|
The Rest of the Story
Keith Dysart wrote:
There was an 'if' there, wasn't there? Do you think the 'if' is satisfied? Or not? The rest is useless without knowing. Under the laws of physics governing transmission lines inserting an ideal 1WL line does not change the steady- state conditions. If you think it does, you have invented some new laws of physics. You still have to explain where this destructive energy is stored for those 90 degrees. Please identify the element and its energy flow as a function of time. Your request is beyond the scope of my Part 1 article. If interference exists at the source resistor, the energy associated with the interference flows to/from the source and/or to/from the load. That condition is NOT covered in my Part 1 article. Please stand by for Part 2 which will explain destructive interference and Part 3 which will explain constructive interference. One advantage of moving the source voltage one wavelength away from the source resistor is that it is impossible for the source to respond instantaneously You have previously claimed that the steady-state conditions are the same (which I agree), Glad you agree so there is nothing stopping you from an analysis of the following example: source---1WL 50 ohm---Rs---1WL 50 ohm---+j50 Pfor1-- Pfor2-- --Pref1 --Pref2 Make Rs a 4-terminal network and a standard s-parameter analysis is possible. but now you have moved to discussing transients, for which the behaviour is quite different. Nope, you are confused. I am saying absolutely nothing about transients. Why do you think an instantaneous power analysis during steady-state is not possible? If you want to claim similarity, then you need to allow the circuit to settle to steady state after any change. Instantaneous response is not required if the analysis is only steady-state. Are you saying that an analysis of instantaneous power does not apply during steady-state? If that is true, then all of your earlier analysis involving transmission lines is bogus. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)
Keith Dysart wrote: On Apr 4, 11:41*am, Roger Sparks wrote: On Fri, 4 Apr 2008 06:30:08 -0700 (PDT) Keith Dysart wrote: On Apr 4, 1:29*am, Roger Sparks wrote: [snip] Another way to figure the power to the source would be by using the voltage and current through the source. * This is how I did it. Taking Esource.50[90..91] = 0.03046 J as an example ... Psource.50[90] = V * I * * * * * * * *= 0.000000 * 0.000000 * * * * * * * *= 0 W Psource.50[91] = -2.468143 * -0.024681 * * * * * * * *= 0.060917 W Using the trapezoid rule for numerical integration, Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval * * * * * * * * * *= ((0+0.060917)/2)*1 * * * * * * * * * *= 0.030459 J The other powers and energies in the spreadsheet are computed similarly. There was an error in the computation of the 'delta's; the sign was wrong. The spreadsheet available athttp://keith.dysart.googlepages.com/radio6 has now been corrected. (And, for Cecil, this spreadsheet no longer has macros so it may be downloadable.) To use this spreadsheet to compute my numbers above, set the formulae for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero. Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J. I could not match to the above data to any rows, so I can't comment. But perhaps the explanation above will correct the discrepancy. I took a look at your revised spreadsheet entitled "Reflected45degrees-1.xls". Using the numbers from row 94 (91 degrees), the voltage developed at the source would be Vs = -2.468143v. *The current folowing through the source would be found from Ig which is 1.389317a in todays version of the spreadsheet (it was 1.439a previously). *The power flowing INTO the source is 2.468143* 1.389317 = 3.429032w. *This is the power Ps found in Column 11. * This returning power is all from the reflected wave. * I would not say this. The power *is* from the line, but this is Pg, and it satisfies the equation Ps(t) = Prs(t) + Pg(t) The imputed power in the reflected wave is Pr.g(t) and is equal to -99.969541 W, at 91 degrees. This can not be accounted for in any combination of Ps(91) (-3.429023 W) and Prs(91) (96.510050 W). And recall that expressing Cecil's claim using instantaneous powers requires that the imputed reflected power be accounted for in the source resistor, and not the source. This is column 26 and would require that Prs(91) equal 100 W (which it does not). Taking these numbers and adding the Psource.50[91] = -2.468143 * -0.024681*= 0.060917 W found previously above, we have Total = 0.060917 + 3.429023 + 96.510050 = 99.99999. Very close to 100w, but I am not sure of exactly what I am adding here. I think you are right, that we should always be able to add the power on the line to 100w if we truely account for the power at any instant. This results from the trig identity that sin(x) + cos(x) = 1. Sin(x+90)= cos(x) so sin(x)+sin(x+90)= 1. This explains why the sum of the forward power and reflected power should always equal 100w. Pg(t) is the result of a standing wave, containing power from Pf(x) and Pr(x+90). Only the power from Pr(x+90) is available to at a later time Prs(t+90+delta). Power from Pf(x+delta) is found in the transmission line. The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. * This is not a good way to describe the source. The ratio of the voltage to the current is 1.776 but this is not a resistor since if circuit conditions were to change, the voltage would stay the same while the current could take on any value; this being the definition of a voltage source. Since the voltage does not change when the current does, deltaV/deltaI is always 0 so the voltage source is more properly described as having an impedance of 0. As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. * The returning reflection is affectively a change in the circuit conditions. Using the source impedance of 0 plus the 50 ohm resistor means the reflection sees 50 ohms, so there is no reflection. Using your approach of computing a resistance from the instantaneous voltage and current yeilds a constantly changing resistance. The reflection would alter this computed resistance. This change in resistance would then alter the reflection which would change the resistance. Would the answer converge? The only approach that works is to use the conventional approach of considering that a voltage source has an impedance of 0. The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. That negates the idea that the source has an impedance of zero when we also assign the source a voltage. Of course the result is a another reflection. *Is this the idea you were trying to communicate Cecil? To me, this is destructive interference at work, so all the power in the reflected wave does not simply disappear into the resistor Rs on the instant basis. * I agree with latter, but not for the reason expressed. Rather, because the imputed power of the reflected wave is a dubious concept. This being because it is impossible to account for this power. Does it help to notice that we are applying the power through two wires/paths. We then divide the power in half and say that the power to the resistor from the close half flows seamlessly from source to resistor. On the other hand, the second half of the source power flows through the transmission line and we say that it may never get to the resistor. Something is wrong with this logic. If we can't account for the power, it is because we are doing the accounting incorrectly. clip Yes. *I don't want to force my conventions onto you, so I am trying to understand yours while insisting that the answers from each convention must be the same. * I think we both agree that the reflections are carrying power now. Not I. Not until the imputed power can be accounted for. ...Keith I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. Maybe if we pursue the trig identity sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers? -- 73, Roger, W7WKB |
The Rest of the Story
correction
On Sat, 5 Apr 2008 07:06:22 -0700 Roger Sparks wrote: On Sat, 5 Apr 2008 03:06:18 -0700 (PDT) Keith Dysart wrote: On Apr 4, 11:41*am, Roger Sparks wrote: On Fri, 4 Apr 2008 06:30:08 -0700 (PDT) Keith Dysart wrote: On Apr 4, 1:29*am, Roger Sparks wrote: [snip] Another way to figure the power to the source would be by using the voltage and current through the source. * This is how I did it. Taking Esource.50[90..91] = 0.03046 J as an example ... Psource.50[90] = V * I * * * * * * * *= 0.000000 * 0.000000 * * * * * * * *= 0 W Psource.50[91] = -2.468143 * -0.024681 * * * * * * * *= 0.060917 W Using the trapezoid rule for numerical integration, Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval * * * * * * * * * *= ((0+0.060917)/2)*1 * * * * * * * * * *= 0.030459 J The other powers and energies in the spreadsheet are computed similarly. There was an error in the computation of the 'delta's; the sign was wrong. The spreadsheet available athttp://keith.dysart.googlepages.com/radio6 has now been corrected. (And, for Cecil, this spreadsheet no longer has macros so it may be downloadable.) To use this spreadsheet to compute my numbers above, set the formulae for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero. Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J. I could not match to the above data to any rows, so I can't comment. But perhaps the explanation above will correct the discrepancy. I took a look at your revised spreadsheet entitled "Reflected45degrees-1.xls". Using the numbers from row 94 (91 degrees), the voltage developed at the source would be Vs = -2.468143v. *The current folowing through the source would be found from Ig which is 1.389317a in todays version of the spreadsheet (it was 1.439a previously). *The power flowing INTO the source is 2.468143* 1.389317 = 3.429032w. *This is the power Ps found in Column 11. * This returning power is all from the reflected wave. * I would not say this. The power *is* from the line, but this is Pg, and it satisfies the equation Ps(t) = Prs(t) + Pg(t) The imputed power in the reflected wave is Pr.g(t) and is equal to -99.969541 W, at 91 degrees. This can not be accounted for in any combination of Ps(91) (-3.429023 W) and Prs(91) (96.510050 W). And recall that expressing Cecil's claim using instantaneous powers requires that the imputed reflected power be accounted for in the source resistor, and not the source. This is column 26 and would require that Prs(91) equal 100 W (which it does not). Taking these numbers and adding the Psource.50[91] = -2.468143 * -0.024681*= 0.060917 W found previously above, we have Total = 0.060917 + 3.429023 + 96.510050 = 99.99999. Very close to 100w, but I am not sure of exactly what I am adding here. I think you are right, that we should always be able to add the power on the line to 100w if we truely account for the power at any instant. This results from the trig identity that sin(x) + cos(x) = 1. Sin(x+90)= cos(x) so sin(x)+sin(x+90)= 1. This explains why the sum of the forward power and reflected power should always equal 100w. Pg(t) is the result of a standing wave, containing power from Pf(x) and Pr(x+90). Only the power from Pr(x+90) is available to Rs at a later time Prs(t+90+delta). Power from Pf(x+delta) is found in the transmission line. I omitted the Rs in "available to Rs at a later time Prs(t+90+delta)" Sorry! The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. * This is not a good way to describe the source. The ratio of the voltage to the current is 1.776 but this is not a resistor since if circuit conditions were to change, the voltage would stay the same while the current could take on any value; this being the definition of a voltage source. Since the voltage does not change when the current does, deltaV/deltaI is always 0 so the voltage source is more properly described as having an impedance of 0. As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. * The returning reflection is affectively a change in the circuit conditions. Using the source impedance of 0 plus the 50 ohm resistor means the reflection sees 50 ohms, so there is no reflection. Using your approach of computing a resistance from the instantaneous voltage and current yeilds a constantly changing resistance. The reflection would alter this computed resistance. This change in resistance would then alter the reflection which would change the resistance. Would the answer converge? The only approach that works is to use the conventional approach of considering that a voltage source has an impedance of 0. The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. That negates the idea that the source has an impedance of zero when we also assign the source a voltage. Of course the result is a another reflection. *Is this the idea you were trying to communicate Cecil? To me, this is destructive interference at work, so all the power in the reflected wave does not simply disappear into the resistor Rs on the instant basis. * I agree with latter, but not for the reason expressed. Rather, because the imputed power of the reflected wave is a dubious concept. This being because it is impossible to account for this power. Does it help to notice that we are applying the power through two wires/paths. We then divide the power in half and say that the power to the resistor from the close half flows seamlessly from source to resistor. On the other hand, the second half of the source power flows through the transmission line and we say that it may never get to the resistor. Something is wrong with this logic. If we can't account for the power, it is because we are doing the accounting incorrectly. clip Yes. *I don't want to force my conventions onto you, so I am trying to understand yours while insisting that the answers from each convention must be the same. * I think we both agree that the reflections are carrying power now. Not I. Not until the imputed power can be accounted for. ...Keith I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. Maybe if we pursue the trig identity sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers? -- 73, Roger, W7WKB -- 73, Roger, W7WKB |
The Rest of the Story
Roger Sparks wrote:
The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. That negates the idea that the source has an impedance of zero when we also assign the source a voltage. Consider that what you are seeing is the flip side of the interference at the source resistor. When a local source is present, it can certainly absorb destructive interference energy and supply constructive interference energy. The following example has identical steady-state conditions but brings Pfor1 and Pref1 into play for the instantaneous values. I suspect that Pref1 is being completely ignored in the present analysis. Vs(t)---1WL 50 ohm---Rs---1WL 50 ohm---+j50 Pfor1-- Pfor2-- --Pref1 --Pref2 If we can't account for the power, it is because we are doing the accounting incorrectly. Try the above example and maybe it will become clear. Pref1 = Pfor1(rho1^2) + Pref2(1-rho2^2) + interference1 Pfor2 = Pfor1(1-rho1^2) + Pref2(rho2^2) + interference2 The source power doesn't appear directly in the equations and need not be considered at all. Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 (all average) I suspect the above equation will account for all the energy components even at the instantaneous level such that: Pfor1(t) + Pref2(t) + P.Rs(t) = Pfor2(t) + Pref1(t) Please note that all of these power components exist when the two transmission lines are removed so this analysis is probably the key to understanding what is wrong with the earlier analysis. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Sat, 05 Apr 2008 10:01:13 -0500
Cecil Moore wrote: Roger Sparks wrote: The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. That negates the idea that the source has an impedance of zero when we also assign the source a voltage. Consider that what you are seeing is the flip side of the interference at the source resistor. When a local source is present, it can certainly absorb destructive interference energy and supply constructive interference energy. The following example has identical steady-state conditions but brings Pfor1 and Pref1 into play for the instantaneous values. I suspect that Pref1 is being completely ignored in the present analysis. Vs(t)---1WL 50 ohm---Rs---1WL 50 ohm---+j50 Pfor1-- Pfor2-- --Pref1 --Pref2 I think this would be right on the average basis. Because any sine waves of identical frequency can ultimately be added to make one wave, we can describe a single sine wave on the source side and another wave on the load/reflection side. There is no need for constructive or destructive interference as part of the final sine wave description. On the other hand, if we want to understand how the final wave is assembled for each side of the resistor, we need the idea of constructive and destructive interference. Your pictorial showed only the single reflection, but as you have explained previously, there are many more reflections between the resistor and +j50 points until the power on the +j50 side finally stabalizes to some power level. It seems to me like my spreadsheet found at http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf captures a description of the forward and reflected waves as a single equation for each side of the resistor. The forward wave is y*sin(t) and reflected wave is y*sin(t+90). If we want to learn how to find out how the source load begins at 100 ohms resistive and changes to 70.7 ohms reactive, we can either notice how the peak current has shifted from resistive to reactive (45 degree) from the combined waves described on my spreadsheet, or we can add all the reflections on each side of the resistor and come to the combined wave after many additions of ever smaller reflections. If we can't account for the power, it is because we are doing the accounting incorrectly. Try the above example and maybe it will become clear. Pref1 = Pfor1(rho1^2) + Pref2(1-rho2^2) + interference1 Pfor2 = Pfor1(1-rho1^2) + Pref2(rho2^2) + interference2 Interference1 and interference2 would be the combined effects of successive ever smaller reflections. Frankly, I don't see that combining the smaller reflections in this way to be of much value. It is like saying that the first reflection is special and the subsequent reflections equally special so we will give subsequent refections a group name, "interference". To me, all the reflections are "interference" and must be either added or subtracted in the same manner. The source power doesn't appear directly in the equations and need not be considered at all. We have to have a starting point, which must be the source power. How could we avoid having a starting point? Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 (all average) I suspect the above equation will account for all the energy components even at the instantaneous level such that: Pfor1(t) + Pref2(t) + P.Rs(t) = Pfor2(t) + Pref1(t) Please note that all of these power components exist when the two transmission lines are removed so this analysis is probably the key to understanding what is wrong with the earlier analysis. -- I think this equation is missing some terms, reflections 2,3,4...n. -- 73, Roger, W7WKB |
The Rest of the Story
Roger Sparks wrote:
We have to have a starting point, which must be the source power. How could we avoid having a starting point? We are analyzing a resistor isolated from the source and load by 1WL of ideal feedline on each side. All we need are the steady-state values of Pfor1, Pref1, Pfor2, and Pref2. It's much like an s-parameter analysis of a 4-terminal box with a 50 ohm series resistor. If we know a1 (Vfor1) and a2 (Vref2), that is all that needs to be known to complete the analysis. We don't actually need to know anything about the source and the load if a1 and a2 are given. I think this equation is missing some terms, reflections 2,3,4...n. The equation already contains all of those "missing' terms because it is steady-state and all of those terms have already been added in to the total. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Sat, 05 Apr 2008 14:56:35 -0500
Cecil Moore wrote: Roger Sparks wrote: We have to have a starting point, which must be the source power. How could we avoid having a starting point? We are analyzing a resistor isolated from the source and load by 1WL of ideal feedline on each side. All we need are the steady-state values of Pfor1, Pref1, Pfor2, and Pref2. It's much like an s-parameter analysis of a 4-terminal box with a 50 ohm series resistor. If we know a1 (Vfor1) and a2 (Vref2), that is all that needs to be known to complete the analysis. We don't actually need to know anything about the source and the load if a1 and a2 are given. I think this equation is missing some terms, reflections 2,3,4...n. The equation already contains all of those "missing' terms because it is steady-state and all of those terms have already been added in to the total. -- 73, Cecil http://www.w5dxp.com From your previous posting, you gave the equation Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 (all average) Where Pfor1 and Pref2 were toward the resistor, and Pfor2 and Pref1 were away from the resistor. This equation seems to assume that the power to the resistor is zero. This is because the same current must flow in Pfor1 and Pfor2, and the same current in Pref2 and Pref1. The voltage drop through the resistor only happens when the circuit is complete on both sides of the resistor. If we try to separate them as you are suggesting, we would have to recognize that it is also correct to say that Pfor1 + Pref2 = Pfor2 + Pref1 + P.Rs (all average) This is true because power only flows when the circuit is complete, and we can not say all the power comes from the left, or from the right. The power only flows when both are connected so the power must come from both sides. There will never be one answer for your equation, unless, as you suggest, a1 and a2 are given. -- 73, Roger, W7WKB |
The Rest of the Story
On Sat, 5 Apr 2008 15:28:59 -0700, Roger Sparks wrote: BIG SNIPPAGE Why don't you two take this to private email? |
The Rest of the Story
Roger Sparks wrote:
Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 (all average) Sorry for the obvious typo. The sign of P.Rs should have been minus. This equation seems to assume that the power to the resistor is zero. No, these are really basic concepts. The energy flow away from a junction is equal to the energy flow into the junction minus the dissipation in the junction. (Pfor1 + Pref2) is the energy flow into the junction P.Rs is the dissipation in Rs (Pfor2 + Pref1) is the energy flow away from the junction. (Pfor1 + Pref2) - P.Rs = (Pfor2 + Pref1) or (Pfor1 + Pref2) = (Pfor2 + Pref1) + P.Rs There will never be one answer for your equation, unless, as you suggest, a1 and a2 are given. a1 and a2 *are* given - at least indirectly so they can be calculated. Since we know Vs, Rs, and ZL, and given the two 1WL feedlines, we can easily calculate a1 and a2. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Lets put the 'Rest of the Story' to bed before it becomes 'The Never Ending
Story' or are we too late already. Mike. VK6MO "Cecil Moore" wrote in message . .. Roger Sparks wrote: Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 (all average) Sorry for the obvious typo. The sign of P.Rs should have been minus. This equation seems to assume that the power to the resistor is zero. No, these are really basic concepts. The energy flow away from a junction is equal to the energy flow into the junction minus the dissipation in the junction. (Pfor1 + Pref2) is the energy flow into the junction P.Rs is the dissipation in Rs (Pfor2 + Pref1) is the energy flow away from the junction. (Pfor1 + Pref2) - P.Rs = (Pfor2 + Pref1) or (Pfor1 + Pref2) = (Pfor2 + Pref1) + P.Rs There will never be one answer for your equation, unless, as you suggest, a1 and a2 are given. a1 and a2 *are* given - at least indirectly so they can be calculated. Since we know Vs, Rs, and ZL, and given the two 1WL feedlines, we can easily calculate a1 and a2. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Apr 6, 7:02*am, "Mike" wrote:
Lets put the 'Rest of the Story' to bed before it becomes 'The Never Ending Story' or are we too late already. Too late. The first reference I could find in which Cecil is chasing the imputed energy in reflected waves is "Is Maxwell wrong?", posted December 31, 1994. Soon to be a decade and a half. ...Keith |
The Rest of the Story
Mike wrote:
Lets put the 'Rest of the Story' to bed before it becomes 'The Never Ending Story' or are we too late already. Do you have anything technical to contribute to resolving the on-topic thread issues? If not, it is very likely to continue until it is resolved. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Keith Dysart wrote:
Soon to be a decade and a half. One would think that by now someone would have figured it out. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Apr 6, 9:42*am, Cecil Moore wrote:
Keith Dysart wrote: Soon to be a decade and a half. One would think that by now someone would have figured it out. Perhaps, like the search for Sasquatch*, doomed to failure because of the object's non-existance. ...Keith * Substitute any of: Yeti, Nessie, Champ, Bigfoot, Memphremagog Monster, Ogopogo, Reflected Wave Power. |
The Rest of the Story
Keith Dysart wrote:
Perhaps, like the search for Sasquatch*, doomed to failure because of the object's non-existance. I agree that the conjugate of Sasquatch is probably non-existent. :-) -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Apr 5, 9:05*am, Cecil Moore wrote:
Keith Dysart wrote: Please expand on what it tells us "about what is wrong with the analysis so far". You have not been able to tell where the instantaneous reflected energy goes. This new example should help you solve your problem. It is not really my problem. It is only a problem for those who expect the imputed power of a partial contributor to superposition to represent a real energy flow. If your steady-state equations for the new example are not identical to the steady- state equations for the previous example, then something is wrong with your previous analysis. I would fully expect the same results. IMO, something is obviously wrong with your previous analysis since it requires reflected waves to contain something other than ExH joules/sec, a violation of the laws of EM wave physics. Not at all a violation. Just as one does not expect the partial values of volts and currents during superposition to produce a power value that represents a real energy flow, one should not expect it from Es and Hs which are the partial values being superposed. * * * * * feedline1 * * * * feedline2 source---1WL 50 ohm---Rs---1WL 50 ohm---+j50 The beauty of this example is that conditions at the source resistor, Rs, are isolated from any source of energy other than the ExH forward wave energy in feedline1 and the ExH reflected wave energy in feedline2. That's all the energy there is available at Rs and that energy cannot be denied. I suppose we shall see when you complete the arithmetic. ...Keith |
The Rest of the Story
On Apr 5, 10:02*am, Cecil Moore wrote:
Keith Dysart wrote: There was an 'if' there, wasn't there? Do you think the 'if' is satisfied? Or not? The rest is useless without knowing. Under the laws of physics governing transmission lines inserting an ideal 1WL line does not change the steady- state conditions. If you think it does, you have invented some new laws of physics. You still have to explain where this destructive energy is stored for those 90 degrees. Please identify the element and its energy flow as a function of time. Your request is beyond the scope of my Part 1 article. If interference exists at the source resistor, the energy associated with the interference flows to/from the source and/or to/from the load. That condition is NOT covered in my Part 1 article. Please stand by for Part 2 which will explain destructive interference and Part 3 which will explain constructive interference. One advantage of moving the source voltage one wavelength away from the source resistor is that it is impossible for the source to respond instantaneously You have previously claimed that the steady-state conditions are the same (which I agree), Glad you agree so there is nothing stopping you from an analysis of the following example: True, but as you say, the results will be the same. source---1WL 50 ohm---Rs---1WL 50 ohm---+j50 * * * * * *Pfor1-- * * * * *Pfor2-- * * * * * *--Pref1 * * * * *--Pref2 Make Rs a 4-terminal network and a standard s-parameter analysis is possible. Yes, but that would be an average analysis and we have already seen how averages mislead. but now you have moved to discussing transients, for which the behaviour is quite different. Nope, you are confused. I am saying absolutely nothing about transients. You did say: "One advantage of moving the source voltage one wavelength away from the source resistor is that it is impossible for the source to respond instantaneously." The words "respond instantaneously" suggested transient, rather than waiting for the system to settle. Why do you think an instantaneous power analysis during steady-state is not possible? Haven't said that. In fact, I think that is what I have been doing. If you want to claim similarity, then you need to allow the circuit to settle to steady state after any change. Instantaneous response is not required if the analysis is only steady-state. ...Keith |
The Rest of the Story
On Apr 5, 10:06*am, Roger Sparks wrote:
On Sat, 5 Apr 2008 03:06:18 -0700 (PDT) Keith Dysart wrote: On Apr 4, 11:41*am, Roger Sparks wrote: On Fri, 4 Apr 2008 06:30:08 -0700 (PDT) Keith Dysart wrote: On Apr 4, 1:29*am, Roger Sparks wrote: [snip] Another way to figure the power to the source would be by using the voltage and current through the source. * This is how I did it. Taking Esource.50[90..91] = 0.03046 J as an example ... Psource.50[90] = V * I * * * * * * * *= 0.000000 * 0.000000 * * * * * * * *= 0 W Psource.50[91] = -2.468143 * -0.024681 * * * * * * * *= 0.060917 W Using the trapezoid rule for numerical integration, Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval * * * * * * * * * *= ((0+0.060917)/2)*1 * * * * * * * * * *= 0.030459 J The other powers and energies in the spreadsheet are computed similarly. There was an error in the computation of the 'delta's; the sign was wrong. The spreadsheet available athttp://keith.dysart.googlepages.com/radio6 has now been corrected. (And, for Cecil, this spreadsheet no longer has macros so it may be downloadable.) To use this spreadsheet to compute my numbers above, set the formulae for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero. Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J. I could not match to the above data to any rows, so I can't comment. But perhaps the explanation above will correct the discrepancy. I took a look at your revised spreadsheet entitled "Reflected45degrees-1.xls". Using the numbers from row 94 (91 degrees), the voltage developed at the source would be Vs = -2.468143v. *The current folowing through the source would be found from Ig which is 1.389317a in todays version of the spreadsheet (it was 1.439a previously). *The power flowing INTO the source is 2.468143* 1.389317 = 3.429032w. *This is the power Ps found in Column 11. * This returning power is all from the reflected wave. * I would not say this. The power *is* from the line, but this is Pg, and it satisfies the equation * Ps(t) = Prs(t) + Pg(t) The imputed power in the reflected wave is Pr.g(t) and is equal to -99.969541 W, at 91 degrees. This can not be accounted for in any combination of Ps(91) (-3.429023 W) and Prs(91) (96.510050 W). And recall that expressing Cecil's claim using instantaneous powers requires that the imputed reflected power be accounted for in the source resistor, and not the source. This is column 26 and would require that Prs(91) equal 100 W (which it does not). Taking these numbers and adding the Psource.50[91] = -2.468143 * -0.024681*= 0.060917 W found previously above, we have Total = 0.060917 + 3.429023 + 96.510050 = 99.99999. *Very close to 100w, but I am not sure of exactly what I am adding here. Using my notation where Ptot = Pfor + Pref we have Pg(t) = Pf.g(t) + Pr.g(t) = 50 + 50 cos(2wt) + (-50 + 50 cow(2wt)) = 100 cos(2wt) If you subtract Pr.g from Pf.g(t) you get 100. If you follow through the math, you will find that this is what you have done. Ps.50[91] + (- Ps.0[91]) + Prs[91] which is 2 * Pf.g[91] + (-Prs[91] - Pg[91]) + Prs[91] which is Pf.g[91] + Pf.g[91] - prs[91] - Pf.g[91] - Pr.g[91] + Prs[91] which is Pf.g[91] - Pr.g[91] which is 100 I think you are right, that we should always be able to add the power on the line to 100w if we truely account for the power at any instant. *This results from the trig identity that sin(x) + cos(x) = 1. *Sin(x+90)= cos(x) so sin(x)+sin(x+90)= 1. *This explains why the sum of the forward power and reflected power should always equal 100w. If I recall correctly, it is sin**2 + cos**2 = 1. Pg(t) is the result of a standing wave, containing *power from Pf(x) and Pr(x+90). * This is one way of thinking of it, but it is less misleading to consider that Pg(t) describes the actual energy flow, just as Vg(t) describes the actual voltage and Ig(t) describes the actual current. Using superposition Vf, If, Vr and Ir can be derived and from these Pf and Pr. Only the power from Pr(x+90) is available to at a later time Prs(t+90+delta). *Power from Pf(x+delta) is found in the transmission line. The source is acting like a resistor with an impedance of 2.468143/1.389317 = 1.776 ohms. * This is not a good way to describe the source. The ratio of the voltage to the current is 1.776 but this is not a resistor since if circuit conditions were to change, the voltage would stay the same while the current could take on any value; this being the definition of a voltage source. Since the voltage does not change when the current does, deltaV/deltaI is always 0 so the voltage source is more properly described as having an impedance of 0. As a result, the returning reflection does not truely see 50 ohms but sees 50 + 1.776 = 51.776 ohms. * The returning reflection is affectively a change in the circuit conditions. Using the source impedance of 0 plus the 50 ohm resistor means the reflection sees 50 ohms, so there is no reflection. Using your approach of computing a resistance from the instantaneous voltage and current yeilds a constantly changing resistance. The reflection would alter this computed resistance. This change in resistance would then alter the reflection which would change the resistance. Would the answer converge? The only approach that works is to use the conventional approach of considering that a voltage source has an impedance of 0. The overriding issue is to account for all the power, which we are having a hard time doing. *The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. *That negates the idea that the source has an impedance of zero when we also assign the source a voltage. When a current flows out of a voltage source, the source is providing energy. When the current flows into it, the voltage source is absorbing energy. A charging battery is reasonable real world example. Of course the result is a another reflection. *Is this the idea you were trying to communicate Cecil? To me, this is destructive interference at work, so all the power in the reflected wave does not simply disappear into the resistor Rs on the instant basis. * I agree with latter, but not for the reason expressed. Rather, because the imputed power of the reflected wave is a dubious concept. This being because it is impossible to account for this power. Does it help to notice that we are applying the power through two wires/paths. *We then divide the power in half and say that the power to the resistor from the close half flows seamlessly from source to resistor. *On the other hand, the second half of the source power flows through the transmission line and we say that it may never get to the resistor. *Something is wrong with this logic. * If we can't account for the power, it is because we are doing the accounting incorrectly. And the error in the accounting may be the expectation that the particular set of powers chosen should balance. Attempting to account for Pr fails when Pr is the imputed power from a partial voltage and current because such computations do not yield powers which exist. clip Yes. *I don't want to force my conventions onto you, so I am trying to understand yours while insisting that the answers from each convention must be the same. * I think we both agree that the reflections are carrying power now. Not I. Not until the imputed power can be accounted for. ...Keith I am convinced that accounting for the power can be done on an instant basis, but neither of us has done it on a mutually acceptable basis yet. *Maybe if we pursue the trig identity sin(x) + sin(x+90) = 1, we can have a constant reference and avoid mixing forward and reflected powers? Except that it is sin squared + cos squared that equals one. ...Keith |
The Rest of the Story
On Apr 5, 11:01*am, Cecil Moore wrote:
Roger Sparks wrote: The overriding issue is to account for all the power, which we are having a hard time doing. The current is flowing the wrong way for the source voltage at Ps(91) so the source is absorbing power. *That negates the idea that the source has an impedance of zero when we also assign the source a voltage. Consider that what you are seeing is the flip side of the interference at the source resistor. When a local source is present, it can certainly absorb destructive interference energy and supply constructive interference energy. The following example has identical steady-state conditions but brings Pfor1 and Pref1 into play for the instantaneous values. I suspect that Pref1 is being completely ignored in the present analysis. Vs(t)---1WL 50 ohm---Rs---1WL 50 ohm---+j50 * * * * * Pfor1-- * * * * * Pfor2-- * * * * * --Pref1 * * * * * --Pref2 If we can't account for the power, it is because we are doing the accounting incorrectly. Try the above example and maybe it will become clear. Pref1 = Pfor1(rho1^2) + Pref2(1-rho2^2) + interference1 Pfor2 = Pfor1(1-rho1^2) + Pref2(rho2^2) + interference2 The source power doesn't appear directly in the equations and need not be considered at all. Pfor1 + Pref2 + P.Rs = Pfor2 + Pref1 * (all average) I suspect the above equation will account for all the energy components even at the instantaneous level such that: Pfor1(t) + Pref2(t) + P.Rs(t) = Pfor2(t) + Pref1(t) Please note that all of these power components exist when the two transmission lines are removed so this analysis is probably the key to understanding what is wrong with the earlier analysis. When the two transmission lines are removed, how can these imputed powers exist? What impedance did you assign to the non-existant section of line to permit you to compute these powers? ...Keith |
The Rest of the Story
Keith Dysart wrote:
Just as one does not expect the partial values of volts and currents during superposition to produce a power value that represents a real energy flow, one should not expect it from Es and Hs which are the partial values being superposed. Given two coherent EM waves, W1 and W2, we know that the power in W1 is E1xH1 and the power in W2 is E2xH2. That is simple physics. No matter what the results of superposition of those two waves, the total energy in both waves must be conserved. That is simple physics. If you would think about what happens to two coherent light waves in free space, you wouldn't be able to justify your assertion above. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Keith Dysart wrote:
When the two transmission lines are removed, how can these imputed powers exist? What impedance did you assign to the non-existant section of line to permit you to compute these powers? I have very deliberately chosen and limited the scope of the examples to a 50 ohm environment. Those imputed powers are therefore whatever an ideal 50 ohm directional wattmeter indicates. The distributed network model also works for circuits as it is a superset of the lumped circuit model. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Apr 7, 8:27*am, Cecil Moore wrote:
Keith Dysart wrote: Just as one does not expect the partial values of volts and currents during superposition to produce a power value that represents a real energy flow, one should not expect it from Es and Hs which are the partial values being superposed. Given two coherent EM waves, W1 and W2, we know that the power in W1 is E1xH1 and the power in W2 is E2xH2. That is simple physics. No matter what the results of superposition of those two waves, the total energy in both waves must be conserved. That is simple physics. If you would think about what happens to two coherent light waves in free space, you wouldn't be able to justify your assertion above. I did not expect you would be happy with the answer. As long as you stick with simple assertions, followed by sentences such as, "That is simple physics.", spoken in a tone which says no further understanding is necessary, you will be locked in the fruitless search for the imputed reflected energy flow. ...Keith |
The Rest of the Story
On Apr 7, 8:48*am, Cecil Moore wrote:
Keith Dysart wrote: When the two transmission lines are removed, how can these imputed powers exist? What impedance did you assign to the non-existant section of line to permit you to compute these powers? I have very deliberately chosen and limited the scope of the examples to a 50 ohm environment. Those imputed powers are therefore whatever an ideal 50 ohm directional wattmeter indicates. You seem to be saying that the answers would be completely different if you chose a different impedance for the non-existant transmission line. Or even for the one wavelength line. That does not make a very robust explanation. ...Keith |
The Rest of the Story
Keith Dysart wrote:
I did not expect you would be happy with the answer. If you ever comprehend that those RF waves in a transmission line could just as easily be light waves in free space, you will realize that your "answer" violates the existing laws of physics, i.e. EM waves whose energy content varies according to your whim. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Sun, 6 Apr 2008 19:21:00 -0700 (PDT)
Keith Dysart wrote: On Apr 5, 10:06*am, Roger Sparks wrote: On Sat, 5 Apr 2008 03:06:18 -0700 (PDT) Keith Dysart wrote: On Apr 4, 11:41*am, Roger Sparks wrote: On Fri, 4 Apr 2008 06:30:08 -0700 (PDT) Keith Dysart wrote: On Apr 4, 1:29*am, Roger Sparks wrote: [snip] Another way to figure the power to the source would be by using the voltage and current through the source. * This is how I did it. Taking Esource.50[90..91] = 0.03046 J as an example ... Psource.50[90] = V * I * * * * * * * *= 0.000000 * 0.000000 * * * * * * * *= 0 W Psource.50[91] = -2.468143 * -0.024681 * * * * * * * *= 0.060917 W Using the trapezoid rule for numerical integration, Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval * * * * * * * * * *= ((0+0.060917)/2)*1 * * * * * * * * * *= 0.030459 J The other powers and energies in the spreadsheet are computed similarly. There was an error in the computation of the 'delta's; the sign was wrong. The spreadsheet available athttp://keith.dysart.googlepages.com/radio6 has now been corrected. (And, for Cecil, this spreadsheet no longer has macros so it may be downloadable.) To use this spreadsheet to compute my numbers above, set the formulae for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero. Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J. I could not match to the above data to any rows, so I can't comment. |
The Rest of the Story
Keith Dysart wrote:
As long as you stick with simple assertions, followed by sentences such as, "That is simple physics.", spoken in a tone which says no further understanding is necessary, you will be locked in the fruitless search for the imputed reflected energy flow. My vision is returning and could turn out to be the best vision that I've ever had in my life. :-) You have been asking for the mechanism for storage and return of the interference energy in the system. That mechanism is standing waves. Are you aware that standing waves store energy and return it to the system every 90 degrees? In the examples being discussed, there are standing waves inside the source. For the 1/8WL shorted line, there appears to be 125 watts of forward power and 25 watts of reflected power at points on each side of the source. With powers given in average values, the circuit that you should be using for your instantaneous power equations is: 50 ohm ----50-ohm----/\/\/\/\----50-ohm---- 125w-- 100w 50w-- --25w --50w I will be very surprised if the instantaneous powers don't balance. Your previous problem is that you were using net power values on one side of Rs and component power values on the other side of Rs. That's apples and oranges, a known no-no. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Roger Sparks wrote:
. . . If we remove the transmission line from the circuit, we have an open circuit with no current. Without current, there can be no power. How can power arrive at Rs if there is no power coming through the transmission line? . . . I regret that I haven't been able to take the time to contribute to or even follow this very interesting thread. So please pardon me if the following comment isn't relevant. The quoted statement reminded me of an earlier thread where Keith asked a similar question, and proposed some interesting alternative explanations. The point in question was a current node in a lossless line which had infinite SWR, e.g., an open circuited line. A look at the energy flow in the vicinity of the node reveals a plausible explanation. It turns out that energy is being stored at the node, in the line's capacitance, to the tune of C*V^2/2. For half the cycle, energy is flowing into the node equally from both directions, so the power measured at that node is zero. During the other half of the cycle, energy is flowing out of the node equally in both directions. So the power remains zero at the node for the entire cycle. This is, of course, consistent with the zero current at the node. The node's energy increases to a maximum and down to zero periodically in step with the square of the node voltage. Now, I don't know of any way to assign "ownership" to bundles of energy. But let's suppose that the energy which flows into the node from the left side during the "inhalation" part of the cycle is the energy which flows out to the right during the "exhalation" part of the cycle, and the energy flowing into the node from the right exits on the left. So now we've managed to get energy past the node going in both directions while maintaining zero power and current at the node and conserving energy as we must. This can be seen graphically with the little TLVis1 program I made up a while back -- see the thread of that name for more info. The phenomenon I'm describing can be seen in demo 4, one cycle from the output (right) end of the line. Roy Lewallen, W7EL |
The Rest of the Story
Roy Lewallen wrote:
Now, I don't know of any way to assign "ownership" to bundles of energy. One way is to add a unique bit of modulation to each bundle of wave energy. I am fond of using a TV signal and observing ghosting on the screen. This, of course, assumes that the modulation stays with the same component wave to which it was originally associated. But let's suppose that the energy which flows into the node from the left side during the "inhalation" part of the cycle is the energy which flows out to the right during the "exhalation" part of the cycle, and the energy flowing into the node from the right exits on the left. So now we've managed to get energy past the node going in both directions while maintaining zero power and current at the node and conserving energy as we must. This agrees with the distributed network model. Since there is no impedance discontinuity and no impedor at the node, there can be no reflections at the node. The forward wave flows unimpeded through the node as does the equal magnitude reflected wave. The net energy flow is zero. The average energy flow is zero. Anyone who believes there is zero energy at a standing- wave current node should touch that point on a transmission line (which just happens to be the same point as the maximum voltage anti-node). One must be careful not to confuse the net signal with the component signals. One must be careful not to confuse the average values with the instantaneous values. This can best be visualized using light waves in free space. Unimpeded EM waves do not bounce off of each other. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Keith Dysart wrote:
You seem to be saying that the answers would be completely different if you chose a different impedance for the non-existant transmission line. There you go again, trying to shove your words into my mouth. (Pattooieee!) Please don't do that. You have apparently done the math and found it to be valid so, once again, you have to change the specified conditions in order to try to make your point. A Z0 of 50 ohms is the *only* characteristic impedance that will meet the specified precondition of zero average interference. Choosing any other characteristic impedance will move the example outside of the scope of my Part 1 article. I understand why you want to do such a thing but obfuscation, diversions, and straw men are not part of the scientific method. My Part 1 article has a very narrow scope. Please abide by it. Killing a snake and using it for the transmission line would likewise violate the same specified preconditions. Please feel free to do that and report the results. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Apr 7, 12:14*pm, Roger Sparks wrote:
On Sun, 6 Apr 2008 19:21:00 -0700 (PDT) Keith Dysart wrote: On Apr 5, 10:06*am, Roger Sparks wrote: Pg(t) is the result of a standing wave, containing *power from Pf(x) and Pr(x+90). * This is one way of thinking of it, but it is less misleading to consider that Pg(t) describes the actual energy flow, just as Vg(t) describes the actual voltage and Ig(t) describes the actual current. Using superposition Vf, If, Vr and Ir can be derived and from these Pf and Pr. Your argument is correct to the extent that the power you describe is passing point Pg(t) at the instant (t). *It is the equivalent statement that an observer watching cars pass on the freeway would make, saying "one blue car moving left and one red car moving right, so two cars are passing". *Not wrong, just "how is the information useful"? Pg(t) is the actual power at that point in the circuit. It can be derived by simply multiplying the direct measurement of the actual voltages and currents at that point in the circuit. One measures the same voltages and currents regardless of whether it is a transmission line to the right of point g, or the equivalent lumped circuit element. While Vf, Vr, etc. can be used to derive the same information and, therefore is arguably just a different point of view, Vf and Vr, If and Ir, etc., must always be used in pairs to arrive at the actual circuit conditions. It is when one starts to look at them separately, as if they individually represent some part of reality, that confusion awaits. Thus I strongly suggest that Vg, Ig, Pg, represent reality. The others are a convenient alternative view for the purposes of solving problems. Typically we see Vg split into Vf and Vr, but why stop at two. Why not 3, or 4? Analyzing a two wire telephone line will use four or more, forward to the east, forward to the west, reflected to east, reflected to the west, and sometimes many different reflections. How do we choose how many? Depends on what is convenient for solving the problem. The power of superposition. But assigning too much reality to the individual contributors can be misleading. If we can't account for the power, it is because we are doing the accounting incorrectly. And the error in the accounting may be the expectation that the particular set of powers chosen should balance. Attempting to account for Pr fails when Pr is the imputed power from a partial voltage and current because such computations do not yield powers which exist. If we remove the transmission line from the circuit, we have an open circuit with no current. *Without current, there can be no power. How can power arrive at Rs if there is no power coming through the transmission line? * There is power coming from the transmission line. Looking at Pg(t), some of the time energy flows into the line, later in the cycle it flows out. The energy transfer would be exactly the same if the transmission line was replaced by a lumped circuit element. And we don't need Pf and Pr for an inductor. But this flow is quite different than the flow suggested by Pf and Pr. These suggest a continuous flow in each direction. It is only when they are summed that it becomes clear that flow is first in one direction and then other. Would it help to consider that before the "reflection from the short" arrives, power arrives via the transmission line path but the impedance is 100 ohms for our example, composed of Rs = 50 ohms and transmission line = 50 ohms? *After the "reflection from the short" arrives, the impedance drops to 70.7 ohms so the power to the circuit goes up (assuming a constant voltage source). *How can this happen if power is not carried via the "reflection from the short"? It goes up because the impedance presented by the transmission changes when the reflection returns. This change in impedance alters the circuit conditions and the power in the various elements change. Depending on the details of the circuit, these powers may go up, or they may go down when the reflection arrives. ...Keith |
The Rest of the Story
On Apr 7, 5:31*pm, Cecil Moore wrote:
Keith Dysart wrote: As long as you stick with simple assertions, followed by sentences such as, "That is simple physics.", spoken in a tone which says no further understanding is necessary, you will be locked in the fruitless search for the imputed reflected energy flow. My vision is returning and could turn out to be the best vision that I've ever had in my life. :-) You have been asking for the mechanism for storage and return of the interference energy in the system. That mechanism is standing waves. Are you aware that standing waves store energy and return it to the system every 90 degrees? More correctly, the energy is stored in the distributed capacitance and inductance of the transmission line. In the examples being discussed, there are standing waves inside the source. There is no capacitance or inductance in the source to store energy. For the 1/8WL shorted line, there appears to be 125 watts of forward power and 25 watts of reflected power at points on each side of the source. Not if there is no transmission line. With powers given in average values, the circuit that you should be using for your instantaneous power equations is: * * * * * * * * 50 ohm ----50-ohm----/\/\/\/\----50-ohm---- * * *125w-- * * 100w * * 50w-- * * *--25w * * * * * * * --50w I will be very surprised if the instantaneous powers don't balance. Perhaps. But I don't need more examples where the powers balance. I already have the one example where they don't. And it only takes one to disprove an hypothesis. Your previous problem is that you were using net power values on one side of Rs and component power values on the other side of Rs. But there are no component powers in the source. It is a simple circuit element. ...Keith |
The Rest of the Story
On Apr 8, 8:51*am, Cecil Moore wrote:
Roy Lewallen wrote: Now, I don't know of any way to assign "ownership" to bundles of energy. One way is to add a unique bit of modulation to each bundle of wave energy. I am fond of using a TV signal and observing ghosting on the screen. This, of course, assumes that the modulation stays with the same component wave to which it was originally associated. But as soon as you modulate, you no longer have sinusoidal steady state. You can split the signal up into its spectral components and, using superposition, analyze the circuit for each spectral component individually, then sum them to obtain the total system response. But chasing the energy with one frequency is hard enough. The conundrums that arise when doing it for several are much worse than the ones here. For your enjoyment consider the composite signal cos(9.95*2*pi*t)+cos(10.05*2*pi*t) working into 1 ohm. The imputed average power for each of the components is 0.5 W. The total average power is 1 W as expected. Consider the 1 second interval from 4.5 to 5.5 seconds. In this second 0.016393 joules flow for an average power of 0.016393 W. But the sum of the imputed power in the two spectral components is 1 W. Where did the missing energy go? Just another example of why assigning too much reality to the imputed powers of the components of superposition is misleading. But let's suppose that the energy which flows into the node from the left side during the "inhalation" part of the cycle is the energy which flows out to the right during the "exhalation" part of the cycle, and the energy flowing into the node from the right exits on the left. So now we've managed to get energy past the node going in both directions while maintaining zero power and current at the node and conserving energy as we must. This agrees with the distributed network model. Since there is no impedance discontinuity and no impedor at the node, there can be no reflections at the node. In other examples, you have suggested inserting a zero length transmission line to aid analysis. Why not insert a zero length transmission line with an impedance to produce the desired reflection? At steady state the reflection cancels but this would be due to the redistribution of energy according to your explanations. The forward wave flows unimpeded through the node as does the equal magnitude reflected wave. The net energy flow is zero. The average energy flow is zero. Anyone who believes there is zero energy at a standing- wave current node should touch that point on a transmission line (which just happens to be the same point as the maximum voltage anti-node). No one has said there is zero energy. Only that there is zero energy flow. For energy flow, one needs simultaneous voltage and current. One must be careful not to confuse the net signal with the component signals. Agreed. Assigning too much reality to component signals is seriously misleading. Now actual voltages, currents and powers, that's a different thing. One must be careful not to confuse the average values with the instantaneous values. This can best be visualized using light waves in free space. Unimpeded EM waves do not bounce off of each other. Until one can grasp the simplicity of a transmission line, moving to the complexity of free space offers nothing but obfuscation. ...Keith |
The Rest of the Story
On Apr 8, 11:44*am, Cecil Moore wrote:
Keith Dysart wrote: You seem to be saying that the answers would be completely different if you chose a different impedance for the non-existant transmission line. There you go again, trying to shove your words into my mouth. (Pattooieee!) Please don't do that. Did you not say that 50 ohms for the non-existant line was the correct impedance because other impedances would yield the 'wrong' answer? You have apparently done the math and found it to be valid so, once again, you have to change the specified conditions in order to try to make your point. I don't recall changing anything. I'm still with Fig 1-1 from your paper, which did not include non-existant transmission lines. A Z0 of 50 ohms is the *only* characteristic impedance that will meet the specified precondition of zero average interference. Choosing any other characteristic impedance will move the example outside of the scope of my Part 1 article. Did you not say that adding 1 wave length of transmission line does not alter the conditions? Are you now saying it does? I understand why you want to do such a thing but obfuscation, diversions, and straw men are not part of the scientific method. My Part 1 article has a very narrow scope. Please abide by it. Yes. That is why I prefer the simplicity of Fig 1-1 without the non-existant transmission lines. ...Keith |
The Rest of the Story
Keith Dysart wrote:
Thus I strongly suggest that Vg, Ig, Pg, represent reality. The others are a convenient alternative view for the purposes of solving problems. Of course they represent *net* reality but we are trying to determine what is happening at a component wave level. Defining the component waves out of existence is an un- acceptable substitute for ascertaining what is happening in reality. Typically we see Vg split into Vf and Vr, but why stop at two. Why not 3, or 4? Because two is what a directional wattmeter reads. The two superposed waves, forward and reverse, can be easily distinguished from one another. Two superposed coherent forward waves cannot be distinguished from each other. That's why we stop at two - because it is foolish to go any farther. There is power coming from the transmission line. Looking at Pg(t), some of the time energy flows into the line, later in the cycle it flows out. The energy transfer would be exactly the same if the transmission line was replaced by a lumped circuit element. And we don't need Pf and Pr for an inductor. OTOH, the distributed network model is a superset of the lumped circuit model so the inadequate lumped circuit model might confuse people. Hint: changing models to make waves disappear from existence doesn't make the waves disappear. The lumped circuit model is adequate for lumped circuits. It is inadequate for a lot of distributed network problems. If the lumped circuit model worked for everything, we wouldn't ever need the distributed network model. I suggest that you take your circuit and apply distributed network modeling techniques to it including reflection coefficients and forward and reflected voltages, currents, and powers at all points in the circuit. Note that the reflections are *same-cycle* reflections. If the lumped circuit model analysis differs from the distributed network model analysis, the lumped circuit analysis is wrong. It goes up because the impedance presented by the transmission changes when the reflection returns. This change in impedance alters the circuit conditions and the power in the various elements change. Depending on the details of the circuit, these powers may go up, or they may go down when the reflection arrives. That is true, but the impedance is *VIRTUAL*, i.e. not an impedor, and is therefore only an *EFFECT* of superposition. We are once again left wondering about the *CAUSE* of the virtual impedance, i.e. the details of the superposition process. Ignoring those details will not solve the problem. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
On Wed, 9 Apr 2008 03:45:19 -0700 (PDT)
Keith Dysart wrote: On Apr 7, 12:14*pm, Roger Sparks wrote: On Sun, 6 Apr 2008 19:21:00 -0700 (PDT) Keith Dysart wrote: On Apr 5, 10:06*am, Roger Sparks wrote: Pg(t) is the result of a standing wave, containing *power from Pf(x) and Pr(x+90). * This is one way of thinking of it, but it is less misleading to consider that Pg(t) describes the actual energy flow, just as Vg(t) describes the actual voltage and Ig(t) describes the actual current. Using superposition Vf, If, Vr and Ir can be derived and from these Pf and Pr. Your argument is correct to the extent that the power you describe is passing point Pg(t) at the instant (t). *It is the equivalent statement that an observer watching cars pass on the freeway would make, saying "one blue car moving left and one red car moving right, so two cars are passing". *Not wrong, just "how is the information useful"? Pg(t) is the actual power at that point in the circuit. It can be derived by simply multiplying the direct measurement of the actual voltages and currents at that point in the circuit. One measures the same voltages and currents regardless of whether it is a transmission line to the right of point g, or the equivalent lumped circuit element. While Vf, Vr, etc. can be used to derive the same information and, therefore is arguably just a different point of view, Vf and Vr, If and Ir, etc., must always be used in pairs to arrive at the actual circuit conditions. It is when one starts to look at them separately, as if they individually represent some part of reality, that confusion awaits. Thus I strongly suggest that Vg, Ig, Pg, represent reality. The others are a convenient alternative view for the purposes of solving problems. Typically we see Vg split into Vf and Vr, but why stop at two. Why not 3, or 4? Analyzing a two wire telephone line will use four or more, forward to the east, forward to the west, reflected to east, reflected to the west, and sometimes many different reflections. How do we choose how many? Depends on what is convenient for solving the problem. The power of superposition. But assigning too much reality to the individual contributors can be misleading. Good thoughts. By breaking Vg into Vf and Vr, we can explain why very long transmission lines, many wavelengths long, have repeating patterns of inductive and capacitive reactance as if they were lumped components. If Vf and Vr work for long lines, they should work for short lines. So far as breaking Vg into many sequential/different Vf and Vr, we usually need to do that. Cecil chose our simple example to prevent re-reflection (reflection of the reflection) but even then it is apparent that the voltage source will have a reactive component. If we can't account for the power, it is because we are doing the accounting incorrectly. And the error in the accounting may be the expectation that the particular set of powers chosen should balance. Attempting to account for Pr fails when Pr is the imputed power from a partial voltage and current because such computations do not yield powers which exist. If we remove the transmission line from the circuit, we have an open circuit with no current. *Without current, there can be no power. How can power arrive at Rs if there is no power coming through the transmission line? * There is power coming from the transmission line. Looking at Pg(t), some of the time energy flows into the line, later in the cycle it flows out. The energy transfer would be exactly the same if the transmission line was replaced by a lumped circuit element. And we don't need Pf and Pr for an inductor. But this flow is quite different than the flow suggested by Pf and Pr. These suggest a continuous flow in each direction. It is only when they are summed that it becomes clear that flow is first in one direction and then other. I understand the delemma here. It is like trying to both fill and empty the bottle at the same time. We can't do that with physical objects and we like to think of energy as if it were a physical object. So how can energy seemingly flow into a line at the same time it flows out? Of course one way would be if Vf actually did reflect from Vr. A reflection beginning at the point Vg, then propagating down the line as an artifact of the original wave. So far as I have been able to figure, the result is the same as when we think of both Vf and Vr as actual waves, which are much easier to follow and calculate. Would it help to consider that before the "reflection from the short" arrives, power arrives via the transmission line path but the impedance is 100 ohms for our example, composed of Rs = 50 ohms and transmission line = 50 ohms? *After the "reflection from the short" arrives, the impedance drops to 70.7 ohms so the power to the circuit goes up (assuming a constant voltage source). *How can this happen if power is not carried via the "reflection from the short"? It goes up because the impedance presented by the transmission changes when the reflection returns. This change in impedance alters the circuit conditions and the power in the various elements change. Depending on the details of the circuit, these powers may go up, or they may go down when the reflection arrives. Your comment almost makes the altered impedance sound like a resistance, probably not quite the picture you want to convey. I think of power to Rs coming via two paths, one longer than the other. In my mind, the changed impedance is the result of two power streams merging back together. The impedance found when the reflection returns is dependent upon the line impedance, length of line, and conditons at the point of reflection. The length of line is measured in terms of time and wave velocity. While this is strong evidence supporting Vf and Vr, it does not rule out reflection between wave components. I don't know how many people have seen an railroad engine starting a train from stop, when there is a small gap between each of the cars. You can hear each of the cars bumping the adjacent car in a chain reaction going from engine to the end of the train. Clearly, the reaction has a velocity of travel. Our EM waves could do the same thing but we would never measure anything except the the resultant wave. -- 73, Roger, W7WKB |
The Rest of the Story
Keith Dysart wrote:
Cecil Moore wrote: There is no capacitance or inductance in the source to store energy. "In" is an oxymoron for the lumped circuit model. The lumped reactance exists *at* the same point as the source because everything is conceptually lumped into a single point. In the real world, circuits are never single points and there exists a frequency at which distributed network effects cannot be ignored. In reality, distributed network effects occur for all real circuits but they can often be ignored as negligible. The two inches of wire connecting the source to the source resistor has a characteristic impedance and is a certain fraction of a wavelength long. If it is not perfectly matched, reflections will occur, i.e. there will exist forward power and reflected power on that two inches of wire. For the 1/8WL shorted line, there appears to be 125 watts of forward power and 25 watts of reflected power at points on each side of the source. Not if there is no transmission line. Aha, there's your error. What would a Bird directional wattmeter read for forward power and reflected power? Consider that short pieces of 50 ohm coax are used to connect the real-world components together. Or chose any characteristic impedance and do the math. You will discover something about the real world, i.e. that you have been seduced by the lumped circuit model. Perhaps. But I don't need more examples where the powers balance. I already have the one example where they don't. And that one example is outside the scope of the preconditions of my Part 1 article. Let me help you out on that one. There are an infinite number of examples where the reflected power is NOT dissipated in the source resistor but none of those examples, including yours, satisfies the preconditions specified in my Part 1 article. Therefore, they are irrelevant to this discussion. But there are no component powers in the source. It is a simple circuit element. No wonder your calculations are in error. Perform your calculations based on the readings of an ideal 50 ohm directional wattmeter and get back to us. Hint: Mismatches cause reflections, even in real-world circuits. The reflections happen to be *same-cycle* reflections. The simplified lumped circuit model, that exists in your head and not in reality, ignores those reflections and thus causes confusion among the uninitiated who do not understand its real-world limitations. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Keith Dysart wrote:
On Apr 8, 8:51 am, Cecil Moore wrote: Roy Lewallen wrote: Now, I don't know of any way to assign "ownership" to bundles of energy. One way is to add a unique bit of modulation to each bundle of wave energy. I am fond of using a TV signal and observing ghosting on the screen. This, of course, assumes that the modulation stays with the same component wave to which it was originally associated. But as soon as you modulate, you no longer have sinusoidal steady state. You know and I know that is a copout diversion to avoid your having to face the technical facts. Consider the 1 second interval from 4.5 to 5.5 seconds. In this second 0.016393 joules flow for an average power of 0.016393 W. But the sum of the imputed power in the two spectral components is 1 W. Where did the missing energy go? Hint: Missing energy is impossible except in your mind. Just because you are ignorant of where the energy goes doesn't mean it is missing. It just means that you fail to understand interference. Have you not read Hecht's Chapter 9 on "Interference"? Obviously, interference is present and there is *NO* missing energy. I have previously listed the possibilities at least four times so will not bother listing them again. Just another example of why assigning too much reality to the imputed powers of the components of superposition is misleading. Just another example of ignorance in action. Waves possess energy that cannot be destroyed. Just because you cannot track it doesn't mean it cannot be tracked. In other examples, you have suggested inserting a zero length transmission line to aid analysis. Why not insert a zero length transmission line with an impedance to produce the desired reflection? What would be the characteristic impedance of a length of transmission that caused a reflection coefficient of 1.0? No one has said there is zero energy. Only that there is zero energy flow. For energy flow, one needs simultaneous voltage and current. Vfor/Ifor = Z0, Vfor*Ifor = Pfor = EforxHfor If an EM wave exists, it is moving at the speed of light and transferring energy. For Z0 purely resistive, Vfor cannot exist without Vfor/Z0 = Ifor. Vfor is always in phase with Ifor. Assigning too much reality to component signals is seriously misleading. Assigning reality to the components of superposition is seriously misleading???? Can we therefore throw out the entire principle of superposition? Until one can grasp the simplicity of a transmission line, moving to the complexity of free space offers nothing but obfuscation. It is obvious that you have many things you desire to hide inside that black transmission line to which we are not even allowed to attach a directional wattmeter. Since you are incapable of explaining what happens in free space for all to see, why should we believe that you have figured out what is happening inside a transmission line where everything is hidden from view? Until one can grasp the transparency of free space, moving to the opaque transmission line where all kinds of important things are hidden from view offers nothing but obfuscation. -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Keith Dysart wrote:
Did you not say that 50 ohms for the non-existant line was the correct impedance because other impedances would yield the 'wrong' answer? ABSOLUTELY NOT! I said that other impedances do not meet the specified preconditions for my Part 1 article so the answers are irrelevant, not wrong. If a 50 ohm Z0 and 50 ohm load is specified, do other Z0s yield wrong answers? Of course not. Those answers are merely irrelevant to the specified preconditions. I don't recall changing anything. I'm still with Fig 1-1 from your paper, which did not include non-existant transmission lines. You have not proved any errors exist in my Part 1 article. You keep trying to change the specified preconditions from average power to instantaneous power but that is simply unethical. Did you not say that adding 1 wave length of transmission line does not alter the conditions? Are you now saying it does? The set(A) conditions are not altered. The set(B) conditions are altered. Exactly what conditions are you referring to? Yes. That is why I prefer the simplicity of Fig 1-1 without the non-existant transmission lines. Of course, you absolutely avoid using any tool that would prove you wrong. So what's new? -- 73, Cecil http://www.w5dxp.com |
The Rest of the Story
Keith Dysart wrote:
On Apr 8, 8:51 am, Cecil Moore wrote: . . . The forward wave flows unimpeded through the node as does the equal magnitude reflected wave. The net energy flow is zero. The average energy flow is zero. Anyone who believes there is zero energy at a standing- wave current node should touch that point on a transmission line (which just happens to be the same point as the maximum voltage anti-node). No one has said there is zero energy. Only that there is zero energy flow. For energy flow, one needs simultaneous voltage and current. . . . In the interesting case of a current node on an infinite-SWR line, it appears we do have energy flow without any current, and therefore without power. Energy flows into the node from both directions in equal amounts at the same time, and out to both directions in equal amounts at the same time. What we don't have is *net* energy flow at the node. Likewise, there's charge flowing into the node from both directions, and out in both directions, which results in the zero net current. I don't believe that's the same as saying there's no energy or charge flow at all, even though the power and current are zero. And it's not necessary to separately consider forward and reverse waves of current, energy, or power in order to observe this -- it can be seen from looking only at the total charge or energy. Roy Lewallen, W7EL |
The Rest of the Story
On Apr 9, 12:51*pm, Roger Sparks wrote:
On Wed, 9 Apr 2008 03:45:19 -0700 (PDT) Keith Dysart wrote: On Apr 7, 12:14*pm, Roger Sparks wrote: On Sun, 6 Apr 2008 19:21:00 -0700 (PDT) Keith Dysart wrote: On Apr 5, 10:06*am, Roger Sparks wrote: Pg(t) is the result of a standing wave, containing *power from Pf(x) and Pr(x+90). * This is one way of thinking of it, but it is less misleading to consider that Pg(t) describes the actual energy flow, just as Vg(t) describes the actual voltage and Ig(t) describes the actual current. Using superposition Vf, If, Vr and Ir can be derived and from these Pf and Pr. Your argument is correct to the extent that the power you describe is passing point Pg(t) at the instant (t). *It is the equivalent statement that an observer watching cars pass on the freeway would make, saying "one blue car moving left and one red car moving right, so two cars are passing". *Not wrong, just "how is the information useful"? Pg(t) is the actual power at that point in the circuit. It can be derived by simply multiplying the direct measurement of the actual voltages and currents at that point in the circuit. One measures the same voltages and currents regardless of whether it is a transmission line to the right of point g, or the equivalent lumped circuit element. While Vf, Vr, etc. can be used to derive the same information and, therefore is arguably just a different point of view, Vf and Vr, If and Ir, etc., must always be used in pairs to arrive at the actual circuit conditions. It is when one starts to look at them separately, as if they individually represent some part of reality, that confusion awaits. Thus I strongly suggest that Vg, Ig, Pg, represent reality. The others are a convenient alternative view for the purposes of solving problems. Typically we see Vg split into Vf and Vr, but why stop at two. Why not 3, or 4? Analyzing a two wire telephone line will use four or more, forward to the east, forward to the west, reflected to east, reflected to the west, and sometimes many different reflections. How do we choose how many? Depends on what is convenient for solving the problem. The power of superposition. But assigning too much reality to the individual contributors can be misleading. Good thoughts. * By breaking Vg into Vf and Vr, we can explain I am not sure that 'explain' is the correct word. It certainly provides a convenient technique for computing the voltate and current along the line, but there are other ways to compute the result; differential equations being one other way, though less convenient. But I am not convinced that a convenient technique for solving the problem is necessarily an 'explanation of why'. why very long transmission lines, many wavelengths long, have repeating patterns of inductive and capacitive reactance as if they were lumped components. *If Vf and Vr work for long lines, they should work for short lines. This is true. But when we descend to zero length lines, as some have done, the rationale becomes quite a bit weaker. So far as breaking Vg into many sequential/different Vf and Vr, we usually need to do that. *Cecil chose our simple example to prevent re-reflection (reflection of the reflection) but even then it is apparent that the voltage source will have a reactive component. I still think of a voltage source as just being a voltage source, not something with resitance, reactance or impedance. If we can't account for the power, it is because we are doing the accounting incorrectly. And the error in the accounting may be the expectation that the particular set of powers chosen should balance. Attempting to account for Pr fails when Pr is the imputed power from a partial voltage and current because such computations do not yield powers which exist. If we remove the transmission line from the circuit, we have an open circuit with no current. *Without current, there can be no power. How can power arrive at Rs if there is no power coming through the transmission line? * There is power coming from the transmission line. Looking at Pg(t), some of the time energy flows into the line, later in the cycle it flows out. The energy transfer would be exactly the same if the transmission line was replaced by a lumped circuit element. And we don't need Pf and Pr for an inductor. But this flow is quite different than the flow suggested by Pf and Pr. These suggest a continuous flow in each direction. It is only when they are summed that it becomes clear that flow is first in one direction and then other. I understand the delemma here. *It is like trying to both fill and empty the bottle at the same time. *We can't do that with physical objects and we like to think of energy as if it were a physical object. *So how can energy seemingly flow into a line at the same time it flows out? Of course one way would be if Vf actually did reflect from Vr. *A reflection beginning at the point Vg, then propagating down the line as an artifact of the original wave. *So far as I have been able to figure, the result is the same as when we think of both Vf and Vr as actual waves, which are much easier to follow and calculate. I agree that the final results are the same. The intermediate results can mislead in different ways. Would it help to consider that before the "reflection from the short" arrives, power arrives via the transmission line path but the impedance is 100 ohms for our example, composed of Rs = 50 ohms and transmission line = 50 ohms? *After the "reflection from the short" arrives, the impedance drops to 70.7 ohms so the power to the circuit goes up (assuming a constant voltage source). *How can this happen if power is not carried via the "reflection from the short"? It goes up because the impedance presented by the transmission changes when the reflection returns. This change in impedance alters the circuit conditions and the power in the various elements change. Depending on the details of the circuit, these powers may go up, or they may go down when the reflection arrives. Your comment almost makes the altered impedance sound like a resistance, Impedance does have some similarity to resistance, but only for single frequency sinusoidal excitation, though I was not trying to say that. probably not quite the picture you want to convey. *I think of power to Rs coming via two paths, one longer than the other. *In my mind, the changed impedance is the result of two power streams merging back together. The impedance found when the reflection returns is dependent upon the line impedance, length of line, and conditons at the point of reflection. *The length of line is measured in terms of time and wave velocity. *While this is strong evidence supporting Vf and Vr, The technique of breaking actual voltage into Vf and Vr certainly works. But I would not say this is evidence for the existance of Vf and Vr, merely agreement with superposition. One of my text books drags the reader through the solution using differential equations, and then introduces Vf and Vr as a simpler way to solve the problem. The student is truly happy from learning that diffyQs will not be required. it does not rule out reflection between wave components. I don't know how many people have seen an railroad engine starting a train from stop, when there is a small gap between each of the cars. *You can hear each of the cars bumping the adjacent car in a chain reaction going from engine to the end of the train. *Clearly, the reaction has a velocity of travel. * Some what like hole flow in semiconductors. Electrons going forward make the holes flow backwards. Our EM waves could do the same thing but we would never measure anything except the the resultant wave. ...Keith |
| All times are GMT +1. The time now is 10:31 PM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com