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Reflection coefficient for total re-reflection
My critics say that a rho = 1.0 cannot be established when the virtual
short is caused by wave interference. A reflection is what happens to a single wave at an impedance discontinuity. The redistribution of energy associated with interference is technically not a reflection since it involves the superposition of two (or more) waves. As a result, reflection mechanics cannot be used to explain interference effects. The "virtual short" is the *result* of the combination of physical reflections plus conditions associated with interference, i.e. the "virtual short" is not the *cause* of anything - it is the *result* of reflections plus interference. Most people deal only with reflected voltages where the interference is transparent. When we start dealing with reflected power, interference is NOT transparent as it must be recognized to be able to track the energy through the system. If we superpose two identical waves of 100 volt at zero degrees, we obviously get 200 volts at zero degrees. However, when we recognize that in a 50 ohm environment some distance away from any source, we have taken two 200 watt waves and created a single 800 watt wave, the question arises: Where did the extra 400 watts come from? The answer is from constructive interference which requires 400 watts of destructive interference in the opposite direction in order to satisfy the conservation of energy principle. When interference is involved the virtual power reflection coefficient may not be the square of the physical voltage reflection coefficient. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 17, 9:32*pm, Cecil Moore wrote:
When interference is involved the virtual power reflection coefficient may not be the square of the physical voltage reflection coefficient. Let me try an example in ASCII to see if it can work. --50 ohm--+--1/2WL 291.3 ohms--50 ohms Given: The steady-state forward power, Pfwd1, on the 50 ohm line is 200 watts. The steady-state reflected power, Pref1, on the 50 ohm line is zero watts. 200 watts of steady-state power is delivered to the 50 ohm load. The power reflection coefficient, rho^2, is 0.5 at point 'x' and at the load. The power transmission coefficient, 1-rho^2, is 0.5 at point 'x' and at the load. The steady-state incident forward power, Pfwd2, at the 50 ohm load is 200/0.5=400 watts. The steady state reflected power, Pref2, at the load is 400(0.5)=200 watts. There are four superposition components: 1. The component power reflected from point 'x' back toward the source is Pfwd1(rho^2)=200(0.5)=100w. 2. The component power transmitted through point 'x' in the direction of the source is Pref2(1-rho^2)=200(0.5)=100w. The fields of these two component waves are 180 degrees out of phase and they are equal in amplitude so they cancel to zero. Note that Pref1=zero. This is 200 watts of total destructive interference. The conservation of energy principle says that, in the absence of a nearby source (which this is), any destructive interference must be offset by constructive interference in a different direction. In a transmission line, there are only two directions, i.e. destructive interference in the direction of the source must necessarily be offset by an equal magnitude of constructive interference toward the load. That's exactly how Z0-matches function from an energy standpoint. Addition of EM wave powers are accomplished by the irradiance equation from the field of optics. Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A) where A is the angle between the two electric fields associated with the two waves. Pref1 = 100w + 100w - 200w = 0, where the -200w is destructive interference (negative sign). 3. The component power transmitted through point 'x' in the direction of the load is Pfwd1(1-rho^2)=200(0.5)=100w. 4. The component power reflected from point 'x' back toward the load is Pref2(rho^2)=200(0.5)=100w. The fields of these two component waves are in phase and they are equal in amplitude so they add to double the voltage and current magnitudes. When we double both the voltages and currents, we multiply the power by a factor of four. Thus, two 100 watt waves result in a single 400 watt wave when they engage in total constructive interference as they do in the above example. Pfor2 = 100w + 100w + 200w = 400w, where the +200w is constructive interference (positive sign). |destructive interference| = |constructive interference| and all is well with the conservation of energy principle. Note that the physical power reflection coefficient is 0.5 everywhere. However, Pref1/Pfwd1=0, i.e. the virtual power reflection coefficient is zero (same as the single-port reflection coefficient). All of the power reflected from the load, Pref2, is redistributed back toward the load at point 'x', i.e. the virtual short, virtual power reflection coefficient is 1.0 even though the physical power reflection coefficient is 0.5 at point 'x', i.e. A single-port analysis yields a power reflection coefficient of 1.0 looking back into point 'x'. Walt is obviously using a single-port analysis where *everything is virtual*. Impedances are V/I ratios, not actual devices. Thus the Pref/ Pfwd power ratios are virtual. The difference between reflected energy and interference energy is the difference between real-world reflection coefficients based on real-world impedance discontinuities and virtual reflection coefficients based on V/I ratios and Pref/Pfwd ratios. Walt, here is my humble opinion of what the problem is: 1. You are considering that virtual impedances and virtual reflection coefficients exist in reality to the point that they can be the cause of something. IMO, everything virtual is a result, not a cause. That's why it is called "virtual". 2. You are considering the redistribution of reflected power back toward the load to be 100% caused by reflections governed by reflection mechanics. IMO, that is OK for a voltage analysis, where energy considerations are completely transparent, but definitely NOT for a power analysis. When power is being considered, the difference between reflected energy and interference energy MUST be taken into account because INTERFERENCE ENERGY DOES NOT OBEY THE RULES OF REFLECTION MECHANICS. It obeys the rules of interference mechanics. 3. The assumption that all sources are non-dissipative is a convention that I was taught in college. My professor emphasized that it is a *convention*, convenient for analysis, but not necessarily true in reality. Thus it is assumed that any reflected power that is absorbed by the source was never generated in the first place. But that is a *convention* to make the math easier and not necessarily what happens in reality. 4. The V/I ratio that is your mismatched source impedance is indeed non-dissipative but is most likely a combination of reflections and interference caused by reflected energy flowing into the source and superposing with the generated wave. IT IS SIMPLY IMPOSSIBLE FOR REFLECTED ENERGY TO SIMULTANEOUSLY BE THE CAUSE OF A VIRTUAL SOURCE IMPEDANCE WHILE BEING RE-REFLECTED FROM THE SOURCE TERMINAL, i.e. it cannot be used at the same time for two completely different purposes at two different locations. Somewhere back inside the source, it is likely that dissipation of reflected energy is occurring in any mismatched system. I have said this before and it fell on deaf ears. Since RF energy cannot travel faster than the speed of light, the only way that a source can know the value of its load impedance is in the form of feedback of *reflected energy* from the load, i.e. unless the load impedance is equal to the source impedance, REFLECTED ENERGY MUST NECESSARILY BE FLOWING INSIDE THE SOURCE FOR THE SOURCE TO DETECT THE VALUE OF ITS LOAD IMPEDANCE. The reason for confusion about this subject is that the lumped circuit model presumes instantaneous, faster than light speed, throughout the universe. This faster than light concept is set in concrete brains throughout the RF engineering world. It is indeed strange that RF engineers using the lumped circuit model, don't realize that it takes about one ns for a source to detect a load that is six inches away. The V/I ratio from the source changes after one ns when the first reflected energy arrives. The source has absolutely no way of detecting its load impedance until the first reflected energy arrives. Any other concept is just magical thinking about the real world. Here is a special case example where all of the steady-state reflected power is dissipated in the source resistor. This particular source was presented by w7el in his food-for-thought series. http://www.w5dxp.com/nointfr.htm -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 13, 11:55*am, 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 resultingreflectioncoefficient. Assume a 50-ohm line terminated in a purely resistive 150-ohm load, yielding a 3:1 mismatch with a voltagereflectioncoefficientVñ 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 voltagereflectioncoefficientVñ 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, cancelingreflectionthat cancels the primary reflected wave generated by the mismatched line termination. Recall the voltagereflectioncoefficientVñ of the line impedance prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current reflectioncoefficientIñ = 0.5 at +120° We now look at the voltagereflectioncoefficientof the stub, which is Vñ = 0.5 at +60°, and the corresponding currentreflection coefficientof 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-reflectionof the two sets of reflected waves at the match point. Now the question--what is the resultant voltagecoefficientof reflectionat the match point? My position is that ñ = 1.0, which indicates totalreflection. 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 18, 5:50*pm, K7ITM wrote:
For some reason, Google Groups won't let me see any postings to this thread past June 13th. Tom, I had the same problem and clicking on "sort by date" solved the problem. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 19, 2:57*pm, walt wrote:
The problem here Cecil, is that you've made the same error as Steve in believing that the reflected voltage adds to the source voltage at the source to establish the forward voltage. I've proven that it does not, but it's power or energy that adds. What you seem to be saying is that the voltage reflection coefficient is NOT the square root of the power reflection coefficient since if the power reflection coefficient is 1.0, the voltage reflection coefficient would also be 1.0. It seems that would be a violation of reflection mechanics. Is reflection mechanics wrong or did you misunderstand what I (and Steve) are saying? Walt, I believe we have proven the same thing. But neither Steve nor I believe that the reflected voltage adds to the source voltage at the source. We had this same conversation years ago when I lived at my last QTH and you still seem to be confused about what Steve was saying. Steve's P1 is *NOT* the source power (as you seem to believe). It is the source power that is transmitted through the impedance discontinuity, i.e. Pfwd1(1-rho^2). His P2 is *NOT* the reflected power (as you seem to believe). It is the reflected power that is re- reflected from the impedance discontinuity, i.e. Pref2(rho^1). It seems you have never understood what Steve was saying. Let's go over it again. The power reflection coefficient at point '+' is 0.5 looking in either direction. 100w Source--50 ohm--+--1/2WL 291.3 ohms--50 ohm load On the 50 ohm line, Pfwd1=100w and Pref1=0w On the 291.3 ohm line, Pfwd2=200w and Pref2=100w Steve's P1 equals Pfwd1(1-rho^2)=100w*0.5=50w That's the component of source power that makes it through the impedance discontinuity. Steve's P2 equals Pref2(rho^2)=100w*0.5=50w That's the component of reflected power that is re-reflected from the impedance discontinuity. The total forward power on the 291.3 ohm feedline then becomes: Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(0) Ptot = 50w + 50w + 2*SQRT(50w*50w) = 200w And that is indeed the forward power on the 291.3 ohm line. Please go back and read what Steve said given that his P1 is NOT the source power and his P2 is NOT the reflected power. His power equation comes directly from any optics physics book and is the irradiance equation for EM waves. Steve did indeed make some errors in part 3 of his articles but the above is not one of them. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 19, 2:57*pm, walt wrote:
The problem here Cecil, is that you've made the same error as Steve in believing that the reflected voltage adds to the source voltage at the source to establish the forward voltage. I've proven that it does not, but it's power or energy that adds. For a voltage analysis, one doesn't need to use power (or energy) at all. Given the proper voltage reflection coefficients, the voltages simply phasor add up to the proper values. The conservation of energy principle is completely transparent during the superposition of voltages. It is the power (energy) analysis that is causing the problems. Walt, you may be right about reflected power incident upon a source being 100% redistributed back toward the load but it is definitely not due to 100% re-reflection. In any power (energy) analysis, interference effects must be included in the analysis. If the source completely rejects incident reflected energy, it is because total destructive interference is occurring within the source which results in constructive interference toward the load. Here's a reference from my "Worldradio" energy article that explains the interference phenomenon. "[8] Maxwell, Walter, Reflections II, (c) 2001 Worldradio Books, ISBN 0-9705206-0-3 page 4-3, "The destructive wave interference between these two complementary waves ... causes a complete cancellation of energy flow in the direction toward the generator. Conversely, the constructive wave interference produces an energy maximum in the direction toward the load, ..." page 23-9, "Consequently, all corresponding voltage and current phasors are 180 degrees out of phase at the matching point. ... With equal magnitudes and opposite phase at the same point (point A, the matching point), the sum of the two (reflected) waves is zero." i.e., a source doesn't have to have a power reflection coefficient of 1.0 in order for 100% of the incident reflected energy to be *redistributed* back toward the load. If total destructive interference is occurring within the source, the power reflection coefficient can have any value between 0.0 and 1.0. Consider a Zg=50 ohm source driving a Z0=50 ohm 1/4WL ideal shorted stub. The voltage reflection coefficient looking back into the source is 0.0 so the re-reflected power is zero. However, zero power is being sourced because total destructive interference is occurring in the source. All of the reflected power is redistributed back into the stub by destructive interference and none of it is re-reflected, i.e., the power reflection coefficient looking back into the source is 0.0 because Zg=Z0. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 18, 6:57*pm, Cecil Moore wrote:
On Jun 18, 5:50*pm, K7ITM wrote: For some reason, Google Groups won't let me see any postings to this thread past June 13th. Tom, I had the same problem and clicking on "sort by date" solved the problem. -- 73, Cecil, w5dxp.com ;-) Well, I'm not sure if you also sent that to my email address; I didn't see it there. And of course, I didn't see it here, since I never "sort by date" -- at least not till now, when I've managed to find a free newsgroup server where I can see the messages. I'm happy to see that Owen has jumped into the thread with his usual well thought out postings. Measuring S-parameters (which includes reflection coefficients) and using them in calculations has been the topic of a great many articles. There are some good Hewlett-Packard notes on them that I believe you can get from the Agilent website, and if not there, from other archives. HP AN95-1 and HP/Agilent AN154 are a couple, but Googling "S Parameter Application Note" will get you lots more. Honestly, I'd have to say they are a better place to try to learn this sort of thing than threads here... Cheers, Tom |
Reflection coefficient for total re-reflection
On Jun 21, 2:58*pm, K7ITM wrote:
;-) *Well, I'm not sure if you also sent that to my email address; Yes, I did Tom, to the email address in your attribution line above. HP AN95-1 Available at: http://www.sss-mag.com/pdf/an-95-1.pdf One interesting thing about the s-parameter equations is that if one squares them, one obtains the irradiance (power density) equation from the field of optics. For instance: b1 = s11*a1 + s12*a2 are the normalized voltages where b1^2 is the reflected power toward the source. Squaring the right side of the equation uncovers the interference term, 2*s11*a1*s12*a2. If that term is positive, the interference is constructive. If that term is negative, the interference is destructive. Zero reflected power toward the source indicates total destructive interference toward the source and that is the goal of tuning a system to zero reflected power incident upon the source. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
Let's return to an earlier example and compare a single-port analysis
with a dual-port analysis. 100w source--50 ohm--+--1/2WL 291.4 ohm--50 ohm load The 50 ohm Z0-match point is at '+'. The forward power on the 50 ohm line is 100 watts and the reflected power on the 50 ohm line is zero watts. The forward power on the 291.4 ohm line is 200 watts and the reflected power on the 291.4 ohm line is 100 watts. 100 watts is being sourced and delivered to the 50 ohm load. The voltage reflection coefficient, rho, at the load is (50-291.4)/ (50+291.4)=0.7071. The power reflection coefficient, rho^2, at the load is 0.5, i.e. half of the power incident upon the load (200w) is reflected (100w). Since the load is a single-port, these parameters are consistent with a single-port analysis. In a single-port analysis, we cannot tell the difference between a virtual reflection coefficient and a physical reflection coefficient. The problem comes when we use a single-port analysis on the Z0-match point. Since the reflected power on the 50 ohm line is zero, a single- port analysis would yield rho=0.0 and rho^2=0.0 when viewing the Z0- match from the source side. When we perform a dual-port analysis, we get different values for rho and rho^2, i.e. we get the complement of the reflection coefficients at the load which is a characteristic of any simple Z0-match similar to the above example. For a dual-port analysis, rho looking into the Z0-match from the source side is (291.4-50)/(291.4+50)=0.7071 and rho^2 looking into the Z0-match from the source side is 0.5, the same as at the load. Looking back into the Z0-match from the load side, the sign of rho is negative just as it is at the load with rho^2=0.5, the same as at the load. Since the two analyses yield different values for the reflection coefficients, which analysis is correct? The answer gives the clue to the resolution of this discussion. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 22, 1:49*pm, Cecil Moore wrote:
Let's return to an earlier example and compare a single-port analysis with a dual-port analysis. 100w source--50 ohm--+--1/2WL 291.4 ohm--50 ohm load The 50 ohm Z0-match point is at '+'. The forward power on the 50 ohm line is 100 watts and the reflected power on the 50 ohm line is zero watts. The forward power on the 291.4 ohm line is 200 watts and the reflected power on the 291.4 ohm line is 100 watts. 100 watts is being sourced and delivered to the 50 ohm load. The voltage reflection coefficient, rho, at the load is (50-291.4)/ (50+291.4)=0.7071. The power reflection coefficient, rho^2, at the load is 0.5, i.e. half of the power incident upon the load (200w) is reflected (100w). Since the load is a single-port, these parameters are consistent with a single-port analysis. In a single-port analysis, we cannot tell the difference between a virtual reflection coefficient and a physical reflection coefficient. The problem comes when we use a single-port analysis on the Z0-match point. Since the reflected power on the 50 ohm line is zero, a single- port analysis would yield rho=0.0 and rho^2=0.0 when viewing the Z0- match from the source side. When we perform a dual-port analysis, we get different values for rho and rho^2, i.e. we get the complement of the reflection coefficients at the load which is a characteristic of any simple Z0-match similar to the above example. For a dual-port analysis, rho looking into the Z0-match from the source side is (291.4-50)/(291.4+50)=0.7071 and rho^2 looking into the Z0-match from the source side is 0.5, the same as at the load. Looking back into the Z0-match from the load side, the sign of rho is negative just as it is at the load with rho^2=0.5, the same as at the load. Since the two analyses yield different values for the reflection coefficients, which analysis is correct? The answer gives the clue to the resolution of this discussion. -- 73, Cecil, w5dxp.com ok, i'm afraid i'm going to have to ask the simple question... if you blackbox the load and stub and look at just the one connection to it and that gives you no reflected power... where do you define the second port, and why? |
Reflection coefficient for total re-reflection
On 6/22/2011 3:24 PM, dave wrote:
On Jun 22, 1:49 pm, Cecil wrote: Let's return to an earlier example and compare a single-port analysis with a dual-port analysis. 100w source--50 ohm--+--1/2WL 291.4 ohm--50 ohm load The 50 ohm Z0-match point is at '+'. The forward power on the 50 ohm line is 100 watts and the reflected power on the 50 ohm line is zero watts. The forward power on the 291.4 ohm line is 200 watts and the reflected power on the 291.4 ohm line is 100 watts. 100 watts is being sourced and delivered to the 50 ohm load. The voltage reflection coefficient, rho, at the load is (50-291.4)/ (50+291.4)=0.7071. The power reflection coefficient, rho^2, at the load is 0.5, i.e. half of the power incident upon the load (200w) is reflected (100w). Since the load is a single-port, these parameters are consistent with a single-port analysis. In a single-port analysis, we cannot tell the difference between a virtual reflection coefficient and a physical reflection coefficient. The problem comes when we use a single-port analysis on the Z0-match point. Since the reflected power on the 50 ohm line is zero, a single- port analysis would yield rho=0.0 and rho^2=0.0 when viewing the Z0- match from the source side. When we perform a dual-port analysis, we get different values for rho and rho^2, i.e. we get the complement of the reflection coefficients at the load which is a characteristic of any simple Z0-match similar to the above example. For a dual-port analysis, rho looking into the Z0-match from the source side is (291.4-50)/(291.4+50)=0.7071 and rho^2 looking into the Z0-match from the source side is 0.5, the same as at the load. Looking back into the Z0-match from the load side, the sign of rho is negative just as it is at the load with rho^2=0.5, the same as at the load. Since the two analyses yield different values for the reflection coefficients, which analysis is correct? The answer gives the clue to the resolution of this discussion. -- 73, Cecil, w5dxp.com ok, i'm afraid i'm going to have to ask the simple question... if you blackbox the load and stub and look at just the one connection to it and that gives you no reflected power... where do you define the second port, and why? Logic, immediately, suggests to me, that varying the frequency and measuring voltage, amperage, and SWR would begin to immediately point the answer(s.) Regards, JS |
Reflection coefficient for total re-reflection
On 6/22/2011 5:50 PM, John Smith wrote:
On 6/22/2011 3:24 PM, dave wrote: On Jun 22, 1:49 pm, Cecil wrote: Let's return to an earlier example and compare a single-port analysis with a dual-port analysis. 100w source--50 ohm--+--1/2WL 291.4 ohm--50 ohm load The 50 ohm Z0-match point is at '+'. The forward power on the 50 ohm line is 100 watts and the reflected power on the 50 ohm line is zero watts. The forward power on the 291.4 ohm line is 200 watts and the reflected power on the 291.4 ohm line is 100 watts. 100 watts is being sourced and delivered to the 50 ohm load. The voltage reflection coefficient, rho, at the load is (50-291.4)/ (50+291.4)=0.7071. The power reflection coefficient, rho^2, at the load is 0.5, i.e. half of the power incident upon the load (200w) is reflected (100w). Since the load is a single-port, these parameters are consistent with a single-port analysis. In a single-port analysis, we cannot tell the difference between a virtual reflection coefficient and a physical reflection coefficient. The problem comes when we use a single-port analysis on the Z0-match point. Since the reflected power on the 50 ohm line is zero, a single- port analysis would yield rho=0.0 and rho^2=0.0 when viewing the Z0- match from the source side. When we perform a dual-port analysis, we get different values for rho and rho^2, i.e. we get the complement of the reflection coefficients at the load which is a characteristic of any simple Z0-match similar to the above example. For a dual-port analysis, rho looking into the Z0-match from the source side is (291.4-50)/(291.4+50)=0.7071 and rho^2 looking into the Z0-match from the source side is 0.5, the same as at the load. Looking back into the Z0-match from the load side, the sign of rho is negative just as it is at the load with rho^2=0.5, the same as at the load. Since the two analyses yield different values for the reflection coefficients, which analysis is correct? The answer gives the clue to the resolution of this discussion. -- 73, Cecil, w5dxp.com ok, i'm afraid i'm going to have to ask the simple question... if you blackbox the load and stub and look at just the one connection to it and that gives you no reflected power... where do you define the second port, and why? Logic, immediately, suggests to me, that varying the frequency and measuring voltage, amperage, and SWR would begin to immediately point the answer(s.) Regards, JS A 50 ohm, non-reactive/carbon load would complicate matters, so naturally, I am assuming that is NOT the case ... Regards, JS |
Reflection coefficient for total re-reflection
On Jun 22, 5:24*pm, dave wrote:
ok, i'm afraid i'm going to have to ask the simple question... if you blackbox the load and stub and look at just the one connection to it and that gives you no reflected power... where do you define the second port, and why? For the two-port analysis, only the impedance discontinuity at point 'x' is in the black box. One port is the source side of the impedance discontinuity. The second port is the load side of the impedance discontinuity. It allows the standard s-parameters to be measured and the standard s-parameter equations to be used. On the source side of the impedance discontinuity: b1 = s11(a1) + s12(a2) On the load side of the impedance discontinuity: b2 = s21(a1) + s22(a2) Those are the normalized voltage equations. Squaring those equations shows what happens to the component powers including interference components. Reference: http://www.sss-mag.com/pdf/an-95-1.pdf -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 22, 7:52*pm, John Smith wrote:
A 50 ohm, non-reactive/carbon load would complicate matters, so naturally, I am assuming that is NOT the case ... IMO, the resistor load simplifies things as a single-port analysis can be used on a resistor because there is only one component of power accepted by the resistor. What would complicate things, IMO, is a 50 ohm antenna feedpoint impedance which is a virtual impedance. Analyzing the antenna as a multiple port device shows where the multiple energy components go which is a complication of the present point I am trying to make about the Z0-match point 'x'. However, the single-port vs dual-port analysis differences at the impedance discontinuity 'x' also apply to the analysis at the load resistor vs an antenna AND at the source where Walt seems to be using a single-port analysis involving a virtual source impedance which depends for its very existence upon forward and reflected energy components flowing in opposite directions at the source impedance point which necessarily causes interference accompanied by a redistribution, not a reflection, of energy components. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On 6/23/2011 5:59 AM, Cecil Moore wrote:
On Jun 22, 7:52 pm, John wrote: A 50 ohm, non-reactive/carbon load would complicate matters, so naturally, I am assuming that is NOT the case ... IMO, the resistor load simplifies things as a single-port analysis can be used on a resistor because there is only one component of power accepted by the resistor. What would complicate things, IMO, is a 50 ohm antenna feedpoint impedance which is a virtual impedance. Analyzing the antenna as a multiple port device shows where the multiple energy components go which is a complication of the present point I am trying to make about the Z0-match point 'x'. However, the single-port vs dual-port analysis differences at the impedance discontinuity 'x' also apply to the analysis at the load resistor vs an antenna AND at the source where Walt seems to be using a single-port analysis involving a virtual source impedance which depends for its very existence upon forward and reflected energy components flowing in opposite directions at the source impedance point which necessarily causes interference accompanied by a redistribution, not a reflection, of energy components. -- 73, Cecil, w5dxp.com But cecil, with a 50 ohm NON-inductive load (not inductive over all freqs of concern), fed by 50 ohm line, from a 50 ohm source ... would find a stub a bit of a problem ... indeed, a shorted stub more so -- choice of proper line length and placement of the stub would allow its' use, on very limited frequencies, with excellent dummy load results, but the need, ever, escapes me! Anyway, I just commented to **** old dave off ... :-) I admit it! Regards, JS |
Reflection coefficient for total re-reflection
On Jun 23, 12:43*pm, Cecil Moore wrote:
On Jun 22, 5:24*pm, dave wrote: ok, i'm afraid i'm going to have to ask the simple question... if you blackbox the load and stub and look at just the one connection to it and that gives you no reflected power... where do you define the second port, and why? For the two-port analysis, only the impedance discontinuity at point 'x' is in the black box. One port is the source side of the impedance discontinuity. The second port is the load side of the impedance discontinuity. It allows the standard s-parameters to be measured and the standard s-parameter equations to be used. On the source side of the impedance discontinuity: b1 = s11(a1) + s12(a2) On the load side of the impedance discontinuity: b2 = s21(a1) + s22(a2) Those are the normalized voltage equations. Squaring those equations shows what happens to the component powers including interference components. Reference: http://www.sss-mag.com/pdf/an-95-1.pdf -- 73, Cecil, w5dxp.com but what is your second source? you can always represent the second source in that case in terms of the transmitter output so the second input can be eliminated giving you a single port model. |
Reflection coefficient for total re-reflection
On Jun 23, 12:43*pm, Cecil Moore wrote:
On Jun 22, 5:24*pm, dave wrote: ok, i'm afraid i'm going to have to ask the simple question... if you blackbox the load and stub and look at just the one connection to it and that gives you no reflected power... where do you define the second port, and why? For the two-port analysis, only the impedance discontinuity at point 'x' is in the black box. One port is the source side of the impedance discontinuity. The second port is the load side of the impedance discontinuity. It allows the standard s-parameters to be measured and the standard s-parameter equations to be used. On the source side of the impedance discontinuity: b1 = s11(a1) + s12(a2) On the load side of the impedance discontinuity: b2 = s21(a1) + s22(a2) Those are the normalized voltage equations. Squaring those equations shows what happens to the component powers including interference components. Reference: http://www.sss-mag.com/pdf/an-95-1.pdf -- 73, Cecil, w5dxp.com p.s. if the separation between the two ports is just the discontinuity connection 'point' then the voltages must be the same and the currents are exact opposites only because of the direction convention defined, there can be no difference measuring on one side of a point to the other. |
Reflection coefficient for total re-reflection
On Jun 23, 4:41*pm, dave wrote:
but what is your second source? *you can always represent the second source in that case in terms of the transmitter output so the second input can be eliminated giving you a single port model. a1 is the normalized forward voltage on the 50 ohm feedline from the source. a2 is the normalized reflected voltage on the 291.4 ohm feedline from the load. Those are the two sources associated with the impedance discontinuity inside the black box. a2 could just as easily be from a second generator instead of a reflection. When the single-port model is used, if the impedance is not an impedor, i.e. if the impedance is virtual, the reflection coefficients are virtual reflection coefficients that do not reflect anything and do not absorb power. I will repeat an earlier assertion: Since a virtual impedance is result of the superposition of a forward wave and a reflected wave, a virtual impedance cannot re-reflect the reflected wave, i.e. one cannot re-reflect the reflected wave while at the same time the reflected wave is being used to generate an impedance. It has to be one or the other. Otherwise, there is a violation of the conservation of energy principle. RF EM ExH energy cannot be used simultaneously to generate a virtual impedance while at the same time being re-reflected. If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. On Jun 24, 6:27 am, dave wrote: p.s. if the separation between the two ports is just the discontinuity connection 'point' then the voltages must be the same and the currents are exact opposites only because of the direction convention defined, there can be no difference measuring on one side of a point to the other. The total voltage and total current on both sides of the impedance discontinuity must be equal. But the superposition components do not have to be equal and, in fact, cannot be equal. In the case of the Z0- matched example, the forward voltage on the 50 ohm side is 70.7 volts while the forward voltage on the 291.4 ohm side is 241.4 volts. In order for the total voltage to be the same, the reflected voltage on the 291.4 ohm side, which is 170.7 volts, must be subtracted from the 241.4 volts of forward voltage to yield a total of 70.7 volts. For the Z0-matched example: Vfwd1 = Vfwd2 - Vref2 70.7v = 241.4v - 170.7v Please note that the Z0-match point is at a voltage minimum on the 291.4 ohm feedline. 1/4WL toward the load, the total voltage is 241.4+170.7=412.1 volts (in a lossless system). -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 24, 1:52*pm, Cecil Moore wrote:
On Jun 23, 4:41*pm, dave wrote: but what is your second source? *you can always represent the second source in that case in terms of the transmitter output so the second input can be eliminated giving you a single port model. a1 is the normalized forward voltage on the 50 ohm feedline from the source. a2 is the normalized reflected voltage on the 291.4 ohm feedline from the load. Those are the two sources associated with the impedance discontinuity inside the black box. a2 could just as easily be from a second generator instead of a reflection. When the single-port model is used, if the impedance is not an impedor, i.e. if the impedance is virtual, the reflection coefficients are virtual reflection coefficients that do not reflect anything and do not absorb power. I will repeat an earlier assertion: Since a virtual impedance is result of the superposition of a forward wave and a reflected wave, a virtual impedance cannot re-reflect the reflected wave, i.e. one cannot re-reflect the reflected wave while at the same time the reflected wave is being used to generate an impedance. It has to be one or the other. Otherwise, there is a violation of the conservation of energy principle. RF EM ExH energy cannot be used simultaneously to generate a virtual impedance while at the same time being re-reflected. If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. On Jun 24, 6:27 am, dave wrote: p.s. if the separation between the two ports is just the discontinuity connection 'point' then the voltages must be the same and the currents are exact opposites only because of the direction convention defined, there can be no difference measuring on one side of a point to the other. The total voltage and total current on both sides of the impedance discontinuity must be equal. But the superposition components do not have to be equal and, in fact, cannot be equal. In the case of the Z0- matched example, the forward voltage on the 50 ohm side is 70.7 volts while the forward voltage on the 291.4 ohm side is 241.4 volts. In order for the total voltage to be the same, the reflected voltage on the 291.4 ohm side, which is 170.7 volts, must be subtracted from the 241.4 volts of forward voltage to yield a total of 70.7 volts. For the Z0-matched example: Vfwd1 = Vfwd2 - Vref2 70.7v = 241.4v - 170.7v Please note that the Z0-match point is at a voltage minimum on the 291.4 ohm feedline. 1/4WL toward the load, the total voltage is 241.4+170.7=412.1 volts (in a lossless system). -- 73, Cecil, w5dxp.com meaningless hair splitting. if i put a meter on one side of the stub connection point i will measure the exact same voltage as on the other side of the connection point. why don't you guys do something practical instead of arguing about split hairs and things that can't be measured? |
Reflection coefficient for total re-reflection
On 6/24/2011 1:24 PM, dave wrote:
On Jun 24, 1:52 pm, Cecil wrote: On Jun 23, 4:41 pm, wrote: but what is your second source? you can always represent the second source in that case in terms of the transmitter output so the second input can be eliminated giving you a single port model. a1 is the normalized forward voltage on the 50 ohm feedline from the source. a2 is the normalized reflected voltage on the 291.4 ohm feedline from the load. Those are the two sources associated with the impedance discontinuity inside the black box. a2 could just as easily be from a second generator instead of a reflection. When the single-port model is used, if the impedance is not an impedor, i.e. if the impedance is virtual, the reflection coefficients are virtual reflection coefficients that do not reflect anything and do not absorb power. I will repeat an earlier assertion: Since a virtual impedance is result of the superposition of a forward wave and a reflected wave, a virtual impedance cannot re-reflect the reflected wave, i.e. one cannot re-reflect the reflected wave while at the same time the reflected wave is being used to generate an impedance. It has to be one or the other. Otherwise, there is a violation of the conservation of energy principle. RF EM ExH energy cannot be used simultaneously to generate a virtual impedance while at the same time being re-reflected. If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. On Jun 24, 6:27 am, wrote: p.s. if the separation between the two ports is just the discontinuity connection 'point' then the voltages must be the same and the currents are exact opposites only because of the direction convention defined, there can be no difference measuring on one side of a point to the other. The total voltage and total current on both sides of the impedance discontinuity must be equal. But the superposition components do not have to be equal and, in fact, cannot be equal. In the case of the Z0- matched example, the forward voltage on the 50 ohm side is 70.7 volts while the forward voltage on the 291.4 ohm side is 241.4 volts. In order for the total voltage to be the same, the reflected voltage on the 291.4 ohm side, which is 170.7 volts, must be subtracted from the 241.4 volts of forward voltage to yield a total of 70.7 volts. For the Z0-matched example: Vfwd1 = Vfwd2 - Vref2 70.7v = 241.4v - 170.7v Please note that the Z0-match point is at a voltage minimum on the 291.4 ohm feedline. 1/4WL toward the load, the total voltage is 241.4+170.7=412.1 volts (in a lossless system). -- 73, Cecil, w5dxp.com meaningless hair splitting. if i put a meter on one side of the stub connection point i will measure the exact same voltage as on the other side of the connection point. why don't you guys do something practical instead of arguing about split hairs and things that can't be measured? It can be measured. Why don't you go to another group or thread? Nobody is forcing you to read this one. |
Reflection coefficient for total re-reflection
On 6/24/2011 8:52 AM, Cecil Moore wrote:
When the single-port model is used, if the impedance is not an impedor, i.e. if the impedance is virtual, the reflection coefficients are virtual reflection coefficients that do not reflect anything and do not absorb power. I will repeat an earlier assertion: Since a virtual impedance is result of the superposition of a forward wave and a reflected wave, a virtual impedance cannot re-reflect the reflected wave, i.e. one cannot re-reflect the reflected wave while at the same time the reflected wave is being used to generate an impedance. It has to be one or the other. Otherwise, there is a violation of the conservation of energy principle. RF EM ExH energy cannot be used simultaneously to generate a virtual impedance while at the same time being re-reflected. If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. I disagree. There are 100W supplied by the source and 100W consumed by the load. There are 200W in the 291.4 ohm line. 100W of that is just "passing through". The other 100W is circulating, that is, stored energy which was put there by the start-up transient. If it is circulating, then it must be reflected from each end of the 291.4 ohm line. Cheers, John |
Reflection coefficient for total re-reflection
On Jun 24, 2:20*pm, John S wrote:
If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. I disagree. There are 100W supplied by the source and 100W consumed by the load. There are 200W in the 291.4 ohm line. 100W of that is just "passing through". The other 100W is circulating, that is, stored energy which was put there by the start-up transient. If it is circulating, then it must be reflected from each end of the 291.4 ohm line. Let's assume that the 100 watts is just "passing through". It would change from 70.7 volts in the 50 ohm environment to 170.7 volts in the 291.4 ohm environment. The reflected power is 100 watts so the reflected voltage is also 170.7 volts. Those two voltages would have to add together to get the forward voltage. The forward voltage is known to be 241.4 volts. Exactly how do you add two 170.7 volt in- phase waves to get a total of 241.4 volts? Here is actually what happens. Since the physical power reflection coefficient at the Z0-match point is 0.5, only 50 watts of the source power makes it through the impedance discontinuity. That voltage is 120.7 volts. The same thing applies to the reflected power - only 50 watts is re-reflected by the 0.5 power reflection coefficient. So the re-reflected voltage is also 120.7 volts. Adding those two voltages together yields 241.4 volts which we know is the correct forward voltage. Now you are going to ask how two 50 watt waves can add up to 200 watts forward power. That's just the nature of constructive interference since power is proportional to voltage squared. If we add two 50 watt waves in phase, we get a 200 watt wave. Where did the extra 100 watts come from? Why, from the 100 watts of destructive interference toward the source that eliminated the reflections toward the source. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 24, 1:24*pm, dave wrote:
meaningless hair splitting. That's my attitude toward religion so I don't frequent any religious newsgroups. That meaningless hair splitting is the answer to the apparent contradiction with which Walt is wrestling. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 24, 7:05*pm, John S wrote:
On 6/24/2011 1:24 PM, dave wrote: On Jun 24, 1:52 pm, Cecil *wrote: On Jun 23, 4:41 pm, *wrote: but what is your second source? *you can always represent the second source in that case in terms of the transmitter output so the second input can be eliminated giving you a single port model. a1 is the normalized forward voltage on the 50 ohm feedline from the source. a2 is the normalized reflected voltage on the 291.4 ohm feedline from the load. Those are the two sources associated with the impedance discontinuity inside the black box. a2 could just as easily be from a second generator instead of a reflection. When the single-port model is used, if the impedance is not an impedor, i.e. if the impedance is virtual, the reflection coefficients are virtual reflection coefficients that do not reflect anything and do not absorb power. I will repeat an earlier assertion: Since a virtual impedance is result of the superposition of a forward wave and a reflected wave, a virtual impedance cannot re-reflect the reflected wave, i.e. one cannot re-reflect the reflected wave while at the same time the reflected wave is being used to generate an impedance. It has to be one or the other. Otherwise, there is a violation of the conservation of energy principle. RF EM ExH energy cannot be used simultaneously to generate a virtual impedance while at the same time being re-reflected. If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. On Jun 24, 6:27 am, *wrote: p.s. if the separation between the two ports is just the discontinuity connection 'point' then the voltages must be the same and the currents are exact opposites only because of the direction convention defined, there can be no difference measuring on one side of a point to the other. The total voltage and total current on both sides of the impedance discontinuity must be equal. But the superposition components do not have to be equal and, in fact, cannot be equal. In the case of the Z0- matched example, the forward voltage on the 50 ohm side is 70.7 volts while the forward voltage on the 291.4 ohm side is 241.4 volts. In order for the total voltage to be the same, the reflected voltage on the 291.4 ohm side, which is 170.7 volts, must be subtracted from the 241.4 volts of forward voltage to yield a total of 70.7 volts. For the Z0-matched example: Vfwd1 = Vfwd2 - Vref2 70.7v = 241.4v - 170.7v Please note that the Z0-match point is at a voltage minimum on the 291.4 ohm feedline. 1/4WL toward the load, the total voltage is 241.4+170.7=412.1 volts (in a lossless system). -- 73, Cecil, w5dxp.com meaningless hair splitting. *if i put a meter on one side of the stub connection point i will measure the exact same voltage as on the other side of the connection point. *why don't you guys do something practical instead of arguing about split hairs and things that can't be measured? It can be measured. Why don't you go to another group or thread? Nobody is forcing you to read this one. try it! you will read the exact same voltage on either side of that connection point! |
Reflection coefficient for total re-reflection
On Jun 25, 5:53*am, dave wrote:
try it! *you will read the exact same voltage on either side of that connection point! I already told you that only applies to the total voltage and total current. You will NOT read the same forward voltage on either side, you will NOT read the same forward current on either side, you will NOT read the same reflected voltage on either side, and you will NOT read the same reflected current on either side. The total voltage and total current are the results of the superposition of the four component voltages and currents that obey the rules of wave reflection mechanics. Recognizing the interference patterns when two phasor voltages are superposed is the key to understanding exactly what is happening to the energy in the waves. At an impedance discontinuity in a transmission line some distance from any active source, the average destructive interference power in one direction MUST equal the average constructive interference power in the opposite direction in order to avoid a violation of the conservation of energy principle. So why isn't the forward current flowing into the impedance discontinuity equal to the forward current flowing out of the impedance discontinuity? The answer to that question will solve Walt's apparent contradiction between voltages and powers. Look at the Z0- match again. source--50 ohm--+--1/2WL Z050 ohm--50 ohm load The total current on the 50 ohm side of point '+' is equal to the total current on the Z050 ohm side but the current on the 50 ohm side is a flat traveling wave *constant* current while the current on the Z050 ohm side is a standing-wave current maximum, i.e. the total current on the Z050 ohm side is a *variable* that changes with a change in the measurement point. A variable current is NOT the same as a constant current. The total voltage on the 50 ohm side is a flat traveling wave *constant* voltage while the voltage on the Z050 ohm side is a standing wave voltage minimum, i.e. the total voltage on the Z050 ohm side is a *variable* that changes with a change in the measurement point. The power on the 50 ohm side is V*I where V and I are constant values. The power on the Z050 ohm side is V*I*cos(A) where A is the angle between the current phasor and the voltage phasor and, because of the standing waves, all three parameters vary with location on the feedline. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On 6/24/2011 7:41 PM, Cecil Moore wrote:
On Jun 24, 2:20 pm, John wrote: If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. I disagree. There are 100W supplied by the source and 100W consumed by the load. There are 200W in the 291.4 ohm line. 100W of that is just "passing through". The other 100W is circulating, that is, stored energy which was put there by the start-up transient. If it is circulating, then it must be reflected from each end of the 291.4 ohm line. Let's assume that the 100 watts is just "passing through". It would change from 70.7 volts in the 50 ohm environment to 170.7 volts in the 291.4 ohm environment. The reflected power is 100 watts so the reflected voltage is also 170.7 volts. Those two voltages would have to add together to get the forward voltage. The forward voltage is known to be 241.4 volts. Exactly how do you add two 170.7 volt in- phase waves to get a total of 241.4 volts? Here is actually what happens. Since the physical power reflection coefficient at the Z0-match point is 0.5, only 50 watts of the source power makes it through the impedance discontinuity. That voltage is 120.7 volts. The same thing applies to the reflected power - only 50 watts is re-reflected by the 0.5 power reflection coefficient. So the re-reflected voltage is also 120.7 volts. Adding those two voltages together yields 241.4 volts which we know is the correct forward voltage. Now you are going to ask how two 50 watt waves can add up to 200 watts forward power. That's just the nature of constructive interference since power is proportional to voltage squared. If we add two 50 watt waves in phase, we get a 200 watt wave. Where did the extra 100 watts come from? Why, from the 100 watts of destructive interference toward the source that eliminated the reflections toward the source. -- 73, Cecil, w5dxp.com If the reflections toward the source is eliminated, how is it that it appears to be 50 ohms at that point rather than 291.4 ohms? John |
Reflection coefficient for total re-reflection
On Jun 25, 12:02*pm, John S wrote:
If the reflections toward the source is eliminated, how is it that it appears to be 50 ohms at that point rather than 291.4 ohms? You answered your own question - if reflections toward the source are eliminated in a Z0=50 ohm environment, the apparent (virtual) impedance cannot be anything except 50 ohms. When you look at yourself in the mirror, your reflected apparent (virtual) distance behind the mirror is the same as your actual distance from the mirror. The reflection is NOT where it appears to be, i.e. it is virtual. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On 6/25/2011 12:41 PM, Cecil Moore wrote:
On Jun 25, 12:02 pm, John wrote: If the reflections toward the source is eliminated, how is it that it appears to be 50 ohms at that point rather than 291.4 ohms? You answered your own question - if reflections toward the source are eliminated in a Z0=50 ohm environment, the apparent (virtual) impedance cannot be anything except 50 ohms You said "Since a virtual impedance is result of the superposition of a forward wave and a reflected wave, a virtual impedance cannot re-reflect the reflected wave, i.e. one cannot re-reflect the reflected wave while at the same time the reflected wave is being used to generate an impedance." But, it does. First, it causes the 50 ohms line (looking into the 291.4 ohms line to see a match due to the reflection. Second, the re-reflection from that discontinuity is half of what maintains the circulating energy on the line. The other half is the discontinuity of the non-virtual load. |
Reflection coefficient for total re-reflection
On Jun 23, 12:50*am, John Smith wrote:
On 6/22/2011 3:24 PM, dave wrote: On Jun 22, 1:49 pm, Cecil *wrote: Let's return to an earlier example and compare a single-port analysis with a dual-port analysis. 100w source--50 ohm--+--1/2WL 291.4 ohm--50 ohm load The 50 ohm Z0-match point is at '+'. The forward power on the 50 ohm line is 100 watts and the reflected power on the 50 ohm line is zero watts. The forward power on the 291.4 ohm line is 200 watts and the reflected power on the 291.4 ohm line is 100 watts. 100 watts is being sourced and delivered to the 50 ohm load. The voltage reflection coefficient, rho, at the load is (50-291.4)/ (50+291.4)=0.7071. The power reflection coefficient, rho^2, at the load is 0.5, i.e. half of the power incident upon the load (200w) is reflected (100w). Since the load is a single-port, these parameters are consistent with a single-port analysis. In a single-port analysis, we cannot tell the difference between a virtual reflection coefficient and a physical reflection coefficient. The problem comes when we use a single-port analysis on the Z0-match point. Since the reflected power on the 50 ohm line is zero, a single- port analysis would yield rho=0.0 and rho^2=0.0 when viewing the Z0- match from the source side. When we perform a dual-port analysis, we get different values for rho and rho^2, i.e. we get the complement of the reflection coefficients at the load which is a characteristic of any simple Z0-match similar to the above example. For a dual-port analysis, rho looking into the Z0-match from the source side is (291.4-50)/(291.4+50)=0.7071 and rho^2 looking into the Z0-match from the source side is 0.5, the same as at the load. Looking back into the Z0-match from the load side, the sign of rho is negative just as it is at the load with rho^2=0.5, the same as at the load. Since the two analyses yield different values for the reflection coefficients, which analysis is correct? The answer gives the clue to the resolution of this discussion. -- 73, Cecil, w5dxp.com ok, i'm afraid i'm going to have to ask the simple question... if you blackbox the load and stub and look at just the one connection to it and that gives you no reflected power... where do you define the second port, and why? Logic, immediately, suggests to me, that varying the frequency and measuring voltage, amperage, and SWR would begin to immediately point the answer(s.) Regards, JS but you can't do that! they don't like real measurements, only arguing about their virtual reflections and the separate currents and voltages. |
Reflection coefficient for total re-reflection
On Jun 25, 1:18*pm, Cecil Moore wrote:
On Jun 25, 5:53*am, dave wrote: try it! *you will read the exact same voltage on either side of that connection point! I already told you that only applies to the total voltage and total current. You will NOT read the same forward voltage on either side, you will NOT read the same forward current on either side, you will NOT read the same reflected voltage on either side, and you will NOT read the same reflected current on either side. when someone gives me a voltmeter i can touch to that connection point and measure the 4 components then we can talk. as far as designing anything i need i can do it without giving those s parameters or your 4 components a second thought. it is very easy to transform and combine the impedances to tell me what the load seen by the transmitter is, or to figure out the needed stub for providing a proper match without all that stuff. |
Reflection coefficient for total re-reflection
On Jun 25, 5:25*pm, John S wrote:
But, it does. First, it causes the 50 ohms line (looking into the 291.4 ohms line to see a match due to the reflection. Second, the re-reflection from that discontinuity is half of what maintains the circulating energy on the line. The other half is the discontinuity of the non-virtual load. You are confusing reflection with wave cancellation (destructive interference). I suggest that you study the separate sections on reflections vs interference in "Optics", by Hecht. Nowhere does any optical textbook indicate that superposition and reflection are the same thing (and they are indeed NOT the same thing). Superposition/interference applies to two or more waves. Reflection applies to a single wave. When a reflected wave is re-reflected, it is always a single wave event. Take a look at one of the s-parameter equations: b1 = s11(a1) + s12(a2) = 0 b1 is the reflected voltage toward the source which is, of course, zero when looking into a Z0-match. s11(a1) is the forward source wave reflected from the physical impedance discontinuity at the Z0-match point. Note: This is a single reflected component of a single forward wave. s12(a2) is the part of the reflected wave from the load that is transmitted back through the impedance discontinuity at the Z0-match point. Note: This is a single transmitted component of a single reflected wave. Reflections are now over and done with. What happens next is superposition/interference, i.e. the phasor addition of the two component reflections. Note: The results of the phasor addition of two component waves is NOT a reflection!!! The zero result of the addition of those two phasors is associated with destructive interference toward the load. That's what causes the 50 ohm Z0-match, not the component reflections. The redistribution of the destructive interference energy back toward the load is NOT a re- reflection and here's why: A re-reflection preserves the modulation content of the re-reflected wave which can be proved by TV ghosting experiments. Wave cancellation due to destructive interference does NOT preserve the modulation. In fact, any differences in modulation between the two superposed wave components would wind up incident upon the source. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 26, 6:43*am, dave wrote:
when someone gives me a voltmeter i can touch to that connection point and measure the 4 components then we can talk. I could design an expensive device that will do exactly that but it is a lot easier to just calculate the values using Mathcad. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On 6/29/2011 11:04 AM, Cecil Moore wrote:
On Jun 25, 5:25 pm, John wrote: But, it does. First, it causes the 50 ohms line (looking into the 291.4 ohms line to see a match due to the reflection. Second, the re-reflection from that discontinuity is half of what maintains the circulating energy on the line. The other half is the discontinuity of the non-virtual load. You are confusing reflection with wave cancellation (destructive interference). I suggest that you study the separate sections on reflections vs interference in "Optics", by Hecht. Nowhere does any optical textbook indicate that superposition and reflection are the same thing (and they are indeed NOT the same thing). You always fall back on the optics thing, don't you? |
Reflection coefficient for total re-reflection
On Jun 24, 1:52*pm, Cecil Moore wrote:
If the reflected wave is re-reflected, it must be by an impedance other than the virtual impedance generated by the reflected wave itself. If the reflected wave is being used to generate a virtual impedance, it cannot at the same time be being re-reflected. I think i have finally seen the confusion point. you can not at the same time talk about reflected waves and re-reflected waves and also use s parameters or other steady state reflection parameters or impedances... this means you can not talk about individual reflections from the stub connection point without also analyzing the traveling wave inside the stub, nor can you track and sum up all the reflections using the steady state impedance of the stub as if it were a lumped value. to do so violates the conditions where the s parameters and reflection coefficients are valid, the sinusoidal steady state realm. |
Reflection coefficient for total re-reflection
On Jun 25, 5:25*pm, John S wrote:
But, it does. First, it causes the 50 ohms line (looking into the 291.4 ohms line to see a match due to the reflection. Second, the re-reflection from that discontinuity is half of what maintains the circulating energy on the line. The other half is the discontinuity of the non-virtual load. John, EM wave reflection is what happens to a single wave. EM wave superposition/interference is what happens between two (or more) waves. They are not the same phenomenon and do NOT obey the same rules of physics. Wave cancellation between two waves is *NOT a reflection* but it does *redistribute* reflected energy back toward the load at the Z0-match point. The destructive interference energy toward the source is redistributed as constructive interference energy toward the load but one can tell from the resulting phase and magnitude that it was NOT a reflection. A good reference on the differences between wave reflection and wave superposition/interference is "Optics", by Eugene Hecht. The international 4th edition is available in paperback for around $20 from Abebooks.com. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 29, 2:13*pm, John S wrote:
You always fall back on the optics thing, don't you? That's because optical physicists seem to be the only technical people who understand interference effects. However, most of what one needs to understand about the subject is contained in my Worldradio energy article: http://www.w5dxp.com/energy.htm -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jun 29, 5:41*pm, K1TTT wrote:
I think i have finally seen the confusion point. you can not at the same time talk about reflected waves and re-reflected waves and also use s parameters or other steady state reflection parameters or impedances... Someone said it a long time ago: You can't have your cake and eat it too. One cannot be using the reflected wave to establish a dynamic virtual short while, at the same time, the virtual short is causing a dynamic re-reflection of the reflected wave. The virtual impedance is (Vfor+Vref)/(Ifor+Iref) but if that is the impedance doing the re-reflection of that same Vref and Iref, then it cannot also be being used to establish the dynamic virtual impedance, i.e. if a virtual impedance is re-reflecting all of the reflected energy, it will necessarily disappear from existence because there is no longer a Vref and Iref to cause it to exist. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
On Jul 1, 7:56*am, W5XP wrote:
My apologies to JAMES W GRIFFITH, W5XP. I apparently made a typo and left out a 'D' from my call. -- 73, Cecil, w5dxp.com |
Reflection coefficient for total re-reflection
"W5DXP" wrote in message ... On Jul 1, 7:56 am, W5XP wrote: My apologies to JAMES W GRIFFITH, W5XP. I apparently made a typo and left out a 'D' from my call. -- 73, Cecil, w5dxp.com **************** - LOL....I saw the callsign, knew something was wrong, but didn't know what. ......that happens at my age --Wayne W5GIE (exiled to W6) |
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