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Reflection coefficient for total re-reflection
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. This is a straight- forward question for which I 'm dead serious. The answer will involve a resulting reflection coefficient. Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Placing a short-circuited stub having an inductive reactance of +j57.735 ohms on the line cancels the line reactance, establishing a new line impedance at this matching point of 50 + j0. However, a lot more is involved than just canceling the reactances to establish the impedance match. To begin, the stub generates a new, secondary, canceling reflection that cancels the primary reflected wave generated by the mismatched line termination. Recall the voltage reflection coefficient Vñ of the line impedance prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current reflection coefficient Iñ = 0.5 at +120° We now look at the voltage reflection coefficient of the stub, which is Vñ = 0.5 at +60°, and the corresponding current reflection coefficient of the stub which is Iñ = 0.5 at -120°. Observe that resultant angle of the voltage coefficients is 0° and resultant angle of the current coefficients is 180°, all with the same magnitude--0.5. It is well known that when these respective coefficients exist at this point, the matching point, this condition establishes a virtual open circuit to rearward traveling waves at this point, prohibiting any further rearward travel of the reflected waves, thus reversing their direction toward the load. Consequently, the result is total re-reflection of the two sets of reflected waves at the match point. Now the question--what is the resultant voltage coefficient of reflection at the match point? My position is that ñ = 1.0, which indicates total reflection. However, I've been criticized on this value, my critics saying that because this virtual open circuit was established by wave interference, ñ cannot equal 1.0. They assert that ñ = 1.0 can be obtained only through a physical open circuit. So I now ask you, am I correct in saying that the reflection coefficient in this situation is 1.0? I'm holding my breath for your answers. Walt |
Reflection coefficient for total re-reflection
On Jun 12, 3:05*pm, walt wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to transmission lines.This is not a trick question. *This is a straight- forward question for which I 'm dead serious. *The answer will involve a resulting reflection coefficient. Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Placing a short-circuited stub having an inductive reactance of +j57.735 ohms on the line cancels the line reactance, establishing a new line impedance at this matching point of 50 + j0. However, a lot more is involved than just canceling the reactances to establish the impedance match. To begin, the stub generates a new, secondary, canceling reflection that cancels the primary reflected wave generated by the mismatched line termination. Recall the voltage reflection coefficient Vñ of the line impedance prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current reflection coefficient Iñ = 0.5 at +120° We now look at the voltage reflection coefficient of the stub, which is Vñ = 0.5 at +60°, and the corresponding current reflection coefficient of the stub which is Iñ = 0.5 at -120°. Observe that resultant angle of the voltage coefficients is 0° and resultant angle of the current coefficients is 180°, all with the same magnitude--0.5. It is well known that when these respective coefficients exist at this point, the matching point, this condition establishes a virtual open circuit to rearward traveling waves at this point, prohibiting any further rearward travel of the reflected waves, thus reversing their direction toward the load. Consequently, the result is total re-reflection of the two sets of reflected waves at the match point. Now the question--what is the resultant voltage coefficient of reflection at the match point? My position is that ñ = 1.0, which indicates total reflection. However, I've been criticized on this value, my critics saying that because this virtual open circuit was established by wave interference, ñ cannot equal 1.0. They assert that *ñ = 1.0 can be obtained only through a physical open circuit. So I now ask you, am I correct in saying that the reflection coefficient in this situation is 1.0? *I'm holding my breath for your answers. Walt Unfortunately, the symbol for rho came thru as 'n' with a diacritical mark over it. That incorrect character should be replace with the Greek symbol rho, which this venue cannot print. Walt |
Reflection coefficient for total re-reflection
On Jun 12, 3:05*pm, walt wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to transmission lines.This is not a trick question. *This is a straight- forward question for which I 'm dead serious. *The answer will involve a resulting reflection coefficient. Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Placing a short-circuited stub having an inductive reactance of +j57.735 ohms on the line cancels the line reactance, establishing a new line impedance at this matching point of 50 + j0. However, a lot more is involved than just canceling the reactances to establish the impedance match. To begin, the stub generates a new, secondary, canceling reflection that cancels the primary reflected wave generated by the mismatched line termination. Recall the voltage reflection coefficient Vñ of the line impedance prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current reflection coefficient Iñ = 0.5 at +120° We now look at the voltage reflection coefficient of the stub, which is Vñ = 0.5 at +60°, and the corresponding current reflection coefficient of the stub which is Iñ = 0.5 at -120°. Observe that resultant angle of the voltage coefficients is 0° and resultant angle of the current coefficients is 180°, all with the same magnitude--0.5. It is well known that when these respective coefficients exist at this point, the matching point, this condition establishes a virtual open circuit to rearward traveling waves at this point, prohibiting any further rearward travel of the reflected waves, thus reversing their direction toward the load. Consequently, the result is total re-reflection of the two sets of reflected waves at the match point. Now the question--what is the resultant voltage coefficient of reflection at the match point? My position is that ñ = 1.0, which indicates total reflection. However, I've been criticized on this value, my critics saying that because this virtual open circuit was established by wave interference, ñ cannot equal 1.0. They assert that *ñ = 1.0 can be obtained only through a physical open circuit. So I now ask you, am I correct in saying that the reflection coefficient in this situation is 1.0? *I'm holding my breath for your answers. Walt Forgot to say, the stub is placed in series with the line, not in shunt. Walt |
Reflection coefficient for total re-reflection
On Jun 12, 3:12*pm, walt wrote:
On Jun 12, 3:05*pm, walt wrote: I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to transmission lines.This is not a trick question. *This is a straight- forward question for which I 'm dead serious. *The answer will involve a resulting reflection coefficient. Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Placing a short-circuited stub having an inductive reactance of +j57.735 ohms on the line cancels the line reactance, establishing a new line impedance at this matching point of 50 + j0. However, a lot more is involved than just canceling the reactances to establish the impedance match. To begin, the stub generates a new, secondary, canceling reflection that cancels the primary reflected wave generated by the mismatched line termination. Recall the voltage reflection coefficient Vñ of the line impedance prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current reflection coefficient Iñ = 0.5 at +120° We now look at the voltage reflection coefficient of the stub, which is Vñ = 0.5 at +60°, and the corresponding current reflection coefficient of the stub which is Iñ = 0.5 at -120°. Observe that resultant angle of the voltage coefficients is 0° and resultant angle of the current coefficients is 180°, all with the same magnitude--0.5. It is well known that when these respective coefficients exist at this point, the matching point, this condition establishes a virtual open circuit to rearward traveling waves at this point, prohibiting any further rearward travel of the reflected waves, thus reversing their direction toward the load. Consequently, the result is total re-reflection of the two sets of reflected waves at the match point. Now the question--what is the resultant voltage coefficient of reflection at the match point? My position is that ñ = 1.0, which indicates total reflection. However, I've been criticized on this value, my critics saying that because this virtual open circuit was established by wave interference, ñ cannot equal 1.0. They assert that *ñ = 1.0 can be obtained only through a physical open circuit. |
Reflection coefficient for total re-reflection
On Sun, 12 Jun 2011 12:12:46 -0700 (PDT), walt wrote:
On Jun 12, 3:05*pm, walt wrote: I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to transmission lines. Hi Walt, What I need to see is a labeled schematic. 73's Richard Clark, KB7QHC |
Reflection coefficient for total re-reflection
On Jun 12, 2:05*pm, walt wrote:
So I now ask you, am I correct in saying that the reflection coefficient in this situation is 1.0? *I'm holding my breath for your answers. Walt, some of us have previously discussed the difference between a one-port s11 (reflection coefficient) and a two-port s11 and I think that difference is what you are seeing. The one-port s11 (rho) is a virtual reflection coefficient based on the available system information. In that case, rho is 1.0. However, if we use the two-port parameters, I believe s11 (rho) will be 0.5. The matching stub complicates the analysis so maybe we can reduce the complexity using a slightly different example about which you can ask the same question and the answer will be a little easier. --50 ohm--+--1/2WL 150 ohm--50 ohm load It's essentially the same problem without the complexity of the stub. There is a 50 ohm Z0-match at point '+'. The one-port s11 (rho) looking into the Z0-match point from the source side is 0.0 because there are no reflections. The one-port analysis is blind to any interference that might be recognized if we were using the two-port analysis. The one-port s11 (rho) looking back into the Z0-match from the load is 1.0, same as your virtual open/short. The one-port analysis used for your virtual open/short concept is unaware of the interference that is visible in a two-port analysis. A one-port analysis cannot tell the difference between an actual reflection and a redistribution of energy associated with superposition interference. However, the two-port s11 (rho) looking into the Z0-match from the source side is 0.5, i.e. (150-50)/(150+50). Interference is not invisible in the two-port analysis. In fact, when the voltage reflected toward the source is zero, there is total destructive interference in the direction of the source. b1 = s11*a1 + s12*a2 = 0 where all math is phasor math. The two terms cancel destructively to zero. The ExH energy components in those two terms are redistributed back toward the load. In a two-port analysis, the reflection coefficient looking back into the Z0-match from the load is called s22. The matching equation is: b2 = s21*a1 + s22*a2 All of the destructive interference energy involved toward the source is redistributed back toward the load as constructive interference. But such is invisible in a one-port analysis such as your virtual open/ short circuit. There is nothing wrong with your one-port virtual open/short circuit analysis. There is also nothing wrong with the more in-depth two-port analysis. It is simply that more information is available during the two-port analysis. Over on another newsgroup, some reported using a one-port analysis almost all of the time. Tell your critics that you are using a one- port analysis where the interference information is simply not available during the analysis. What is the one-port reflection coefficient looking into a Z0-match from the source side? 0.0 What is the one-port reflection coefficient looking into a Z0-match from the load side? plus or minus 1.0 That is what you are doing with your virtual open/short and there's nothing wrong with a one-port analysis. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 12, 5:46*pm, Cecil Moore wrote:
On Jun 12, 2:05*pm, walt wrote: So I now ask you, am I correct in saying that the reflection coefficient in this situation is 1.0? *I'm holding my breath for your answers. Walt, some of us have previously discussed the difference between a one-port s11 (reflection coefficient) and a two-port s11 and I think that difference is what you are seeing. The one-port s11 (rho) is a virtual reflection coefficient based on the available system information. In that case, rho is 1.0. However, if we use the two-port parameters, I believe s11 (rho) will be 0.5. The matching stub complicates the analysis so maybe we can reduce the complexity using a slightly different example about which you can ask the same question and the answer will be a little easier. --50 ohm--+--1/2WL 150 ohm--50 ohm load It's essentially the same problem without the complexity of the stub. There is a 50 ohm Z0-match at point '+'. The one-port s11 (rho) looking into the Z0-match point from the source side is 0.0 because there are no reflections. The one-port analysis is blind to any interference that might be recognized if we were using the two-port analysis. The one-port s11 (rho) looking back into the Z0-match from the load is 1.0, same as your virtual open/short. The one-port analysis used for your virtual open/short concept is unaware of the interference that is visible in a two-port analysis. A one-port analysis cannot tell the difference between an actual reflection and a redistribution of energy associated with superposition interference. However, the two-port s11 (rho) looking into the Z0-match from the source side is 0.5, i.e. (150-50)/(150+50). Interference is not invisible in the two-port analysis. In fact, when the voltage reflected toward the source is zero, there is total destructive interference in the direction of the source. b1 = s11*a1 + s12*a2 = 0 where all math is phasor math. The two terms cancel destructively to zero. The ExH energy components in those two terms are redistributed back toward the load. In a two-port analysis, the reflection coefficient looking back into the Z0-match from the load is called s22. The matching equation is: b2 = s21*a1 + s22*a2 All of the destructive interference energy involved toward the source is redistributed back toward the load as constructive interference. But such is invisible in a one-port analysis such as your virtual open/ short circuit. There is nothing wrong with your one-port virtual open/short circuit analysis. There is also nothing wrong with the more in-depth two-port analysis. It is simply that more information is available during the two-port analysis. Over on another newsgroup, some reported using a one-port analysis almost all of the time. Tell your critics that you are using a one- port analysis where the interference information is simply not available during the analysis. What is the one-port reflection coefficient looking into a Z0-match from the source side? 0.0 What is the one-port reflection coefficient looking into a Z0-match from the load side? plus or minus 1.0 That is what you are doing with your virtual open/short and there's nothing wrong with a one-port analysis. -- 73, Cecil, w5dxp.com Hi Cecil and Richard, Thank you for the insightful response, Cecil. However, when you go to the 2-port version I’m unable to correlate that configuration with my stubbing problem. Here’s what my problem really involves. Referring to Eq 8 in the first part of Steve Best’s three-part article published in QEX, the equation is as follows: The section within the parentheses is the voltage- increase factor when all reflections have been re-reflected. That section is 1/1 - (rho_’a’ x rho_’s’), where ‘a’ is the mismatch at the antenna (the load) and ‘s’ is the mismatch at the source. We’re considering the source to be an RF power amp, where we know the output source resistance is non-dissipative, thus re-reflects all reflected power incident on it. I maintain that the reflection coefficient at the source is 1.0 because of the total re-reflection there. Now consider working with the output of the pi-network of the RF anp, where I believe the reflection coefficient is 1.0, and the 3:1 antenna mismatch that yields a reflection coefficient of 0.5. Plugging these coefficients into the equation section yields 1/1- (1 x 0.5), which equals 2.0, which is incorrect, making that equation invalid, IMHO. The correct value with these coefficients should be 1.1547, not 2.0. However, mathematical experts say that the equation is correct, saying that rho_¬’s’ cannot be equal to 1.0, because the virtual open circuit was established by wave interference, not a physical open circuit. |
Reflection coefficient for total re-reflection
On Jun 12, 8:16*pm, walt wrote:
Thank you for the insightful response, Cecil. However, when you go to the 2-port version I’m unable to correlate that configuration with my stubbing problem. The reason that I didn't say anything about the stub example is because I cannot comprehend it without a schematic. That's why I changed examples. Do you agree with what I said about my example? Could you post a schematic of your first example? It is the "series stub" part that I don't understand. Such is usually called a "series section" because a stub is usually a parallel dead end open or short circuit. It is also difficult to comprehend how a two-port analysis could be done at the stub connection point. Wouldn't that require a three-port analysis? We’re considering the source to be an RF power amp, where we know the output source resistance is non-dissipative, thus re-reflects all reflected power incident on it. I maintain that the reflection coefficient at the source is 1.0 because of the total re-reflection there. How's about we limit the *initial* discussion and examples to a source with zero incident reflected power so the source impedance doesn't matter? IMO, a two-port analysis of a Z0-match point will reveal the main ingredients of the energy flow. However, mathematical experts say that the equation is correct, saying that rho_¬’s’ cannot be equal to 1.0, because the virtual open circuit was established by wave interference, not a physical open circuit. A one-port analysis cannot tell the difference between wave interference and reflections. You are correct that the reflection coefficients are not necessarily the same between a one-port analysis and a two-port analysis. Your "mathematical experts" don't seem to understand the limitations of a one-port analysis. It's akin to not knowing what is inside a black box, i.e. one cannot tell the difference between a resistor and a virtual resistance. However, with a two-port analysis, one can tell the difference. It appears that your "mathematical experts" are insisting on a two-port analysis such as provided by the s-parameter equations: b1 = s11*a1 + s12*a2 b2 = s21*a1 + s22*a2 http://www.sss-mag.com/pdf/an-95-1.pdf -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 12, 9:45*pm, Cecil Moore wrote:
On Jun 12, 8:16*pm, walt wrote: Thank you for the insightful response, Cecil. However, when you go to the 2-port version I’m unable to correlate that configuration with my stubbing problem. The reason that I didn't say anything about the stub example is because I cannot comprehend it without a schematic. That's why I changed examples. Do you agree with what I said about my example? Could you post a schematic of your first example? It is the "series stub" part that I don't understand. Such is usually called a "series section" because a stub is usually a parallel dead end open or short circuit. It is also difficult to comprehend how a two-port analysis could be done at the stub connection point. Wouldn't that require a three-port analysis? We’re considering the source to be an RF power amp, where we know the output source resistance is non-dissipative, thus re-reflects all reflected power incident on it. I maintain that the reflection coefficient at the source is 1.0 because of the total re-reflection there. How's about we limit the *initial* discussion and examples to a source with zero incident reflected power so the source impedance doesn't matter? IMO, a two-port analysis of a Z0-match point will reveal the main ingredients of the energy flow. However, mathematical experts say that the equation is correct, saying that rho_¬’s’ cannot be equal to 1.0, because the virtual open circuit was established by wave interference, not a physical open circuit. A one-port analysis cannot tell the difference between wave interference and reflections. You are correct that the reflection coefficients are not necessarily the same between a one-port analysis and a two-port analysis. Your "mathematical experts" don't seem to understand the limitations of a one-port analysis. It's akin to not knowing what is inside a black box, i.e. one cannot tell the difference between a resistor and a virtual resistance. However, with a two-port analysis, one can tell the difference. It appears that your "mathematical experts" are insisting on a two-port analysis such as provided by the s-parameter equations: b1 = s11*a1 + s12*a2 b2 = s21*a1 + s22*a2 http://www.sss-mag.com/pdf/an-95-1.pdf -- 73, Cecil, w5dxp.com Cecil, the series stubbing appears in Reflections, Chapter 23, with the same values as I presented above, with detailed diagrams shown in each step in the progression of the explanation. I hope these diagrams can help. As I said in the previous post, the experts were referring to the output of the RF amp as not establishing a reflection coefficient rho = 1.0, which has put me in a corner. Sorry Cecil, I can't correlate a two-port configuration using S parameters with the problem I have. Thanks a million for the discourse. Walt |
Reflection coefficient for total re-reflection
On Sun, 12 Jun 2011 19:09:29 -0700 (PDT), walt wrote:
the series stubbing appears in Reflections, Chapter 23, with the same values as I presented above, with detailed diagrams shown in each step in the progression of the explanation. I hope these diagrams can help. In other words, consult: http://www.w2du.com/Chapter%2023.pdf Figures 1 through 5 As I said in the previous post, the experts were referring to the output of the RF amp as not establishing a reflection coefficient rho = 1.0, which has put me in a corner. Hi Walt, How so? (What is the corner?) 73's Richard Clark, KB7QHC |
Reflection coefficient for total re-reflection
On Jun 12, 11:14*pm, Richard Clark wrote:
On Sun, 12 Jun 2011 19:09:29 -0700 (PDT), walt wrote: the series stubbing appears in Reflections, Chapter 23, with the same values as I presented above, with detailed diagrams shown in each step in the progression of the explanation. I hope these diagrams can help. In other words, consult:http://www.w2du.com/Chapter%2023.pdf Figures 1 through 5 As I said in the previous post, the experts were referring to the output of the RF amp as not establishing a reflection coefficient rho = 1.0, which has put me in a corner. Hi Walt, How so? *(What is the corner?) 73's Richard Clark, KB7QHC The corner I'm in, Richard, is that In Reflections 3, Chapter 25, I assert that Steve Best's Eq 8 in the first part of his three-part article appearing in QEX is invalid, because it gives incorrect answers when I plug in what I believe are correct values of reflection coefficients. Yet his equation agrees with that of Johnson on Page 100 of his "Transmission Lines" text book. In addition, a mathematics expert whom I respect says Best's equation is correct. So I've got to make the decision whether to delete my criticism of his equation or leave it in and be accused of criticizing him incorrectly. What to do! Walt |
Reflection coefficient for total re-reflection
On Sun, 12 Jun 2011 20:31:42 -0700 (PDT), walt wrote:
On Jun 12, 11:14*pm, Richard Clark wrote: On Sun, 12 Jun 2011 19:09:29 -0700 (PDT), walt wrote: the series stubbing appears in Reflections, Chapter 23, with the same values as I presented above, with detailed diagrams shown in each step in the progression of the explanation. I hope these diagrams can help. In other words, consult:http://www.w2du.com/Chapter%2023.pdf Figures 1 through 5 As I said in the previous post, the experts were referring to the output of the RF amp as not establishing a reflection coefficient rho = 1.0, which has put me in a corner. Hi Walt, How so? *(What is the corner?) 73's Richard Clark, KB7QHC The corner I'm in, Richard, is that In Reflections 3, Chapter 25, In other words, consult: http://www.w2du.com/r3ch25.pdf I assert that Steve Best's Eq 8 in the first part of his three-part article appearing in QEX is invalid, because it gives incorrect answers when I plug in what I believe are correct values of reflection coefficients. Yet his equation agrees with that of Johnson on Page 100 of his "Transmission Lines" text book. In addition, a mathematics expert whom I respect says Best's equation is correct. So I've got to make the decision whether to delete my criticism of his equation or leave it in and be accused of criticizing him incorrectly. What to do! Walt Hi Walt, So this is not only double-deep, through your work to Steve's, but triple deep then through Steve to Johnson. Lacking the necessary, culminating edition of Johnson's, I still don't know what the corner is. Lacking the complete math from all sides of the argument (not somewhere I would like to go), and noting that many authors (not making attributions here) frequently ignore some relatively basic mandates where they don't matter, to then expand into situations where they do matter; then I don't really trust heavily editorialized math analysis. I note your summary statement for Steve that you find contentious, viz. "A total re-reflection of power at the match point is not necessary for the impedance match to occur." is one where I would agree with Steve; but not necessarily for reasons brought forward. What is worse, this simple statement may mean three things to two people. 73's Richard Clark, KB7QHC |
Reflection coefficient for total re-reflection
Thoughts on voltage vs power, reflection vs interference:
What most RF people do is deal exclusively with voltages. Power is considered only at the beginning and end of a voltage analysis, NOT during the voltage analysis. If joules/second are to be tracked seamlessly throughout the analysis, a working knowledge of the effects of superposition/interference is absolutely necessary. Optical physicists do not have the luxury of working exclusively with voltages, as we do in RF, so they must necessarily understand superposition/interference and be able to track every component of irradiance (power density). I took a look at Johnson and he is dealing with voltage, not power, and certainly not with dissipationless resistances as part of the generator source impedance. He uses 'k' sub-script 'g' as the symbol for the voltage reflection coefficient. I'm going to use 'rho' for his 'k' with braces {g} indicating subscripts. His *voltage* reflection coefficient at the generator is: rho{g} = (Zg-Z0)/(Zg+Z0) which is just standard *voltage* wave reflection mechanics. What happens to the energy (power) in superposed waves is completely transparent when superposing voltages. For instance, let's say we have two 200 watt waves in a 50 ohm environment which makes each of their voltage magnitudes equal to 100 volts RMS. The electric fields of the two waves are 120 degrees apart. What happens when we superpose 100 volts at +60 degrees with 100 volts at -60 degrees? Every student of three-phase power systems knows the result will be 100 volts at zero degrees. All is well until we take a look at the energy in those two superposed waves. Each wave is associated with an ExH amount of power, V^2/Z0=200w, for a total of 400 watts in the two waves. The resultant (total?) superposed wave contains 200 watts of ExH power. Most people don't give this idea a second thought but where did the other 200 watts go? To answer the question, one must understand destructive/constructive interference. In the above example, there is 200 watts of destructive interference present so the resulting "total" voltage is not the only component of superposition. If the above occurs in a transmission line, the amount of destructive interference energy that is lost in the direction of superposition, e.g. toward the load, is redistributed in the only other direction possible, i.e. toward the source. There is a second 200w wave generated that travels toward the source but that fact is not covered when voltage superposition is involved. Note that it is a reverse- traveling wave but it is technically not a reflection of a single wave as it is the result of superposition of two waves. Voltage superposition takes care of itself and everyone believes in the conservation of energy principle which is probably why very few people ask, "Where does the power go?" It is only when we are trying to track energy throughout the system that we are forced to understand the effects associated with interference. Thoughts on one-port analysis vs two-port analysis. Sources are necessarily treated as single-port devices. We know we often get completely different reflection coefficients when treating something as a single-port device vs as a dual-port device. For instance, most of us treat a dipole feedpoint as a single-port device when it is actually far from being a single-port device. In reality, many other things besides a single reflection, are happening at a dipole's feedpoint. The actual physical reflection coefficient at the feedpoint of a "50 ohm" dipole fed with 50 ohm coax is around 0.845 because the characteristic impedance of a #14 wire 30 feet above ground is around 600 ohms. Proof: Eliminate the reflections from the ends of the dipole by terminating the ends of the inv-V dipole to ground through 600 ohm resistors and the SWR on the 50 ohm feedline goes to 12:1. Because of reflections from the ends of the dipole, a lot of interference is happening at the feedpoint which results in a *virtual* reflection coefficient of 0.0 only because of the single- port analysis that is ordinarily used. IMO, a virtual reflection coefficient is a *result* and cannot cause anything including reflections. IMO, only physical reflection coefficients, i.e. physical impedance discontinuities, can *cause* reflections. Much of what we consider to be reflections are the result of interference. Seems that something similar, but more complicated, is happening inside a source where there is an active-source component in the mix. IMO, what is happening to the energy inside a source cannot possibly be understood without taking the effects associated with interference into account. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 13, 1:55*pm, K7ITM wrote:
*It doesn't make any sense to me to put a shorted section of line in series with another line, so my confusion starts. Tom, I didn't know initially that the example was in "Reflections III". A series stub can be used instead of a loading coil on a wire antenna. I had never seen a series stub used in such a manner on a transmission line and that's why I was confused. I'm assuming that the center conductor is broken and one side is connected to the inner conductor and one side is connected to the braid on a stub, but I am not sure that is correct. There's got to be a less complicated example that we can use. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On 6/13/2011 12:34 PM, Cecil Moore wrote:
On Jun 13, 1:55 pm, wrote: It doesn't make any sense to me to put a shorted section of line in series with another line, so my confusion starts. Tom, I didn't know initially that the example was in "Reflections III". A series stub can be used instead of a loading coil on a wire antenna. I had never seen a series stub used in such a manner on a transmission line and that's why I was confused. I'm assuming that the center conductor is broken and one side is connected to the inner conductor and one side is connected to the braid on a stub, but I am not sure that is correct. There's got to be a less complicated example that we can use. -- 73, Cecil, w5dxp.com Here is a small handout, with a smith chart example! -- Regards, JS “The Constitution is not an instrument for the government to restrain the people, it’s an instrument for the people to restrain the government.” -- Patrick Henry |
Reflection coefficient for total re-reflection
On 6/13/2011 12:34 PM, Cecil Moore wrote:
On Jun 13, 1:55 pm, wrote: It doesn't make any sense to me to put a shorted section of line in series with another line, so my confusion starts. Tom, I didn't know initially that the example was in "Reflections III". A series stub can be used instead of a loading coil on a wire antenna. I had never seen a series stub used in such a manner on a transmission line and that's why I was confused. I'm assuming that the center conductor is broken and one side is connected to the inner conductor and one side is connected to the braid on a stub, but I am not sure that is correct. There's got to be a less complicated example that we can use. -- 73, Cecil, w5dxp.com Larger example: http://personal.ee.surrey.ac.uk/Pers...ries-stub.html -- Regards, JS “The Constitution is not an instrument for the government to restrain the people, it’s an instrument for the people to restrain the government.” -- Patrick Henry |
Reflection coefficient for total re-reflection
On 6/13/2011 12:34 PM, Cecil Moore wrote:
On Jun 13, 1:55 pm, wrote: It doesn't make any sense to me to put a shorted section of line in series with another line, so my confusion starts. Tom, I didn't know initially that the example was in "Reflections III". A series stub can be used instead of a loading coil on a wire antenna. I had never seen a series stub used in such a manner on a transmission line and that's why I was confused. I'm assuming that the center conductor is broken and one side is connected to the inner conductor and one side is connected to the braid on a stub, but I am not sure that is correct. There's got to be a less complicated example that we can use. -- 73, Cecil, w5dxp.com Small handout with example and smith chart: http://www.ittc.ku.edu/~jstiles/723/...b%20Tuning.pdf -- Regards, JS “The Constitution is not an instrument for the government to restrain the people, it’s an instrument for the people to restrain the government.” -- Patrick Henry |
Reflection coefficient for total re-reflection
On Jun 13, 2:46*pm, John Smith wrote:
http://personal.ee.surrey.ac.uk/Pers...ries-stub.html Thanks John, I can't help but point out that such a "series stub" unbalances the transmission line currents resulting in common-mode currents on the outside braid and is probably a deviation from the original design specifications. There has got to be a simpler example that illustrates the question/answer. I thought I had it with my Z0- match example but maybe not. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On 6/13/2011 1:22 PM, Cecil Moore wrote:
On Jun 13, 2:46 pm, John wrote: http://personal.ee.surrey.ac.uk/Pers...ries-stub.html Thanks John, I can't help but point out that such a "series stub" unbalances the transmission line currents resulting in common-mode currents on the outside braid and is probably a deviation from the original design specifications. There has got to be a simpler example that illustrates the question/answer. I thought I had it with my Z0- match example but maybe not. -- 73, Cecil, w5dxp.com Quite possible ... -- Regards, JS “The Constitution is not an instrument for the government to restrain the people, it’s an instrument for the people to restrain the government.” -- Patrick Henry |
Reflection coefficient for total re-reflection
On Jun 13, 3:08*pm, John Smith wrote:
http://www.ittc.ku.edu/~jstiles/723/...b%20Tuning.pdf Thanks John, but I would like to see the schematic for coax. Shirley, one doesn't cut the braid and add a stub? -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 13, 2:55*pm, K7ITM wrote:
walt wrote: I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to transmission lines.This is not a trick question. *This is a straight- forward question for which I 'm dead serious. *The answer will involve a resulting reflection coefficient. Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Placing a short-circuited stub having an inductive reactance of +j57.735 ohms on the line cancels the line reactance, establishing a new line impedance at this matching point of 50 + j0. However, a lot more is involved than just canceling the reactances to establish the impedance match. To begin, the stub generates a new, secondary, canceling reflection that cancels the primary reflected wave generated by the mismatched line termination. Recall the voltage reflection coefficient Vñ of the line impedance prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current reflection coefficient Iñ = 0.5 at +120° We now look at the voltage reflection coefficient of the stub, which is Vñ = 0.5 at +60°, and the corresponding current reflection coefficient of the stub which is Iñ = 0.5 at -120°. Observe that resultant angle of the voltage coefficients is 0° and resultant angle of the current coefficients is 180°, all with the same magnitude--0.5. It is well known that when these respective coefficients exist at this point, the matching point, this condition establishes a virtual open circuit to rearward traveling waves at this point, prohibiting any further rearward travel of the reflected waves, thus reversing their direction toward the load. Consequently, the result is total re-reflection of the two sets of reflected waves at the match point. Now the question--what is the resultant voltage coefficient of reflection at the match point? My position is that ñ = 1.0, which indicates total reflection. However, I've been criticized on this value, my critics saying that because this virtual open circuit was established by wave interference, ñ cannot equal 1.0. They assert that *ñ = 1.0 can be obtained only through a physical open circuit. |
Reflection coefficient for total re-reflection
On Jun 12, 9:09*pm, walt wrote:
Sorry Cecil, I can't correlate a two-port configuration using S parameters with the problem I have. In a one-port analysis, virtual impedances cause reflections. In a two- port analysis, only physical impedance discontinuities cause reflections. As I see it, that is the entire problem in a nutshell. You are using a one-port analysis for virtually :-) all of your analyses, including your source analysis. Your detractors are trying to talk you into using a two-port analysis. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 13, 6:17*pm, Cecil Moore wrote:
On Jun 12, 9:09*pm, walt wrote: Sorry Cecil, I can't correlate a two-port configuration using S parameters with the problem I have. In a one-port analysis, virtual impedances cause reflections. In a two- port analysis, only physical impedance discontinuities cause reflections. As I see it, that is the entire problem in a nutshell. You are using a one-port analysis for virtually :-) all of your analyses, including your source analysis. Your detractors are trying to talk you into using a two-port analysis. -- 73, Cecil, w5dxp.com Cecil, as I said earlier, I can't see any relation between a two-port configuration and the tank circuit of an RF power amp. Why are you pushing it? Can't you just tell me whether you agree that the reflection coefficient at the output of the tank circuit is rho = 1.0 or not. If you don't believe it does, even though it re-reflects all the reflected power incident on it, then please explain what you believe the reflection coefficient is at this point. Walt, W2DU |
Reflection coefficient for total re-reflection
Cecil Moore wrote: On Jun 13, 1:55*pm, K7ITM wrote: *It doesn't make any sense to me to put a shorted section of line in series with another line, so my confusion starts. Tom, I didn't know initially that the example was in "Reflections III". A series stub can be used instead of a loading coil on a wire antenna. I had never seen a series stub used in such a manner on a transmission line and that's why I was confused. I'm assuming that the center conductor is broken and one side is connected to the inner conductor and one side is connected to the braid on a stub, but I am not sure that is correct. There's got to be a less complicated example that we can use. -- 73, Cecil, w5dxp.com Ah, OK. Now I understand what Walt meant... Though it's possible to use a series stub on a transmission line, as you say, it's not all that common in practice. I suppose that's why my mind wasn't going there. Mea culpa. Perhaps now I can go back and look at Walt's original question and make more sense out of it. When a series stub is used in an antenna (as in a quarter-wave stub coupling colinear half-waves), King points out that coupling from the antenna fields to the wires in the stub, when the stub is an open-wire line perpendicular to the axis of the half-waves, is an important factor in how the stub operates to establish in-phase currents on the adjacent half-waves... I see you made a comment about antenna currents on the stub in John Smith's example, too. Cheers, Tom |
Reflection coefficient for total re-reflection
On 6/13/2011 3:05 PM, Cecil Moore wrote:
On Jun 13, 3:08 pm, John wrote: http://www.ittc.ku.edu/~jstiles/723/...b%20Tuning.pdf Thanks John, but I would like to see the schematic for coax. Shirley, one doesn't cut the braid and add a stub? -- 73, Cecil, w5dxp.com LOL! Never lose that sense of humor, love it! -- Regards, JS “The Constitution is not an instrument for the government to restrain the people, it’s an instrument for the people to restrain the government.” -- Patrick Henry |
Reflection coefficient for total re-reflection
walt wrote: I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to transmission lines.This is not a trick question. This is a straight- forward question for which I 'm dead serious. The answer will involve a resulting reflection coefficient. Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Placing a short-circuited stub having an inductive reactance of +j57.735 ohms on the line cancels the line reactance, establishing a new line impedance at this matching point of 50 + j0. However, a lot more is involved than just canceling the reactances to establish the impedance match. To begin, the stub generates a new, secondary, canceling reflection that cancels the primary reflected wave generated by the mismatched line termination. Recall the voltage reflection coefficient Vñ of the line impedance prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current reflection coefficient Iñ = 0.5 at +120° We now look at the voltage reflection coefficient of the stub, which is Vñ = 0.5 at +60°, and the corresponding current reflection coefficient of the stub which is Iñ = 0.5 at -120°. Observe that resultant angle of the voltage coefficients is 0° and resultant angle of the current coefficients is 180°, all with the same magnitude--0.5. It is well known that when these respective coefficients exist at this point, the matching point, this condition establishes a virtual open circuit to rearward traveling waves at this point, prohibiting any further rearward travel of the reflected waves, thus reversing their direction toward the load. Consequently, the result is total re-reflection of the two sets of reflected waves at the match point. Now the question--what is the resultant voltage coefficient of reflection at the match point? My position is that ñ = 1.0, which indicates total reflection. However, I've been criticized on this value, my critics saying that because this virtual open circuit was established by wave interference, ñ cannot equal 1.0. They assert that ñ = 1.0 can be obtained only through a physical open circuit. So I now ask you, am I correct in saying that the reflection coefficient in this situation is 1.0? I'm holding my breath for your answers. Walt OK, now that I understand the series stub configuration, I have to ask: "the" reflection coefficient looking into WHAT? Looking back toward the source in its feedline? Looking back toward the source but including the series stub? Looking toward the load, including the series stub? The way I look at this problem is as a measurements person: once I know where I'm looking, how will I measure the value of the reflection coefficient? The way I would normally do it is with a network analyzer. If I want to do it in the presence of significant power coming from a source, I can set up directional couplers to allow me to introduce a test signal (that I must be able to distinguish from the power coming from the source) and observe the reflection. You can also replace each piece of transmission line (including the piece that's the stub) with an S-parameter two-port, and the source and load as S-parameter one-ports, and apply standard S-parameter math to see what the reflection coefficient is looking into any particular port. I suppose this is going to give you an answer different from the one you're hoping for...but I believe it's consistent with the definitions for linear system analysis that I've always used. Consider the case of two identical sources generating the same signal but 180 degrees out of phase. Each source is connected to a length of transmission line, and the two lines are identical. Just for fun, the line impedance equals the source impedance. The free ends of the lines are connected in parallel and to a resistive load. Of course, we find that the signals cancel and there is no power dissipated in the load. Would we say that the reflection coefficient magnitude looking back into either line toward the source is unity? I would not! I would say it is zero, since any signal I send in that direction is absorbed in the source. Note in this example that you can disconnect the load and not change anything. Then you simply have the signal from each source traveling to the other source and being absorbed; if you put the sources on two different frequencies, you'll see no standing wave for either frequency (though you'll have to average for a while to be sure about that, if the frequencies are close together). And note, please, that "absorbed" is not in general the same thing as "dissipated." Cheers, Tom |
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Reflection coefficient for total re-reflection
On Jun 12, 10:31*pm, walt wrote:
So I've got to make the decision whether to delete my criticism of his equation or leave it in and be accused of criticizing him incorrectly. What to do! Personally, I didn't find anything wrong with parts 1 and 2. It's in part 3 where Dr. Best gets lost. In an exchange on this newsgroup before the article was published, he asserted that there was no interference occurring at an impedance discontinuity on a transmission line. When he published his component powers, he took into account P1and P2 while completely ignoring (what I have dubbed) the component powers, P3 and P4 in the opposite direction. He even invented something like two phantom waves traveling forever in the direction of total destructive interference while transporting energy but completely canceling each other in the process, obviously a physical impossibility. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 14, 12:25*am, K7ITM wrote:
Cecil Moore wrote: On Jun 13, 1:55*pm, K7ITM wrote: *It doesn't make any sense to me to put a shorted section of line in series with another line, so my confusion starts. Tom, I didn't know initially that the example was in "Reflections III". A series stub can be used instead of a loading coil on a wire antenna. I had never seen a series stub used in such a manner on a transmission line and that's why I was confused. I'm assuming that the center conductor is broken and one side is connected to the inner conductor and one side is connected to the braid on a stub, but I am not sure that is correct. There's got to be a less complicated example that we can use. -- 73, Cecil, w5dxp.com Ah, OK. *Now I understand what Walt meant... *Though it's possible to use a series stub on a transmission line, as you say, it's not all that common in practice. *I suppose that's why my mind wasn't going there. *Mea culpa. *Perhaps now I can go back and look at Walt's original question and make more sense out of it. When a series stub is used in an antenna (as in a quarter-wave stub coupling colinear half-waves), King points out that coupling from the antenna fields to the wires in the stub, when the stub is an open-wire line perpendicular to the axis of the half-waves, is an important factor in how the stub operates to establish in-phase currents on the adjacent half-waves... *I see you made a comment about antenna currents on the stub in John Smith's example, too. Cheers, Tom Tom and Cecil, My use of the series stub wasn't meant to be of practical use. I used it only as a tool to explain a simple matching procedure. I used series rather than parallel, because I wanted it to be simple to use for those who are not as accustomed to admittance procedure as they are with impedance. Sorry if I misled you. Walt |
Reflection coefficient for total re-reflection
On Jun 14, 12:25*am, K7ITM wrote:
Cecil Moore wrote: On Jun 13, 1:55*pm, K7ITM wrote: *It doesn't make any sense to me to put a shorted section of line in series with another line, so my confusion starts. Tom, I didn't know initially that the example was in "Reflections III". A series stub can be used instead of a loading coil on a wire antenna. I had never seen a series stub used in such a manner on a transmission line and that's why I was confused. I'm assuming that the center conductor is broken and one side is connected to the inner conductor and one side is connected to the braid on a stub, but I am not sure that is correct. There's got to be a less complicated example that we can use. -- 73, Cecil, w5dxp.com Ah, OK. *Now I understand what Walt meant... *Though it's possible to use a series stub on a transmission line, as you say, it's not all that common in practice. *I suppose that's why my mind wasn't going there. *Mea culpa. *Perhaps now I can go back and look at Walt's original question and make more sense out of it. When a series stub is used in an antenna (as in a quarter-wave stub coupling colinear half-waves), King points out that coupling from the antenna fields to the wires in the stub, when the stub is an open-wire line perpendicular to the axis of the half-waves, is an important factor in how the stub operates to establish in-phase currents on the adjacent half-waves... *I see you made a comment about antenna currents on the stub in John Smith's example, too. Cheers, Tom I'm sorry to have misled you in bringing up the series stub problem. What I'm really concerned about is the reflection coefficient at the output of the tank circuit of the RF power amp. Being non-dissipative it cannot absorb the reflected power, so it must re-reflect it. Therefore, my position is that its reflection coefficient rho = 1.0, My critics say that a rho = 1.0 cannot be established when the virtual short is caused by wave interference. However, if Best's Eq 8 is valid, then why does it yield an incorrect answer when I plug in rho s = 1, and rho a = 0.5, which gives 2 as the answer, instead of the correct 1.1547 as the correct answer. Now that I've narrowed the problem down, what do you believe is the answer? Your answer will determine whether I need to delete the pertinent paragraph from Chapter 25 in Reflections 3, in which I have declared Best's Eq 8 invalid. Walt Walt |
Reflection coefficient for total re-reflection
On Tue, 14 Jun 2011 11:26:41 -0700 (PDT), walt wrote:
Hi Walt, The focus seems to have shifted away from the matching hardware previously described to: What I'm really concerned about is the reflection coefficient at the output of the tank circuit of the RF power amp. It includes a point argued by many, and which you make a condition of success for your argument: Being non-dissipative it cannot absorb the reflected power, so it must re-reflect it. Your argument now applied at two points, one not illustrated in your references (the source): Therefore, my position is that its reflection coefficient rho = 1.0, The counter argument: My critics say that a rho = 1.0 cannot be established when the virtual short is caused by wave interference. which they express NOT as a condition of the tank circuit, but of the matching hardware. However, if Best's Eq 8 is valid, You have acknowledge that experts in math whom you trust say it is valid. This yields a paradox. By all the shifts in focus, it is apparent that: 1. You are both right; 2. You are both wrong. 3. One of you is right, and one of you is wrong; more likely given the drift of focus: 4. You are at cross purposes and arguing two different things. Now that I've narrowed the problem down, what do you believe is the answer? You have not narrowed the problem down at all. It is composed of a succession of premises that all have arguments for-and-against them. This compounds the uncertainty. It also brings the hazard of any single premise being rejected bringing down the entire goal you are pursuing. * * * * * * * * * * * * * * * * * * * To put this bluntly, let's test the assertion repeated he output of the tank circuit of the RF power amp. Being non-dissipative it cannot absorb the reflected power, so it must re-reflect it. Odds are, you will lose support from at least half of the readership for this statement alone - and it isn't even in your written work that I've provided links for, and it certainly doesn't appear in Johnson. It is not discussed by the experts in math either. It is a flyer, a leap of faith whose validity is now emeshed with the outcome of another problematic discussion. Steve's opinion: viz. "A total re-reflection of power at the match point is not necessary for the impedance match to occur." offers NOTHING to the matter of non-dissipation. Odds are, Steve could easily lose support from at least half of the readership for this statement alone. Unfortunately, that doesn't leverage your position one iota. There are too many irons in the fire. 73's Richard Clark, KB7QHC |
Reflection coefficient for total re-reflection
On Jun 14, 2:58*pm, Richard Clark wrote:
On Tue, 14 Jun 2011 11:26:41 -0700 (PDT), walt wrote: Hi Walt, The focus seems to have shifted away from the matching hardware previously described to: What I'm really concerned about is the reflection coefficient at the output of the tank circuit of the RF power amp. It includes a point argued by many, and which you make a condition of success for your argument: Being non-dissipative it cannot absorb the reflected power, so it must re-reflect it. Your argument now applied at two points, one not illustrated in your references (the source): Therefore, my position is that its reflection coefficient rho = 1.0, The counter argument:My critics say that a rho = 1.0 cannot be established when the virtual short is caused by wave interference. which they express NOT as a condition of the tank circuit, but of the matching hardware. However, if *Best's Eq 8 is valid, You have acknowledge that experts in math whom you trust say it is valid. *This yields a paradox. *By all the shifts in focus, it is apparent that: 1. *You are both right; 2. *You are both wrong. 3. *One of you is right, and one of you is wrong; more likely given the drift of focus: 4. *You are at cross purposes and arguing two different things. Now that I've narrowed the problem down, what do you believe is the answer? You have not narrowed the problem down at all. *It is composed of a succession of premises that all have arguments for-and-against them. This compounds the uncertainty. *It also brings the hazard of any single premise being rejected bringing down the entire goal you are pursuing. * * * * * * * * * * * * * * * * * * * To put this bluntly, let's test the assertion repeated he output of the tank circuit of the RF power amp. Being non-dissipative it cannot absorb the reflected power, so it must re-reflect it. Odds are, you will lose support from at least half of the readership for this statement alone - and it isn't even in your written work that I've provided links for, and it certainly doesn't appear in Johnson. It is not discussed by the experts in math either. *It is a flyer, a leap of faith whose validity is now emeshed with the outcome of another problematic discussion. Steve's opinion:viz. * *"A total re-reflection of power at the match * *point is not necessary for the impedance * *match to occur." offers NOTHING to the matter of non-dissipation. Odds are, Steve could easily lose support from at least half of the readership for this statement alone. *Unfortunately, that doesn't leverage your position one iota. There are too many irons in the fire. 73's Richard Clark, KB7QHC Thank you for your insightful response, Richard. However, I must plod on, and introduce the results of my own measurements, the data of which appear in Chapter 19, Sec 19.14 of Reflections 3. Using a Kenwood TS-830S I measured the parameters required to prove the output source resistance of the tank circuit of its RF power amp is non- dissipative. The measurements also prove that the reflected power incident on its output is totally re-reflected. From reading an earlier post of yours, Richard, I know you have read this portion of Chapter 19, and you agreed with it. My measured data conflicts with Best's Eq 8, so there's got to be a valid answer to this dilemma, but where? Walt |
Reflection coefficient for total re-reflection
On Tue, 14 Jun 2011 12:25:33 -0700 (PDT), walt wrote:
Thank you for your insightful response, Richard. However, I must plod on, and introduce the results of my own measurements, the data of which appear in Chapter 19, Sec 19.14 of Reflections 3. Using a Kenwood TS-830S I measured the parameters required to prove the output source resistance of the tank circuit of its RF power amp is non- dissipative. The measurements also prove that the reflected power incident on its output is totally re-reflected. From reading an earlier post of yours, Richard, I know you have read this portion of Chapter 19, and you agreed with it. My measured data conflicts with Best's Eq 8, so there's got to be a valid answer to this dilemma, but where? Walt Hi Walt, Yes, I am very familiar with your bench work. I am very annoyed by those who trivialize it. I am further annoyed by those who patronize you to then shift the parameters to explain something else - completely abandoning you. However, I also acknowledge that I can hurt your feelings about where I stand on these matters. Yes, to a limited degree, I do agree with much of your assessment, but I do not follow it into the arena of where you express it as being "non-dissipative." However, my denying this proposition does NOT mean I dispute your other proposition of complete reflection. This cross connection of several topics is where things get overly complex and gives the entire discussion the appearance of a house of cards. I cannot tell what your agenda is, but the material you quote as source, and for which I have provided links to in this thread - all seem a subtext to non-dissipative sources. Attention directed towards this pursuit of matching points rendered in stubs seems like tea leaf reading to resolve a larger issue. As a student, tech, and engineer employing a LOT of precision waveguide technology, I am quite comfortable with the components and topologies you describe. Arguments for wave interferences being rendered into useful circuit constructs (that is, constructed on the basis of wave interference) has been with us easily since the early 40s. One component, the ATR tube, is one I have handled and replaced to insure the proper operation of RF paths being steered in waveguides. One proviso, however = for any of these "virtualizations" to work, they require a physical (and dissipative) element as the initiator. Waves do not interfere without the presence of a physical element (often some form of detector, or minimally a load). 73's Richard Clark, KB7QHC |
Reflection coefficient for total re-reflection
On Jun 14, 3:53*pm, Richard Clark wrote:
On Tue, 14 Jun 2011 12:25:33 -0700 (PDT), walt wrote: Thank you for your insightful response, Richard. However, I must plod on, and introduce the results of my own measurements, the data of which appear in Chapter 19, Sec 19.14 of Reflections 3. Using a Kenwood TS-830S I measured the parameters required to prove the output source resistance of the tank circuit of its RF power amp is non- dissipative. The measurements also prove that the reflected power incident on its output is totally re-reflected. From reading an earlier post of yours, Richard, I know you have read this portion of Chapter 19, and you agreed with it. My measured data conflicts with Best's Eq 8, so there's got to be a valid answer to this dilemma, but where? Walt Hi Walt, Yes, I am very familiar with your bench work. *I am very annoyed by those who trivialize it. *I am further annoyed by those who patronize you to then shift the parameters to explain something else - completely abandoning you. However, I also acknowledge that I can hurt your feelings about where I stand on these matters. *Yes, to a limited degree, I do agree with much of your assessment, but I do not follow it into the arena of where you express it as being "non-dissipative." * However, my denying this proposition does NOT mean I dispute your other proposition of complete reflection. *This cross connection of several topics is where things get overly complex and gives the entire discussion the appearance of a house of cards. I cannot tell what your agenda is, but the material you quote as source, and for which I have provided links to in this thread - all seem a subtext to non-dissipative sources. *Attention directed towards this pursuit of matching points rendered in stubs seems like tea leaf reading to resolve a larger issue. As a student, tech, and engineer employing a LOT of precision waveguide technology, I am quite comfortable with the components and topologies you describe. *Arguments for wave interferences being rendered into useful circuit constructs (that is, constructed on the basis of wave interference) has been with us easily since the early 40s. *One component, the ATR tube, is one I have handled and replaced to insure the proper operation of RF paths being steered in waveguides. *One proviso, however = for any of these "virtualizations" to work, they require a physical (and dissipative) element as the initiator. *Waves do not interfere without the presence of a physical element (often some form of detector, or minimally a load). 73's Richard Clark, KB7QHC Thanks again Richard, for your insightful response. And goodness no, Richard, you won't hurt my feelings. I like being correct, but if I'm not I surely want to be told about it. This situation is no different. However, if you're not following me on the 'non-dissipative' path, I'd like you to review the last portion of the TS-830S experiment, and follow the numbers. If you don't have that material in front of you, you can find it again on my web page at www.w2du.com. Click on 'Preview Chapters from Reflections 3', and then click on Chapter 19A, which is a part of Chapter 19 in the 3rd edition. Observe that when the load impedance is changed from 50+j0 to the complex impedance 17.98 + j8.77 ohms, the plate current rose expectedly from 260ma to 290ma. This change occurred because the amp is now mismatched, and the pi-network is also detuned from resonance. The unwashed would conclude that the reflected power caused the increase in plate current. However, one will also observe that after the pi-network has been retuned and adjusted to again deliver all the available power into the otherwise 'mismatched' load (which is now matched to the source), the plate current went back to it's original value, 260ma. In addition, the amp returned to deliver 100w into the complex impedance at the line input, with the 30.6w of reflected power in the line, and adding to the 100w of source power, making 130.6w incident on the mismatched load, absorbing 100w and reflecting 30.6w. IMHO, these data prove beyond doubt that the pi-network not only didn't absorb any of the reflected power, but totally re-reflected it. In my book this says the pi-network reflection coefficient rho = 1.0. How can anyone disagree with this? Walt |
Reflection coefficient for total re-reflection
On Tue, 14 Jun 2011 14:18:53 -0700 (PDT), walt wrote:
Thanks again Richard, for your insightful response. And goodness no, Richard, you won't hurt my feelings. I like being correct, but if I'm not I surely want to be told about it. This situation is no different. However, if you're not following me on the 'non-dissipative' path, I'd like you to review the last portion of the TS-830S experiment, and follow the numbers. If you don't have that material in front of you, you can find it again on my web page at www.w2du.com. Click on 'Preview Chapters from Reflections 3', and then click on Chapter 19A, which is a part of Chapter 19 in the 3rd edition. Observe that when the load impedance is changed from 50+j0 to the complex impedance 17.98 + j8.77 ohms, the plate current rose expectedly from 260ma to 290ma. This change occurred because the amp is now mismatched, and the pi-network is also detuned from resonance. The unwashed would conclude that the reflected power caused the increase in plate current. However, one will also observe that after the pi-network has been retuned and adjusted to again deliver all the available power into the otherwise 'mismatched' load (which is now matched to the source), the plate current went back to it's original value, 260ma. In addition, the amp returned to deliver 100w into the complex impedance at the line input, with the 30.6w of reflected power in the line, and adding to the 100w of source power, making 130.6w incident on the mismatched load, absorbing 100w and reflecting 30.6w. Hi Walt, There is absolutely NOTHING that I can dispute in your numbers or method. However, as to "dissipation" this says nothing. As for what it says about the reflection coefficient, I would agree with you. However, the math and the math experts you speak of - they are not answering the model we are examining above. You distracted them with stubs and reflecting waves. If you cast this agreement in the reflection coefficient back into effects purported to exist in the matching with stubs model, we may yet argue. IMHO, these data prove beyond doubt that the pi-network not only didn't absorb any of the reflected power, but totally re-reflected it. In my book this says the pi-network reflection coefficient rho = 1.0. How can anyone disagree with this? *** Nothing further useful about the topic is to be found below *** There is a curious side bar to this found at: http://www.w5big.com/purchase4170c.htm Observe the second data screen that purports to examine a mismatch through a length of both RG58 and RG59 terminated with 100 Ohms. Of particular note is the distinct transition from the Zc of 50 Ohms to the Zc of 75 Ohms and the similarly distinct transition from the Zc of 75 Ohms to the termination R of 100 Ohms. Note that the transitions both span a distance of 2 feet. 2 feet is not insubstantial compared to connection technology that spans, probably, no more than one tenth that distance. Whence the extra 11 inches on both sides of the connection plane of either connection? In the physics I've been studying for the past 10 years, near fields, this would not be unusual. In fact it would be expected. There is a transition zone (in waveguide design, there would be a taper or a sweep section to anticipate this and they would be physically large in terms of wavelength) not a transition cliff. Similarly, at the connector to your TS-830S, there is a zone that surrounds it that only approximates the 50 Ohm which occurs (in the terms of this link's demonstration) some distance away, deep inside your TS-830S. To me, dissipation inhabits this zone (certainly large enough for the tube and tank to occupy) and embraces the match with a complex addition of phases that could result in loss or even gain. This is also to say that I do not subscribe to dissipation being all about loss - especially when the hand of man is on the tuning knob instead of letting the chips fall where they may. 73's Richard Clark, KB7QHC |
Reflection coefficient for total re-reflection
On Jun 15, 8:22*am, "J.B. Wood" wrote:
On 06/12/2011 03:05 PM, walt wrote: Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Walt, *I can only get your 1.0 - j1.1547 normalized impedance value (looking into the line towards the normalized load of 3.0) if I move 60 degrees, not 30, from the load towards the generator. *I'm also assuming a lossless line. *Sincerely, and 73s from N4GGO, -- J. B. Wood * * * * * * * * *e-mail: Hello JB, thank you for the response. I'm sure I understand what's going on. When you view the Smith Chart at the unity resistance circle with a 3:1 mismatch the normalized impedance there is 1.0 - j1.1547, as you stated. However, a radius through that point yields a voltage reflection coefficient rho = 0.5 @ -60°. In other words, the radial line intersects the periphery of the Smith Chart at -60°. As I'm sure you know, reflection degrees equals two electrical degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs at 30° rearward of the load. That OK? In the absence of a Smith Chart here's an easy way to determine the reactance appearing at the unity resistance circle for any given degree of mismatch: 1/(sqrt SWR/ SWR-1) Walt, W2DU |
Reflection coefficient for total re-reflection
On Jun 15, 10:57*am, walt wrote:
On Jun 15, 8:22*am, "J.B. Wood" wrote: On 06/12/2011 03:05 PM, walt wrote: Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal to 0.5 at 0°. We want to place a stub on the line at the position relating to the unit-resistance circle on the Smith Chart. For a 3:1 mismatch this position is exactly 30° rearward from the load, and has a voltage reflection coefficient Vñ equal to 0.5 at -60°. The normalized line resistance at this point is 1.1547 chart ohms, or 57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735 ohms. Walt, *I can only get your 1.0 - j1.1547 normalized impedance value (looking into the line towards the normalized load of 3.0) if I move 60 degrees, not 30, from the load towards the generator. *I'm also assuming a lossless line. *Sincerely, and 73s from N4GGO, -- J. B. Wood * * * * * * * * *e-mail: Hello JB, thank you for the response. I'm sure I understand what's going on. When you view the Smith Chart at the unity resistance circle with a 3:1 mismatch the normalized impedance there is 1.0 - j1.1547, as you stated. However, a radius through that point yields a voltage reflection coefficient rho = 0.5 @ -60°. In other words, the radial line intersects the periphery of the Smith Chart at -60°. As I'm sure you know, reflection degrees equals two electrical degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs at 30° rearward of the load. That *OK? In the absence of a Smith Chart here's an easy way to determine the reactance appearing at the unity resistance circle for any given degree of mismatch: * * * * * * * * * * * * * * * * * * * * * * * * * * * * *1/(sqrt SWR/ SWR-1) Walt, W2DU Fixing the equation: 1/(sqrt SWR/SWR-1) Walt |
Reflection coefficient for total re-reflection
On 06/15/2011 10:57 AM, walt wrote:
As I'm sure you know, reflection degrees equals two electrical degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs at 30° rearward of the load. That OK? Well, I'm not clear what you mean by "rearward of the load". At 60 degrees toward the generator the normalized impedance looking into the line is 1.0 - j1.1547 ohms and that corresponds to a reflection coefficient of -j1.1547/(2-j1.1547) or alternatively, 0.5 @ -60 degrees (sorry my char set doesn't do Steinmetz notation). So I think we're in agreement here. What more needs to be said? Now put in a shorted stub (of appropriate length at the desired frequency) in series to tune out (cancel) the reactive part of the line impedance. The combination of the series stub and the line will appear as 1.0 + j0 to the generator/source and the reflection coefficient is now 0. How could it be anything else (at the desired frequency) in this scenario? You don't need a discussion of travelling waves to arrive at the correct answer. The reflection coefficient corresponding to a normalized (say to 50 ohms) impedance zx = Zx/50 is rho = (zx - 1)/(zx +1). Clearly a values of rho = 1 + j0 or -1 + j0 correspond to a zx values of infinity (open circuit) or zero (short circuit), respectively. You might want to think of the shorted line stub as inside one box and the transmission line + 150 ohm load in another black box. Let's assume someone hands you these two boxes and you have no idea what's inside but you do have access to set of terminals (input port) on each box from which you can measure steady-state impedance of each. At some frequency you find the normalized input impedance on one box is +j1.1547 ohms and on the other box at that same frequency it is 1.0 - j1.1547 ohms. Now you connect the boxes is series and measure the impedance of the series combination again at that frequency. The point I'm making here is that at the terminals of either of the boxes you measured the impedance at some frequency - it doesn't matter what's inside the box. (Of course the impedance behavior over a range of frequencies certainly is dependent on the circuit topology within the box.) Sincerely, -- J. B. Wood e-mail: |
Reflection coefficient for total re-reflection
On Jun 15, 2:55*pm, "J.B. Wood" wrote:
On 06/15/2011 10:57 AM, walt wrote: As I'm sure you know, reflection degrees equals two electrical degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs at 30 rearward of the load. That *OK? Well, I'm not clear what you mean by "rearward of the load". *At 60 degrees toward the generator the normalized impedance looking into the line is 1.0 - j1.1547 ohms and that corresponds to a reflection coefficient of -j1.1547/(2-j1.1547) or alternatively, 0.5 @ -60 degrees (sorry my char set doesn't do Steinmetz notation). *So I think we're in agreement here. *What more needs to be said? *Now put in a shorted stub (of appropriate length at the desired frequency) in series to tune out (cancel) the reactive part of the line impedance. *The combination of the series stub and the line will appear as 1.0 + j0 to the generator/source and the reflection coefficient is now 0. *How could it be anything else (at the desired frequency) in this scenario? *You don't need a discussion of travelling waves to arrive at the correct answer. The reflection coefficient corresponding to a normalized (say to 50 ohms) impedance zx = Zx/50 is rho = (zx - 1)/(zx +1). *Clearly a values of rho = 1 + j0 or -1 + j0 correspond to a zx values of infinity (open circuit) or zero (short circuit), respectively. You might want to think of the shorted line stub as inside one box and the transmission line + 150 ohm load in another black box. *Let's assume someone hands you these two boxes and you have no idea what's inside but you do have access to set of terminals (input port) on each box from which you can measure steady-state impedance of each. *At some frequency you find the normalized input impedance on one box is +j1.1547 ohms and on the other box at that same frequency it is 1.0 - j1.1547 ohms. *Now you connect the boxes is series and measure the impedance of the series combination again at that frequency. *The point I'm making here is that at the terminals of either of the boxes you measured the impedance at some frequency - it doesn't matter what's inside the box. *(Of course the impedance behavior over a range of frequencies certainly is dependent on the circuit topology within the box.) *Sincerely, -- J. B. Wood * * * * * * * * *e-mail: Thanks again for the response, JB. What I mean by rearward of the load is simply going toward the source from the load. But are we in agreement? We probably are as long as we agree on terminology regarding the Smith Chart. The unity resistance circle for a 3:1 mismatch yields a normalized impedance of 1 - j1.1547 as you know. As I said in my previous post, a radial line passing through this point on the unity-resistance circle joins the periphery of the Chart at -60°. It important to note that this -60° is reflection degrees, which is two-times greater than degrees along the transmission line. Therefore, the normalized line impedance of 1 -j1.1547 appears at 30° from the load toward the source on the xmsn line, not at -60°. Hope we're in agreement on this. Walt |
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