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"Reg Edwards" wrote in message ...
"Dr. Slick" wrote What about the ARRL? ================================ Dear Slick, you must be new round this neck of the woods. Don't you realise the ARRL bibles are written by the same sort of people who haggle with you on this newsgroup? Well, yeah, the "A" stands for amateur, right? But it seems the hams don't even trust one another, which really, as we have shown here, can lead to new insights. Mistrust is actually good science. Slick |
"Tarmo Tammaru" wrote in message ...
"Dr. Slick" wrote in message om... What exactly do you mean by Zr at point z=0? i don't fully understand the page you sent, and neither do you obviously. Lower case z is distance, with the load at z=0 If the power RC is the square of the MAGNITUDE of the voltage RC, then a voltage RC 1 will lead to a power RC 1. He squares it to get the magnitude of the vector. There is still a phase angle How do you get more reflected power than incident power into a passive network, praytell?? You don't. at gamma =2.41, the phase angle is about 65 degrees, and the real part of gamma =1.0 What??!? if gamma, or rho, is greater than one, the reflected power is definitely greater than the incident! Now try this: using the conjugate formula, calculate gamma for the case where the line is terminated in a short circuit, and tell us how that meets the boundary condition. Tam/WB2TT Now try this: understand the page you sent me before you attempt to discuss it with others! Slick |
Do I have this right?
Dr Slick examined the generally accepted formula for rho and learned that its magnitude can be greater than one. This appears to imply that reflected power is greater than incident; something that would violate various conservation of energy laws. Dr Slick has therefore rejected the generally accepted formula and produced one which does not result in more power being reflected than is incident, thus satisfying various conservation of energy laws. Many people took issue with this redefinition of rho and attempted to show why the generally accepted formula is correct. But that does not address the issue with the generally accepted formula; how can reflected power be greater than incident? A clear explanation of why rho greater than one does not violate conservation of energy would seem to remove Dr Slick's objection to the generally accepted formula and then everyone could agree on the formula. I doubt that any proof of the correctness of the generally accepted formula will convince Dr Slick until it is shown why it does not violate conservation of energy. ....Keith |
Reg Edwards wrote:
Given a line's primary characteristics, R,L,C,G, length, or it's secondary characteristics Zo, dB, phase angle, plus the line's terminatiing impedance it is possible to calculate, by classical methods, all other quantities of engineering interest - WITHOUT ANY REFERENCE TO REFLECTION COEFFICIENT OR SWR which are mere man-made notions supposed to assist understanding of what goes on in the real world but, as exchanges on this newsgroup show, are just a pair of bloody useless nuisances. Nevertheless, the outer circle of the Smith chart is *always* the locus of zero positive resistance and infinite SWR, and a rho vector cannot terminate on, or cross over, this circle when a load R0 is present, regardless of the rest of the circuit, including any possible combination of resistances and reactances and complex Z0. One can argue "ignore rho=1 and just jump over it". This cannot be done in good mathematics. Dismissing rho and SWR as "contrived nuisances" is a convenient way to get rid of this problem, but it does not "wash". Rho and SWR are fundamental properties of transmission lines that do not go away, and a non-zero R precludes rho=1.0. Any attempt to circumvent (bypass) these small inconveniences is doomed to failure, regardless of the analytic geometry considerations. Bill W0IYH |
William E. Sabin wrote:
Reg Edwards wrote: Given a line's primary characteristics, R,L,C,G, length, or it's secondary characteristics Zo, dB, phase angle, plus the line's terminatiing impedance it is possible to calculate, by classical methods, all other quantities of engineering interest - WITHOUT ANY REFERENCE TO REFLECTION COEFFICIENT OR SWR which are mere man-made notions supposed to assist understanding of what goes on in the real world but, as exchanges on this newsgroup show, are just a pair of bloody useless nuisances. Nevertheless, the outer circle of the Smith chart is *always* the locus of zero positive resistance and infinite SWR, and a rho vector cannot terminate on, or cross over, this circle when a load R0 is present, regardless of the rest of the circuit, including any possible combination of resistances and reactances and complex Z0. One can argue "ignore rho=1 and just jump over it". This cannot be done in good mathematics. Dismissing rho and SWR as "contrived nuisances" is a convenient way to get rid of this problem, but it does not "wash". Rho and SWR are fundamental properties of transmission lines that do not go away, and a non-zero R precludes rho=1.0. Any attempt to circumvent (bypass) these small inconveniences is doomed to failure, regardless of the analytic geometry considerations. Bill W0IYH Power wave theory avoids the Smith chart, since there are no transmission lines. Scattering matrices are used instead. Nevertheless, rho is still an important parameter, but it does not involve distance separation between generator and load as a parameter. Bill W0IYH |
Richard Clark wrote:
W5DXP wrote: So how do you get the reflections in a single source system to be incoherent? Two reflective interfaces with an aperiodic distance between. That won't do it unless the distance between them is somehow dynamically changing. For fixed distances, steady-state signals will be coherent. The example of the challenge serves to illuminate (pun intended) the logical shortfall of those here who insist that a Transmitter exhibits no Z, or that it is unknowable (to them, in other words), or that it reflects all power that returns to it (to bolster their equally absurd notion that the Transmitter does not absorb that power). This is a convenient rule-of-thumb, nothing more. It solves the problem of something being unknowable. A source obeys the rules of the wave reflection model. Unfortunately, we don't usually know the exact value of source impedance seen by the reflected waves. Thus, the rule-of-thumb. Engineers and scientists simply converse with the tacit agreement that the source matches the line when going into the discussion of SWR (and why Chapman plainly says this up front on the page quoted earlier). This is so commonplace that literalists who lack the background (and skim read) fall into a trap of asserting some pretty absurd things. It follows that for these same literalists, any evidence to the contrary is anathema, heresy, or insanity - people start wanting to "help" you :-P I agree with Chipman on that. Now, be advised that when I say "accurately" that this is of concern only to those who care for accuracy. That's the part I don't understand. You can assume a whole range of impedances for the source while the forward power and reflected power remain the same. Is "accuracy" somehow involved with efficiency? -- 73, Cecil, W5DXP |
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William E. Sabin wrote:
William E. Sabin wrote: Reg Edwards wrote: Given a line's primary characteristics, R,L,C,G, length, or it's secondary characteristics Zo, dB, phase angle, plus the line's terminatiing impedance it is possible to calculate, by classical methods, all other quantities of engineering interest - WITHOUT ANY REFERENCE TO REFLECTION COEFFICIENT OR SWR which are mere man-made notions supposed to assist understanding of what goes on in the real world but, as exchanges on this newsgroup show, are just a pair of bloody useless nuisances. Nevertheless, the outer circle of the Smith chart is *always* the locus of zero positive resistance and infinite SWR, and a rho vector cannot terminate on, or cross over, this circle when a load R0 is present, regardless of the rest of the circuit, including any possible combination of resistances and reactances and complex Z0. One can argue "ignore rho=1 and just jump over it". This cannot be done in good mathematics. Dismissing rho and SWR as "contrived nuisances" is a convenient way to get rid of this problem, but it does not "wash". Rho and SWR are fundamental properties of transmission lines that do not go away, and a non-zero R precludes rho=1.0. Any attempt to circumvent (bypass) these small inconveniences is doomed to failure, regardless of the analytic geometry considerations. Bill W0IYH Power wave theory avoids the Smith chart, since there are no transmission lines. Scattering matrices are used instead. Nevertheless, rho is still an important parameter, but it does not involve distance separation between generator and load as a parameter. Bill W0IYH I am not satisfied with this post. I will try to improve it a little later. Bill W0IYH |
I get it. The challenge is to figure out what you're talking about!!
What do I win if I can do it? ;-) AC6XG Richard Clark wrote: On Thu, 28 Aug 2003 21:01:54 -0500, W5DXP wrote: Richard Clark wrote: To put it ironically, the challenge I offer is deliberately incoherent to give that math a deliberate solution that is other than the result of simple addition or subtraction. So how do you get the reflections in a single source system to be incoherent? Hi Cecil, Two reflective interfaces with an aperiodic distance between. The cable (or any transmission line) falls in between. So does most instrumentation to measure power. All fall prey to this indeterminacy (unless, of course, it is made determinant through the specification of distance, which it is for the challenge). As I offered, this challenge is not my own hodge-podge of boundary conditions, it was literally drawn from a standard text many here have - hence the quote marks that attend its publication by me. I am not surprised no one has caught on, I also pointed out this discussion is covered in the parts of Chapman that no one reads. Whatchagonnado? The example of the challenge serves to illuminate (pun intended) the logical shortfall of those here who insist that a Transmitter exhibits no Z, or that it is unknowable (to them, in other words), or that it reflects all power that returns to it (to bolster their equally absurd notion that the Transmitter does not absorb that power). Chapman is quite clear to this last piece of fluff science - specifically and to the very wording. Engineers and scientists simply converse with the tacit agreement that the source matches the line when going into the discussion of SWR (and why Chapman plainly says this up front on the page quoted earlier). This is so commonplace that literalists who lack the background (and skim read) fall into a trap of asserting some pretty absurd things. It follows that for these same literalists, any evidence to the contrary is anathema, heresy, or insanity - people start wanting to "help" you :-P Ian grasped at the straw that the discussion simply peters out by the steady state and wholly disregards the compelling evidence (and further elaboration of Chapman to this, but he lacks another voice, the same Chapman, to accept it) with a forced mismatch at both ends of the line. It is impossible to accurately describe the power delivered to the load without knowing all parameters, the most overlooked is distances traversed by the power (total phase in the solution for interference). I put the challenge up to illustrate where the heat goes (the line); and it is well into the steady state, as I am sure no one could argue, but could easily gust "t'ain't so!" At least I saved them from the prospect of strangling on their own spit sputtering "shades of conjugation." [Another topic that barely goes a sentence without being corrupted with a Z-match characteristic.] Using this example for the challenge forces out the canards that the source is adjusting to the load (in fact, the challenge presents no such change in the first place) and dB cares not a whit what power is applied unless we have suddenly entered a non-linear physics. None have gone that far as they have already fallen off the edge earlier. Now, be advised that when I say "accurately" that this is of concern only to those who care for accuracy. Between mild mismatches the error is hardly catastrophic, and yet with the argument that the Transmitter is wholly reflective, it becomes catastrophic. The lack of catastrophe does not reject the math, it rejects the notion of the Transmitter being wholly reflective. This discussion in their terms merely drives a stake through their zombie theories. I would add there has been another voice to hear in this matter. The same literalist skim readers suffer the same shortfall of perception. We both enjoy the zen-cartwheels so excellently exhibited by the drill team of naysayers. ;-) 73's Richard Clark, KB7QHC |
William E. Sabin wrote:
William E. Sabin wrote: Reg Edwards wrote: Given a line's primary characteristics, R,L,C,G, length, or it's secondary characteristics Zo, dB, phase angle, plus the line's terminatiing impedance it is possible to calculate, by classical methods, all other quantities of engineering interest - WITHOUT ANY REFERENCE TO REFLECTION COEFFICIENT OR SWR which are mere man-made notions supposed to assist understanding of what goes on in the real world but, as exchanges on this newsgroup show, are just a pair of bloody useless nuisances. Nevertheless, the outer circle of the Smith chart is *always* the locus of zero positive resistance and infinite SWR, and a rho vector cannot terminate on, or cross over, this circle when a load R0 is present, regardless of the rest of the circuit, including any possible combination of resistances and reactances and complex Z0. One can argue "ignore rho=1 and just jump over it". This cannot be done in good mathematics. Dismissing rho and SWR as "contrived nuisances" is a convenient way to get rid of this problem, but it does not "wash". Rho and SWR are fundamental properties of transmission lines that do not go away, and a non-zero R precludes rho=1.0. Any attempt to circumvent (bypass) these small inconveniences is doomed to failure, regardless of the analytic geometry considerations. Bill W0IYH Power wave theory avoids the Smith chart, since there are no transmission lines. Scattering matrices are used instead. Nevertheless, rho is still an important parameter, but it does not involve distance separation between generator and load as a parameter. Bill W0IYH I am not satisfied with this post. I will try to improve it a little later. Bill W0IYH |
On Fri, 29 Aug 2003 08:25:06 -0700, W5DXP
wrote: Now, be advised that when I say "accurately" that this is of concern only to those who care for accuracy. That's the part I don't understand. You can assume a whole range of impedances for the source while the forward power and reflected power remain the same. Is "accuracy" somehow involved with efficiency? -- Hi Cecil, This is a real (the example of the challenge) issue of the Mismatch Uncertainty. It is only uncertain insofar as most folks are unaware of the contribution of error the source presents, and doubly unaware of the phase distances between the interfaces. Accuracy is unrelated to efficiency, both valuations exist distinct from the other. You can be efficient but express it with poor accuracy or good accuracy, efficiency cares not a whit about your poor situation to resolve it. You can also be pristine accurate and horribly inefficient, here again, accuracy is not a function of efficiency, accuracy cares not a whit about your waste. As for your conjecture You can assume a whole range of impedances for the source while the forward power and reflected power remain the same. I assume nothing, but I can portray the range through the same challenge's example. It merely involves moving the SWR measurement. I have already demonstrated that. 73's Richard Clark, KB7QHC |
On Fri, 29 Aug 2003 08:27:09 -0700, W5DXP
wrote: Richard Clark wrote: You have expended far more time climbing Everest in search of the guru, than simply doing the challenge at the bench. I don't understand the challenge. Until I do, I would be wasting my time on the bench. I'm sure I'm not the only one who doesn't understand what you are trying to say. Hi Cecil, Yes, it is terribly cryptic: how much heat is lost in the line? It requires advanced techniques in measuring power lost over a known stretch of transmission line. This is an art unknown to many here and baffles them in how to proceed when written directions are provided. Even if they reject those methods and directions, they struggle through massive tomes to come to no conclusion. What's more, the example resides in one of their cherished scriptures that many would swear by and would scream outrage at the suggestion that they might read it. The irony is delicious. 73's Richard Clark, KB7QHC |
On Fri, 29 Aug 2003 12:00:10 -0700, W5DXP
wrote: Richard Clark wrote: Accuracy is unrelated to efficiency, ... This is what I thought. I just wanted to be sure you were thinking the same way. I assume nothing, but I can portray the range through the same challenge's example. It merely involves moving the SWR measurement. I have already demonstrated that. But: "Every time you make a measurement, you make an error." So said my EE prof almost half a century ago. I assume it's still true. :-) Hi Cecil, Yes. 73's Richard Clark, KB7QHC |
"Tarmo Tammaru" wrote in message ...
Why didn't you do the gamma for a shorted line using your formula? I think you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes to -Z0*/Z0, which, for Zo having phase angle b equates to -1 at angle(-2b). Agreed. This doesn't prove anything, does it? This is in more detail of what ACF do: 1. Normalize the RC to (Zn -1)/(Zn +1) 2. Let's say we agree that for biggest magnitude, this has to be in the left hand plane. 3. Draw a vector for Zn = a at -135 deg 4. Draw vectors for Zn - 1 and Zn +1. Note that these 3 vectors have the same y coordinate. 5. You now can draw two triangles with the corners at 0, Zn, and Zn -1 for one, and 0, Zn, and Zn+1 for the other. 6. Solve for Zn+/-1 in terms of "a" 7. By plane geometry the magnitude of, of Zn -1= SQRT(1 + a^2 + aSQRT(2)) 8. The magnitude of Zn +1 is SQRT(1 + a^2 - aSQRT(2)) 9. Square both sides of the equation, and gamma^2= (1 + a^2 + aSQRT(2))/(1 + a^2 - aSQRT(2)) 10 This equates to 1 + (2SQRT(2))/((a + 1/a) - SQRT(2)) 11 |Gamma| max occurs when (a + 1/a) is a minimum, which is 2 at a=1. 12|Gamma| max is 1 + SQRT(2) They anticipate people being concerned about |Gamma| 1 and later come up with a formula for time average power. I don't know that looking at it is going to give anybody any insight, but for this is what they end up with ( I am typing CM for Gamma): P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)] Remember, z is distance from the load. Tam/WB2TT ok, you've done a nice job of copying the text you sent me. They also mention that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr because Zo is complex in the general case. "They anticipate people being concerned about |Gamma| 1 and later come up with a formula for time average power." Really? Could you tell us more about that? Could you email me more pages, especially the one that has Eq. 5.2b? Slick |
"William E. Sabin" sabinw@mwci-news wrote in message ...
Reg Edwards wrote: Given a line's primary characteristics, R,L,C,G, length, or it's secondary characteristics Zo, dB, phase angle, plus the line's terminatiing impedance it is possible to calculate, by classical methods, all other quantities of engineering interest - WITHOUT ANY REFERENCE TO REFLECTION COEFFICIENT OR SWR which are mere man-made notions supposed to assist understanding of what goes on in the real world but, as exchanges on this newsgroup show, are just a pair of bloody useless nuisances. Nevertheless, the outer circle of the Smith chart is *always* the locus of zero positive resistance and infinite SWR, and a rho vector cannot terminate on, or cross over, this circle when a load R0 is present, regardless of the rest of the circuit, including any possible combination of resistances and reactances and complex Z0. "gasp!" ...saved by a lifesaver of logic while drowning in the Sea of Ignorance! Thank you Bill. This is probably why they say that an oscillator has negative resistance (postitive feedback), and also why the stability circles have their centers located outside the unit 1 Radius. Clearly, areas outside the rho=1 circle on the Smith, are reserved for active networks. Slick One can argue "ignore rho=1 and just jump over it". This cannot be done in good mathematics. Dismissing rho and SWR as "contrived nuisances" is a convenient way to get rid of this problem, but it does not "wash". Rho and SWR are fundamental properties of transmission lines that do not go away, and a non-zero R precludes rho=1.0. Any attempt to circumvent (bypass) these small inconveniences is doomed to failure, regardless of the analytic geometry considerations. Bill W0IYH |
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Bill:
[snip] In the case of rather lossy cable, it is best to abandon rho and SWR before they become "bloody useless nuisances", and use the computer for everything. Bill W0IYH [snip] Amen Brother! -- Peter K1PO Indialatic By-the-Sea, FL. |
"Dr. Slick" wrote in message om... "Tarmo Tammaru" wrote in message ... Why didn't you do the gamma for a shorted line using your formula? I think you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes to -Z0*/Z0, which, for Zo having phase angle b equates to -1 at angle(-2b). Agreed. This doesn't prove anything, does it? How does this allow for the sum of V+ and V- to be 0? That is what you have across a short. .......................................... P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)] Remember, z is distance from the load. Tam/WB2TT ok, you've done a nice job of copying the text you sent me. As I recall, you said you were not familiar with these diagrams and did not understand them. What do you want me to do? derive it in a different way? They also mention that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr because Zo is complex in the general case. "They anticipate people being concerned about |Gamma| 1 and later come up with a formula for time average power." Really? Could you tell us more about that? Could you email me more pages, especially the one that has Eq. 5.2b? Slick I can't scan the whole chapter, and part of the next. You should be able to get your library to borrow the book for you. Or, look at some other college textbook under lossy lines. 5.2B is straightforward enough: Zo=(1/Yo)=SQRT[(R+jwL)/(G+jwC)]=Ro+jXo Tam/WB2TT |
"W5DXP" wrote in message ... Consider the following two examples: 1. The source is generating 100 watts and 20 watts of the 50 watts of incident reflected power is being re-reflected in phase from the source. The forward power meter reads 120 watts. The reflected power meter reads 50 watts. The conventional rule-of-thumb has the source generating (100-30)= 70 watts since it is dissipating 30 watts of reflected power. 2. The source is generating 110 watts and 10 watts of the 50 watts of incident reflected power is being re-reflected in phase from the source. The forward power meter reads 120 watts. The reflected power meter reads 50 watts. The conventional rule-of-thumb has the source generating (110-40)= 70 watts since it is dissipating 40 watts of reflected power. How can you possibly distinguish between the above two identical conditions caused by different source impedances? -- 73, Cecil, W5DXP Cecil, I actually did a test along these lines. Believe we discussed that some weeks ago before I did it. 1/4 wave 50 Ohm line shorted at the end. Freq ~ 2 MHz 50 Ohm generator in series with a 75 ohm resistor at the other end. Looked at both ends of the 75 Ohms, both to ground, and differential. At resonance, all voltages were in phase. Unfortunately I used very low loss line, but to a good approximatio, the only power supplied by the source could be explained by cable loss, and all reflected power was re reflected. It would be interesting to do this with a total source impedance of 50. BTW, I also did the thing with measuring VSWR and then changing the source impedance. Power changed, VSWR did not, for either 1:1 or 1.7:1. Tam/WB2TT |
In the case of rather lossy cable, it is best to
abandon rho and SWR before they become "bloody useless nuisances", and use the computer for everything. Bill W0IYH [snip] Amen Brother! -- Peter K1PO Indialatic By-the-Sea, FL. ========================== Amen ? Peter, does "Amen" imply worship of Terman and the ARRL Handbook has now been heretically switched to a mere box of electronics ? Heaven help us! ---- Reg, G4FGQ |
Richard Clark wrote in message . ..
Thank you Bill. This is probably why they say that an oscillator has negative resistance (postitive feedback), and also why the stability circles have their centers located outside the unit 1 Radius. Clearly, areas outside the rho=1 circle on the Smith, are reserved for active networks. Slick Hi OM, This does not follow logically. Active networks exhibit this behavior and offer a library full of applications. Strained passive networks, as evidenced by other correspondence, also reside in this "forbidden region" but their number are significantly smaller and for good reason. Please give an example. Slick |
"William E. Sabin" sabinw@mwci-news wrote in message ...
For a moderately complex Z0 (moderately lossy cable), the conventional formula for rho can have magnitude values a little greater than 1.0. The boundary of the Smith chart represents rho=1.0, but a complex Z0 can push rho a little beyond the circumference. Because SWR is extremely high at the outer circle of the Smith chart, we should avoid that region in our work with the Smith chart. Incorrect, Bill. Rho greater than one is for active networks only. Slick |
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On Fri, 29 Aug 2003 23:36:02 -0500, W5DXP
wrote: Richard Clark wrote: W5DXP wrote: But: "Every time you make a measurement, you make an error." So said my EE prof almost half a century ago. I assume it's still true. :-) Yes. What can I assume from the fact that you ignored the tough questions in the rest of my posting? Hi Cecil, I have no idea what you can assume. You generally express it freely anyway. How do you measure the exact amount of reflected power re-reflected from the source? Dan's spreadsheet work seems like a good place to start. It is of no particular interest to me and doesn't respond to the challenge in and of itself. You might find an exact answer there. Simply going to the bench would be adequate for amateurs, guessing would eclipse the efforts of everyone else. ;-) It should seem obvious that the power reflected at the source interface is governed by the standard mechanics no one wishes to impart to the source. In other words the power returning to the source will be reflected through the same mechanics it met at the mismatched load, with the "exact" value varying only by the reflection coefficient of the source mismatch and the power incident upon it. As the line presents a media of 50 Ohms, and the source presents a 100 Ohm discontinuity, the portion of power reflected is a rather trivial computation - except for those who apparently employ "first principles" in one direction only. Gad, I love that bit of irony! And they do it for the sake of educating lurkers too! Hi Ian, George, you will find this in chapter 9 from Chipman (got it right that time George ;-). It also accounts for the increased loss exhibited by the example of the Challenge that so baffles everyone. I wonder what those so bench-shy would attribute that additional loss to? Would their astrologers suggest the opposition of Mars? Hint: the answer works at all aspects of any planets. For those who closely follow their astrologer's advice, I once again suggest this only matters to those interested in accuracy. With equipment available across the counter in exchange for a credit card charge, their purchase is not likely to suffer any issues brought to the forum here and they can rest assured their SWR meters will work as advertised given the likely source Z of being 50 Ohms or insignificantly off from that value. (Many will gladly suffer 2:1 mismatch straight from the antenna connector so they can worry it at the antenna.) Further, for inferior purchases at similar cost (how would they know?), they will still be unaware barring some change in the length of cabling that will have them muttering a moment or two before they shrug it off anyway. 73's Richard Clark, KB7QHC |
"W5DXP" wrote in message ... Tarmo Tammaru wrote: ...and all reflected power was re reflected. Since the re-reflected waves are coherent with the source waves, how do you know that? Would the amplitude of ghosting in a TV signal support that assertion? -- 73, Cecil http://www.qsl.net/w5dxp There was no actual power generated by the source. Some people seem to think that *power* dissipated by the resistor due to current flowing in one direction and *power* caused by current in the other direction cancel. Of course, the experiment was moot, or 1/4 wave stubs wouldn't work. Ghosts live in the realm of pulses, where things always work unambiguously. In logic design, you often terminate the source, so there is only one reflection. You can't series terminate the load, because the input impedance is several K; you can't shunt terminate the load because the source can't deliver enough current, although I have done things like shunt terminating the load with a Zo resistor in series with 10PF or so. Tam/WB2TT |
Ian White, G3SEK wrote:
I believe you have specified more numbers than you can actually know. That's the whole point, Ian. Exactly how are those things knowable if they give identical readings? I could argue that either of those conditions exist and you cannot prove otherwise. There are probably an infinite number of configurations that will yield identical measurements. What I don't understand is how any of them can be knowable. -- 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! =----- |
Tarmo Tammaru wrote:
There was no actual power generated by the source. How do you know? Was it 100% efficient and consumed no power? -- 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! =----- |
W5DXP wrote:
Ian White, G3SEK wrote: [Re-inserting my quote from Cecil] 1. The source is generating 100 watts and 20 watts of the 50 watts of incident reflected power is being re-reflected in phase from the source. The forward power meter reads 120 watts. The reflected power meter reads 50 watts. The conventional rule-of-thumb has the source generating (100-30)= 70 watts since it is dissipating 30 watts of reflected power. 2. The source is generating 110 watts and 10 watts of the 50 watts of incident reflected power is being re-reflected in phase from the source. The forward power meter reads 120 watts. The reflected power meter reads 50 watts. The conventional rule-of-thumb has the source generating (110-40)= 70 watts since it is dissipating 40 watts of reflected power. How can you possibly distinguish between the above two identical conditions caused by different source impedances? I believe you have specified more numbers than you can actually know. That's the whole point, Ian. Exactly how are those things knowable if they give identical readings? I could argue that either of those conditions exist and you cannot prove otherwise. There are probably an infinite number of configurations that will yield identical measurements. What I don't understand is how any of them can be knowable. Sorry, I didn't make myself clear. I believe you have specified more numbers than you can know, in *each* example. Therefore trying to distinguish between them is double-doomed :-)) -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
On Sat, 30 Aug 2003 11:58:55 -0500, W5DXP
wrote: Richard Clark wrote: Simply going to the bench would be adequate for amateurs, guessing would eclipse the efforts of everyone else. ;-) I'm ready to go to the bench but I don't know how to separate the incident reflected power that is re-reflected from the source from the generated power. That's what I am asking. How does one separate those two coherent superposed forward waves? It should seem obvious that the power reflected at the source interface is governed by the standard mechanics no one wishes to impart to the source. I agree that the source obeys the wave reflection model rules. The fly in the ointment is the unknowable source impedance encountered by the incident reflected waves. Worry about it if the example of the challenge presents more than 3+ dB loss. If not, there's hardly any point is there? We can work out the differences then, but only if you care about accuracy and the lack of confirming this exotic characteristic. I can appreciate that this is not everyone's interest. In other words the power returning to the source will be reflected through the same mechanics it met at the mismatched load, with the "exact" value varying only by the reflection coefficient of the source mismatch and the power incident upon it. As the line presents a media of 50 Ohms, and the source presents a 100 Ohm discontinuity, ... I must have missed how you know the source presents a 100 ohm impedance to incident reflected waves. Do you have a 100 ohm pad between the transmitter output and the transmission line? Hi Cecil, I didn't say I did the example, it is drawn from a reference. We've already been through the mechanics of how to do it employing a variable transmission line. Consult our correspondence for specific details, you already offered that your equipment could tolerate that mismatch, however THAT discussion is separate and distinct from the example of the challenge. The challenge merely offers another approach. You can add a 50 Ohm Dummy Load in series to the output of your rig (this, of course presumes it presents 50 Ohms characteristic, but as many declaim that specification perhaps they could offer another value - eh, unlikely). Another issue of adding a series resistor is one of shielding and common mode issues. You asked me in that earlier correspondence if I considered this, and yes I did. That is why I used massive parallel loads that insured an entirely shielded system. This is one of those methods that one correspondent pondered: There is no institutionalized ignorance, just a lot of skepticism regarding the reliability of the analysis methods and the measurement methods. which is understandable from those not trained in the art of designing measurement scenarios. I am trained but those still caught in the quandary are immobilized by rejecting every method (institutionalized ignorance). The measure of RF power is not simple by any means so it is best left to those who are serious about accuracy - hardly an amateur pursuit, and it hardly matters anyway as it is not a problem with modern equipment. I offered that you already had the tools to perform a simple first pass approximation, you really need to consult that thread of correspondence again. None of this with forced mismatches is all that hard in the first place. The greater the mismatch at each end, the more compelling the evidence. The simple fact of the matter is that most rigs conform to 50 Ohm source Z and do not exhibit this issue. If those who held to their cherished fantasy of source Z being other than 50 Ohms, then they should be able to ace this test from the beginning (the lack of their correspondence reveals the invalidity of their claims). I'm working on two repeater systems today (10M and GMRS), so enjoy. 73's Richard Clark, KB7QHC |
"W5DXP" wrote in message ... Tarmo Tammaru wrote: There was no actual power generated by the source. How do you know? Was it 100% efficient and consumed no power? -- 73, Cecil http://www.qsl.net/w5dxp The output rose to the open circuit value. Actually, I had thought of building a very efficient class C amplifier and calibrating the output power in terms of DC collector current, but think it is a waste of time. Tam/WB2TT -----= 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! =----- |
Tarmo Tammaru wrote:
The output rose to the open circuit value. Assuming you are talking about voltage, for a Norton equivalent, that's a very lossy situation. What was your source? -- 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! =----- |
"Cecil Moore" wrote in message ... Tarmo Tammaru wrote: The output rose to the open circuit value. Assuming you are talking about voltage, for a Norton equivalent, that's a very lossy situation. What was your source? -- 73, Cecil http://www.qsl.net/w5dxp The source was an MFJ269, and as I mentioned, the external series resistor was 75 Ohms. I should try it with something like 1K. Coax was ~100 feet of 9913. My scope is only good to 100 MHz; so, I have to keep the frequency down. BTW, I have met Ed Norton. Tam/WB2TT |
"Tarmo Tammaru" wrote in message ...
you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes to -Z0*/Z0, which, for Zo having phase angle b equates to -1 at angle(-2b). Agreed. This doesn't prove anything, does it? How does this allow for the sum of V+ and V- to be 0? That is what you have across a short. This is the correct answer, you just interpret it incorrectly. The reflection coefficient is defined as the reflected voltage divided by the incident voltage. Sure, the voltage is zero right at the short, but there is a reflected voltage wave that moves back towards the generator. When you have a short, the phase is flipped 180 degrees, which is exactly what the -1 means. Notice if you had a Zl=infinity, that the RC would be +1, which would be full reflections INPHASE with the generator, or in phase with the incident voltage wave. It's very simple stuff, but many people don't understand this. ......................................... P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)] Remember, z is distance from the load. Tam/WB2TT ok, you've done a nice job of copying the text you sent me. As I recall, you said you were not familiar with these diagrams and did not understand them. What do you want me to do? derive it in a different way? I want you to tell me the significance of the fact that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr because Zo is complex in the general case. And how this may make this example incorrect for this discussion. Slick |
"Dr. Slick" wrote in message om... "Tarmo Tammaru" wrote in message ... you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes to -Z0*/Z0, which, for Zo having phase angle b equates to -1 at angle(-2b). Agreed. This doesn't prove anything, does it? How does this allow for the sum of V+ and V- to be 0? That is what you have across a short. This is the correct answer, you just interpret it incorrectly. The reflection coefficient is defined as the reflected voltage divided by the incident voltage. Sure, the voltage is zero right at the short, but there is a reflected voltage wave that moves back towards the generator. When you have a short, the phase is flipped 180 degrees, which is exactly what the -1 means. Notice if you had a Zl=infinity, that the RC would be +1, which would be full reflections INPHASE with the generator, or in phase with the incident voltage wave. It's very simple stuff, but many people don't understand this. That's the whole point. By the conjugate formula RC is *not* -1. It is -1 with a phase angle. I agree it works for an open circuit, since you can divide both sides by Zl, and anything divided by infinity is 0 ......................................... P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)] Remember, z is distance from the load. Tam/WB2TT ok, you've done a nice job of copying the text you sent me. As I recall, you said you were not familiar with these diagrams and did not understand them. What do you want me to do? derive it in a different way? I want you to tell me the significance of the fact that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr Obviously true because Zo is complex in the general case. And how this may make this example incorrect for this discussion. Slick I am not sure which example you mean, but A*/A does not have 0 phase angle, unless the phase angle of A is 0. PS I scanned in the other stuff. Compressed TIF Tam |
"Tarmo Tammaru" wrote in message ...
I can't scan the whole chapter, and part of the next. You should be able to get your library to borrow the book for you. Or, look at some other college textbook under lossy lines. 5.2B is straightforward enough: Zo=(1/Yo)=SQRT[(R+jwL)/(G+jwC)]=Ro+jXo Tam/WB2TT I've looked at the 9 pages you sent me, and I'm not certain what to tell you. They say, "the fact that the rho may exceed unity on a dissipative line does not violate a condition of power conservation (as it would on a lossless structure)." On the one hand, I give them creedence for addressing the concept...it's what i've been saying all along. However, they don't explain why a lossy line can INCREASE the reflected power! The lossless line would not attenuate the reflected wave at all! They also mention that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr because Zo is complex in the general case. This may or may not make their example moot. And i don't trust their Smith Chart extended out to 1+sqrt(2) for a dissipative line. Maybe for an active network, but not a passive one. Also, they go from equation 5.12 to 5.13 without showing us how they got there. Perhaps they just copied it out of another book and used the formula incorrectly. Certainly text books can disagree, just as the "experts" often do. Maybe this book is hard to find because it's out of print because nobody trusted it. Slick |
"Dr. Slick" wrote in message om... However, they don't explain why a lossy line can INCREASE the reflected power! The lossless line would not attenuate the reflected wave at all! They make a pont of the fact that they are *not* violating the concept of conservation of energy They also mention that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr because Zo is complex in the general case. This may or may not make their example moot. I don't see the problem. 100 /_30 degrees divided by 2/_5 degrees is 50/_15 degrees. Different phase angle. By general case they mean not the lossless case. And i don't trust their Smith Chart extended out to 1+sqrt(2) for a dissipative line. Maybe for an active network, but not a passive one. No idea. Never had to extend a Smith chart Also, they go from equation 5.12 to 5.13 without showing us how they got there. They use the identity e**jx = cos x + jsin x Perhaps they just copied it out of another book and used the formula incorrectly. I don't think a technical textbook is supposed to introduce new concepts unless they are labeled as such Certainly text books can disagree, just as the "experts" often do. Maybe this book is hard to find because it's out of print because nobody trusted it. Slick As I said out front. The book is copyrighted 1960. There is a certain life to these things. Tam |
"Tarmo Tammaru" wrote in message ...
How does this allow for the sum of V+ and V- to be 0? That is what you have across a short. This is the correct answer, you just interpret it incorrectly. The reflection coefficient is defined as the reflected voltage divided by the incident voltage. Sure, the voltage is zero right at the short, but there is a reflected voltage wave that moves back towards the generator. When you have a short, the phase is flipped 180 degrees, which is exactly what the -1 means. Notice if you had a Zl=infinity, that the RC would be +1, which would be full reflections INPHASE with the generator, or in phase with the incident voltage wave. It's very simple stuff, but many people don't understand this. That's the whole point. By the conjugate formula RC is *not* -1. It is -1 with a phase angle. I agree it works for an open circuit, since you can divide both sides by Zl, and anything divided by infinity is 0 It was clear that you misunderstood what the -1 meant. Ok, -1 with a phase angle. I believe it, if the Zo is complex. The point is that the ratio of the reflected to incident voltage is one, and has nothing to with the DC voltage measured at the short itself. -1 with a phase angle would simply be 180 degrees phase shift, plus or minus and additional angle. I want you to tell me the significance of the fact that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr because Zo is complex in the general case. And how the example you sent me may make be incorrect for this discussion. Slick |
"Tarmo Tammaru" wrote in message ...
"Dr. Slick" wrote in message om... However, they don't explain why a lossy line can INCREASE the reflected power! The lossless line would not attenuate the reflected wave at all! They make a pont of the fact that they are *not* violating the concept of conservation of energy But they never explain WHY a lossy line can INCREASE the reflected power! The lossless line would not attenuate the reflected wave at all! I don't trust their claims on this. If you get more power reflected than you send into a passive network, you are getting energy from nowhere, and are thus violating conservation of energy. They also mention that the normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr because Zo is complex in the general case. This may or may not make their example moot. I don't see the problem. 100 /_30 degrees divided by 2/_5 degrees is 50/_15 degrees. Different phase angle. By general case they mean not the lossless case. I believe you mean 50 @ 25 degrees. And i don't trust their Smith Chart extended out to 1+sqrt(2) for a dissipative line. Maybe for an active network, but not a passive one. No idea. Never had to extend a Smith chart Do some research, and you will never see an "extended Smith Chart" for a passive network. Oh, certainly for a active device, for stability circles and such, but passive networks can never have a rho greater than 1. Also, they go from equation 5.12 to 5.13 without showing us how they got there. They use the identity e**jx = cos x + jsin x Yes? And? How did they get the Zo=(Zn-1)/(Zn+1) from this? As I said out front. The book is copyrighted 1960. There is a certain life to these things. Tam But it seems to be out of print, perhaps with good reason... Slick |
Opps!
I meant, how did they derive Reflection Coefficient=(Zn-1)/Zn+1)? They certainly don't show us! This would be key to answering our questions. Slick |
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