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#171




For Vfwd1 = 1 volt or the value of your choice at the input end of the
line, fill in the following blanks: At the input end of the lossy line: Vref1 = volts Vfwd2 = volts Vref2 = volts Ifwd1 = amps Iref1 = amps Ifwd2 = amps Iref2 = amps V = volts I = amps P = watts At the output end of the lossy line: Vfwd1 = volts Vref1 = volts Vfwd2 = volts Vref2 = volts Ifwd1 = amps Iref1 = amps Ifwd2 = amps Iref2 = amps V = volts I = amps P = watts Feel free to toss in values for as many Pf's and Pr's as you think there are, at both ends of the line. That's what you're going to have to do to get me to pay attention to you. Volts, amps, watts. If it's beneath your dignity, too complicated, too simple, too much work, too boring, or any other excuse, that's perfectly ok. I'll just continue ignoring your handwaving. (Which is, come to think of it, probably a great relief anyway.) Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: Frankly, I haven't paid any attention to your ducking, dodging, and handwaving. You haven't been able to produce an analysis showing the voltages, currents, and powers in the same simple circuit I analyzed. As far as I'm concerned, nothing you've posted constitutes a proof of anything. If one leads a horse to water and it refuses to drink, digging another lake is just a waste of time. Analyze the following and see if you get the same results. I'm betting you will catch your error. Z0=68j39 lossy feedline+1WL 50 ohm lossless line10+j50 load Vfwd1 Vfwd2 Vref1 Vref2 Vfwd1 = Vfwd1*rho1 + Vfwd1*tau1 one forward, one reflected component Vref2 = Vref2*rho2 + Vref2*tau2 one forward, one reflected component Vref1 = Vfwd1*rho1 + Vref2*tau2 both reflected components added together Vfwd2 = Vfwd1*tau1 + Vref2*rho2 both forward components added together Note that every voltage has two components. You chose only one component for your Vfwd. You ignored the other component of Vfwd. Hint: you cannot have voltages left over from calculating the total forward voltage and the total reflected voltage. 
#172




On Fri, 05 Sep 2003 00:28:43 0500, Cecil Moore
wrote: If one leads a horse to water and it refuses to drink, digging another lake is just a waste of time. Hi you two, The Navy had a way of dealing with these academic issues that put both of you to shame. I once offered that stale observation of "leading a horse to water..." to a Chief Bos'un's Mate. He, understandably, was not amused by the lack of imagination, insight, invention, intelligence, or whatever was suppose to be the hallmark of my specialty and simply replied with a hohum: "You hold its head under water and suck on its ass." Seems no one is interested in actually getting anything accomplished when trading complaints is obviously more fun. ;) 73's Richard Clark, KB7QHC 
#173




Roy Lewallen wrote:
That's what you're going to have to do to get me to pay attention to you. Volts, amps, watts. Roy, you are the one who made the mistake. I have told you exactly what your mistake is. The rest is your problem, not mine. Once again there are four component voltages and four component currents. Forward voltage and current contain TWO terms, not one term as you assert. Reflected voltage and current contain TWO terms, not one term as you assert. I'm at work and don't remember all the values in the experiment but here is something similar which I posted yesterday. lossy feedline+1WL 50 ohm lossless feedline10+j60 load Pfwd1 Pfwd2 Pref1 Pref2 The ratio of Pref2 to Pfwd2 is 0.7225 on the 50 ohm line. (Pfwd1  Pref1) = (Pfwd2  Pref2) I don't remember the Z0 of the lossy line but the reflection coefficient can be calculated to solve the problem. The main thing to realize from the above is Pfwd2 Pref2. Therefore, Pfwd1 Pref1, i.e. total reflected power is ALWAYS less than total forward power when dealing with passive loads. You have terms left over. There should be no terms left over. There should be one forward voltage, one forward current, one reflected voltage, and one reflected current. Until you collect all the terms, your analysis will contain an error. Why are you so dead set against collecting like terms?  73, Cecil, W5DXP 
#174




No, I don't have a problem, nor did I make a mistake. I presented an
example with voltages, currents, and powers that are all selfconsistent, obey Ohm's and Kirchoff's laws, and don't violate any physical laws. It simply uses those laws, equations that can be found in nearly any transmission line text, and arithmetic. The analysis is correct as written. You've demonstrated that you're unable to produce a similar analysis which can be fit into your conception of how things work. If your problem is dyslexia or unfamiliarity with complex arithmetic, all you have to do is say so. Neither is anything to be ashamed of. I now return the readers to the standard flurry of handwaving, objecting, dodging, and excuses. But don't expect an analysis, derivation, or proof. And I'll be back outta here. Roy Lewallen, W7EL W5DXP wrote: Roy, you are the one who made the mistake. I have told you exactly what your mistake is. The rest is your problem, not mine. . . . 
#175




Roy Lewallen wrote:
No, I don't have a problem, nor did I make a mistake. I presented an example with voltages, currents, and powers that are all selfconsistent, obey Ohm's and Kirchoff's laws, and don't violate any physical laws. It simply uses those laws, equations that can be found in nearly any transmission line text, and arithmetic. The analysis is correct as written. Nope, it isn't and after following all those laws, you violated one you should have learned in the 4th grade, i.e. to collect like terms. You've demonstrated that you're unable to produce a similar analysis which can be fit into your conception of how things work. No I haven't. I have produced a simple logical analysis that proved yours to be wrong. Do you really believe yourself incapable of making a conceptual error? In the following: lossy linex1WL 50 ohm lossless line10+j50 load Pfwd1 Pfwd2 = 30.53W rho=0.82 at 88.9 deg Pref1 Pref2 = 20.53W I measure ten volts across the ten ohm resistor. I measure 30.53W forward on the 50 ohm line and 20.53W reflected on the 50 ohm line. Since the 50 ohm line is lossless, the forward voltage is in phase with the forward current and the reflected voltage is in phase with the reflected current. The values of voltages and currents on the 50 ohm line are easy to calculate. The load reflection coefficient is easy to calculate. Analysis from 'x' to the load is a nobrainer. The forward power on the 50 ohm line is 10W less than the reflected power on the 50 ohm line, i.e. Pfwd2Pref12 = 10W just as it should. At point 'x', on the source side of 'x', Pfwd1Pref1 MUST equal that same 10W. It doesn't matter what happens between 'x' and the source. At point 'x', Pfwd1 simply cannot be less than Pref1. Conditions are the same whether the 1WL of lossless 50 ohm feedline is in the circuit or out of the circuit. You have not taken all the forward and reflected terms into account. You have violated something you should have learned in the 4th grade, i.e. to collect like terms. Instead, you threw away half of the forward terms and half of the reflected terms. No wonder you got it wrong.  73, Cecil http://www.qsl.net/w5dxp = Posted via Newsfeeds.Com, Uncensored Usenet News = http://www.newsfeeds.com  The #1 Newsgroup Service in the World! == Over 100,000 Newsgroups  19 Different Servers! = 
#176




Cecil Moore wrote:
Roy Lewallen wrote: No, I don't have a problem, nor did I make a mistake. I presented an example with voltages, currents, and powers that are all selfconsistent, obey Ohm's and Kirchoff's laws, and don't violate any physical laws. It simply uses those laws, equations that can be found in nearly any transmission line text, and arithmetic. The analysis is correct as written. Nope, it isn't and after following all those laws, you violated one you should have learned in the 4th grade, i.e. to collect like terms. You've demonstrated that you're unable to produce a similar analysis which can be fit into your conception of how things work. No I haven't. I have produced a simple logical analysis that proved yours to be wrong. Do you really believe yourself incapable of making a conceptual error? In the following: lossy linex1WL 50 ohm lossless line10+j50 load Pfwd1 Pfwd2 = 30.53W rho=0.82 at 88.9 deg Pref1 Pref2 = 20.53W I measure ten volts across the ten ohm resistor. I measure 30.53W forward on the 50 ohm line and 20.53W reflected on the 50 ohm line. Since the 50 ohm line is lossless, the forward voltage is in phase with the forward current and the reflected voltage is in phase with the reflected current. The values of voltages and currents on the 50 ohm line are easy to calculate. The load reflection coefficient is easy to calculate. Analysis from 'x' to the load is a nobrainer. The forward power on the 50 ohm line is 10W less than the reflected power on the 50 ohm line, i.e. Pfwd2Pref12 = 10W just as it should. At point 'x', on the source side of 'x', Pfwd1Pref1 MUST equal that same 10W. It doesn't matter what happens between 'x' and the source. At point 'x', Pfwd1 simply cannot be less than Pref1. I observe that you make the assertion, but fail to do the arithmetic to demonstrate the assertion, while Roy does the arithmetic which seems to contradict your assertion. There is a strong suspicion that the reason you don't do the arithmetic (since you do seem fond of numerical examples) in this case is that you have been unable to make arithmetic produce the desired result. Note that a requirement for Vr to be greater than Vf is that the reactive component of the load has a different sign than the reactive component of the line. So the problem specification is incomplete. Since the reactances on each side of 'x' have different signs, the circuit looks like it might be somewhat resonant and it should not be a surprise when resonant circuits produce higher voltages. ....Keith 
#178




wrote:
Perhaps completing the arithmetic would demonstrate whether the definitions you think represent Pfwd and Prev actually do. It ain't rocket science. Everything flowing toward the load is Pfwd. Everything flowing toward the source is Pref. There's no third term because there's no third direction. The directional possibilities of power in a transmission line = binary. There's no magic third term pointing toward Alpha Centauri.  73, Cecil http://www.qsl.net/w5dxp = Posted via Newsfeeds.Com, Uncensored Usenet News = http://www.newsfeeds.com  The #1 Newsgroup Service in the World! == Over 100,000 Newsgroups  19 Different Servers! = 
#179




I'm trying to stay out of this, but can't avoid mentioning that although
Cecil insists on only one Pref and one Pfwd, he has different criteria for voltage waves, and requires multiple Vfwd and Vref. So there are four directions for voltage, but only two for power? Roy Lewallen, W7EL Cecil Moore wrote: wrote: Perhaps completing the arithmetic would demonstrate whether the definitions you think represent Pfwd and Prev actually do. It ain't rocket science. Everything flowing toward the load is Pfwd. Everything flowing toward the source is Pref. There's no third term because there's no third direction. The directional possibilities of power in a transmission line = binary. There's no magic third term pointing toward Alpha Centauri. 
#180




Cecil Moore wrote:
wrote: Perhaps completing the arithmetic would demonstrate whether the definitions you think represent Pfwd and Prev actually do. It ain't rocket science. Everything flowing toward the load is Pfwd. Everything flowing toward the source is Pref. Well, I suppose. If that is the extent of the concept of Pfwd and Prev its pretty simple. But without some equations defining Pfwd and Prev, the concept is not particularly useful since there are no predictions which you can make from it. ....Keith 
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