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#91




Hey, I've got it. It's sort of a "virtual reflection coefficient". Now I
understand fully. Thanks! Roy Lewallen, W7EL W5DXP wrote: . . . The energy analysis on my web page deals only with physical reflection coefficients. If 'rho' is not a physical reflection coefficient, then it is the END RESULT of a mathematical calculation and is not the CAUSE of anything. If a source doesn't "see" a physical impedance discontinuity, it doesn't "see" anything except forward and reflected waves. Coherent waves traveling in opposite directions are "unaware" of each other. Coherent waves traveling in the same direction merge, lose their separate identies, and become indistinguishable from one another. 
#92




Roy Lewallen wrote:
If people want to argue about the reflection of power waves, I'll gladly bow out and let Cecil and his colleagues resume their interminable arguments without me. I don't think there's anything to argue about. From _Optics_, by Hecht: "We define the reflectance R to be the ratio of the reflected power to the incident power." "When we talk about the 'amount' of light illuminating a surface, we are referring to something called the irradiance, denoted by I  the average energy per unit area per unit time." Light and RF are both EM waves. These scientific facts concerning light have only been known to physics for about 300 years.  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! = 
#93




Peter O. Brackett wrote:
But I maintain that is "unatural" and "Mother Nature" naturally likes the classical rho = (Z  Zo)/(Z + Zo) Thoughts, comments? I agree with you, Peter. That's what I have been calling the physical reflection coefficient. At the junction of two transmission lines of different characteristic impedances, it becomes s11=(Z02Z01)/(Z02+Z01) and is usually not equal to rho=Sqrt(Pref/Pfwd).  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! = 
#94




I'm trying to follow this, but have gone astray on the first couple of
steps. Peter O. Brackett wrote: The definition of the reflection coefficient is dependent upon what you define the "reflected voltage" to be. Consider the classical bridge circuit for measuring reflection coefficients. If Z is an unknown load presented to a generator of internal impedance R and that internal impedance R is used as the "reference" impedance to observe/measure the reflected voltage b and that reflected voltage b is calculated/measured by observation of the voltage v across the load Z and the current i through the load Z then the classical definition of a reflected voltage would be calculated as: b = v  Ri = Zi  Ri = (Z  R)i Volts. First of all, you're speaking of a circuit with a source impedance R and load impedance Z, rather than a terminated transmission line. Forward and reflected wave terminology is widely used in S parameter analysis, which also uses this model, so I'll be glad to follow along to see if and how S parameter terminology differs from the transmission line terminology we've been discussing so far. Please correct me where my assumptions diverge from yours. Your "classical definition" of b isn't one familiar to me. v + Ri would of course be the source voltage (which I'll call Vs). So v  Ri is Vs  2*Ri. Where does this come from and what does it mean? From your equation, and given source voltage Vs, i = Vs/(R+Z). Therefore, your "classical definition" of reflected voltage b is, in terms of Vs, Vs*((ZR)/(Z+R)). and the incident voltage a would be the Thevinins equivalent voltage across the sum of Z and R, i.e. a = (Z + R)i Since i = Vs(Z+R), you're saying that a = the source voltage Vs (from your two equations). So what you're calling the "incident voltage" is simply the source voltage Vs. Let's do a consistency check. The voltage at the load should be a + b = Vs + Vs*((ZR)/(Z+R)) = Vs*2Z/(Z+R). Inspection of the circuit as I understand it shows that the voltage at the load should be half this value. So, we already diverge. Which is true: 1. I've goofed up my algebra (a definite possibility) 2. I've misinterpreted your circuit, or 3. The voltage at the load is not equal to the sum of the forward and reflected voltages a and b, as you use the terms "forward voltage" and "reflected voltage". If v isn't equal to a + b, then what is the relationship between v, a, and b, and what are the physical meanings of the forward and reflected voltages? I'd like to continue with the remainder of the analysis, but can't proceed until this problem is cleared up. . . . Roy Lewallen, W7EL 
#95




Roy Lewallen wrote:
As soon as anyone starts arguing about average power waves, I'm outta here. I'd just as soon argue about the temperature of ghosts. You probably don't discuss light waves all that often, eh? :)  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! = 
#96




Thanks for the suggestion, Reg, and it's absolutely true. But even more
information is thrown away when you take the average of the power. Remember the statistician who drowned when crossing the creek whose average depth is only three feet. As soon as anyone starts arguing about average power waves, I'm outta here. I'd just as soon argue about the temperature of ghosts. Roy Lewallen, W7EL Reg Edwards wrote: Dear Roy, To reduce the amount of bafflegab to an absolute minimum why don't you just say that immediately the value of the reflection coefficient is squared (to allow 'power' to be introduced) half of the information it contains is tossed into the nearest garbage heap. Any conclusions drawn from following calculations are inevitably ambiguous and highly suspicious to say the least. To mention one wellknown example, that's why it is impossible to deduce the value of the lineterminating impedance from the calculated SWR. Ignorance is the root cause of these silly, timewasting arguments. Get back to basics, erase Smith Charts, misquoted worshipped idols, and illconceived inventions from your minds and start afresh from R, L, C, G, F and little t. Fortunately, the success of proposed missions to Mars does not depend on the deliberations of this newsgroup. Cec, by the way, are there any vinyards in Texas? It *is* as big as France.  Yours, Reg, G4FGQ 
#97




"David Robbins" wrote in message ...
Incorrect. You need the conjugate in the numerator if the Zo is complex. If it is purely real, WHICH MOST TEXTS ASSUME, then you can use the normal equation. sorry, the derivation for the table in the book i sent before is for the general case of a complex Zo. they then go on to simplify for an ideal line and for a nearly ideal line... nowhere does a conjugate show up. Please post this derivation again. When they say "ideal line" do they mean purely real? and that reference you give is not for a load on a transmission line, it is talking about a generator supplying power to a load... a completely different animal. Not at all really. The impedance seen by the load can be from either a source or a source hooked up with a transmission line. It doesn't matter with this equation. As Reg points out about the normal equation: "Dear Dr Slick, it's very easy. Take a real, long telephone line with Zo = 300  j250 ohms at 1000 Hz. Load it with a real resistor of 10 ohms in series with a real inductance of 40 millihenrys. The inductance has a reactance of 250 ohms at 1000 Hz. If you agree with the following formula, Magnitude of Reflection Coefficient of the load, ZL, relative to line impedance = ( ZL  Zo ) / ( ZL + Zo ) = 1.865 which exceeds unity, and has an angle of 59.9 degrees. The resulting standing waves may also be calculated. Are you happy now ?"  Reg, G4FGQ Well, i wasn't happy, because how can you have a R.C. greater than one into a passive network??? Quite impossible. But, if you use the conjugate formula, the R.C. will indeed be less than one. Convince yourself. Slick 
#98




(Tdonaly) wrote in message ... Very impressive. You've designed 5/8s vertical ground planes, 1/4 wavelength [something or others, I guess] and dipoles. Where are you working now? Did you go to Lowell? 73, Tom Donaly, KA6RUH Did you skip the part about U. C. Davis? I'm working parttime in the RF field, after being laid off among what seems like everyone else. Gives me time to paint my next masterpiece! Tit for Tat, maybe you can tell us something about you, Tom. last school attended? Job responsibilities? Slick My pitifully inadequate education could be of no interest to you, Garvin; I'm just a humble ham. (This is an _amateur_ newsgroup after all.) It is interesting to me, though, that a person of your age and attainments would pose as a pottymouth little blackfaced god whenever someone disagreed with you about something as abstruse as the reflection coefficient on a transmission line. I can only suppose that your social education was deficient, or that you really do want your name to be the most popular in the group's collective killfile. Anyway, you're wasting your time with the infantile behavior. Most of the fellows on this group are old men who gave up that form of discourse when they learned to talk. By the way, some of your art isn't half bad and shows the influence of some training. Did you have an art minor in college? 73, Tom Donaly, KA6RUH 
#99




W5DXP wrote:
Nope, they're not Reg. For Z0matched systems (which most ham systems are), rho^2 is all you need to know along with the forward and reflected powers to completely solve the voltage, current, and phase conditions at the impedance discontinuity. There in one sentence is the whole problem with the "power" approach. For a complete solution including phase conditions, you have to assume a Z0match, and even Cecil acknowledges that is only true for "most ham systems". As Reg says, this is because the power approach throws away the phase information at the start, and if you want it back again, you have to make assumptions. So the problem is not that the "power" approach cannot give a complete solution, but that it cannot do it for all cases. In other words, it isn't completely general  and that flaw is fatal. The conventional approach based on voltage (or current) waves doesn't discard the phase information, but uses it to give a complete solution for every case. It is completely general, and that's precisely why engineers use it.  73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek 
#100




Roy:
[snip] b = v  Ri = Zi  Ri = (Z  R)i Volts. First of all, you're speaking of a circuit with a source impedance R and load impedance Z, rather than a terminated transmission line. Forward and reflected wave terminology is widely used in S parameter analysis, which also uses this model, so I'll be glad to follow along to see if and how S parameter terminology differs from the transmission line terminology we've been discussing so far. Please correct me where my assumptions diverge from yours. Your "classical definition" of b isn't one familiar to me. v + Ri would of course be the source voltage (which I'll call Vs). So v  Ri is Vs  2*Ri. Where does this come from and what does it mean? [snip] Yes it should be familiar to you because it is the most common definition and one you seem to agree with. I presume that you are not used to seeing the use of the symbols "a" and "b" for those quantities. The use of "a" and "b" is widely used in Scattering Formalism and is less confusing to many than using subscripts. In your terminology above the "Vs" symbol is nothing more than the incident voltage usually given the symbol "a" in the Scattering Formalism, or the symbol V with a "+" sign subscript in many developments. Often you will find authors use a V with a plus sign "+" as a subscript to indicate the "a" voltage and a V with a minus sign "" subscript to idicate the "b" voltage. Personally I find the use of math symbols "" and "+" or other subscripts to variables to be confusing, I much prefer the use of "a" and "b" for forward or incident and reflected voltages. Simply put, if a generator with "open circuit" voltage "a" and "internal impedance" R is driving a load Z [Z could be a transmission line driving point impedance, for instance Z would be the characteristic or surge impedance Zo of a transmission line if the generator was driving a semiinfinite line.] then v is the voltage drop across Z and I is the current through Z, and so... a = v + Ri = Zi  Ri = (Z  R)i is simply the [usual] forward voltage or incident voltage applied by the generator to the to the load Z, which may be a lumped element load or if you prefer to talk about transmission lines, Z can be just the driving point impedance of a transmsision line, whatever you wish. Then its'just a simple application of Ohms Law tosee that b = v  Ri = Zi  Ri = (Z  R)i is the [usual] reflected voltage. b is just the difference between the voltage across Z which is calculated as Zi and the voltage that would be across Z if Z was actually equal to R. i.e. the reflected voltage b is just the voltage that would exist across Z if there was an "image match" between Z and R. [If Z is the Zo of a semiinfinite transmsision line you could call this a Zo match]. Taking the ratio of "b" to "a" just yeilds the [usual] reflection coefficient as b/a = (Z  R)i/(Z + R)i = (Z  R)/(Z + R). A well known result. Simple? [snip] From your equation, and given source voltage Vs, i = Vs/(R+Z). Therefore, your "classical definition" of reflected voltage b is, in terms of Vs, Vs*((ZR)/(Z+R)). and the incident voltage a would be the Thevinins equivalent voltage across the sum of Z and R, i.e. a = (Z + R)i [snip] Yep you got it all right! [snip] Since i = Vs(Z+R), you're saying that a = the source voltage Vs (from your two equations). So what you're calling the "incident voltage" is simply the source voltage Vs. [snip] Yes, mathematically "a" = "Vs", what else would it be? Nothing mysterious about that. The incident voltage is always simply the open circuit voltage of the source. In words a is not the source voltage because the source is a Thevinin equivalent made up of the ideal voltage generator Vs = a behind the "internal" source impedance R. A better way to describe Vs = a in words would be the incident voltage a is the "open circuit source voltage". [snip] Let's do a consistency check. The voltage at the load should be a + b = Vs + Vs*((ZR)/(Z+R)) = Vs*2Z/(Z+R). Inspection of the circuit as I understand it shows that the voltage at the load should be half this value. So, we already diverge. Which is true: [snip] No, the voltage at the load is not (a + b) rather it is [the quite obvious by Ohms Law] v = Zi. and the sum of the incident and reflected voltage is simply a + b = (v + Ri) + (v  Ri) = 2v = 2Zi Now if there is an "image match" and the "unknown" Z is actually equal to R, i.e. let Z = R in all of the above, then... a = Vs b = 0 a + b = 2Ri and i = Vs/2R = a/2R. [snip] 1. I've goofed up my algebra (a definite possibility) [snip] Only a little :) [snip] 2. I've misinterpreted your circuit, or [snip] No you have it correct! [snip] 3. The voltage at the load is not equal to the sum of the forward and reflected voltages a and b, as you use the terms "forward voltage" and "reflected voltage". If v isn't equal to a + b, then what is the relationship between v, a, and b, and what are the physical meanings of the forward and reflected voltages? [snip] I showed those relationships above. There is nothing new here... these are the [most] widely accepted definitions of incident and reflected voltages. [snip] I'd like to continue with the remainder of the analysis, but can't proceed until this problem is cleared up. Roy Lewallen, W7EL [snip] OK, let's carry on.  Peter K1PO Indialantic BytheSea, FL. 
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