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Where does it go? (mismatched power)
On Jun 16, 6:04*pm, K1TTT wrote:
would you care to provide us with the general equation for A given a complex load impedance and a line length other than a multiple of 90 degrees? It's simply the relative phase angle between the forward voltage and reflected voltage at the source resistor. I trust that you can manage that without my help. -- 73, Cecil, w5dxp.com |
#2
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Where does it go? (mismatched power)
On Jun 16, 11:16*pm, Cecil Moore wrote:
On Jun 16, 6:04*pm, K1TTT wrote: would you care to provide us with the general equation for A given a complex load impedance and a line length other than a multiple of 90 degrees? It's simply the relative phase angle between the forward voltage and reflected voltage at the source resistor. I trust that you can manage that without my help. -- 73, Cecil, w5dxp.com ah, so you do have to calculate the voltages, which would requires two transformations of the length of the coax plus calculating the magnitude and angle of the complex reflection coefficient then adding them all up. still sounds like more work than necessary. |
#3
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Where does it go? (mismatched power)
On 16 jun, 20:50, K1TTT wrote:
On Jun 16, 11:16*pm, Cecil Moore wrote: On Jun 16, 6:04*pm, K1TTT wrote: would you care to provide us with the general equation for A given a complex load impedance and a line length other than a multiple of 90 degrees? It's simply the relative phase angle between the forward voltage and reflected voltage at the source resistor. I trust that you can manage that without my help. -- 73, Cecil, w5dxp.com ah, so you do have to calculate the voltages, which would requires two transformations of the length of the coax plus calculating the magnitude and angle of the complex reflection coefficient then adding them all up. *still sounds like more work than necessary. Hello all If I do not make any mistake, my numbers agree with yours Cecil, it is a pleasure (my procedure was the old plain and simple standard) :) I know, 1000 examples probe nothing, only one popperianists man can bring a black swan at any time and falsify the induction :). but now seems more clear to me you have a predictive model that render identical numbers of my classical one, at least in basic tests, it is a step forward to better considerate your efforts and with goodwill begin to establish basic points of agreement. I recognize it is a more simple approach the classic method (as said K1TTT) to me too, but also I think not always one good method or model it is more convenient to understand another useful view of phenomena. Phasorial solutions are good, practical and likely complete electric solutions but in my opinion they do not married very well with other more general electromagnetic and physics models (wave guides perhaps? I have not experience); unification has its advantages too (However my favourite answer was definitely the Roy's one in the fourth post of the thread, hi hi...) 73 - Miguel - LU6ETJ PS: Owen do not be upset with me :) most of my available newsgroup time it is spent in translate to english without flaws that may induce to misinterpretations (all of you are very demanding with precise wording and exact definitions), one mistake and I will need three or four painly translations more to clarify :D Today I tested your interesting formula with a Half wave 50 ohms TL, loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give 2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving Vf=50 V and Vr=16,6 V aprox. for both loads. I am using (Pf=Pnet / (1-Rho^2) and Pr=Pnet / ((1/1/Rho^2) -1) formula which not gives different sign to Vr, in such case applying to PRs=(((Vs/2)-Vr)^2)/Rs = ((50/2)-16.66)^2/50. I get PRs=22,2 with both loads because the sign of Vf (always +) from simple Pr and Pf formulas, changing the sign of Pr render Rs=88,8 W for Prs (OK). I have in my disk a very descriptive, advisable and friendly article downloaded from: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf In page number 88 there is a agreement with your formula in "Is zo of aa HF ham tx typycally 50+j0?". |
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Where does it go? (mismatched power)
lu6etj wrote in
: .... PS: Owen do not be upset with me :) most of my available newsgroup time it is spent in translate to english without flaws that may induce to misinterpretations (all of you are very demanding with precise wording and exact definitions), one mistake and I will need three or four painly translations more to clarify :D Today I tested your interesting formula with a Half wave 50 ohms TL, loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give 2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving Vf=50 V and Vr=16,6 V aprox. for both loads. I posted a correction to the formula to properly account for the fact that Vr is a complex quantity. The corrected expression is Prs=|(Vs/2-Vr)|^2/Rs . Apologies if you missed it. Of course, the V quantities are RMS, it is a bit of a botch of RMS with phase as we often do in thinking... but I see you using the same shorthand in your calcs. Although we are using RMS values, don't overlook that where they add (eg Vf+Vr) you must properly account for the phase. For your cases: For the 25+j0 load, Vr=-16.67+j0, so Prs=|50--16.67|^2/50=88.9W. For the 100+j0 load, Vr=16.67+j0, so Prs=|50-16.67|^2/50=22.2W. Of course, for a 50+j0 load, Vr=0, so Prs=|50-0|^2/50=50.0W. As you can see, the first two cases have the same |Vr| (though different phase), the same 'reflected power', and yet Prs is very different. Consider an extreme case, Zl=1e6, VSWR is extreme, almost infinity, Vref= 50+j0, so Prs=|50-50|^2/50=0.0W. Here, your 'reflected power' is as large as it gets, but the power dissipated in the source is zero. The notion that reflected power is simply and always absorbed in the real source resistance is quite wrong. Sure you can build special cases where that might happen, but there is more to it. Thinking of the reflected wave as 'reflected power' leads to some of the misconception. Owen |
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Where does it go? (mismatched power)
On Jun 17, 12:38*am, Owen Duffy wrote:
The notion that reflected power is simply and always absorbed in the real source resistance is quite wrong. Sure you can build special cases where that might happen, but there is more to it. Thinking of the reflected wave as 'reflected power' leads to some of the misconception. Again, exactly who proposed that notion? The "power" in a reflected wave is actually reflected energy measured at some point on the transmission line. Reflected EM waves cannot exist without ExH energy per unit time per unit area. The special case where 100% of the reflected power incident from a Z0=50 ohm transmission line is dissipated in the 50 ohm source resistor happens when there is zero interference between the forward wave and reflected wave. Since it is not true when interference exists, it is logical to conclude that interference has something to do with the reflected power results deviating from the zero interference case - and so it does. The interference term in the power density equation indicates what kind of interference exists and what happens to the interference energy. Roy's experiment was designed to eliminate re-reflections from the source and it does exactly that. What that means is that all of the variations in power distribution are associated with interference, the other mechanism by which RF wave energy can be redistributed. Contrary to w7el's assumption, EM wave reflection is NOT the only mechanism for redistributing RF energy in a transmission line. That glaring fact of physics is what most of the RF experts are missing. -- 73, Cecil, w5dxp.com |
#6
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Where does it go? (mismatched power)
On Jun 17, 1:36*pm, Cecil Moore wrote:
On Jun 17, 12:38*am, Owen Duffy wrote: The notion that reflected power is simply and always absorbed in the real source resistance is quite wrong. Sure you can build special cases where that might happen, but there is more to it. Thinking of the reflected wave as 'reflected power' leads to some of the misconception. Again, exactly who proposed that notion? The "power" in a reflected wave is actually reflected energy measured at some point on the transmission line. Reflected EM waves cannot exist without ExH energy per unit time per unit area. The special case where 100% of the reflected power incident from a Z0=50 ohm transmission line is dissipated in the 50 ohm source resistor happens when there is zero interference between the forward wave and reflected wave. Since it is not true when interference exists, it is logical to conclude that interference has something to do with the reflected power results deviating from the zero interference case - and so it does. The interference term in the power density equation indicates what kind of interference exists and what happens to the interference energy. Roy's experiment was designed to eliminate re-reflections from the source and it does exactly that. What that means is that all of the variations in power distribution are associated with interference, the other mechanism by which RF wave energy can be redistributed. Contrary to w7el's assumption, EM wave reflection is NOT the only mechanism for redistributing RF energy in a transmission line. That glaring fact of physics is what most of the RF experts are missing. -- 73, Cecil, w5dxp.com but you have already admitted that it is the only mechanism... though you will now undoubtedly argue against yourself. em wave reflection IS the only mechanism for redistributing rf energy, you admitted that when you agreed that superposition of the waves is the mechanism that causes the interference in the first place. since interference is caused by superposition then wave cancelation and this so called redistribution of energy is also caused by the superposition of the reflected wave with the incident wave. RF experts are not missing anything, they just don't talk in your fancy energy and power terms all the time or figure how to calculate power superpositions... read the 500 pages BEFORE the s-parameter discussion in your version of the fields and waves book and learn the simple way to handle the single wave components that makes all that stuff you preach look over complicated. |
#7
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Where does it go? (mismatched power)
On Jun 17, 5:08*pm, K1TTT wrote:
but you have already admitted that it is the only mechanism... though you will now undoubtedly argue against yourself. * em wave reflection IS the only mechanism for redistributing rf energy, you admitted that when you agreed that superposition of the waves is the mechanism that causes the interference in the first place. Superposition and reflection are NOT the same mechanism. You are totally confused about what I have said. Wave REFLECTION (of one wave) is not caused by SUPERPOSITION (of two waves) and vice versa. Wave reflection happens to a single wave when it encounters an impedance discontinuity. Superposition requires two or more waves. They are clearly two completely different mechanisms. So I will repeat what I said befo There are two mechanisms for redistributing the reflected energy back toward the load. 1. The re-reflection of the SINGLE reflected wave from the load at an impedance discontinuity associated with the power reflection coefficient, e.g. 0.5 in my earlier example. Thus in that earlier example, 1/2 of the reflected energy from the load is re-reflected back toward the load and joins the forward wave toward the load. That leaves 1/2 of the reflected energy that is transmitted through the impedance discontinuity toward the source without being reflected. The second energy redistribution mechanism occurs associated with superposition of MULTIPLE WAVES. 2. The percentage of the reflected energy from the load that is transmitted through the impedance discontinuity toward the source superposes with the reflection of the source forward wave from the impedance discontinuity. In my earlier example, the following two wavefronts superpose in the direction of the source. Pfor1(rho^2) = 50w and Pref2(1-rho^2) = 50w Pref1 = 50w + 50w - 2*SQRT(50w*50w) = 0 This is the second mechanism (wave cancellation) that redistributes the energy in the canceled wavefronts back toward the load. This step 2 is technically NOT a reflection since it involves two waves. This is the step that the RF gurus are missing. I really wish you understood the s-parameter equations which are easier to discuss than the above RF equations. In the s-parameter equation: b2 = s21*a1 + s22*a2 the term, s22*a2, is the SINGLE WAVE re-reflection term, i.e. a2 is the reflected voltage from the load. In the other s-parameter equation: b1 = s11*a1 + s12*a2 = 0 s11*a1 and s12*a2 are the TWO WAVEFRONTS that superpose to zero, i.e. engage in wave cancellation. -- 73, Cecil, w5dxp.com |
#8
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Where does it go? (mismatched power)
On Jun 16, 11:47*pm, lu6etj wrote:
If I do not make any mistake, my numbers agree with yours Cecil, it is a pleasure (my procedure was the old plain and simple standard) *:) If, instead of a voltage analysis, one does an energy analysis, what happens to the reflected energy becomes obvious. I posted an earlier example that nobody solved. So let me repeat it here. ------Z01------+------Z02------ The power reflection coefficient at point '+' is rho^2 = 0.5. The power transmission coefficient is (1-rho^2) = 0.5 Pfor1 on the Z01 line is 100w Pref1 on the Z01 line is 0w What are Pfor2 and Pref2 on the Z02 line? What is the SWR on the Z02 line? The power reflected back toward the source at point '+' is Pfor1(rho^2) = 100(0.5) = 50w We know that Pref1 is zero, so Pref2(1-rho^2) = 50w Since 1-rho^2 is 0.5, Pref2 must be 100w and Pref2(rho^2) = 50w We also know that Pfor1(1-rho^2) = 50w So we have two 50w waves trying to flow toward the source and two 50w waves trying to flow toward the load. Do we have 100 watts flowing in both directions? No, that would be superposition of power which is a no-no. Using the power density equation indicates what is happening. Pref1 = 0 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive interference at the Z0-match Pref1 = 50 + 50 - 2*SQRT(50*50) = 0 Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus total constructive interference Pfor2 = 50 + 50 + 2*SQRT(50*50) = 200w Since Pref2/Pfor2 = 0.5, the SWR on the Z02 line is 5.83:1 Note that we have solved the problem without knowing the value of Vfor1, Vfor2, and Vref2. But let's assume Z01 is 50 ohms which makes Z02 equal to 291.5 ohms. Now we can calculate the voltages assuming Vfor1 at '+' is our phase reference: Vfor1 = 70.7 volts at zero degrees at '+' Vref1 = 0 Vfor2 = 241.4 volts at zero degrees at '+' Vref2 = 170.7 volts at 180 deg at '+' Vfor1(rho1) = 50v at zero deg Vref2(tau2) = 50v at 180 deg Vfor1(tau1) = 120.7v at zero deg Vref2(rho2) = 120.7v at zero deg There's the interference, voltage style. Two voltages superpose to zero toward the source while engaged in total destructive interference. What happens to the energy components that are supporting those two voltages? There's only one thing that can happen. They are obviously not flowing in their original directions so they must necessarily flow in the only other direction possible. Total destructive interference toward the source results in total constructive interference toward the load. Someone earlier remarked about the fact that when two 50w waves combine, the result is one 200w wave. That is total constructive interference in action and that extra energy had to come from somewhere. -- 73, Cecil, w5dxp.com |
#9
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Where does it go? (mismatched power)
On 17 jun, 11:30, Cecil Moore wrote:
On Jun 16, 11:47*pm, lu6etj wrote: If I do not make any mistake, my numbers agree with yours Cecil, it is a pleasure (my procedure was the old plain and simple standard) *:) If, instead of a voltage analysis, one does an energy analysis, what happens to the reflected energy becomes obvious. I posted an earlier example that nobody solved. So let me repeat it here. ------Z01------+------Z02------ The power reflection coefficient at point '+' is rho^2 = 0.5. The power transmission coefficient is (1-rho^2) = 0.5 Pfor1 on the Z01 line is 100w Pref1 on the Z01 line is 0w What are Pfor2 and Pref2 on the Z02 line? What is the SWR on the Z02 line? The power reflected back toward the source at point '+' is Pfor1(rho^2) = 100(0.5) = 50w We know that Pref1 is zero, so Pref2(1-rho^2) = 50w Since 1-rho^2 is 0.5, Pref2 must be 100w and Pref2(rho^2) = 50w We also know that Pfor1(1-rho^2) = 50w So we have two 50w waves trying to flow toward the source and two 50w waves trying to flow toward the load. Do we have 100 watts flowing in both directions? No, that would be superposition of power which is a no-no. Using the power density equation indicates what is happening. Pref1 = 0 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive interference at the Z0-match Pref1 = 50 + 50 - 2*SQRT(50*50) = 0 Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus total constructive interference Pfor2 = 50 + 50 + 2*SQRT(50*50) = 200w Since Pref2/Pfor2 = 0.5, the SWR on the Z02 line is 5.83:1 Note that we have solved the problem without knowing the value of Vfor1, Vfor2, and Vref2. But let's assume Z01 is 50 ohms which makes Z02 equal to 291.5 ohms. Now we can calculate the voltages assuming Vfor1 at '+' is our phase reference: Vfor1 = 70.7 volts at zero degrees at '+' Vref1 = 0 Vfor2 = 241.4 volts at zero degrees at '+' Vref2 = 170.7 volts at 180 deg at '+' Vfor1(rho1) = 50v at zero deg Vref2(tau2) = 50v at 180 deg Vfor1(tau1) = 120.7v at zero deg Vref2(rho2) = 120.7v at zero deg There's the interference, voltage style. Two voltages superpose to zero toward the source while engaged in total destructive interference. What happens to the energy components that are supporting those two voltages? There's only one thing that can happen. They are obviously not flowing in their original directions so they must necessarily flow in the only other direction possible. Total destructive interference toward the source results in total constructive interference toward the load. Someone earlier remarked about the fact that when two 50w waves combine, the result is one 200w wave. That is total constructive interference in action and that extra energy had to come from somewhere. -- 73, Cecil, w5dxp.com Good day: wave as 'reflected power' leads to some of the misconception. Ugh! The slippery word again...! Please Owen, remember me what power definition are you using here and expand the sentence idea. EM waves cannot exist without ExH energy per unit time per unit area. This is what teachers taught me, but when the EM TL waves reachs a "target" seems the issue arise in the newsgroup. Puzzle to me why we can (want? :) ) not reconcile those physics concepts with usual electricity concepts if today they arise from the same place ultimately? Certainly in electromagnetism we deal with vectors E,D,B,H,S and in electricity with scalars V(t), I(t) or phasors, this is for convenience and simplicity, but I accept we should be able to understand them in both forms without contradiction. 73 - Miguel Ghezzi - LU6ETJ |
#10
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Where does it go? (mismatched power)
On Jun 17, 9:28*pm, lu6etj wrote:
On 17 jun, 11:30, Cecil Moore wrote: On Jun 16, 11:47*pm, lu6etj wrote: If I do not make any mistake, my numbers agree with yours Cecil, it is a pleasure (my procedure was the old plain and simple standard) *:) If, instead of a voltage analysis, one does an energy analysis, what happens to the reflected energy becomes obvious. I posted an earlier example that nobody solved. So let me repeat it here. ------Z01------+------Z02------ The power reflection coefficient at point '+' is rho^2 = 0.5. The power transmission coefficient is (1-rho^2) = 0.5 Pfor1 on the Z01 line is 100w Pref1 on the Z01 line is 0w What are Pfor2 and Pref2 on the Z02 line? What is the SWR on the Z02 line? The power reflected back toward the source at point '+' is Pfor1(rho^2) = 100(0.5) = 50w We know that Pref1 is zero, so Pref2(1-rho^2) = 50w Since 1-rho^2 is 0.5, Pref2 must be 100w and Pref2(rho^2) = 50w We also know that Pfor1(1-rho^2) = 50w So we have two 50w waves trying to flow toward the source and two 50w waves trying to flow toward the load. Do we have 100 watts flowing in both directions? No, that would be superposition of power which is a no-no. Using the power density equation indicates what is happening. Pref1 = 0 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive interference at the Z0-match Pref1 = 50 + 50 - 2*SQRT(50*50) = 0 Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus total constructive interference Pfor2 = 50 + 50 + 2*SQRT(50*50) = 200w Since Pref2/Pfor2 = 0.5, the SWR on the Z02 line is 5.83:1 Note that we have solved the problem without knowing the value of Vfor1, Vfor2, and Vref2. But let's assume Z01 is 50 ohms which makes Z02 equal to 291.5 ohms. Now we can calculate the voltages assuming Vfor1 at '+' is our phase reference: Vfor1 = 70.7 volts at zero degrees at '+' Vref1 = 0 Vfor2 = 241.4 volts at zero degrees at '+' Vref2 = 170.7 volts at 180 deg at '+' Vfor1(rho1) = 50v at zero deg Vref2(tau2) = 50v at 180 deg Vfor1(tau1) = 120.7v at zero deg Vref2(rho2) = 120.7v at zero deg There's the interference, voltage style. Two voltages superpose to zero toward the source while engaged in total destructive interference. What happens to the energy components that are supporting those two voltages? There's only one thing that can happen. They are obviously not flowing in their original directions so they must necessarily flow in the only other direction possible. Total destructive interference toward the source results in total constructive interference toward the load. Someone earlier remarked about the fact that when two 50w waves combine, the result is one 200w wave. That is total constructive interference in action and that extra energy had to come from somewhere. -- 73, Cecil, w5dxp.com Good day: wave as 'reflected power' leads to some of the misconception. Ugh! *The slippery word again...! Please Owen, remember me what power definition are you using here and expand the sentence idea. EM waves cannot exist without ExH energy per unit time per unit area. This is what teachers taught me, but when the EM TL waves reachs a "target" seems the issue arise in the newsgroup. Puzzle to me why we can (want? :) ) not reconcile those physics concepts with usual electricity concepts if today they arise from the same place ultimately? Certainly in electromagnetism we deal with vectors E,D,B,H,S and in electricity with scalars V(t), I(t) or phasors, this is for convenience and simplicity, but I accept we should be able to understand them in both forms without contradiction. 73 - Miguel Ghezzi - LU6ETJ except for the one guy in the other thread that likes his dc 'waves' i think most everyone else gets the same answers but won't acknowledge that the other peoples methods also work. even my simple transformed impedance method shows the same results, you just have to look at what it really means. the whole key is that it doesn't really matter, we know how to use voltage/current, E/H, powers, s-parameters, etc to design circuits, and as long as we know their limitations and stay within them everything should work the same way no matter how its designed on paper. |
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