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Roy:
[snip] "Roy Lewallen" wrote in message ... I'm sorry I'm having such a hard time communicating, and maybe the trouble is that I've been misunderstanding what has been said. Here's how it looks to me. [snip] Heh, heh... don't feel bad, that is common in any discourse using the medium of USENET Newsgroup postings! [snip] So what's bothering me is that I'm completely unable to start with what I believe to be true about the relationships between voltages, currents, and characteristic impedance of a transmission line (itemized as three assumptions in a recent posting), and arrive at anything but the classical equation which is universally used for transmission line analysis. [snip] I'm with you man! I like and, only use, the classical definitions of rho and the Scattering Matrix wich correspond to the "actual" waves supported by the wave equation [The Telegraphist's Equations] on passive transmission lines. I simply don't like the version of rho/Scattering matrix which uses the complex conjugate of the "reference impedance" or what ever else you wish to call the charateristic or surge impedance simply because the transition between Z and conj(Z) represents an impedance discontinuity, a big one in fact, and physical intuition tells me that there will alwasy be some effect on waves which have to transit such a discontinuity. I am not a bigot about this use of the conj(Z) and I can quote several excellent authoritative references that do so. I don't however agree with Slick's use of conj(Z) in the numerator and just Z in the denominator, I believe that to be the result of some typographical error or complete errors in misunderstanding. I do understand that for special kinds of problems it may be convenient to adopt different definitions for rho and the Scattering Matrix, and that is fine just as long as the definers remain consistent thereafter and indicate their departure from the more common usual definitions. Confusion sets in when there are proponents of the different definitions discussing/arguing about results without first agreeing on the definitions. [snip] I've posted both a detailed derivation of the formula for reflection coefficient and a numerical example of a terminated transmission line with complex Z0. [snip] You do good work Roy. [snip] elements or "semi-infinite" transmission lines -- or no derivation at all. The alternate dervivations, it seems, can't for some reason withstand the condition that a simple, finite length transmission line be involved. Rather than presenting a simple analysis of a system with a transmission line (as I did), all I've seen in response is lumped element or "semi-infinite" transmission line analyses with indignant protestations if I question the possibility that the presented models don't accurately represent a finite transmission line. [snip] Because Roy, the use of semi-infinite transmission lines in a simple "theoretical" derivation is easier to communicate in short NG postings and easier to understand and manipulate than complex long "arithmetical" developments, no matter how practical. I too can give the group numerous practical Engineering design calculations of real problems that many have solved over the years in the communications industry, not made up problems like yours, but real ones involving complex Zo transmission lines with highly variable Zo's of lengths up to 18,000 feet operating with very little "talker echo" [reflection coefficient] over frequency ranges of a handful of decades, using very economical lumped approximations to Zo in the balancing networks. There are several patents issued in this area. But I am sure that most would get glassy eyed with those detailed Engineering calculations. Which would you rather have, some detailed long drawn out Engineering calculations or a simple two line theoretical proof that using Zo and not conj(Zo) results in no talker echo only for an image matched line. The theory of "image matching" was developed by Campbell and Zobel in the time frame of the early 1920's, why are we trying to prove it again now using arithmetic? [snip] I really think that by now, anyone who is able to benefit from the discussion has done so, and those who remain will continue to hold their views no matter what. So I won't further waste the time of either group by continuing to question the issue. I hope that the derivations I've posted are helpful to those people who are interested in seeing where the common formulas and equations come from. [snip] Thanks for all of your efforts Roy. As you know I agree completely with your results and conclusions. But I must re-iterate I don't agree with your methods. The use of arithmetic where a little algebra and mathematical theory of trasmission lines will suffice. is not my ideal of good communications. Kraus, Balmain and all of your "other heros" make use of semi-infinite lines in their developments and descriptions of the meaning of Zo, I just don't see what *your* problem is with that approach. I can assure you that Maxwell did not use "arithmetic" in the development of his equations, nor did the first person to "define" a reflection coefficient! I am perplexed by your approach to say the least, but perhaps I just don't understand... -- Peter K1PO Indialantic By-the-Sea, FL. |
Slick:
[snip] If you believe that there are theoretically no reflections in a conjugate match, then with Zl=50+j10 and Zo=50-j10, the conjugate equation correctly cancels the reactances giving no reflections, while the non-conjugate still incorrectly gives a magitude (non zero) for rho. Slick [snip] Oh yes here are voltage reflections at a conjugate match! Simply put as waves pass across the transition from an impedance of Zo to an impedance of conj(Zo) they are crossing a boundary with an impedance discontinuity. Zo on one side and conj(Zo) on the other side is definitely discontinuous! Unless of course, Zo = conj(Zo) which occurs only when Zo is real. There will always reflections at such an impedance discontinuity where an impedance faces its' conjugate. If an impedance Zo faces itself Zo there will be no "voltage reflections", this phenomena was called an "image match" by Campbell and Zobel way back around 1920 or so, it represents the basis for the image match design of filters and transmission systems. Do any reader's here recall image parameter filter design? I designed more than a few that way myself. If an impedance Zo faces its' conjugate conj(Zo) then there will be no "power reflections", but there will in general be voltage reflections, and this is called a "conjugate match" and represents the basis for "insertion loss" design of filters and transmission systems. Darlington and Cauer introduced the insertion loss design of filters which eventually supplanted the older Campbell, Zobel image parameter design. The "classical definition of voltage reflection coefficient from theoretical physics and the solutions to the wave equations is the one that does not use the conjugate. This classical definition corresponds to the calculations made by a reflectometer configured to measure the reflected voltage at the boundary between Z and Zo such that if Z=Zo the reflected voltage (often called the "talker echo") will be zero. This will not correspond to maximum power transfer in the general case. i.e. "classical" rho = (Z - Zo)/(Z + Zo) is not zero at a conjugate match. I also believe that this is "Mother Nature's" reflection coefficient for it is exactly what she uses as she lets the waves propagate down her lines of surge impedance Zo following her partial differential equations at every point along the way. At every infinitesimal length of line all along it's length the waves are passing from a infinitesimal region of surge impedance Zo to the next infinitesimal region of surge impedance Zo and there are no voltage reflections anywhere along that [uniform] line, although if the line is not lossless there will be energy lost as the wave progresses. Slick... On another whole level it simply does not matter which defiinition of the reflection coefficient one uses to make design calculations though, as long as the definition is used consistently throughout any calculations. One can convert any results based on the non-conjugate version of rho to results based on the conjugate version of rho and vice versa. In other words, neither version is "RIGHT" or "WRONG" as long as the results from using that particular definition are interperted correctly in terms of the original definition. In fact one can cook up an [almost] completely ficticious reference impedance, one which has no relation whatsoever to the Zo of the line. Just call the ficticious, perhaps complex reference impedance R [I like to use R since in my mind it stands for "Reference", and it need not be a pure resistance.] and use this R in a defintition of rho and/or a Scattering Matrix and then any and all subsequent calculations afterwords will be correct as long as this ficticious R is used consistently with the definitions in terms of the electrical port vectors the voltage v, current i and wave vector comprising the "waves" a and b. rho = (Z - R)/(Z + R) = b/a = (v - Ri)/(v + Ri) This is true simply because the "waves" a and b are mathematically just simple linear combinations of the voltage v and current i! Look at how simple the relationship is... [a, b] and [i, v] are simply related by a simple transformation matrix as follows: a = v + Ri b = v - Ri or wave vector = matrix * electrical vector And the matrix, in this case is: | 1 R | | 1 -R| As long as you choose a reference impedance R such that the transformation matrix is non-singular then you can go back and forth from a, and b to v and i any time, any where, etc... So... who gives a damm about the defintion of rho as long as you are consistent. Unless of course, like Roy and others, you insist that the "waves" correspond to some preconceived notion of what waves really are... ;-) Thoughts, comments, -- Peter K1PO Indialantic By-the-Sea, FL |
"Peter O. Brackett" wrote in message thlink.net...
Slick: [snip] If you believe that there are theoretically no reflections in a conjugate match, then with Zl=50+j10 and Zo=50-j10, the conjugate equation correctly cancels the reactances giving no reflections, while the non-conjugate still incorrectly gives a magitude (non zero) for rho. Slick [snip] Oh yes here are voltage reflections at a conjugate match! Simply put as waves pass across the transition from an impedance of Zo to an impedance of conj(Zo) they are crossing a boundary with an impedance discontinuity. Zo on one side and conj(Zo) on the other side is definitely discontinuous! Unless of course, Zo = conj(Zo) which occurs only when Zo is real. There will always reflections at such an impedance discontinuity where an impedance faces its' conjugate. I disagree completely. The theoretical impedance of a resonant series L and C (which is lossless) is zero. So in a conjugate match, where they cancel out, in an ideal loss-less world, it would be equivalent to the series C and L not being there at all, with the source and load 50 ohms free to pass max. power delivered to the load. If an impedance Zo faces its' conjugate conj(Zo) then there will be no "power reflections", but there will in general be voltage reflections, ??? if the square of the magnitude of the voltage RC is the power RC, then your statement is incorrect. And rho (magnitude of Voltage RC) is the square root of the Power RC. i.e. "classical" rho = (Z - Zo)/(Z + Zo) is not zero at a conjugate match. That's why it is incorrect for complex Zo. I also believe that this is "Mother Nature's" reflection coefficient for it is exactly what she uses as she lets the waves propagate down her lines of surge impedance Zo following her partial differential equations at every point along the way. At every infinitesimal length of line all along it's length the waves are passing from a infinitesimal region of surge impedance Zo to the next infinitesimal region of surge impedance Zo and there are no voltage reflections anywhere along that [uniform] line, although if the line is not lossless there will be energy lost as the wave progresses. Maybe "Mother Nature" should take a Les Besser course... :) Slick... On another whole level it simply does not matter which defiinition of the reflection coefficient one uses to make design calculations though, as long as the definition is used consistently throughout any calculations. I totally disagree again: Did you read Williams' data? The data follows: Note: |rho1*| is conjugated rho1, SWR1 is for |rho1*|, |rho2| is not conjugated and SWR2 applies to |rho2| X0.......|rho1*|..SWR1.....|rho2|..SWR2 -250..... 0.935...30.0.....1.865...-3.30 -200..... 0.937...30.8.....1.705...-3.80 -150..... 0.942...33.3.....1.517...-4.87 -100..... 0.948...37.5.....1.320...-7.25 -050..... 0.955...43.3.....1.131...-16.3 -020..... 0.959...47.6.....1.030...-76.5 -015..... 0.960...48.4.....1.010...-204 -012..... 0.960...48.9.....0.997....+/- infinity -010..... 0.960...49.2.....0.990....+305 -004..... 0.961...76.3.....0.974....+76.3 0000..... 0.961...50.9.....0.961....+50.9 The numbers for not-conjugate rho are all over the place and lead to ridiculous numbers for SWR. It is also obvious that for a low-loss line it doesn't matter much. But values of rho greater than 1.0, on a Smith chart correspond to negative values of resistance (see the data). Excellent work William. You are also showing how a rho1 leads to ridiculous numbers for the equation: SWR = (1 + rho)/(1 - rho) The non-conjugate equation simply cannot handle complex Zo. Some people think we should throw out the SWR formula completely, but this is complete nonsense, of course. SWR = (1 + rho)/(1 - rho) works for 0=rho=1, for very good reason, as it applies to passive networks only. And the conjugate will always give 0=rho=1, even with a complex Zo. Slick |
Slick:
[snip] I disagree completely. The theoretical impedance of a resonant series L and C (which is lossless) is zero. So in a conjugate match, where they cancel out, in an ideal loss-less world, it would be equivalent to the series C and L not being there at all, with the source and load 50 ohms free to pass max. power delivered to the load. [snip] Which is exactly what happens for all energy passing through at the resonant frequency of the series LC! And for instance if you are testing with a sinusoidal generator at that frequency that is exactly what you will observe. Of course if you are testing with a broad band signal rather than a sinusoidal signal lots of much more interesting stuff happens. That all can be calculated simply by using the full functional descriptions of the network/transmission system, i.e. assuming Z = Z(p) where p = s + jw, etc, etc... [snip] ??? if the square of the magnitude of the voltage RC is the power RC, then your statement is incorrect. [snip] To which voltage reflection coefficient do you refer? :-) No! The square of the magnitude of the voltage reflection coefficient is not, in general, equal to the power reflection coefficient. [snip] That's why it is incorrect for complex Zo. [snip] Slick, no "correctly" defined reflection coefficient is "incorrect". There can easily be an infinity of different "correct" reflection coefficients so defined, and none is "incorrect" so long as no incorrect conclusions are drawn from their use. The only "incorrect" ones are the reflection coefficients that are not defined based upon simple non-singular linear combinations of the electrical variables i and v. Slick, your view of the reflection coefficient world is far too narrow! Widen your horizons, there is more than one way to go to hell, and chosing a particular definition of a reflection coefficient and forcing all others to believe in it is nothing short of bigotry! [snip] Maybe "Mother Nature" should take a Les Besser course... :) [snip] I am sure that Dr. Besser is an honorable and accomplished man despite his obviously narrow views of "waves". [snip] I totally disagree again: Did you read Williams' data? [snip] Yes I "scanned" it and lost interest quickly, because of the gratuitous use of mind boggling numerical tables in ASCII text on a newsgroup posting! I am sure that William did a lot of work whilst typing in those long strings of numbers without error. Good work William! Hey I'll scan in and post a listing of a couple of thousand lines of the Zo versus frequency of 18,000 feet of plastic insulated AWG 24 wire if that will help. I've got hundreds and hundreds of pages of such data! :-) [snip] Excellent work William. You are also showing how a rho1 leads to ridiculous numbers for the equation: SWR = (1 + rho)/(1 - rho) The non-conjugate equation simply cannot handle complex Zo. Some people think we should throw out the SWR formula completely, but this is complete nonsense, of course. SWR = (1 + rho)/(1 - rho) works for 0=rho=1, for very good reason, as it applies to passive networks only. And the conjugate will always give 0=rho=1, even with a complex Zo. [snip] Hey, again your view of rho and VSWR is too narrow. Ask yourself what is the meaning of SWR in that formula when rho is complex and SWR is complex! Actually if you let your mind expand a little beyond your narrow view of things you will find that complex SWR can have a physical and useful meaning as well. -- Peter K1PO Indialantic By-the-Sea, FL |
"Peter O. Brackett" wrote in message link.net... Slick: Hey, again your view of rho and VSWR is too narrow. Ask yourself what is the meaning of SWR in that formula when rho is complex and SWR is complex! how can swr be complex... in my book it is: SWR=(1+|rho|)/(1-|rho|) so swr can't be complex. |
Dr. Slick wrote:
"Peter O. Brackett" wrote: There will always reflections at such an impedance discontinuity where an impedance faces its' conjugate. I disagree completely. The theoretical impedance of a resonant series L and C (which is lossless) is zero. So in a conjugate match, where they cancel out, in an ideal loss-less world, it would be equivalent to the series C and L not being there at all, with the source and load 50 ohms free to pass max. power delivered to the load. Better be careful. Did you just assert that you can change the SWR on a feedline by forming a conjugate match at the source? All Peter is saying is that the VSWR on the feedline will not be 1:1 if Z-complex-load differs from the purely resistive Z0 of a lossless line. For a lossless line, there is nothing you can do at the source to change the SWR at the load. However, if the line is lossless, you can achieve maximum power transfer anyway even in the face of a high SWR. -- 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! =----- |
"Peter O. Brackett" wrote in message hlink.net...
Slick: [snip] I disagree completely. The theoretical impedance of a resonant series L and C (which is lossless) is zero. So in a conjugate match, where they cancel out, in an ideal loss-less world, it would be equivalent to the series C and L not being there at all, with the source and load 50 ohms free to pass max. power delivered to the load. [snip] Which is exactly what happens for all energy passing through at the resonant frequency of the series LC! And for instance if you are testing with a sinusoidal generator at that frequency that is exactly what you will observe. Of course if you are testing with a broad band signal rather than a sinusoidal signal lots of much more interesting stuff happens. That all can be calculated simply by using the full functional descriptions of the network/transmission system, i.e. assuming Z = Z(p) where p = s + jw, etc, etc... Well, of course i assume the conjugate match to occur at ONE frequency, and with a small signal sine wave. [snip] ??? if the square of the magnitude of the voltage RC is the power RC, then your statement is incorrect. [snip] To which voltage reflection coefficient do you refer? :-) No! The square of the magnitude of the voltage reflection coefficient is not, in general, equal to the power reflection coefficient. Nope! Page 16-2 of the 1993 ARRL: rho=sqrt(Preflected/Pforward) Look it up yourself, don't take my word for it. The only "incorrect" ones are the reflection coefficients that are not defined based upon simple non-singular linear combinations of the electrical variables i and v. I don't think you know what you are typing about here.. Slick, your view of the reflection coefficient world is far too narrow! Widen your horizons, there is more than one way to go to hell, and chosing a particular definition of a reflection coefficient and forcing all others to believe in it is nothing short of bigotry! Believe what you will... I ain't forcing anyone to accept anything! I will tell you what respected authorities have written, though. Jesus, dude. Do you want me to agree with you even when i think you are incorrect? Or would you prefer me to be honest? [snip] Maybe "Mother Nature" should take a Les Besser course... :) [snip] I am sure that Dr. Besser is an honorable and accomplished man despite his obviously narrow views of "waves". YOU are the one with the narrow views. Besser's courses are like $1,200 a head. Do you think companies would pay him to steer them wrong? Please! Yes I "scanned" it and lost interest quickly, because of the gratuitous use of mind boggling numerical tables in ASCII text on a newsgroup posting! I am sure that William did a lot of work whilst typing in those long strings of numbers without error. Good work William! Lost interest, or don't want to look at information that you disagree with? [snip] Excellent work William. You are also showing how a rho1 leads to ridiculous numbers for the equation: SWR = (1 + rho)/(1 - rho) The non-conjugate equation simply cannot handle complex Zo. Some people think we should throw out the SWR formula completely, but this is complete nonsense, of course. SWR = (1 + rho)/(1 - rho) works for 0=rho=1, for very good reason, as it applies to passive networks only. And the conjugate will always give 0=rho=1, even with a complex Zo. [snip] Hey, again your view of rho and VSWR is too narrow. Ask yourself what is the meaning of SWR in that formula when rho is complex and SWR is complex! Sigh... rho is the MAGNITUDE of the RC, so it isn't complex. And SWR is never complex! And a negative SWR is pretty meaningless! If you want to rewrite the RF books, good luck. Cheers, Slick |
Cecil Moore wrote in message ...
I disagree completely. The theoretical impedance of a resonant series L and C (which is lossless) is zero. So in a conjugate match, where they cancel out, in an ideal loss-less world, it would be equivalent to the series C and L not being there at all, with the source and load 50 ohms free to pass max. power delivered to the load. Better be careful. Did you just assert that you can change the SWR on a feedline by forming a conjugate match at the source? All Peter is saying is that the VSWR on the feedline will not be 1:1 if Z-complex-load differs from the purely resistive Z0 of a lossless line. For a lossless line, there is nothing you can do at the source to change the SWR at the load. As usual, your sentences don't make too much sense, which is probably why you go one with your record-breaking threads. Maybe you actually agree with people when you argue with them... well, we could all be accused of that one. However, if the line is lossless, you can achieve maximum power transfer anyway even in the face of a high SWR. If Zo=50-j5 and Zload=50+j5, you will have a conjugate match, and max power delivered to the load. Slick |
Frankly, I haven't paid any attention to your ducking, dodging, and
hand-waving. 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. One thing I have gotten from your postings, though, is an appreciation for what you said about your alma mater being a military school. They obviously taught you to always present a moving target, and you learned the lesson well. I return the readers now to tau, s11, n-port networks, optics, virtual photons, and whatever else can be produced to avoid directly facing the stark reality of Ohm's and Kirchoff's laws. Enjoy! Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: The calculation used for reflection coefficient is based on its definition, namely reflected voltage divided by forward voltage. Unfortunately, you did not correctly identify the forward voltage and reflected voltage. V1*tau is only one of the forward voltage components. There is another one, V2*rho. Same for current. Did you see my example where by adding one wavelength of lossless feedline, it can be proven that reflected power can never be greater than forward power? |
Roy Lewallen wrote:
Frankly, I haven't paid any attention to your ducking, dodging, and hand-waving. 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=68-j39 ---lossy feedline---+---1WL 50 ohm lossless line---10+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. -- 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! =----- |
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 hand-waving. (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 hand-waving. 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=68-j39 ---lossy feedline---+---1WL 50 ohm lossless line---10+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. |
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 what-ever was suppose to be the hallmark of my specialty and simply replied with a ho-hum: "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 |
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 feedline---10+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 |
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 self-consistent, 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 hand-waving, 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. . . . |
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 self-consistent, 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 line---x---1WL 50 ohm lossless line---10+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 no-brainer. The forward power on the 50 ohm line is 10W less than the reflected power on the 50 ohm line, i.e. Pfwd2-Pref12 = 10W just as it should. At point 'x', on the source side of 'x', Pfwd1-Pref1 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! =----- |
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 self-consistent, 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 line---x---1WL 50 ohm lossless line---10+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 no-brainer. The forward power on the 50 ohm line is 10W less than the reflected power on the 50 ohm line, i.e. Pfwd2-Pref12 = 10W just as it should. At point 'x', on the source side of 'x', Pfwd1-Pref1 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 |
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! =----- |
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. |
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 |
Roy Lewallen wrote:
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? There are four voltage components flowing in two directions. Add up the two voltages flowing in the forward direction and you have total forward voltage. Add up the two voltages flowing in the reverse direction and you have total reverse voltage. From the total forward voltage, you can calculate the total forward power. From the total reverse voltage, you can calculate the total reverse power. There are no magic third terms after superposition. Vfwd2 = Vfwd1(tau1) + Vref2(rho2) Vref1 = Vfwd1(rho1) + Vref2(tau2) Four voltage components, two total voltages, two directions. -- 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! =----- |
wrote:
Just to make sure I understand correctly, are you saying that Vrev^2/Z0 can never be greater than Vfwd^2/Z0 (these being the common definitions of Prev and Pfwd) or did you have another set of defintions in mind for Prev and Pfwd when you state that Prev can never be greater than Pfwd? Assuming a single source, single feedline, a passive load, RMS voltage, and a real Z0, yes, average Prev can never be greater than average Pfwd, i.e. the total Poynting vector can never point away from the load. Such would be a violation of the conservation of energy principle. The fact that the Poynting vector still points toward the load in Roy's example means that reverse power cannot be greater than forward 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! =----- |
Cecil Moore wrote:
wrote: Just to make sure I understand correctly, are you saying that Vrev^2/Z0 can never be greater than Vfwd^2/Z0 (these being the common definitions of Prev and Pfwd) or did you have another set of defintions in mind for Prev and Pfwd when you state that Prev can never be greater than Pfwd? Assuming a single source, single feedline, a passive load, RMS voltage, and a real Z0, yes, average Prev can never be greater than average Pfwd, i.e. the total Poynting vector can never point away from the load. Such would be a violation of the conservation of energy principle. Yes indeed. But in the original example Z0 was complex. So once again, with the clarification that the following question was phrased in general terms and not constrained to lines where Z0 is real... Just to make sure I understand correctly, are you saying that Vrev^2/Z0 can NEVER be greater than Vfwd^2/Z0 (these being the common definitions of Prev and Pfwd) or did you have another set of defintions in mind for Prev and Pfwd when you state that Prev can NEVER be greater than Pfwd? ....Keith |
Cecil Moore wrote:
wrote: 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. The equations are to be found in Dr. Best's Nov/Dec 2001 QEX article, "Wave Mechanics of Transmission Lines: Part 3" and in _Optics_, by Hecht. When the characteristic impedances are real: Do recall that this discussion started with a line whose characteristic was complex not real, so all that follows here is moot. V1 = Vfwd1(tau1) V2 = Vref2(rho2) V3 = Vfwd1(rho1) V4 = Vref2(tau2) Vfwd2 = V1 + V2 Vref1 = V3 + V4 P1 = Pfwd1(1-|rho|^2) P2 = Pref2(|rho|^2) forward components P3 = Pfwd1(|rho|^2) P4 = Pref2(1-|rho|^2) reverse components Pfwd2 = P1 + P2 + 2*Sqrt(P1*P2)cos(delta-1) Pref1 = P3 + P4 + 2*Sqrt(P3*P4)cos(delta-2) where (delta-1) is the angle between V1 and V2, (delta-2) is the angle between V3 and V4, and that last term is called the interference term. In _Optics_, average power is called irradiance (I) and the equation is: I1 + I2 + 2*Sqrt(I1*I2)cos(delta) For further information, check out my Part 1 energy analysis article on my web page. Part 2 will appear shortly. But, going back to the original question, is Pfwd and Prev on a line with complex Z0? ....Keith |
wrote:
Cecil Moore wrote: Assuming a single source, single feedline, a passive load, RMS voltage, and a real Z0, yes, average Prev can never be greater than average Pfwd, i.e. the total Poynting vector can never point away from the load. Such would be a violation of the conservation of energy principle. Yes indeed. But in the original example Z0 was complex. So once again, with the clarification that the following question was phrased in general terms and not constrained to lines where Z0 is real... I believe my statement holds true for any possible Z0. Here's the logical proof. Replace the following: source---Z01 lossy line---x--passive load Pfwd1-- --Pref1 with an extra 1WL of lossless line that doesn't change anything. The lossy line still sees the same impedance looking into point 'x'. source---Z01 lossy line---x---1WL lossless Z02---+--passive load Pfwd1-- Pfwd2-- --Pref1 --Pref2 Pfwd1 and Pref1 exist just to the left of point 'x'. We know that (Pfwd1-Pref1) has to equal (Pfwd2-Pref2) which equals the power delivered to the load. Everything to the right of point 'x' is easy to analyze. We know that Pfwd2 Pref2 if there is any resistance in the load. If resistance in the load is zero, Pfwd2 = Pref2. In any possible case of a passive load, Pref2 cannot be greater than Pfwd2. Therefore, just to the left of point 'x', Pref1 cannot be greater than Pfwd1. Backtracking from the load, the lossy line really doesn't enter into the discussion at all. Note that Pfwd1 will not equal Pfwd2 and Pref1 will not equal Pref2 but the difference between forward power and reflected power must be equal to the power delivered to the load in order to satisfy the conservation of energy principle. -- 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! =----- |
wrote:
But, going back to the original question, is Pfwd and Prev on a line with complex Z0? See my logical proof in another posting. -- 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! =----- |
wrote in message ... Vrev^2/Z0 can NEVER be greater than Vfwd^2/Z0 (these being the common definitions of Prev and Pfwd) or did you have another set of defintions in mind for Prev and Pfwd this is another of the basic misconceptions that is being applied over and over on there. note from my big message about powers a couple days ago that from phasor notation there are really only two types of power that can be discussed: average power: Pav= .5 |I|^2 Re[Z] = .5 |V|^2 Re[Y] and complex power. P=.5 V I* now, note that given Pav you can NOT recover enough information about V to know its proper phase and magnitude in order to be able to compare it with anything properly. Complex power on the other had requires the conjugate of the current, which when expanded gives an equation like: P=.5 |V| |I| cos(/_V-/_I) + j.5 |V| |I| sin(/_V-/_I) where again, it is not possible to back up from the power to the full phasor description of the voltage to be able to use it in calculations or comparisons. now, you CAN do this if you restrict Z0 to be purely real, you can NOT do it if you use a complex Z0 because you can't extract the imaginary term back out of a power calculation that has specifically removed it. so all these places where you guys start doing sqrt(Pref/Pfwd)=Vfwd/Vref or Vrev^2/Z0/Vfwd^2/Z0 are worthless in the general case. |
Cecil Moore wrote:
wrote: Cecil Moore wrote: Assuming a single source, single feedline, a passive load, RMS voltage, and a real Z0, yes, average Prev can never be greater than average Pfwd, i.e. the total Poynting vector can never point away from the load. Such would be a violation of the conservation of energy principle. Yes indeed. But in the original example Z0 was complex. So once again, with the clarification that the following question was phrased in general terms and not constrained to lines where Z0 is real... I believe my statement holds true for any possible Z0. Here's the logical proof. Replace the following: source---Z01 lossy line---x--passive load Pfwd1-- --Pref1 with an extra 1WL of lossless line that doesn't change anything. The lossy line still sees the same impedance looking into point 'x'. source---Z01 lossy line---x---1WL lossless Z02---+--passive load Pfwd1-- Pfwd2-- --Pref1 --Pref2 Pfwd1 and Pref1 exist just to the left of point 'x'. We know that (Pfwd1-Pref1) has to equal (Pfwd2-Pref2) which equals the power delivered to the load. While Pnet = Pfwd - Prev using the classic definitions Pfwd = Vi^2/R0 and Prev = Vr^2/R0 works for lossless line, we do not yet have an equivalent set of definitions for Pfwd and Prev on a lossy line. When we do find appropriate definitions for Prev and Pfwd on a lossy line such that Pnet = Pfwd - Prev, then Prev/Pfwd will not be greater than one and it definitely will not be equal to rho^2 (which can be greater than one). So we are still looking for appropriate definitions of Pfwd and Prev on a lossy line. Or we could just throw this whole power analysis thing away and stick with voltage analysis which works just fine and lets us compute everything of interest. ....Keith |
|
wrote:
So we have the following requirements Pload = Pfwd - Prev Prho = Prev/Pfwd There is still a very large selection of pairs which will satisfy these constraints. Nope, there are not. Assuming Pload is known and Prho is known, you have two equations and two unknowns which can be solved for with unique solutions. For instance, in my previous example: Pload = 100 watts, Prho = 0.25. Prho can be calculated. Prev = Pfwd*Prho and substituting, we have: 100W = Pfwd - 0.25*Pfwd 100W = 0.75*Pfwd, therefore Pfwd = 133.33W, a unique solution. Incidentally, if you want a reference for losses in low-loss lines, _Fields_and_Waves_... by Ramo, Whinnery, and Van Duzer has a section 1.22 on page 41 titled, "Physical Approximations for Low-Loss Lines". They assume an average loss per unit length. I won't even try to reproduce the equations in ASCII. -- 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! =----- |
I have not followed the whole thread, but at the bottom line, the impact
of a complex Zo on power is an attenuation factor. It can be shown as follows. The general equations a Vf = [[(R+jwL)/(G+jwC)]^0.5]*If Vr = [[(R+1wL)/(G+jwC)]^0.5]*Ir. A numerical example for typical 50 ohm cable at 30 MHZ follows: Assume C = 30E-12 F/ft. Therefore L = 0.075E-6 H/ft [50 ohm lossless]. Assume 1/R [G] 1E-8 [100 megohms][lossy dielectric] Finally, assume R = 10 ohm [moderate rf resistance]. The resulting Zo for the assumed conditions is [[(10+jw7.5E-8)/(1E-8+jw30E-12)]^0.5] ohms. |Zo| = [[(10+j14.137)/(1E-8+j0.00565)]^0.5] = 55.360 ohms. Determine the relative phase shift as follows: atan[10+j14.137] = 54.7257 degrees. atan[1E-8+j0.00565] = 89.99989859 degrees [approximately 90]. The relative phase shift for the assumed complex Zo at 30 MHZ is 54.7257 - 89.99989 degrees [35.274 degrees]. Result: Zo = 55.360 ohms @ 35.274 degrees. Therefore Vf = If*[55.360 degrees] Therefore Pf(load) = Vf(source)*If(source)*cos(35.274) = 81.63% of maximum. The transmission line does not care if the signal is left to right or right to left; or forward and reflected. The effect is to introduce a mirror image to both the Vr and Ir terms. I hope I have not been redundant to other posts on this thread! Deacon Dave, W1MCE |
"Richard Clark" wrote in message ...
| | Hi Fellows, | | Such intense interest and no comments to SV7DMC for his work to this | matter? | Dear Mr. Richard Clark, Thank you very much for your kind interest! You have absolutely right. It's a pity indeed but unfortunately there are but very few comments for our point of view. However at least, all of them which are centered on the heard of the matter, I have a feeling that, are rather positive. I hope. Perhaps we have to put the blame on this damned language barrier. Who knows? Too much silence... Sincerely yours, pez SV7BAX P.S. Permit me please to correct you on that: It is not "his work" but "her work". |
Dear Mr. Richard Clark,
Here it is, once again! I do my best but it happens constantly. As I just told you for the Language Barrier, now it is "The heart of the matter", not "The heard of the matter". But I keep try... Sincerely yours, pez SV7BAX |
Dr. Slick wrote:
Cecil Moore wrote: Yes, the pair that satisfies the conservation of energy principle including source power and losses in the feedline. For instance, if a load is accepting 100 watts when the power reflection coefficient is 0.25, Pfwd(0.75) = 100W, so Pfwd = 133.33W, Pref = 33.33W. From Pozar: For a lossless 2-port network, [s11]**2 + [s21]**2 = 1 0.25 + 0.75 = 1 So hopefully this would be a lossless, passive network. Yes, the above is from an earlier lossless Z0-matched example. If any heat was generated in this passive network (NOT the load, the network itself), then the sum of the power reflection coefficient and the power transmission coefficient would be less than one. My example was the impedance discontinuity point between 50 ohm feedline and 150 ohm feedline. Not much room for losses in 0.1" of copper. -- 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! =----- |
On Mon, 8 Sep 2003 08:16:34 +0300, "pez" wrote:
P.S. Permit me please to correct you on that: It is not "his work" but "her work". Hi, I am at a loss for the equivalent to OM in your regard, and initials do mask you further. I should have been able to surmise this essential difference from your listing on Buckmaster. Another essential difference is that you performed the work that others simply meander around in the hopes of being mistaken as workers. Myself, I cannot task myself to publish complete derivations for two reasons. One, very few here are here to resolve anything but a sagging ego - hence the lack of standards in maintaining a progressive dialog. Two, I abide by a maxim that if one cannot express the key argument to the limited confines of one screen, then that argument lacks focus. I hope you do not take this as a diminution of your work, but more as a statement of stylistic variation. Keep up your effort, you are doing well. 73's Richard Clark, KB7QHC |
pez, your English is no barrier. It's better than many Americans posting
on this and other newsgroups. The "language" of the "language barrier" here is the language of mathematics. Some of the posters who have been the most verbal and insistent don't seem to able to deal with the math necessary to understand the concepts involved, so they're reduced to arguments which can't be mathematically demonstrated. Roy Lewallen, W7EL pez wrote: "Richard Clark" wrote in message ... | | Hi Fellows, | | Such intense interest and no comments to SV7DMC for his work to this | matter? | Dear Mr. Richard Clark, Thank you very much for your kind interest! You have absolutely right. It's a pity indeed but unfortunately there are but very few comments for our point of view. However at least, all of them which are centered on the heard of the matter, I have a feeling that, are rather positive. I hope. Perhaps we have to put the blame on this damned language barrier. Who knows? Too much silence... Sincerely yours, pez SV7BAX P.S. Permit me please to correct you on that: It is not "his work" but "her work". |
On Thu, 11 Sep 2003 12:13:44 +0300, "pez" wrote:
Dear Mr. Richard Clark, Thank you for the encouragement and your kind words. but I am not so sure that we deserve them (*). Hello, Your manners are superlative, and far better than my own manners (according to others, and that's fine as this is a characteristic of my aggressive style). Yes, you have absolute right about the initials, etc. But this became the motivation to discover, with a great deal of surprise I admit, that someone has given a bunch of inaccurate details to this site. We do not have something to hide You offer your call sign, that is far more unique than any signature that may be applied (like Dr. Slick). We have (as I am sure your part of the world does also) a history of pseudonyms or nom-de-plume for writers who wish to obscure their identity. I take care to write everything such that I can accept responsibility - or take responsibility. Many here do so too. Others do not, but in this day and age, anonymity is not guaranteed (again, taking the example of our same Dr. Slick) nor is it an affront if the writer commits to dialog and not simply slander or attack. Such occurrences of attack are rare here, and generally ignored. therefore we certainly do not belong to those with an extremely sensitive interest on private data. But I have to wonder: is there any responsibility by the site owner to guard the amateurs' personal data from anybody's will? Well, if by "site owner" you mean this newsgroup, then there is no "owner." There are nearly 40000 newsgroups that simply exist by nature of a combination of news-servers that are interconnected and sharing a protocol called NNTP. This protocol has been around since the beginning of the Internet (long before the WWW). It is like the public square where people meet to talk and listen and as in that square, if you are recognized, others will call you by name and perhaps comment about your ancestors. Such is privacy. Just as anything may be written here for free, anything written here has no intrinsic value. Value comes through association to the writer and that writer's continuity of thought and logic. You are already establishing a very good continuity and achieving a good association - to those who care! Finally, I do share your point of view for the length of a message and I try to keep them as short as possible but anyhow, take a look please on their rejection (*) in the related thread of our lengthy, indeed, derivations... do;^) Sincerely yours, pez SV7BAX Rejection is the forge of ideas. It does not condemn your work, it merely marks the critic by the nature of their criticism. If that criticism is weak, so is the critic. If that criticism is strong (not abuse, not refusal), then your argument is weak. Being technically and mathematically correct is not necessarily strength in an argument. You may be quite accurate, but the message is obscured by the bulk of presentation. A Canadian social scientist, Marshall McCluhan, demonstrated that identical material presented through different media become different messages. For instance, if you were to present you material to a class, they would accept it and work through it for understanding. If you were to present it to that same group of students in the public square, some might scoff, others might wander off with indifference. If you were to present it on TV, you probably wouldn't even make it to the first commercial. If you presented it on radio, you might have a highly interested audience. Same material, different venues, different responses. The same goes here. The nature of newsgroups is with dialog and hopefully you exchange correspondence with those who will allow you to develop your idea and stick with you to the end. If you attempt to make one presentation that answers all questions in one place, it violates a social contract that excludes them - except for them to accept or reject your work. They are not here to submit. This means you have to allow a style that is dialog in nature that allows their questions along the way to the end, where you do not find yourself wandering down the wrong path. (Think of the Socratic or Platonic methods of argument.) Unfortunately there are those who will manipulate discussion to the wrong path, or confusion. Dialog over time will reveal these individuals. This is why "lurking" is advised to newcomers so that they recognize personality with the signatures. One last point. Nothing is done once here. Nothing is solved. Nothing is fixed. The message will need repeating if only because new participants have not read the first message. The message may need repeating because you did not convince your critics, but they have had more time to ponder. This means that you sometimes have to vary your style and presentation so that repeating is not ignored. If you absolutely need the lengthy, mathematical treatment, then publish it at a web site and make a reference to it in your discussion. Offering the complete mathematical treatment each time you write here automatically limits you and reduces your audience. The craft of writing is knowing what to throw away. I generally discard 3/4ths of what I write before I hit the send button (some may groan not enough was thrown away here ;-) 73's Richard Clark, KB7QHC |
Dear Mr. Richard Clark,
What can I say? This is the fourth time, I have to call on my inadequate English to justify a misunderstanding. And this time it is related to the Buckmaster site mentioned by you and that's all. As for the rest three quarters of your message, I don't feel that I am in position to make any comment since they are looking to me as advises to successful advertisers. But, there is one point which, remarkably enough, is referenced by you in both of your letters and about the bottom line of them. It has to do with the matter of "maths" publication. Noticeably, you show about this matter a strong predilection to web site publication and a rejection of any inclusion into newsgroup messages. Although for the moment is seems that the right is on your side, at the same time, a recall to the need for direct communication surpasses any other argument. After all this is the deepest reason for newsgroups existence, I believe. Anyway. Thank you very much for your time. Sincerely yours, pez SV7BAX "Richard Clark" wrote in message ... | On Thu, 11 Sep 2003 12:13:44 +0300, "pez" wrote: | | Dear Mr. Richard Clark, | | Thank you for the encouragement and your kind words. | but I am not so sure that we deserve them (*). | | Hello, | | Your manners are superlative, and far better than my own manners | (according to others, and that's fine as this is a characteristic of | my aggressive style). | | Yes, you have absolute right about the initials, etc. | But this became the motivation | to discover, with a great deal of surprise I admit, | that someone has given a bunch | of inaccurate details to this site. | We do not have something to hide | | You offer your call sign, that is far more unique than any signature | that may be applied (like Dr. Slick). We have (as I am sure your part | of the world does also) a history of pseudonyms or nom-de-plume for | writers who wish to obscure their identity. I take care to write | everything such that I can accept responsibility - or take | responsibility. Many here do so too. Others do not, but in this day | and age, anonymity is not guaranteed (again, taking the example of our | same Dr. Slick) nor is it an affront if the writer commits to dialog | and not simply slander or attack. Such occurrences of attack are rare | here, and generally ignored. | | therefore we certainly do not belong to those | with an extremely sensitive interest | on private data. | But I have to wonder: | is there any responsibility | by the site owner | to guard the amateurs' personal data | from anybody's will? | | Well, if by "site owner" you mean this newsgroup, then there is no | "owner." There are nearly 40000 newsgroups that simply exist by | nature of a combination of news-servers that are interconnected and | sharing a protocol called NNTP. This protocol has been around since | the beginning of the Internet (long before the WWW). It is like the | public square where people meet to talk and listen and as in that | square, if you are recognized, others will call you by name and | perhaps comment about your ancestors. Such is privacy. | | Just as anything may be written here for free, anything written here | has no intrinsic value. Value comes through association to the writer | and that writer's continuity of thought and logic. | | You are already establishing a very good continuity and achieving a | good association - to those who care! | | Finally, | I do share your point of view for the length of a message | and I try to keep them as short as possible | but anyhow, | take a look please | on their rejection (*) | in the related thread | of our lengthy, indeed, derivations... | | do;^) | | Sincerely yours, | | pez | SV7BAX | | Rejection is the forge of ideas. It does not condemn your work, it | merely marks the critic by the nature of their criticism. If that | criticism is weak, so is the critic. If that criticism is strong (not | abuse, not refusal), then your argument is weak. Being technically | and mathematically correct is not necessarily strength in an argument. | You may be quite accurate, but the message is obscured by the bulk of | presentation. | | A Canadian social scientist, Marshall McCluhan, demonstrated that | identical material presented through different media become different | messages. For instance, if you were to present you material to a | class, they would accept it and work through it for understanding. If | you were to present it to that same group of students in the public | square, some might scoff, others might wander off with indifference. | If you were to present it on TV, you probably wouldn't even make it to | the first commercial. If you presented it on radio, you might have a | highly interested audience. Same material, different venues, | different responses. | | The same goes here. The nature of newsgroups is with dialog and | hopefully you exchange correspondence with those who will allow you to | develop your idea and stick with you to the end. If you attempt to | make one presentation that answers all questions in one place, it | violates a social contract that excludes them - except for them to | accept or reject your work. They are not here to submit. This means | you have to allow a style that is dialog in nature that allows their | questions along the way to the end, where you do not find yourself | wandering down the wrong path. (Think of the Socratic or Platonic | methods of argument.) Unfortunately there are those who will | manipulate discussion to the wrong path, or confusion. Dialog over | time will reveal these individuals. This is why "lurking" is advised | to newcomers so that they recognize personality with the signatures. | | One last point. Nothing is done once here. Nothing is solved. | Nothing is fixed. The message will need repeating if only because new | participants have not read the first message. The message may need | repeating because you did not convince your critics, but they have had | more time to ponder. This means that you sometimes have to vary your | style and presentation so that repeating is not ignored. If you | absolutely need the lengthy, mathematical treatment, then publish it | at a web site and make a reference to it in your discussion. Offering | the complete mathematical treatment each time you write here | automatically limits you and reduces your audience. The craft of | writing is knowing what to throw away. I generally discard 3/4ths of | what I write before I hit the send button (some may groan not enough | was thrown away here ;-) | | 73's | Richard Clark, KB7QHC |
On Fri, 12 Sep 2003 17:37:26 +0300, "pez" wrote:
And this time it is related to the Buckmaster site mentioned by you and that's all. Hello, Buckmaster makes available information that is already part of the public record. That is the point of licensing so that your activities are not shrouded in secrecy; but it also confirms you have certain rights to engage in activities others would not be allowed to do. This, of course, has nothing to do with posting to a newsgroup, but it was also your choice to declare your call sign which is related to the activities here, but it is not necessary to reveal to participate. 73's Richard Clark, KB7QHC |
On Fri, 12 Sep 2003 22:11:59 +0300, "pez" wrote:
Dear Mr. Richard Clark, Do you realize that, this site is unguarded and open to anyone who would like to modify your personal data, even if it is "only" for 48 hours at most, as it claims? That is the point. All of the rest are redundant. Sincerely yours, pez SV7BAX Hello, Well that is news, yes. Perhaps you should write them (Buckmaster) to suggest that they track the IP of the persons responsible for making those changes. This is a trivial, technical modification to their server software (I've written such software for Web servers - probably no more than 5 lines of code and a database call). Then ask them for that IP number so that you can register formal complaints against that individual. There are any number of simple and effective methods to prevent what you describe, nothing is technically difficult nor impossible. 48 Hours is not defensible with today's attention growing against identity theft. 73's Richard Clark, KB7QHC |
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