energy, it can remove it. Without understanding these basics of the ideal voltage source, it will be impossible to correctly analyze circuits that include them. Perhaps this has been the root cause of the misunderstandings. ...Keith This WIKI article mentions the ability of an ideal voltage source to absorb power. http://en.wikipedia.org/wiki/Voltage_source -- 73, Roger, W7WKB |
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Keith Dysart wrote:
Q1. Where does this [source] energy come from? An ideal source simply supplies a fixed voltage devoid of any concern for efficiency or where the energy comes from. This results in an average steady-state number of joules being supplied to the closed system per second. Q2. Where does this [reverse] energy go? Once introduced into a closed system, it obeys the laws of physics including the conservation of energy/momentum principles and the principle of superposition of forward and reverse electromagnetic fields. You can use an ideal directional wattmeter to track the forward and reverse energy flows through the ideal source. Answer to both. We do not know and we do not care. I have been telling you that your concepts violate the conservation of energy principle and now you have essentially admitted it. But just as easily as it can supply energy, it can remove it. I'm afraid you will find that once the ideal source has supplied the energy to a closed system, that energy cannot be destroyed. If you are allowed to willy-nilly suspend the conservation of energy principle, then any magical event is possible and there is no valid reason to even try to track the energy. It is one thing to assume an introduction of steady-state power to a closed system from an ideal source. It is another thing entirely to allow destruction of that energy once it has been introduced. -- 73, Cecil http://www.w5dxp.com |
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Roger Sparks wrote:
This WIKI article mentions the ability of an ideal voltage source to absorb power. It says: "A primary voltage source can supply (or absorb) energy ..." That's easy to comprehend for a battery source. Not so easy for an ideal RF source with a zero series impedance. If we define an RF source as a coherent RF battery, anything is possible (at least in our minds). Which of the following makes more sense? 1. Destructive interference energy is stored somewhere in the system and delivered back to the system 90 degrees later in the cycle just as it is by a physical inductor or capacitor. 2. An RF battery inside the ideal source stores the extra energy in coherent RF form and delivers it back to the system as needed. http://en.wikipedia.org/wiki/Voltage_source It also says: "The internal resistance of an ideal voltage source is zero;" So exactly how does something with an internal resistance of zero absorb any power? -- 73, Cecil http://www.w5dxp.com |
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On Wed, 16 Apr 2008 09:31:32 -0500
Cecil Moore wrote: Roger Sparks wrote: This WIKI article mentions the ability of an ideal voltage source to absorb power. It says: "A primary voltage source can supply (or absorb) energy ..." That's easy to comprehend for a battery source. Not so easy for an ideal RF source with a zero series impedance. Unless we allow it absorb energy with the same ease that it emits energy. If we define an RF source as a coherent RF battery, anything is possible (at least in our minds). Which of the following makes more sense? 1. Destructive interference energy is stored somewhere in the system and delivered back to the system 90 degrees later in the cycle just as it is by a physical inductor or capacitor. No, the problem here is that the individual switched batteries that would make up this system would not be equally discharged. Some would actually gain charge, and never deliver power to the system. So I don't like this choice. 2. An RF battery inside the ideal source stores the extra energy in coherent RF form and delivers it back to the system as needed. This sounds identical to the 1st choice. You can make such a circuit with many individually switched batteries so it becomes a good test of the theory. Again, some of the batteries would actually gain energy. http://en.wikipedia.org/wiki/Voltage_source It also says: "The internal resistance of an ideal voltage source is zero;" So exactly how does something with an internal resistance of zero absorb any power? How do we deal with the I x E power when the reflected voltage is greater than the forward voltage from the source? One way is to ignore it and only deal with averages. Another is to allow the ideal voltage source to absorb power. A third way is to allow the energy to be stored in constructive and destructive interference some place in the system. A fourth way is to allow reflections from the ideal voltage source with the result that power will build internally and the phase shift of the resultant system wave will shift farther away from the driving voltage, until some stable power input to disipation rate is reached. I like your RF battery idea. It could actually be built to a close approximation of an actual sine wave. -- 73, Roger, W7WKB |
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Roger Sparks wrote:
Cecil Moore wrote: That's easy to comprehend for a battery source. Not so easy for an ideal RF source with a zero series impedance. Unless we allow it absorb energy with the same ease that it emits energy. How can a device with a zero impedance, i.e. zero resistance, zero capacitance, and zero inductance, absorb energy? We can certainly allow it to magically absorb energy but of what use is that? A third way is to allow the energy to be stored in constructive and destructive interference some place in the system. As you know, this is the one I prefer. Another thing that neither you nor Keith has done is to account for the reverse-flowing energy through the source. I suspect if that was done, every thimbleful of energy would be accounted for. So far, net energy calculations have been used on one side of Rs and component energy calculations on the other. That would work only if Rs was not dissipating power. -- 73, Cecil http://www.w5dxp.com |
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On Apr 16, 10:04*am, Cecil Moore wrote:
Keith Dysart wrote: Q1. Where does this [source] energy come from? An ideal source simply supplies a fixed voltage devoid of any concern for efficiency or where the energy comes from. This results in an average steady-state number of joules being supplied to the closed system per second. Q2. Where does this [reverse] energy go? You have morphed the questions. Let us try again. Try two ideal voltage sources arranged in the circuit below. 5 ohms +----------\/\/\/\/-----------+ +| +| Vsl=10 VDC Vsr=5 VDC | | +-----------------------------+ Using the circuit analysis technique of your choice you should find that 1 amp is flowing through the resistor. The ideal voltage source on the left is providing 10 joules/second to the circuit. Q1. Where does this energy come from? The ideal voltage source on the right is absorbing 5 joules/second from the circuit. Q2. Where does this energy go? Answer to both. We do not know and we do not care. An ideal voltage source can deliver energy to a circuit and it can remove energy from a circuit; that is part of the definition of an ideal voltage source. It does not matter how it does it. But just as easily as it can supply energy, it can remove it. Without understanding these basics of the ideal voltage source, it will be impossible to correctly analyze circuits that include them. This has been the root cause of the misunderstandings. ...Keith |
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On Apr 16, 10:31*am, Cecil Moore wrote:
Roger Sparks wrote: This WIKI article mentions the ability of an ideal voltage source to absorb power. It says: "A primary voltage source can supply (or absorb) energy ..." That's easy to comprehend for a battery source. Not so easy for an ideal RF source with a zero series impedance. If we define an RF source as a coherent RF battery, anything is possible (at least in our minds). Which of the following makes more sense? Neither. See 3. 1. Destructive interference energy is stored somewhere in the system and delivered back to the system 90 degrees later in the cycle just as it is by a physical inductor or capacitor. 2. An RF battery inside the ideal source stores the extra energy in coherent RF form and delivers it back to the system as needed. 3. An ideal source provides or absorbs energy to satisfy its basic function which is to hold the voltage across its terminals at the desired value. When it is providing energy we do not know where this energy comes from and when it is absorbing energy we do not know where this energy goes. http://en.wikipedia.org/wiki/Voltage_source It also says: "The internal resistance of an ideal voltage source is zero;" So exactly how does something with an internal resistance of zero absorb any power? Unknown. Just as we do not know where the energy that an ideal voltage source sometimes provides comes from. But you are invited to speculate on the many ways that it might be done. Just as you might speculate on where the ideal voltage source gets its energy when it is providing energy to the circuit. Just as there are real devices which approximate ideal voltage sources delivering energy to a circuit (commonly called power supplies), there are devices which approximate ideal voltage sources that remove energy from a circuit. Check the Agilent catalog for "DC Loads". Perhaps the schematic will provide what you seek. Of course power supplies and DC loads are typically one quadrant devices, but with a little ingenuity you can combine them to form a two quadrant device which would be an even better approximation to an ideal voltage source. ...Keith |
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Keith Dysart wrote:
Q1. Where does this energy come from? Q2. Where does this energy go? Answer to both. We do not know and we do not care. Well, that certainly puts an end to this discussion. -- 73, Cecil http://www.w5dxp.com |
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On Apr 16, 11:56*pm, Cecil Moore wrote:
Keith Dysart wrote: Q1. Where does this energy come from? Q2. Where does this energy go? Answer to both. We do not know and we do not care. Well, that certainly puts an end to this discussion. An intriguing response. It was such a good example that it just stopped you in your tracks. But was this because you have learned that you were in error and now better understand the behaviour of sources when they are absorbing energy? Or was it because you have realized the contradictions inherent in your previous explanations but can not work out how to resolve them? Or was it because you have detected just a hint of the contradictions to come and want to give up before having to realize their full impact? ...Keith |
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This discussion about ideal sources and power absorption would be rather
amazing if I hadn't seen so much of the long sad history of this whole thread. It's just another diversion to deflect the discussion away from some of the sticky problems with alternative theories. It looks like Cecil could benefit from momentarily abandoning his power waves, virtual reflections, photons, s parameters, and constructive interference, and go back to basic first or second semester electric circuit theory. Connect an ideal voltage source to a series RL or RC -- all elements lumped, not distributed. Find the power P(t) at each of the three nodes. (It's easy. Just calculate v(t) and i(t) and multiply the two.) Remember that the sign of P(t) shows the direction of energy flow. If I had the patience and self-control to participate in this endless thread (and I admittedly have neither), I'd refuse to say another word about it until Cecil shows that he knows what the power waveforms are at those three nodes. Why argue with someone about distributed networks who hasn't mastered lumped circuit analysis? Roy Lewallen, W7EL |
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Correction: Finding the voltage requires a reference node. So the power
needs to be found at only two of the nodes, with the third being the reference (ground) node. Roy Lewallen, W7EL Roy Lewallen wrote: This discussion about ideal sources and power absorption would be rather amazing if I hadn't seen so much of the long sad history of this whole thread. It's just another diversion to deflect the discussion away from some of the sticky problems with alternative theories. It looks like Cecil could benefit from momentarily abandoning his power waves, virtual reflections, photons, s parameters, and constructive interference, and go back to basic first or second semester electric circuit theory. Connect an ideal voltage source to a series RL or RC -- all elements lumped, not distributed. Find the power P(t) at each of the three nodes. (It's easy. Just calculate v(t) and i(t) and multiply the two.) Remember that the sign of P(t) shows the direction of energy flow. If I had the patience and self-control to participate in this endless thread (and I admittedly have neither), I'd refuse to say another word about it until Cecil shows that he knows what the power waveforms are at those three nodes. Why argue with someone about distributed networks who hasn't mastered lumped circuit analysis? Roy Lewallen, W7EL |
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In article vfqdnbEBq5VjiZrVnZ2dnUVZ_gudnZ2d@easystreetonline , Roy
Lewallen wrote: It looks like Cecil could benefit from momentarily abandoning his power waves, virtual reflections, photons, s parameters, and constructive interference, and go back to basic first or second semester electric circuit theory. Hello, all, and then there's that circuit topology problem found in some circuit theory textbooks: There is an infinite planar square lattice of connected 1-ohm resistors that extends to infinity in all directions. A DC ohmmeter placed across any resistor would register what value? The answer is 0.5 ohm but how do you go about solving for it? (Hint: There is an easy way and a much more difficult (general) way to attack this problem. This problem is a discrete example that in the limit would indicate the distribution of a DC electric field (and currents) in a conducting, isotropic metal plate as one moves away from the source of excitation. Sincerely, and 73s from N4GGO, John Wood (Code 5550) e-mail: Naval Research Laboratory 4555 Overlook Avenue, SW Washington, DC 20375-5337 |
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Keith Dysart wrote:
It was such a good example that it just stopped you in your tracks. Not at all. Everything I have said applies to a distributed network. Since you insist on using a lumped circuit source in a distributed network example, further discussion is futile because you are mixing apples and oranges. What you perceive as the source absorbing energy is the reverse traveling wave energy flowing back through the source unimpeded. After all, how much energy can be absorbed by 0+j0 ohms? The illusion of energy absorption is the direct result of you refusing to deal with the component wave energies. -- 73, Cecil http://www.w5dxp.com |
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Roy Lewallen wrote:
Why argue with someone about distributed networks who hasn't mastered lumped circuit analysis? Roy, after your use of standing-wave current with its unchanging reference phase (as reported by EZNEC) to measure the delay through a 75m bugcatcher coil as virtually zero delay, it is obvious that your "lumped circuit analysis" cannot be trusted. Want to read about "... the Failure of Lumped-Element Circuit Theory" and diagnose your errors regarding those loading coil measurements? http://www.ttr.com/corum/ -- 73, Cecil http://www.w5dxp.com |
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On Wed, 16 Apr 2008 13:36:05 -0500
Cecil Moore wrote: Roger Sparks wrote: Cecil Moore wrote: That's easy to comprehend for a battery source. Not so easy for an ideal RF source with a zero series impedance. Unless we allow it absorb energy with the same ease that it emits energy. How can a device with a zero impedance, i.e. zero resistance, zero capacitance, and zero inductance, absorb energy? We can certainly allow it to magically absorb energy but of what use is that? It gives a tool to check our work inside the system. We think we know what happens once the energy arrives into the known components, but we don't know what exactly happens in the voltage source itself. We solve that by equating "energy in equals energy out". Then we refine that equation to "energy in equals energy returned plus energy disipated plus energy stored during some time period". Ein = Er + Ed + Es This is the "conservation of energy" principle at work. A third way is to allow the energy to be stored in constructive and destructive interference some place in the system. "A third way is to allow the energy to be stored in constructive and destructive interference SOME PLACE IN THE SYSTEM." In this quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need to understand where interference is located. If interference is located within the voltage source, it is really located *OUTSIDE* the system because we do not completely understand the voltage source itself. As you know, this is the one I prefer. Another thing that neither you nor Keith has done is to account for the reverse-flowing energy through the source. I suspect if that was done, every thimbleful of energy would be accounted for. So far, net energy calculations have been used on one side of Rs and component energy calculations on the other. That would work only if Rs was not dissipating power. Are you expecting additional reflections? Any additonal reflections from the source would only be mechanical copies of the reflection at the short, and would add or subtract to the forward wave following the rules of sine wave addition. But how can we have a source with zero resistance, zero capacitance, and zero inductance because in the real world, any source has impedance? The short has "zero resistance, zero capacitance, and zero inductance but it does not emit energy nor have a reverese voltage, both properties of the voltage source. It is not reasonable to assign the properties of the short to the voltage source, ignoring the reverse voltage situation, and expect reflectons from the source to be identical to reflections from a short. -- 73, Roger, W7WKB |
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Roger Sparks wrote:
"A third way is to allow the energy to be stored in constructive and destructive interference SOME PLACE IN THE SYSTEM." In this quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need to understand where interference is located. Actually, all we need to realize is that, in a distributed network, the current through the source is NOT equal to the current being provided by the source. It is the source current superposed with reflected current and re-reflected current. The fact that the *NET* power is negative is NOT because the source is sinking power. It is caused by the other "sources" of energy in the system, i.e. the energy in the reflected waves. Are you expecting additional reflections? Once steady-state is reached, the magnitude of the reflections are also steady-state. The total energy flowing toward the load is the forward energy. The total energy flowing in the other direction is the reflected energy. Both of those magnitudes are fixed during steady-state. I have asked multiple times that this simple mental experiment be performed and my request has been ignored. Please install an ideal 50 ohm directional wattmeter at point 'x' and tell us what are those readings for forward power and reflected power. GND--Vs--x--Rs--------45deg 50 ohm----short 100v 50ohm Once the forward and reverse energy flows at point 'x' are understood, it will become clear that all of the net current through the source is NOT being provided by the source. That the source is sinking energy is an illusion caused by reflections. -- 73, Cecil http://www.w5dxp.com |
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On Thu, 17 Apr 2008 09:46:55 -0500
Cecil Moore wrote: Roger Sparks wrote: "A third way is to allow the energy to be stored in constructive and destructive interference SOME PLACE IN THE SYSTEM." In this quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need to understand where interference is located. Actually, all we need to realize is that, in a distributed network, the current through the source is NOT equal to the current being provided by the source. It is the source current superposed with reflected current and re-reflected current. The fact that the *NET* power is negative is NOT because the source is sinking power. It is caused by the other "sources" of energy in the system, i.e. the energy in the reflected waves. Yes, in nature the returning power will affect the frequency of the source. The ideal voltage source absorbs the reflected power rather than allowing a change in frequency. Are you expecting additional reflections? Once steady-state is reached, the magnitude of the reflections are also steady-state. The total energy flowing toward the load is the forward energy. The total energy flowing in the other direction is the reflected energy. Both of those magnitudes are fixed during steady-state. I have asked multiple times that this simple mental experiment be performed and my request has been ignored. Please install an ideal 50 ohm directional wattmeter at point 'x' and tell us what are those readings for forward power and reflected power. GND--Vs--x--Rs--------45deg 50 ohm----short 100v 50ohm Once the forward and reverse energy flows at point 'x' are understood, it will become clear that all of the net current through the source is NOT being provided by the source. That the source is sinking energy is an illusion caused by reflections. This circuit has an impedance of 73.5 + J44.1 ohms at the point x. Energy sinking into the source occurs during the cycle. It is not an illusion. Think of things this way. Any signal must contain energy. A signal without energy is not a signal. Nature must have some method of signaling between source and load when the load does not match the impedance of power flow. This is simple logic based on the known fact that the apparent impedance of a transmission line depends upon the load at the end of the transmission line. Power returning to the source is the signal nature requires to throttle the output of the source. Power returning to the source will change the frequency of the source in nature, but in our ideal voltage source, nature is suspended and the frequency is held constant. Our problem as engineers is to isolate each effect and to make valid predictions so that we can build more intelligently. -- 73, Roger, W7WKB |
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Keith Dysart wrote:
Q1. Where does this energy come from? Q2. Where does this energy go? Answer to both. We do not know and we do not care. Cecil Moore wrote: Well, that certainly puts an end to this discussion. Then, Roy Lewallen wrote: ... It looks like Cecil could benefit from momentarily abandoning his power waves, virtual reflections, photons, s parameters, and constructive interference, and go back to basic first or second semester electric circuit theory..... Just when you think you're out of it, they pull you back in. K7JEB |
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Roger Sparks wrote:
Yes, in nature the returning power will affect the frequency of the source. If the reflected wave is a sine wave coherent with the forward wave, why would the frequency change when they are superposed? GND--Vs--x--Rs--------45deg 50 ohm----short 100v 50ohm This circuit has an impedance of 73.5 + J44.1 ohms at the point x. Energy sinking into the source occurs during the cycle. It is not an illusion. I got a different impedance value but for purposes of discussion, all that matters is that the impedance is NOT 50 ohms. Let's assume your impedance is correct. That means the ratio of reflected power to forward power at point 'x' is 0.1452 in the 50 ohm environment. Why would reflected energy flowing back through the source be considered "energy sinking". Why cannot the reflected energy flow back through the source unimpeded by the 0+j0 impedance and be reflected by the short to ground on the other side? That same thing happens all the time to standing wave current on the outside of the coax braid. Why do the laws of physics change inside a source? In this example, we are delivering 100 watts to Rs. With a power reflection coefficient of 0.1452, the forward power has to be 117 watts making the reflected power equal to 17 watts at point 'x'. That reflected power of 17 watts is not absorbed by the source. It flows back through the source, unimpeded by the 0+j0 impedance, and is reflected by the short to ground. It joins the 100 watts being generated by the source to become the forward wave of 117 watts. No absorbing of energy required - just good old distributed network physics. Once this 17 watt wave is tracked back through the source, reflected, and superposed with the 100 watts being generated by the source, all will become clear without any power absorption by the source being necessary. Here's a diagram of what's happening. 100w 100w GND--------Vs----------Rs--------45deg----------short 17w-- 117w-- 50w-- --17w --17w --50w All energy flows balance perfectly and there is no absorption by the source. -- 73, Cecil http://www.w5dxp.com |
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On Apr 17, 7:35*am, Cecil Moore wrote:
Keith Dysart wrote: It was such a good example that it just stopped you in your tracks. Not at all. Everything I have said applies to a distributed network. Since you insist on using a lumped circuit source in a distributed network example, further discussion is futile because you are mixing apples and oranges. So you are saying that ideal voltage sources work differently in distributed networks than they do in lumped circuits. Perhaps you could expand on the differences using the following two circuits. This one is lumped. 50 ohms +----------\/\/\/\/-----------+ +| +| Vsl=100 VDC Vsr=50 VDC | | +-----------------------------+ And this one is similar but includes a transmission line. 50 ohms +----------\/\/\/\/----+----------------+-------+ +| arbitrary +| Vsl=100 VDC length, any Vsr=50 VDC | impedance line | +----------------------+----------------+-------+ After the circuit has settled how will the ideal voltage sources on the right be behaving differently in the two examples? If you think it matters, try 50 ohm line. Where does the energy being absorbed by these ideal voltage sources go? What you perceive as the source absorbing energy is the reverse traveling wave energy flowing back through the source unimpeded. After all, how much energy can be absorbed by 0+j0 ohms? None. But pushing a current against a voltage (regardless of the cause of the voltage) will definitely use energy. The illusion of energy absorption is the direct result of you refusing to deal with the component wave energies. These circuits are DC, but Vf and Vr still work. The line settles to a constant (over length) 50 V regardless of the line impedance, but the value of Vf and Vr will depend on the impedance of the line. Why does it settle to 50 V? What else can it do? It is connected to a 50 VDC ideal voltage source. Using a line impedance of 50 ohms, compute Vr. Ooopppps. Vr is equal to 0: no reflected power. The ideal voltage source on the right, after the circuits settle, will be absorbing 50 joules/s in both cases. ...Keith |
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