|
what happens to reflected energy ?
On Jun 27, 5:05*pm, lu6etj wrote:
Have you P1, P2, P3 and P4, for your 100 W example, to clear it? ) 100w---50 ohm---+---1/2WL 291.5 ohm---50 ohm load At '+', rho = 0.707, rho^2 = 0.5 P1 = Pfor1(1-rho^2) = 100w(0.5) = 50w P2 = Pref2(rho^2) = 100w(0.5) = 50w P3 = Pfor1(rho^2) = 100w(0.5) = 50w P4 = Pref2(1-rho^2) = 100w(0.5) = 50w Pref1 = P3 + P4 - 2*SQRT(P3*P4) = 0w Pfor2 = P1 + P2 + 2*SQRT(P1*P2) = 200w How's that? -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 27, 6:20*pm, lu6etj wrote:
Or if you prefer, tell me if in your article: P1=48.98 W; P2=53.15 W; P3=51.02 W; P4=51.02 W- P1+P3 = Pfor1 = 100w P2+P4 = Pref2 = 104.1w Yes, those look like close to the correct values. I guess I should add them to my article. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 27, 11:59*pm, Keith Dysart wrote:
On Jun 27, 4:27*pm, Cecil Moore wrote: On Jun 27, 2:23*pm, Keith Dysart wrote: Example 1: Step function applied to a transmission line. After the * * * * * *line settles, a forward and reflected voltage wave * * * * * *continue on the line but no energy is being transferred. As far as I am concerned, if Maxwell's equations don't work on an example, it might as well be ignored. There is nothing during DC steady-state that allows Maxwell's equations to work because there are no EM waves during DC steady-state. Why don't you already know that? I always thought that Maxwell's equations were more complete than that and worked all the way down to DC. Two of them do not even include time and nothing says that a derivative with respect to time can't be 0. of course they do constant electric field and constant magnetic fields work just fine and go out to infinity in the dc steady state case. this of course means there can be no moving electrons, therefor no current, no d/dt terms, etc... which means no waves. I can take your approach and do you one better. Please prove that you exist. If you cannot prove that you exist, then nothing you say is of any consequence. See, I can do it also. From the above, you have proved that I exist. Thank you. i would debate the existence of a vacuum. Example 2: On a line with infinite VSWR no energy crosses a * * * * * *voltage minimum or maximum. Completely false assumption. You are back to asserting that since the north-bound traffic equals the south-bound traffic on the Golden Gate Bridge that there is no traffic and no bridge maintenance is required. When are you going to give up on that irrational wet dream of yours? No *NET* energy crosses at a voltage zero or current zero point. That doesn't make the north-bound energy equal to zero and doesn't make the south-bound energy equal to zero. It just makes them equal. Just because there is no NET traffic flow on the Golden Gate Bridge doesn't mean there is zero traffic flow in both directions. Please stop clowning around with such absurb notions. I suppose, but then you have to give up on P(t)=V(t)*I(t), generally considered to be a rather fundamental equation. very fundamental, and very restricted. only good for one point in space at one time, and for one pair of voltage and current measurements... can not be applied to separate waves that are superimposed, only to the final total voltage and current at the measurement point at that instant. Example 3: With the 1/8 wavelength line described in * * * * * *http://www.w5dxp.com/nointfr.htmtheenergycan not be * * * * * *properly accounted for on a moment by moment basis. There is no conservation of power principle. There is no mention of power above; simply energy. Are you saying that conservation of energy only applies some of the time? If you would track the RF joules and the conversion of RF joules to heat instead of the joules/ second, everything would become clear to you. As it is, you are laboring under some serious misconceptions about the laws of physics. Power simply doesn't balance within a single cycle - because it doesn't have to - because there is no conservation of power principle. In your example, the RF energy does seem to disappear and re-appear, when tracked on a moment by moment basis. when doing conservation of energy you must include the WHOLE system! it doesn't work on one section of a transmission line any more than it works for the infamous undergraduate teaser: take a refrigerator, put it in a perfectly insulated room, and then open the doors... what happens to the temperature in the room? People who don't learn from their mistakes are doomed to commit the same mistakes over and over. Keith, you seem to be all output and no input. Please enable your input channels for a change. Well, it would help if you could actually find and articulate a flaw inhttp://sites.google.com/site/keithdysart/radio6. ...Keith that site is rather worthless... you say Vs can be used to get the time reference for the other signals, but time is a variable, as is space. you seem to have a snapshot of a bunch of sine waves on an angular scale, but is that scale time or distance? |
what happens to reflected energy ?
On Jun 27, 11:59*pm, Keith Dysart wrote:
On Jun 27, 4:27*pm, Cecil Moore wrote: On Jun 27, 2:23*pm, Keith Dysart wrote: Example 1: Step function applied to a transmission line. After the * * * * * *line settles, a forward and reflected voltage wave * * * * * *continue on the line but no energy is being transferred. As far as I am concerned, if Maxwell's equations don't work on an example, it might as well be ignored. There is nothing during DC steady-state that allows Maxwell's equations to work because there are no EM waves during DC steady-state. Why don't you already know that? I always thought that Maxwell's equations were more complete than that and worked all the way down to DC. Two of them do not even include time and nothing says that a derivative with respect to time can't be 0. of course they do constant electric field and constant magnetic fields work just fine and go out to infinity in the dc steady state case. this of course means there can be no moving electrons, therefor no current, no d/dt terms, etc... which means no waves. I can take your approach and do you one better. Please prove that you exist. If you cannot prove that you exist, then nothing you say is of any consequence. See, I can do it also. From the above, you have proved that I exist. Thank you. i would debate the existence of a vacuum. Example 2: On a line with infinite VSWR no energy crosses a * * * * * *voltage minimum or maximum. Completely false assumption. You are back to asserting that since the north-bound traffic equals the south-bound traffic on the Golden Gate Bridge that there is no traffic and no bridge maintenance is required. When are you going to give up on that irrational wet dream of yours? No *NET* energy crosses at a voltage zero or current zero point. That doesn't make the north-bound energy equal to zero and doesn't make the south-bound energy equal to zero. It just makes them equal. Just because there is no NET traffic flow on the Golden Gate Bridge doesn't mean there is zero traffic flow in both directions. Please stop clowning around with such absurb notions. I suppose, but then you have to give up on P(t)=V(t)*I(t), generally considered to be a rather fundamental equation. very fundamental, and very restricted. only good for one point in space at one time, and for one pair of voltage and current measurements... can not be applied to separate waves that are superimposed, only to the final total voltage and current at the measurement point at that instant. Example 3: With the 1/8 wavelength line described in * * * * * *http://www.w5dxp.com/nointfr.htmtheenergycan not be * * * * * *properly accounted for on a moment by moment basis. There is no conservation of power principle. There is no mention of power above; simply energy. Are you saying that conservation of energy only applies some of the time? If you would track the RF joules and the conversion of RF joules to heat instead of the joules/ second, everything would become clear to you. As it is, you are laboring under some serious misconceptions about the laws of physics. Power simply doesn't balance within a single cycle - because it doesn't have to - because there is no conservation of power principle. In your example, the RF energy does seem to disappear and re-appear, when tracked on a moment by moment basis. when doing conservation of energy you must include the WHOLE system! it doesn't work on one section of a transmission line any more than it works for the infamous undergraduate teaser: take a refrigerator, put it in a perfectly insulated room, and then open the doors... what happens to the temperature in the room? People who don't learn from their mistakes are doomed to commit the same mistakes over and over. Keith, you seem to be all output and no input. Please enable your input channels for a change. Well, it would help if you could actually find and articulate a flaw inhttp://sites.google.com/site/keithdysart/radio6. ...Keith that site is rather worthless... you say Vs can be used to get the time reference for the other signals, but time is a variable, as is space. you seem to have a snapshot of a bunch of sine waves on an angular scale, but is that scale time or distance? |
what happens to reflected energy ?
On Jun 27, 9:16*pm, Cecil Moore wrote:
On Jun 27, 4:42*pm, Keith Dysart wrote: It is not my insistence. It follows from the math. Unfortunately for your arguments, math models do not dictate reality. If the math model doesn't match reality, it is invalid. Your math models obviously do not match reality. And yet, you have not located any errors in the math or the models. ....Keith |
what happens to reflected energy ?
On Jun 27, 9:48*pm, Cecil Moore wrote:
On Jun 27, 6:59*pm, Keith Dysart wrote: I suppose, but then you have to give up on P(t)=V(t)*I(t), generally considered to be a rather fundamental equation. I have absolutely no problem with giving up on the conservation of power principle in which no rational technical person can possibly believe. There again, a non-sequitor non-answer. Do you reject P(t)=V(t)*I(t) ? Are you saying that conservation of energy only applies some of the time? No, I am saying that if you cannot balance the energy equation at all times, you have made a mistake. I agree completely. And my analysis does successfully track all the energy at all times. And after averaging, it even agrees with your analysis. Methinks that you are perturbed that it demonstrates that your analysis does not track all the energy all the time, but only succeeds with averages. ....Keith |
what happens to reflected energy ?
On Jun 28, 5:11 pm, K1TTT wrote:
On Jun 27, 11:59 pm, Keith Dysart wrote: I suppose, but then you have to give up on P(t)=V(t)*I(t), generally considered to be a rather fundamental equation. very fundamental, and very restricted. only good for one point in space at one time, and for one pair of voltage and current measurements... can not be applied to separate waves that are superimposed, only to the final total voltage and current at the measurement point at that instant. True, but others reject it completely. In your example, the RF energy does seem to disappear and re-appear, when tracked on a moment by moment basis. when doing conservation of energy you must include the WHOLE system! it doesn't work on one section of a transmission line any more than it works for the infamous undergraduate teaser: take a refrigerator, put it in a perfectly insulated room, and then open the doors... what happens to the temperature in the room? The teaser is amusing, but hardly relevant. In my example, all of the energy is tracked. Or, I invite you to point out that which was overlooked. Cecil has not found any and would rather prattle on about the difference between energy and power than actually understand. Well, it would help if you could actually find and articulate a flaw inhttp://sites.google.com/site/keithdysart/radio6. ...Keith that site is rather worthless... you say Vs can be used to get the time reference for the other signals, but time is a variable, as is space. you seem to have a snapshot of a bunch of sine waves on an angular scale, but is that scale time or distance? Time, of course. I agree, though, it is not as clear as it could have been. It helps a bit if you look at Cecil’s schematic. Still, it is complicated and will probably take some effort to understand. It would probably be better to start with the step wave example offered previously in another post and copied below for convenience: example I am not sure where you think there is an error. Perhaps you can point them out in the following example: Generator: - 100V step in to an open circuit - 50 ohm source impedance Line: - 50 ohm - open circuit Generator is commanded to produce a step. This will produce 50 V and 1 A at the line input which will propagate down the line. The open end of the line has a reflection co-efficient of 1.0. Just before the 50 V step reaches the end of the line, the whole line will be at 50 V and 1 A will be flowing everywhere. The 50 V step hits the end and is reflected, producing a 50 V step (on top of the 50V already there) which propagates back to the generator. In front of the 50 V step, the current is still 1 A (which provides the charge necessary to produce the reverse propagating 50 V step. Behind the step, the current is 0. When the reverse 50 V step (which is actually a step from 50V to 100V) reaches the generator, the source impedance matches the line impedance so there is no further reflection. The line state is now 100V and 0A all along its length. The settling time was one round-trip. The generator is still producing the step, so the forward step voltage wave is still 'flowing' and being reflected so there is still a reflected step voltage wave, each of 50 V. Since the generator open circuit voltage is 100 V and the line voltage is now 100 V, current is no longer flowing from the generator to the line. Does this agree with your understanding? /example ....Keith |
what happens to reflected energy ?
On Jun 28, 10:50*pm, Keith Dysart wrote:
On Jun 28, 5:11 pm, K1TTT wrote: On Jun 27, 11:59 pm, Keith Dysart wrote: I suppose, but then you have to give up on P(t)=V(t)*I(t), generally considered to be a rather fundamental equation. very fundamental, and very restricted. *only good for one point in space at one time, and for one pair of voltage and current measurements... can not be applied to separate waves that are superimposed, only to the final total voltage and current at the measurement point at that instant. True, but others reject it completely. In your example, the RF energy does seem to disappear and re-appear, when tracked on a moment by moment basis. when doing conservation of energy you must include the WHOLE system! it doesn't work on one section of a transmission line any more than it works for the infamous undergraduate teaser: take a refrigerator, put it in a perfectly insulated room, and then open the doors... what happens to the temperature in the room? The teaser is amusing, but hardly relevant. In my example, all of the energy is tracked. Or, I invite you to point out that which was overlooked. Cecil has not found any and would rather prattle on about the difference between energy and power than actually understand. of course its relevant... so what happens to the temperature? Well, it would help if you could actually find and articulate a flaw inhttp://sites.google.com/site/keithdysart/radio6. ...Keith that site is rather worthless... you say Vs can be used to get the time reference for the other signals, but time is a variable, as is space. *you seem to have a snapshot of a bunch of sine waves on an angular scale, but is that scale time or distance? Time, of course. I agree, though, it is not as clear as it could have been. It helps a bit if you look at Cecil’s schematic. Still, it is complicated and will probably take some effort to understand. It would probably be better to start with the step wave example offered previously in another post and copied below for convenience: example I am not sure where you think there is an error. Perhaps you can point them out in the following example: Generator: - 100V step in to an open circuit - 50 ohm source impedance Line: - 50 ohm - open circuit Generator is commanded to produce a step. This will produce 50 V and 1 A at the line input which will propagate down the line. The open end of the line has a reflection co-efficient of 1.0. Just before the 50 V step reaches the end of the line, the whole line will be at 50 V and 1 A will be flowing everywhere. The 50 V step hits the end and is reflected, producing a 50 V step (on top of the 50V already there) which propagates back to the generator. In front of the 50 V step, the current is still 1 A (which provides the charge necessary to produce the reverse propagating 50 V step. Behind the step, the current is 0. When the reverse 50 V step (which is actually a step from 50V to 100V) reaches the generator, the source impedance matches the line impedance so there is no further reflection. The line state is now 100V and 0A all along its length. The settling time was one round-trip. The generator is still producing the step, so the forward step voltage wave is still 'flowing' and being reflected so there is still a reflected step voltage wave, each of 50 V. Since the generator open circuit voltage is 100 V and the line voltage is now 100 V, current is no longer flowing from the generator to the line. Does this agree with your understanding? /example ...Keith this is different than what you claimed before. on june 17th you claimed: If(t) = 50/50 = 1a Ir(t) = 50/50 = 1a now you sy current is no longer flowing. i will object to you saying the voltage 'wave' is still flowing. The line is at a constant 100v, there is no current, there can be no em wave without current AND voltage, therefore voltage can not be flowing. |
what happens to reflected energy ?
On Jun 28, 5:27*pm, Keith Dysart wrote:
And yet, you have not located any errors in the math or the models. Superposition of power *IS* an error! You add and subtract powers willy-nilly as if that mathematical step were valid which it is not. Two coherent 50w waves do not add up to a 100w wave, even using average powers, except for the special case of zero interference where the waves are 90 degrees out of phase with each other. In order to use power as an energy tracking tool, we must be very careful to ensure that there is a one-to-one correspondence between energy and power, i.e. every joule passing a point in one second must result in one watt of power (no VARs allowed). If that one-to-one correspondence doesn't exist, no valid conclusion can be drawn from tracking the power and any valid conclusion must be based on tracking the energy which is no small task. The key is that there is no such thing as imaginary energy. All energy is real. Some "power" is not real. A one-to-one correspondence does not exist in a standing wave. Therefore, tracking power as if it were equivalent to energy in standing waves is invalid. You have made that error for years. One-to-one correspondence also does not exist over a fraction of a wave. Therefore, instantaneous power is irrevelent in tracking the energy. That's your latest error which is the same conceptual error as before. In general, average power in the traveling waves over at least one complete cycle (or over many cycles) has a one-to-one correspondence to the average energy in the traveling waves. But that one-to-one correspondence is more often than not violated within a fraction of each cycle. Here's a quote from "Optics", by Hecht, concerning power density (irradiance). "If however, the 'T' is now divided out, a highly practical quantity results, one that corresponds to the average energy per unit area per unit time, namely 'I'." - where 'I' is the irradiance (*AVERAGE* power density). If I calculate the Z0 of a 1/4WL transformer, I get two roots when I take the square root of R1*R2. One of the roots is negative. If I ask you to prove something is in error with the math that yielded a negative Z0, could you find the math error? If not, does it follow that you can find the transmission line with the negative characteristic impedance existing in reality? That's your argument in a nutshell. There may (or may not) be an error in your math but it doesn't matter either way. The conclusions that you reach from your math do not match reality so your math is a moot point, i.e. there is no one-to-one correspondence between your math and the real world. If you have forgotten the importance of the one-to-one correspondence concept in mathematics, now would be a good time to review that concept. Without a one-to-one correspondence to reality, math is just fantasy existing only in your mind. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 28, 5:33*pm, Keith Dysart wrote:
Do you reject *P(t)=V(t)*I(t) ? I certainly reject as a moronic method for attempting to track instantaneous energy. I agree completely. And my analysis does successfully track all the energy at all times. That is obviously false. You have tracked the *power* after assuming a one-to-one correspondence between power (watts) and energy (joules). Your assumption is most likely false. You usually cannot use instantaneous watts to track joules within a fraction of a cycle. If the voltage and current are out of phase, some of the joules are occupied as reactive power, and not available as watts of real power. Why do you think the power companies spend so much money trying to balance the power factor? Methinks that you are perturbed that it demonstrates that your analysis does not track all the energy all the time, but only succeeds with averages. What bothers me is that, "Figures don't lie, but liers figure." :-) I have made no assertion or effort to track instantaneous energy. I would have to review a lot of physics to even remember how. What I do know is that you have been tracking power, not energy, and therefore any conclusion that you reach is probably invalid. You first must prove a one-to-one correspondence at all delta-t sections between the joules at that point and the watts at that point. We know that integrating over a complete second will make the joules equal to the watts for a traveling wave (but not for a standing wave). Now you must integrate over every partial cycle to prove that there are the exact number of joules in that delta-t of time to support the exact number of watts in that delta-t of time. I remember being warned by my professors more than half a century ago that it was a "fool's errand". -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 28, 5:50*pm, Keith Dysart wrote:
Cecil has not found any and would rather prattle on about the difference between energy and power than actually understand. I have listed a number of the laws of physics that you are violating. Of course, energy and power are different. Consider an ideal LC oscillator at the instant of time when the voltage on the capacitor is maximum and the current is zero. The energy is certainly not zero and can be calculated knowing the voltage and capacitance (assume 1000 volts and 1uf). Yet the instantaneous power is zero because the instantaneous current is zero. energy = (V^2*C)/2 = 0.5 joule power = V(t)*I(t) = 0 Please prove a one-to-one correspondence between energy and power. How can you possibly say that you are tracking all the energy when the power equals zero and the energy does not? Good grief! I am not sure where you think there is an error. Again, you need to cut off your output and enable your inputs. There are no waves during DC steady-state. Therefore, there are no forward waves and no reflected waves. Therefore, you basic assumptions are invalid. For waves to exist there must be acceleration and deceleration of carrier electrons and such does not exist during DC steady-state. Why don't you know that? -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
"Cecil Moore" wrote ... Again, you need to cut off your output and enable your inputs. There are no waves during DC steady-state. Therefore, there are no forward waves and no reflected waves. Therefore, you basic assumptions are invalid. For waves to exist there must be acceleration and deceleration of carrier electrons and such does not exist during DC steady-state. Why don't you know that? Why don't you know that oscillating current goes into the displacement current. In EM no reflections. You should decide: EM or electrons. The mixture is fun. S* |
what happens to reflected energy ?
On Jun 29, 11:44*am, "Szczepan Bialek" wrote:
You should decide: EM or electrons. The mixture is fun. That's an interesting idea to which I don't know the answer. Let me rephrase the question: Although it is known that electrons cannot flow through the dielectric of an ideal capacitor, how about photons? Can RF photons flow directly through the dielectric layer of a capacitor? If they can, it would explain a lot of things. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
Cecil Moore wrote:
On Jun 29, 11:44 am, "Szczepan Bialek" wrote: You should decide: EM or electrons. The mixture is fun. That's an interesting idea to which I don't know the answer. Let me rephrase the question: Although it is known that electrons cannot flow through the dielectric of an ideal capacitor, how about photons? Can RF photons flow directly through the dielectric layer of a capacitor? If they can, it would explain a lot of things. -- 73, Cecil, w5dxp.com photons can flow through a dielectric.. isn't that what EM propagation is, after all? |
what happens to reflected energy ?
On Jun 29, 12:54*pm, Jim Lux wrote:
photons can flow through a dielectric.. isn't that what EM propagation is, after all? Yes, after I posted it, I realized that it was a rhetorical question. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On 29 jun, 15:08, Cecil Moore wrote:
On Jun 29, 12:54*pm, Jim Lux wrote: photons can flow through a dielectric.. isn't that what EM propagation is, after all? Yes, after I posted it, I realized that it was a rhetorical question. -- 73, Cecil, w5dxp.com Dear friends, I follow with interest your interesting digressions, however in various different posts about differents matters, I notice discussion arises about what is "real" and what is not. IMO that contributes to the solution goes away from us (I remember making this comment in a previous post). In this sense respecto to energy I would like to quote a great physics: "...there is a certain quantity, which we call energy, that does not change in the manifold changes which nature undergoes. That is a most abstract idea, because it is a mathematical principle; it says that there is a numerical quantity which does not change when something happens. It is not a description of a mechanism, or anything concrete; it is just a strange fact that we can calculate some number and when we finish watching nature go through her tricks and calculate the number again, it is the same." "It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and when we add it all together it gives "28"'— always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas." From: Richard Feynman. "Six easy pieces" Miguel LU6ETJ |
what happens to reflected energy ?
On 29 jun, 20:57, lu6etj wrote:
On 29 jun, 15:08, Cecil Moore wrote: On Jun 29, 12:54*pm, Jim Lux wrote: photons can flow through a dielectric.. isn't that what EM propagation is, after all? Yes, after I posted it, I realized that it was a rhetorical question. -- 73, Cecil, w5dxp.com Dear friends, I follow with interest your interesting digressions, however in various different posts about differents matters, I notice discussion arises about what is "real" and what is not. IMO that contributes to the solution goes away from us (I remember making this comment in a previous post). In this sense respecto to energy I would like to quote a great physics: "...there is a certain quantity, which we call energy, that does not change in the manifold changes which nature undergoes. That is a most abstract idea, because it is a mathematical principle; it says that there is a numerical quantity which does not change when something happens. It is not a description of a mechanism, or anything concrete; it is just a strange fact that we can calculate some number and when we finish watching nature go through her tricks and calculate the number again, it is the same." "It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and when we add it all together it gives "28"'— always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas." From: Richard Feynman. "Six easy pieces" Miguel LU6ETJ Sorry, I forget to made clear that my comment not reference Cecil recent post mentioning "real power" in mathematical sense, referencing complex numbers. :) Miguel |
what happens to reflected energy ?
On 29 jun, 15:08, Cecil Moore wrote:
On Jun 29, 12:54*pm, Jim Lux wrote: photons can flow through a dielectric.. isn't that what EM propagation is, after all? Yes, after I posted it, I realized that it was a rhetorical question. -- 73, Cecil, w5dxp.com I learnt displacement current inside a condenser it was = eo* d(phi E)/ dt no EM radiation inside the condenser to made that current possible, in any case EM radiation in physical condenser will come out from condenser to the rest of the universe :). I also learnt photons was necessary to explain certain energy interchange phenomena such as fotoelectric effect or subatomic particle interactions, wave-particle duality for me means "duality", not "wave kaput" :) to account for EM wave well explainable phenomenom. As it was taught to me (I am not physicist), quantum nature of a 80 m wavelenght energy it is useless for calculations and invisible to our instrument resolution because its immensely large quantic number. Is it wrong? Miguel LU6ETJ |
what happens to reflected energy ?
"Cecil Moore" wrote ... On Jun 29, 11:44 am, "Szczepan Bialek" wrote: You should decide: EM or electrons. The mixture is fun. That's an interesting idea to which I don't know the answer. Let me rephrase the question: Although it is known that electrons cannot flow through the dielectric of an ideal capacitor, how about photons? Can RF photons flow directly through the dielectric layer of a capacitor? If they can, it would explain a lot of things. Photons are in the real light. The natural light is not coherent. It is emitted in the portions (packets). Radio waves are emitted continously. Radar waves are in the portions. Radio waves and the light photons flow directly through the dielectric layer of a capacitor. The difference between EM and electrons is in compressibility. Electrons are compressed in the ends of the open circuit (condenser and antenna). There the voltage is doubled (at least). Alternate electric field is created. It is radio wave. The longitudinal electric wave. In Heaviside EM (hydraulic analogy) incompressible current (real + displacement) create the magnetic whirl. The oscillating magnetic whirl is the transverse wave. Electrons are simple. Heaviside did not understand Maxwell and made a complete mess. S* |
what happens to reflected energy ?
On Jun 30, 1:30*am, "Szczepan Bialek" wrote:
The difference between EM and electrons is in compressibility. Electrons are compressed in the ends of the open circuit (condenser and antenna). There the voltage is doubled (at least). Alternate electric field is created. It is radio wave. The longitudinal electric wave. Then shouldn't we all expect a condensor to be as good a radiator of EM waves as a dipole? |
what happens to reflected energy ?
On Jun 28, 7:20 pm, K1TTT wrote:
On Jun 28, 10:50 pm, Keith Dysart wrote: On Jun 28, 5:11 pm, K1TTT wrote: On Jun 27, 11:59 pm, Keith Dysart wrote: I suppose, but then you have to give up on P(t)=V(t)*I(t), generally considered to be a rather fundamental equation. very fundamental, and very restricted. only good for one point in space at one time, and for one pair of voltage and current measurements... can not be applied to separate waves that are superimposed, only to the final total voltage and current at the measurement point at that instant. True, but others reject it completely. In your example, the RF energy does seem to disappear and re-appear, when tracked on a moment by moment basis. when doing conservation of energy you must include the WHOLE system! it doesn't work on one section of a transmission line any more than it works for the infamous undergraduate teaser: take a refrigerator, put it in a perfectly insulated room, and then open the doors... what happens to the temperature in the room? The teaser is amusing, but hardly relevant. In my example, all of the energy is tracked. Or, I invite you to point out that which was overlooked. Cecil has not found any and would rather prattle on about the difference between energy and power than actually understand. of course its relevant... so what happens to the temperature? Any energy that might raise the temperature is accounted for in the source and load resistors, so I still suggest that no energy sources or sinks have been missed. Well, it would help if you could actually find and articulate a flaw inhttp://sites.google.com/site/keithdysart/radio6. ...Keith that site is rather worthless... you say Vs can be used to get the time reference for the other signals, but time is a variable, as is space. you seem to have a snapshot of a bunch of sine waves on an angular scale, but is that scale time or distance? Time, of course. I agree, though, it is not as clear as it could have been. It helps a bit if you look at Cecil’s schematic. Still, it is complicated and will probably take some effort to understand. It would probably be better to start with the step wave example offered previously in another post and copied below for convenience: example I am not sure where you think there is an error. Perhaps you can point them out in the following example: Generator: - 100V step in to an open circuit - 50 ohm source impedance Line: - 50 ohm - open circuit Generator is commanded to produce a step. This will produce 50 V and 1 A at the line input which will propagate down the line. The open end of the line has a reflection co-efficient of 1.0. Just before the 50 V step reaches the end of the line, the whole line will be at 50 V and 1 A will be flowing everywhere. The 50 V step hits the end and is reflected, producing a 50 V step (on top of the 50V already there) which propagates back to the generator. In front of the 50 V step, the current is still 1 A (which provides the charge necessary to produce the reverse propagating 50 V step. Behind the step, the current is 0. When the reverse 50 V step (which is actually a step from 50V to 100V) reaches the generator, the source impedance matches the line impedance so there is no further reflection. The line state is now 100V and 0A all along its length. The settling time was one round-trip. The generator is still producing the step, so the forward step voltage wave is still 'flowing' and being reflected so there is still a reflected step voltage wave, each of 50 V. Since the generator open circuit voltage is 100 V and the line voltage is now 100 V, current is no longer flowing from the generator to the line. Does this agree with your understanding? /example ...Keith this is different than what you claimed before. on june 17th you claimed: If(t) = 50/50 = 1a Ir(t) = 50/50 = 1a now you say current is no longer flowing. i will object to you saying the voltage 'wave' is still flowing. The line is at a constant 100v, there is no current, there can be no em wave without current AND voltage, therefore voltage can not be flowing. I think I detect a bit of a problem originating with definitions. Start by considering that old stalwart from the text books: the infinite transmission line. For convenience, the characteristic impedance is 50 ohms and at the left end is attached a generator with a 50 ohm source impedance. The transmission line extends to the right forever, and this time the generator produces a sinusoid. Turn on the generator and it starts to produce a sinusoidal wave. For greater certainty in the description let us say that the function which describes the wave is: V(t) = 100 cos(wt) where w is the ascii excuse for omega, the angular frequency. Because the line is accepting this signal and transporting it away from the generator, the voltage distribution on the line is also a sinusoid, but spatially distributed along the line. We can visualise this spatially distributed sine wave as moving along the line and thus we have the term travelling wave. If we run fast enough along the line we can get to a point where the signal has not yet arrived and the line is at 0V, and if we wait, the sine wave will eventually arrive and we will observe the same sinusoidal spatially distributed voltage pattern moving along the line. We can run the experiment with the generator producing other functions of time and we would observe the same function spatially distributed along the line and moving. A generator producing a square wave will produce a square wave voltage distribution moving along the line. A generator producing pulses will result in pulses moving along the line. A voice signal will result in a complicated voltage pattern moving along the line. This holds for any wave, V(t), produced by the generator. But most lines are not infinite. What happens when the line is terminated? Most texts will explain that if a section of line is terminated in its characteristic impedance, that section will behave as if it was connected to a line that continued forever. That is, the observations on that section will be no different than if the line went on and on and on. Thus in that section we would observe the moving sine, square, pulse, or whatever wave, but, we can no longer run far down the line to find a place where the wave has not yet arrived, since the line has been truncated. Now what happens if our section of line is open-circuited? The wave reflection model allows us to compute that the reflection coefficient (rc) will be 1. This means that if a forward travelling voltage wave, Vf(t), encounters the open circuit, it will be reflected as a reverse wave, Vr(t). Mathematically Vr(t) = rc * Vf(t) Going back to our waveforms, this is most readily visualised with pulses where the repetition rate is low enough that only one pulse is on the line at a time. The pulse can be seen to travel down the line, hit the end and travel in the reverse direction back to the generator. What happens when the pulse gets back to the generator? Well the generator was specified to have a source impedance that was the same as the line characteristic impedance, so the behaviour on the section of line will be exactly the same as if the line continued forever. There is no reflection, and we can forget about the pulse. The wave reflection model also tells us that the voltage at any point on the line will be V(t) = Vf(t) + Vr(t) where Vf(t) is the voltage from the forward wave at that point on the line and Vr(t) is the voltage from the reverse wave at that point on the line. For pulses, where there is only one pulse on the line at a time, this is easy to visualise, but for continuous signals, the voltage distribution on the line is now the sum of the forward and reverse travelling waves and for the most part, the forward and reverse travelling waves are no longer easily visible. For example, with a sine wave, the resulting voltage distribution on the line is the time varying, so called, standing wave. It is important to note that even though the original travelling waves are no longer visible in the spatial distribution of the voltage on the line, the travelling wave continues to be provided by the generator, it travels to the end of the line where it is reflected and this reverse wave travels back to the generator where it behaves as if the line continued forever because the line is terminated by its characteristic impedance. Recall that all of this holds for any Vf(t) produced by the generator. So what happens with that original sticking point, the step function? Consider the infinite line again. The step is moving along the line. If we run far enough ahead, we can watch the step pass, but behind the step the voltage is constant. This makes it hard to visualise as a moving wave because there are no peaks and valleys to observe, but it is still moving. If in doubt, just run further down the line to observe the step pass again. Now if we terminate the line, we can no longer run on ahead to find the step, but the wave is still moving; we just can’t see it because of the constant nature of the wave after the step. With the open circuited line, the step function is reflected and travels as a reverse wave back to the generator. Once the step reaches the generator, just as happens for any other V(t) wave, the wave continues to move; it just is not visible. And the voltage at any point on the line can be correctly computed from V(t) = Vf(t) + Vr(t) Another way to think of this is to produce the step wave in a different manner. Start with a pulse wave. Then increase the pulse width until it approaches the pulse period. The variation in voltage can still be observed on the line. Then slowly increase the width until it is equal to the period. There is now no longer any variation in the wave to be observed, but the wave is still moving just as it was when the width was infinitesimally shorter than the period. Another view is to start with a square wave, then increase the period until it is much longer than the duration of the experiment. Over the duration of the experiment, no change in voltage will be observed, but it is a square wave. One could perform a Fourier transform on the square wave, treat all the constituent sine waves as travelling waves, work out the reflections for each, and sum them back. Exactly the same results will be obtained as will be for treating the step as a moving wave, even though there are no peaks and valleys to aid the visualisation. So I hope that I have persuaded you that it is completely valid to view a step function as a wave that continues to move (though it is hard to see) even after the step has passed. If not, re-read the above to locate the flaw in my argument. If you have been convinced, then we can get back to If(t) = 50/50 = 1a Ir(t) = 50/50 = 1a Consider a generator, 50 ohms, that produces a 50 V step in to 50 ohms. For this Vf(t) = 50V The line has a 50 ohm impedance, so If(t) = 50/50 = 1A This is the voltage and current that would be observed on the infinite or terminated line after the step passes. With the open circuit line, where the reflection coefficient is 1, the voltage will be reflected so that Vr(t) = 50V after the reverse step passes and Ir(t) = 50/50 = 1A Recalling that at any point on the line V(t) = Vf(t) + Vr(t) and I(t) = If(t) – Ir(t) we have, after the reverse step has returned to the generator, everywhere on the line: V(t) = 50 + 50 = 100V I(t) = 1 – 1 = 0A This is exactly as expected for a generator with an open circuit voltage of 100V and an open circuit line that has no current flowing. So the wave reflection model, with continuously forward and reverse waves, accurately describes the resulting observations on the line, even for a step wave. ....Keith |
what happens to reflected energy ?
On Jun 29, 9:46*am, Cecil Moore wrote:
On Jun 28, 5:27*pm, Keith Dysart wrote: And yet, you have not located any errors in the math or the models. Superposition of power *IS* an error! You have really latched on to that haven't you. I remember the good ol' days when you were superposing powers and the very same people with whom you are arguing today were trying to convince you that superposing power was an invalid operation. Now you have bought in to the rule and apply it even where it does not apply. Just because a sum is being computed does not mean that superposition is at work. When you add up the energies to validate that energy is not created or destroyed, you are not superposing energy. When I add up the energy flows for the same purpose, that does not mean that superposition is at work. From a basic calculus, if the energy balances, so must the energy flow: f(t), g(t), etc. are functions that describes the energy in individual entities of a system with respect to time. Then f(t) + g(t) + h(t) = k is the expression that says the sum of all the energies must be a constant, for energy is neither created nor destroyed. From basic calculus d(f(t) + g(t) + h(t))/dt = d(k)/dt d(f(t) + g(t) + h(t))/dt = 0 d(f(t))/dt + d(g(t))/dt + d(h(t))/dt = 0 d(f(t))/dt is the derivitive with respect to time of the energy (i.e. energy flow, or power) for the entity. So it is completely valid to sum these energy flows to track the energy within the system; as my example does. You add and subtract powers willy-nilly as if that mathematical step were valid which it is not. Two coherent 50w waves do not add up to a 100w wave, even using average powers, except for the special case of zero interference where the waves are 90 degrees out of phase with each other. True. You must always use the actual energy flows in the entity and not the powers that are not real. In order to use power as an energy tracking tool, we must be very careful to ensure that there is a one-to-one correspondence between energy and power, i.e. every joule passing a point in one second must result in one watt of power (no VARs allowed). Also true. And as is done in my example. If that one-to-one correspondence doesn't exist, no valid conclusion can be drawn from tracking the power and any valid conclusion must be based on tracking the energy which is no small task. The key is that there is no such thing as imaginary energy. All energy is real. Some "power" is not real. Agreed. Especially the power in the waves that are being superposed. That is why you need a fudge factor when you sum your partial powers from your waves. I carefully do not sum partial powers from superposed waves. A one-to-one correspondence does not exist in a standing wave. Therefore, tracking power as if it were equivalent to energy in standing waves is invalid. You have made that error for years. Please provide an example of such an error, preferably from one of my expositions. One-to-one correspondence also does not exist over a fraction of a wave. Of course it does. Energy must always balance; not just at the end of the cycle. Therefore, instantaneous power is irrevelent in tracking the energy. That's your latest error which is the same conceptual error as before. In general, average power in the traveling waves over at least one complete cycle (or over many cycles) has a one-to-one correspondence to the average energy in the traveling waves. But that one-to-one correspondence is more often than not violated within a fraction of each cycle. Now you are allowing us to create and destroy energy as long as it balances at the end of the cycle? Are you sure you want to do this? It does not align with the well understood principle of conservation of energy. If I follow that rule, I just make the cycle long enough, say, my lifetime, and I am golden. But I do not hold out any hope for this. Here's a quote from "Optics", by Hecht, concerning power density (irradiance). "If however, the 'T' is now divided out, a highly practical quantity results, one that corresponds to the average energy per unit area per unit time, namely 'I'." - where 'I' is the irradiance (*AVERAGE* power density). There you go again, assuming that the 'practical' challenges of optics also apply to transmission lines. On a transmission line, we can measure the voltage and the current. We do not suffer from the same constraints as optics. Use the tools that are available for transmission lines. Don't tie two hands behind your back just because those tools do not work in the optical domain. ....Keith |
what happens to reflected energy ?
On Jun 29, 10:13*am, Cecil Moore wrote:
On Jun 28, 5:33*pm, Keith Dysart wrote: Do you reject *P(t)=V(t)*I(t) ? I certainly reject as a moronic method for attempting to track instantaneous energy. Dodging the question again! It is not that hard. I agree completely. And my analysis does successfully track all the energy at all times. That is obviously false. You have tracked the *power* after assuming a one-to-one correspondence between power (watts) and energy (joules). Your assumption is most likely false. You usually cannot use instantaneous watts to track joules within a fraction of a cycle. Of course you can. It is just like water flowing from multiple pipes in to a tank. The sum of the instantaneous flows is exactly equal to the rate at which the volume of water is increasing in the tank. Not hard at all. If the voltage and current are out of phase, some of the joules are occupied as reactive power, and not available as watts of real power. Mostly true. But this is accurately accounted for by using the instantaneous voltage and current and not the RMS. This is an important difference expressed in P(t) = V(t) * I(t) and why I do not use Pavg = Vrms * Irms which has all the problems you mention. Why do you think the power companies spend so much money trying to balance the power factor? Not for this reason. The energy meter at my house accurately measures the power (using P(t) = V(t) * I(t)) and integrates so that they bill me for the energy I have used. And so they do. Poor power factor does not affect this, but increases the losses experienced by the power company when delivering the energy to me. They want the customer to pay for these losses so they monitor VAR. ....Keith |
what happens to reflected energy ?
On Jun 29, 11:02*am, Cecil Moore wrote:
On Jun 28, 5:50*pm, Keith Dysart wrote: Cecil has not found any and would rather prattle on about the difference between energy and power than actually understand. I have listed a number of the laws of physics that you are violating. Of course, energy and power are different. Consider an ideal LC oscillator at the instant of time when the voltage on the capacitor is maximum and the current is zero. The energy is certainly not zero and can be calculated knowing the voltage and capacitance (assume 1000 volts and 1uf). Yet the instantaneous power is zero because the instantaneous current is zero. energy = (V^2*C)/2 = 0.5 joule power = V(t)*I(t) = 0 Indeed. Power and energy are related in that power is the derivitive with respect to time of energy. They are very closely related. See my other post for the derivation. Please prove a one-to-one correspondence between energy and power. How can you possibly say that you are tracking all the energy when the power equals zero and the energy does not? Good grief! 'cause that is how calculus works!? If one tracks all the flows, then one knows where the energy is because it can only move by flowing. Think water pipes and tanks for an analogy. I am not sure where you think there is an error. Again, you need to cut off your output and enable your inputs. There are no waves during DC steady-state. Please see my other post where this is covered. Therefore, there are no forward waves and no reflected waves. Therefore, you basic assumptions are invalid. For waves to exist there must be acceleration and deceleration of carrier electrons and such does not exist during DC steady-state. Why don't you know that? You, yourself, explained once how you would use the Fourier transform to analyze such an example. After you have transformed the signal to the frequency domain, there are enough sinusoidal waves to keep anyone happy. Or are you saying that a step function does not have a Fourier transform? ....Keith |
what happens to reflected energy ?
On Jun 29, 8:41*pm, lu6etj wrote:
I learnt displacement current inside a condenser it was = eo* d(phi E)/ dt no EM radiation inside the condenser to made that current possible, in any case EM radiation in physical condenser will come out from condenser to the rest of the universe :). It depends on your definition of "radiation". In an ideal transmission line, energy is not lost to radiation but photons (EM fields and waves) necessarily exist all up and down the line. In an ideal coaxial transmission line, the photons (EM fields and waves) are confined to the dielectric. Electrons cannot travel at the speed of light. EM waves travel at the speed of light. Therefore EM waves are photons. Given the physical nature of a capacitor, refraction would be the primary mechanism for losing energy to radiation and there's probably very, very little refraction in the capacitor dielectric. If electrons are being acelerated and decelerated in the capacitor, photons will be emitted. It seems obvious now that when electrons are decelerated on one capacitor plate, photons are emitted that propogate across the capacitor dielectric and are absorbed by electrons on the opposite plate. When the concept of displacement current was invented, nobody knew that RF fields were actually made up of particles (photons) but now we do know. Displacement current seems only to be EM radiation from one capacitor plate to the other. As it was taught to me (I am not physicist), quantum nature of a 80 m wavelenght energy it is useless for calculations and invisible to our instrument resolution because its immensely large quantic number. Is it wrong? The quantized nature of a single RF photon is no longer open to argument and the energy in that single photon can be calculated, (h*f), where h is Planck's constant, 6.626 x 10^-34 J*s. Whether there are any instruments sensitive enough to detect a single 80m photon is a moot point that does not change the nature of RF fields and waves. What is important is that it is impossible to radiate a signal level less than (h*f). If anyone asserts that RF fields and waves can violate the laws of physics regarding photons, that person is wrong and delusional. (Light left over from the time when the universe became transparent is today red-shifted down to RF microwave frequencies and called background radiation.) -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 30, 1:30*am, "Szczepan Bialek" wrote:
Photons are in the real light. The natural light is not coherent. It is emitted in the portions (packets). Radio waves are emitted continously. Radar waves are in the portions. Laser light can usually be considered to be coherent light as are waves generated by an RF transmitter. But of course, nothing is perfect. Radio waves and the light photons flow directly through the dielectric layer of a capacitor. Energy flow, just not as "current". Electrons are compressed in the ends of the open circuit (condenser and antenna). "Compressed" may be overstating the phenomenon. Have you ever calculated exactly how far an electron moves at 10 MHz? Electrons are simple. Yes, too simple to move at the speed of light. |
what happens to reflected energy ?
On Jun 30, 6:43*am, Keith Dysart wrote:
You have really latched on to that haven't you. I remember the good ol' days when you were superposing powers and the very same people with whom you are arguing today were trying to convince you that superposing power was an invalid operation. And I remember the good ol' days when you were still messing your diapers. Do you still do that? I probably did try to superpose power back when I was young and foolish but I learned the error of my ways and corrected it. When I add up the energy flows for the same purpose, that does not mean that superposition is at work. But you are NOT adding up the energy flows - you are adding up the power. That's superposition of power and is a no-no. Power does not obey any conservation of power principle. The instantaneous maximum charge on a capacitor contains any amount of energy while power equals zero. What is it about that concept that you don't understand? True. You must always use the actual energy flows in the entity and not the powers that are not real. What happens when energy = 1 joule, and de/dt = 0 watts. This happens all the time during an RF cycle so you are not using actual energy flows. You are using power which goes to zero even when maximum energy is still present. There is obviously NOT a one-to-one correspondence between power and energy. YOU CANNOT USE WATTS TO TRACK ENERGY UNLESS THERE IS A ONE-TO-ONE CORRESPONDENCE BETWEEN WATTS AND JOULES. Also true. And as is done in my example. Actually false, as I have proved above. Please plot the joules, not the watts. Please provide an example of such an error, preferably from one of my expositions. At a current node in a standing wave, you claim there is zero power which is true. You claim that since there is zero power, there is also zero energy which is absolutely false since that is a voltage maximum point and you will fry yourself if you grab it. The energy in the voltage maximum is unrelated to the power being zero - just one more proof that energy does not correspond to power. At the zero power points, there is a maximum of energy in one of the fields which you are ignoring. Of course it does. Energy must always balance; not just at the end of the cycle. Energy must balance but power doesn't have to balance and usually does not balance. Power only balances for special cases. You are using power in the general case as an example so anything you say is invalid. Now you are allowing us to create and destroy energy as long as it balances at the end of the cycle? No, energy cannot be created or destroyed. But power is created and destroyed all the time because there is no conservation of power principle. Your glaring error is that you are using power as if it were energy and it is not. I have given numerous examples. In an LC oscillator, when all of the non-destructable energy is stored in the capacitor, there is ZERO power, i.e. power has been destroyed. Why do you refuse to discuss that fact of physics? 90 degrees later in the cycle, power has been created. Every time one of your instantaneous power curves crosses the zero axis, power has been destroyed. Every time one of your instantaneous power curves reaches a peak, power has been created. We do not suffer from the same constraints as optics. You suffer from the delusion that power obeys the same laws of physics that energy does. You willy-nilly interchange power and energy. Until you admit the error of your ways, there is little that can be done to alleviate your ignorance. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 30, 7:24*am, Keith Dysart wrote:
Or are you saying that a step function does not have a Fourier transform? Your example has the step function disappearing during steady-state. DC steady-state has no Fourier transform since omega equals zero. A Fourier analysis requires a recurring function of the sum of sinusoids involving integer multiples of omega*time. Your single step does not recur so your conclusions are invalid. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Jun 30, 10:11*am, Cecil Moore wrote:
On Jun 30, 7:24*am, Keith Dysart wrote: Or are you saying that a step function does not have a Fourier transform? Your example has the step function disappearing during steady-state. DC steady-state has no Fourier transform since omega equals zero. A Fourier analysis requires a recurring function of the sum of sinusoids involving integer multiples of omega*time. Your single step does not recur so your conclusions are invalid. Check the a0 coefficient in the Fourier transform. This represents the DC component of the signal. It is perfectly valid for it to be the only non-zero coefficient. Without this, how would you deal with a signal such as V(t) = 10 + 2 cos(3t) Alternatively, one can use the standard trick for dealing with non-repetitive waveforms: choose an arbitrary period. 24 hours would probably be suitable for these examples and transform from there. Still, you will have zero frequency component to deal with, but there will be some at higher frequencies (if you choose your function to make it so). ....Keith |
what happens to reflected energy ?
On Jun 30, 10:03*am, Cecil Moore wrote:
On Jun 30, 6:43*am, Keith Dysart wrote: You have really latched on to that haven't you. I remember the good ol' days when you were superposing powers and the very same people with whom you are arguing today were trying to convince you that superposing power was an invalid operation. And I remember the good ol' days when you were still messing your diapers. Do you still do that? Warning! Warning! Descent in to scatology. I probably did try to superpose power back when I was young and foolish but I learned the error of my ways and corrected it. Learning is good. But now you want to over apply the rules. When I add up the energy flows for the same purpose, that does not mean that superposition is at work. But you are NOT adding up the energy flows - you are adding up the power. Ummm. Energy flow is power. Joules/s! If it helps, any place I have written 'power', please replace with 'energy flow'. That's superposition of power and is a no-no. Power does not obey any conservation of power principle. Try the multiple pipe, water container, water flow example to understand how flows must also be conserved if matter (or energy) is not to be created or destroyed. The instantaneous maximum charge on a capacitor contains any amount of energy while power equals zero. What is it about that concept that you don't understand? Excellent concept. There is no conflict with conserving flows, or energy for that matter. True. You must always use the actual energy flows in the entity and not the powers that are not real. What happens when energy = 1 joule, and de/dt = 0 watts. This happens all the time during an RF cycle so you are not using actual energy flows. You are using power which goes to zero even when maximum energy is still present. Yes, indeed. That is a fundamental possibility and occurs on transmission lines with infinite VSWR. There is obviously NOT a one-to-one correspondence between power and energy. Correct. Power is the time derivitive of energy. They are related but definitely not one-to-one. YOU CANNOT USE WATTS TO TRACK ENERGY UNLESS THERE IS A ONE-TO-ONE CORRESPONDENCE BETWEEN WATTS AND JOULES. This is quite incorrect. Energy flows must balance, otherwise energy is being created or destroyed to sustain a difference in flow. Also true. And as is done in my example. Actually false, as I have proved above. Please plot the joules, not the watts. If it makes you happier, since it is a discrete simulation, you can simply substitute joules per degree wherever there is a curve labeled Watts. You need to scale, of course, between degrees and seconds, but the shape and sums will be identical. Please provide an example of such an error, preferably from one of my expositions. At a current node in a standing wave, you claim there is zero power which is true. Excellent. I am glad you have come to accept that. You claim that since there is zero power, there is also zero energy I have never claimed that. In fact, the minima and maxima are where the energy peaks will be found. I have claimed there is no energy flow across (i.e. power) these points. which is absolutely false since that is a voltage maximum point and you will fry yourself if you grab it. The energy in the voltage maximum is unrelated to the power being zero Not quite. It follows from the functions for each. Since power is the derivitive of energy, it should come as no surprise that maximum energy occurs at the point of minimum power. In basic calculus looking for the zero in the derivitive is how you locate the maximum value of the curve. - just one more proof that energy does not correspond to power. At the zero power points, there is a maximum of energy in one of the fields which you are ignoring. Of course it does. Energy must always balance; not just at the end of the cycle. Energy must balance but power doesn't have to balance and usually does not balance. Power only balances for special cases. You are using power in the general case as an example so anything you say is invalid. Unfortunately wrong. Energy flows must balance as well. Otherwise, energy is coming from nowhere to sustain the flow. Now you are allowing us to create and destroy energy as long as it balances at the end of the cycle? No, energy cannot be created or destroyed. But power is created and destroyed all the time because there is no conservation of power principle. Your glaring error is that you are using power as if it were energy and it is not. I have given numerous examples. In an LC oscillator, when all of the non-destructable energy is stored in the capacitor, there is ZERO power, i.e. power has been destroyed. Yes, indeed. At that instant, zero energy is flowing from the inductor to the capacitor. But very soon, energy will be flowing from the capacitor to the inductor. The balance is that the energy flowing out of the capacitor is always and exactly equal to the energy flowing in to the inductor. That is the energy flow balance. The only way for this not to be true is for energy to be created or destroyed. Why do you refuse to discuss that fact of physics? 90 degrees later in the cycle, power has been created. One way of thinking about it, I suppose. But not too useful. Instead, think that at every instant, the energy flow between the entities in the experiment must balance. Every time one of your instantaneous power curves crosses the zero axis, power has been destroyed. Every time one of your instantaneous power curves reaches a peak, power has been created. I think you may be confused because you are only looking at the flow in and out of a single entity. This is clearly not conserved. Nor for that matter is the energy within that entity. It is the total energy within the system that is conserved, just as it is the total of the flows of energy between the entities within the system that must be conserved. Put more strictly: The sum of all the energy flows in to all of the entities within the system must equal the energy flow in to the system. We do not suffer from the same constraints as optics. You suffer from the delusion that power obeys the same laws of physics that energy does. You willy-nilly interchange power and energy. Until you admit the error of your ways, there is little that can be done to alleviate your ignorance. An intriguing accusation, but you do need to provide a concrete example of where my analysis has gone wrong. ....Keith |
what happens to reflected energy ?
"Richard Fry" wrote ... On Jun 30, 1:30 am, "Szczepan Bialek" wrote: The difference between EM and electrons is in compressibility. Electrons are compressed in the ends of the open circuit (condenser and antenna). There the voltage is doubled (at least). Alternate electric field is created. It is radio wave. The longitudinal electric wave. Then shouldn't we all expect a condensor to be as good a radiator of EM waves as a dipole? All radio people know that: http://en.wikipedia.org/wiki/File:Dipolentstehung.gif S* |
what happens to reflected energy ?
On Tue, 29 Jun 2010 18:41:52 -0700 (PDT), lu6etj
wrote: As it was taught to me (I am not physicist), quantum nature of a 80 m wavelenght energy it is useless for calculations and invisible to our instrument resolution because its immensely large quantic number. Is it wrong? Yes. We experience 80M activity every day irrespective of it being Newtonian or Quantum. All it reveals is that something with a very, very, very low energy is still quite measurable. However, you "can" deliberately choose the wrong instrument to measure the energy. That instrument reveals more about the choice-maker than the energy. For instance, a 1KW 80M energy source presents a near 0 degree absolute temperature. A fever thermometer is not going to register that energy. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On 30 jun, 16:00, Richard Clark wrote:
On Tue, 29 Jun 2010 18:41:52 -0700 (PDT), lu6etj wrote: As it was taught to me (I am not physicist), quantum nature of a 80 m wavelenght energy it is useless for calculations and invisible to our instrument resolution because its immensely large quantic number. Is it wrong? Yes. We experience 80M activity every day irrespective of it being Newtonian or Quantum. *All it reveals is that something with a very, very, very low energy is still quite measurable. * However, you "can" deliberately choose the wrong instrument to measure the energy. *That instrument reveals more about the choice-maker than the energy. For instance, a 1KW 80M energy source presents a near 0 degree absolute temperature. *A fever thermometer is not going to register that energy. 73's Richard Clark, KB7QHC Dear Richard: What I said is what my physics book says, I swear there no creation of mine... :) (I have not any authority on this matter). I was thinking in quantic number describing the energy of a typical 100 W 80 m oscillator devolped in one second, representing a quantic number n = 4.3 * 10^28; we know our quanta represents the minimun possible energy of a 80 m radiation AND the minimun "delta" Energy possible for a given oscillator, energy difference between (among?) one quanta an two quanta of 80 m radiation is 2.3 * 10^ -27 J, that difference (my physic book say) it is unmeasurable experimentally (this energy leap (skip?, hop?) it is in the order of 10^-8 smaller that green light leap (in reality my book -Resnick Halliday- give a moving dust particle example with quantic number very much lower than my 80 m example yet = n = 3 * 10^14, they said "we can not distiguish energy difference among n = 3 * 10^14 and n = [3 * 10^14] +1") What it is the ohysical sense of working with magnitudes we can not measure? Nobody (as we know) use (or need) quantum mechanics to deal with (or explain) locomotive movement :) Cecil said "electrons can not travel at light speed, photon yes therefore EM waves are photons", well... EM CAN travel at light speed, then photons are EM waves :D :D As I know duality (particles can behave as waves and vice versa) have not dead yet (or he died and I found out?). Cecil said: "When the concept of displacement current was invented, nobody knew that RF fields were actually made up of particles (photons) but now we do know". Cecil seem to me as Zarathustra has declared: "¡Wave is dead!" :) Yes, yes, I know some people bring very strange ideas into the forums, but I think it is not necessary argue them with exotic others -even if they are true- because the partner will double the bet and will bring other even more bizarre yet...! :) Well, dont be bothered by my comments, I am joking a little... 73 - Miguel - LU6ETJ |
what happens to reflected energy ?
Keith obviously understands the requirements for energy flow, but a
casual reader might draw the wrong conclusions. . . Keith Dysart wrote: . . . This is quite incorrect. Energy flows must balance, otherwise energy is being created or destroyed to sustain a difference in flow. On the average, yes. But not moment-by-moment. Energy can be stored and retrieved from storage, resulting in unequal energy flow (power) into and out of a point. For a while. This was explicitly stated earlier when discussing a capacitor, but I think it's important to make the distinction here between instantaneous and average requirements. In steady state, the average condition (energy flow balance) must be met each cycle. That is, the total energy into a node over a cycle has to equal the energy out of a node over a cycle. . . . Unfortunately wrong. Energy flows must balance as well. Otherwise, energy is coming from nowhere to sustain the flow. Ditto. . . . Yes, indeed. At that instant, zero energy is flowing from the inductor to the capacitor. But very soon, energy will be flowing from the capacitor to the inductor. The balance is that the energy flowing out of the capacitor is always and exactly equal to the energy flowing in to the inductor. That is the energy flow balance. The only way for this not to be true is for energy to be created or destroyed. Ditto. . . . Instead, think that at every instant, the energy flow between the entities in the experiment must balance. No, it doesn't, unless I'm misunderstanding the statement. At a given instant, more energy can flow into a component (e.g., a capacitor or inductor) than is flowing out, or vice-versa. But in steady state, whatever flows in during one part of the cycle must flow out during the remainder of the cycle. Every time one of your instantaneous power curves crosses the zero axis, power has been destroyed. Every time one of your instantaneous power curves reaches a peak, power has been created. I think you may be confused because you are only looking at the flow in and out of a single entity. This is clearly not conserved. Nor for that matter is the energy within that entity. It is the total energy within the system that is conserved, just as it is the total of the flows of energy between the entities within the system that must be conserved. Put more strictly: The sum of all the energy flows in to all of the entities within the system must equal the energy flow in to the system. Again, only on an average or steady-state cycle-by-cycle basis. Great inequalities can exist for shorter periods. . . . Like Keith, I firmly believe that an instantaneous time-domain analysis is essential in understanding what really happens to the energy in an AC system. Averaging reduces the amount of information you have -- if all you know is the average value of a waveform, you have no way of going back and finding out what the waveform was, out of an infinite number of possibilities. If averaging is to be done, it should be done after you calculate and understand what's going on at each instant, not before you begin the analysis. But it's also essential to make absolutely clear what conditions must be met every instant, such as p(t) = v(t) * i(t), and which must be met only on the average, such as energy in = energy out. Roy Lewallen, W7EL |
what happens to reflected energy ?
lu6etj wrote:
On 29 jun, 15:08, Cecil Moore wrote: On Jun 29, 12:54 pm, Jim Lux wrote: photons can flow through a dielectric.. isn't that what EM propagation is, after all? Yes, after I posted it, I realized that it was a rhetorical question. -- 73, Cecil, w5dxp.com I learnt displacement current inside a condenser it was = eo* d(phi E)/ dt no EM radiation inside the condenser to made that current possible, in any case EM radiation in physical condenser will come out from condenser to the rest of the universe :). I also learnt photons was necessary to explain certain energy interchange phenomena such as fotoelectric effect or subatomic particle interactions, wave-particle duality for me means "duality", not "wave kaput" :) to account for EM wave well explainable phenomenom. As it was taught to me (I am not physicist), quantum nature of a 80 m wavelenght energy it is useless for calculations and invisible to our instrument resolution because its immensely large quantic number. Is it wrong? Miguel LU6ETJ Photons are very useful in the analysis of transmission lines. They can be brought into the discussion to divert it from taking a path that makes a participant uncomfortable. If unable to answer a question logically, simply toss photons, optics, quantum mechanics, aether, and other confounding factors in, and presto, people will begin arguing about the spurious concepts and forget that you've avoided answering the difficult question. It's called misdirection, a time-honored technique used by politicians and prestidigitators as well as promoters of pseudoscience. Roy Lewallen, W7EL |
what happens to reflected energy ?
On Wed, 30 Jun 2010 14:02:14 -0700 (PDT), lu6etj
wrote: I was thinking in quantic number describing the energy of a typical 100 W 80 m oscillator devolped in one second, Hi Miguel, Power? Energy? One second? Choose one to talk about, and perhaps the mystery of numbers might clear up. one quanta an two quanta Quanta? Two Quanta? We are now up to four intermixed terms. Simplify. Choose one thing. of 80 m radiation is 2.3 * 10^ -27 J, that difference (my physic book say) it is unmeasurable experimentally (this energy leap (skip?, hop?) it is in the order of 10^-8 smaller that green light leap True, but immaterial. You are confusing wavelength and quanta (no surprise given the blearing of topic). Compare Green and IR. Is there a correlation on a scale of two that predicts out to a scale of 10^8? Compare Green and deep IR. Is there a correlation on a scale of ten that predicts out to a scale of 10^8? Compare Green and the Sub-millimeter band. Is there a correlation on a scale of 100 that predicts out to a scale of 10^8? (in reality my book -Resnick Halliday- give a moving dust particle example with quantic number very much lower than my 80 m example yet = n = 3 * 10^14, they said "we can not distiguish energy difference among n = 3 * 10^14 and n = [3 * 10^14] +1") So a quantum of smaller energy of a dust particle is measureable but 80M transmission is not? Common sense is wheezing in this dust. OK, so they are talking about the difference in quantum, not energy. Would it surprise you that you cannot even tell the difference between one quanta of green light and two with conventional detecting technology? What it is the ohysical sense of working with magnitudes we can not measure? Nobody (as we know) use (or need) quantum mechanics to deal with (or explain) locomotive movement :) The limitation is called Quatum Efficiency and the human eye is vastly superior (to all but $1,000,000 components) at rougly QE = 50%. Cecil said Cecil said Yes, yes, I know some people bring very strange ideas into the forums, Indeed. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
Considering the steady state... If we accept the P(t) is the product of instantaneous voltage and current, then there will be some points on any mismatched line where P(t) is always positive. In between those points, P(t) will have positive and negative excursions. I think that it is a reasonable interpretation that at those points where P(t) is always positive, then there is never at any instant, a flow of energy away from the load, energy is never exchanged during a cycle across those points, it always flows from source to load. It may be that energy is exhanged during a cycle at the load end of the line, and it may be that energy is exchanged during a cycle at the source end of the line, but if the line is sufficiently long, there will exist points where instantanous power is always positive, and therefore, energy always flows in the load to source direction at those points. The notion that a reflected wave in general conveys power over the entire path from load to source is not consistent with the above. This notion is emboddied in common language when talking about 'reflected power', but the language belies the actual phenomena. Owen |
what happens to reflected energy ?
On 30 jun, 19:16, Richard Clark wrote:
On Wed, 30 Jun 2010 14:02:14 -0700 (PDT), lu6etj wrote: I was thinking in quantic number describing the energy of a typical 100 W 80 m oscillator devolped in one second, Hi Miguel, Power? *Energy? *One second? *Choose one to talk about, and perhaps the mystery of numbers might clear up. one quanta an two quanta Quanta? *Two Quanta? *We are now up to four intermixed terms. Simplify. *Choose one thing. of 80 m radiation is 2.3 * 10^ -27 J, that difference (my physic book say) it is unmeasurable experimentally (this energy leap (skip?, hop?) it is in the order of 10^-8 smaller that green light leap True, but immaterial. *You are confusing wavelength and quanta (no surprise given the blearing of topic). *Compare Green and IR. *Is there a correlation on a scale of two that predicts out to a scale of 10^8? * Compare Green and deep IR. *Is there a correlation on a scale of ten that predicts out to a scale of 10^8? * Compare Green and the Sub-millimeter band. *Is there a correlation on a scale of 100 that predicts out to a scale of 10^8? (in reality my book -Resnick Halliday- give a moving dust particle example with quantic number very much lower than my 80 m example yet *= n = 3 * 10^14, they said "we can not distiguish energy difference among n = 3 * 10^14 and n = [3 * 10^14] +1") So a quantum of smaller energy of a dust particle is measureable but 80M transmission is not? *Common sense is wheezing in this dust. OK, so they are talking about the difference in quantum, not energy. Would it surprise you that you cannot even tell the difference between one quanta of green light and two with conventional detecting technology? What it is the ohysical sense of working with magnitudes we can not measure? Nobody (as we know) use (or need) quantum mechanics to deal with (or explain) locomotive movement *:) The limitation is called Quatum Efficiency and the human eye is vastly superior (to all but $1,000,000 components) at rougly QE = 50%. Cecil said Cecil said Yes, yes, I know some people bring very strange ideas into the forums, Indeed. 73's Richard Clark, KB7QHC Dear Richard: On examples we usually start with a visible common data, here I made with a 100 W TX power during one second to gives certain amount of energy, this amount of energy stored in a system (for example a LC tank) gives the quantic number of the system. well... Energy it is Power * Time and n=E/h*v, it easy, it is an electrical cuasi identical example as page 1616 part II Spanih translated Resnick & Halliday book. if for Resnick & Halliday guys is a good example for me is good too :) I do not confussing wavelengh with quanta!, quantized energy it is E=nhv and v it is 1/lambda, how do you calculate E without v in such equation? I don not believe my translations are too wrong! I wrote what book say = "they said "we CAN NOT DISTINGUISH energy difference among n = 3 * 10^14 and n = [3 * 10^14] +1"), (page 1652 op.cit.); where you read: "quantum of smaller energy of a dust particle is measureable"? my text says just the opposite! You say: "OK, so they are talking about the difference in quantum, not energy" I do not know if I am translating well your sentence... perhaps you refer to my missuse of the latin word quanta (plural) instead "quantum" (singular) (in spanish we usually say "cuanto/ cuantos" -not latin-, in english I believe you use latin, sorry by my translating error), but I think not is that. Quantum in this context is "energy quantum", they are talking about difference of energy, that difference it is not continuos but quantized, and each energy quantum is 2.3 * 10^ -27 J, one quantum, two quantum... n*quantum, n*quantum in the system = E (op. cit. page 1615), what is wrong? I am talking about 80 m technically useles quantum treatment, and you say to me: "The limitation is called Quatum Efficiency and the human eye is vastly superior (to all but $1,000,000 components) at rougly QE = 50%." What sort of human eye we use to see 80 m "light"? :) I did not want go out off topic, I claimed quantum mechanics do not help so much to solve TL related problems and give some reasons for that. I am not an expert in quantum physics and I am not going further that my elementary physic book examples. Are they wrong? well... then, I am wrong too :) PSE do not argue with me, I am innocent of charges, read the references... 73 - Miguel - LU6ETJ |
what happens to reflected energy ?
Owen Duffy wrote:
Considering the steady state... If we accept the P(t) is the product of instantaneous voltage and current, then there will be some points on any mismatched line where P(t) is always positive. In between those points, P(t) will have positive and negative excursions. I think that it is a reasonable interpretation that at those points where P(t) is always positive, then there is never at any instant, a flow of energy away from the load, energy is never exchanged during a cycle across those points, it always flows from source to load. I'd make a small addition, that . . .there is never at any instant a *net* flow of energy away from the load. . . The problem is that I don't know of any way to keep track of a particular bundle of energy -- it gets mixed together. So you could have energy constantly flowing both ways through a point while maintaining a net flow (power) in one direction and it would look just the same as energy going only one way. Keith's DC thought experiments illustrate these different approaches and some of their logical -- and illogical -- consequences. Quite some time ago I wrote and made available a little graphic program showing the voltage, current, power, and energy on lines under several conditions. When a complete standing wave exists, there are points of zero voltage and current and hence zero power. For one half the cycle you can see energy moving into those points equally from both directions (obviously being stored at the node), and during the other half, energy is moving out of those points in both directions (being retrieved from storage). One interpretation is that the energy arriving from the left exits to the right, and vice-versa, and that fits neatly into the concept of waves of energy simultaneously moving in both directions. Or you can decide that the energy which came in from the right exits to the right, and in from the left exits to the left. If that's your interpretation, then you conclude that no energy ever crosses the boundary. I think this is the genesis of Cecil's view that energy waves somehow bounce off the standing wave node. Both interpretations fit equally well with the observed net flow of energy but, like Keith's DC thought experiments and Cecil's writings show, take you down quite different paths when trying to divine some concept of what's fundamentally happening. . . . Roy Lewallen, W7EL |
what happens to reflected energy ?
On Jun 30, 10:03*pm, Roy Lewallen wrote:
lu6etj wrote: On 29 jun, 15:08, Cecil Moore wrote: On Jun 29, 12:54 pm, Jim Lux wrote: photons can flow through a dielectric.. isn't that what EM propagation is, after all? Yes, after I posted it, I realized that it was a rhetorical question. -- 73, Cecil, w5dxp.com I learnt displacement current inside a condenser it was = eo* d(phi E)/ dt no EM radiation inside the condenser to made that current possible, in any case EM radiation in physical condenser will come out from condenser to the rest of the universe :). I also learnt photons was necessary to explain certain energy interchange phenomena such as fotoelectric effect or subatomic particle interactions, wave-particle duality for me means "duality", not "wave kaput" :) to account for EM wave well explainable phenomenom. As it was taught to me (I am not physicist), quantum nature of a 80 m wavelenght energy it is useless for calculations and invisible to our instrument resolution because its immensely large quantic number. Is it wrong? Miguel LU6ETJ Photons are very useful in the analysis of transmission lines. They can be brought into the discussion to divert it from taking a path that makes a participant uncomfortable. If unable to answer a question logically, simply toss photons, optics, quantum mechanics, aether, and other confounding factors in, and presto, people will begin arguing about the spurious concepts and forget that you've avoided answering the difficult question. It's called misdirection, a time-honored technique used by politicians and prestidigitators as well as promoters of pseudoscience. Roy Lewallen, W7EL yeah, ain't it great fun! |
| All times are GMT +1. The time now is 03:21 AM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com