The Dark Side of the Conjugate Match

Discussion in 'General Technical Questions and Answers' started by KL7AJ, Mar 12, 2010.

Thread Status:
Not open for further replies.
ad: L-HROutlet
ad: l-rl
ad: FBNews-1
ad: Subscribe
ad: L-MFJ
ad: Left-3
ad: Left-2
  1. W5DXP

    W5DXP Ham Member QRZ Page

    Consider the following tube transmitter with an adjustable pi-net output. It will be easier to understand if we assume class-A operation.

    [​IMG]

    Assume the tube has a preferred load-line, the one for which it was designed. Let's call that load-line a Zg-match. Let's call any other load-line a Zg-mismatch.

    For the sake of simplicity, let's assume ideal conditions. At key-down, the tube sends a signal from point 'X' to point 'Y' which takes one nanosecond to get there. If the impedance at point 'Y' equals Zg-match, there will be no reflections from point 'Y' back to point 'X' and the tube continues with the Zg-match load-line. No feedback from point 'Y' causes the tube to continue with the same Zg-match load-line.

    If the impedance at point 'Y' is not equal Zg-match, how does the tube know that fact? It can only know what the impedance at point 'Y' is if the tube gets some sort of feedback from point 'Y' and that feedback cannot reach point 'X' for another one nanosecond. What is the nature of the feedback that tells the tube that point 'Y' is not equal to Zg-match and is instead equal to Zg-mismatch so that the load-line can be changed from Zg-match to Zg-mismatch? What else could it possibly be except reflected voltage/current? Please be specific.

    I am surprised that the lumped-circuit model has caused so many otherwise intelligent people to completely miss this simple concept.
     
  2. W5DXP

    W5DXP Ham Member QRZ Page

    OK, given the location of that impedance, how is the magnitude/phase of that impedance communicated over the physical distance back to the tube at no faster than the speed of light so that the tube can adjust its load-line? Please be specific.

    You misunderstand what I have said. I didn't say all of the reflected energy reaches the source under mismatched conditions, just that enough reflected energy reaches the source under mismatched conditions to cause the load-line to change. Your above statement is simply false. The SWR on the line is 100% dictated by the Z0 of the line and the Z of the load. Whether reflected energy is incident upon the source or not has ZERO effect on the SWR on the transmission line.

    rho = (Zload-Z0)/(Zload+Z0)

    SWR = (1-rho)/(1+rho)

    Absolutely no mention of a source or power levels.

    OTOH, if zero reflections reach the source, as you assert, there is nothing to cause a change in the load-line away from the matched condition, but we know a change in the load-line happens as a result of a mismatch. So exactly how does the source know to change its load-line from matched to mismatched if it gets zero feedback from the mismatched load? Please be specific.

    I agree and nothing that I have said contradicts your statement above. Given all of that happening, there is nothing to prohibit some steady-state reflected energy from reaching the source and affecting the load-line. Exactly what else can change the load-line besides reflected voltage/current? How else does the source know to change the load-line besides reflected voltage/current traveling at the speed of light back from the mismatched load which is some distance away. Are you suggesting "spooky action at a distance" or magic or what?

    I agree and you continue to miss the point. For the tube load-line to make those discrete changes, it must have feedback from the changing impedance that it is driving. Otherwise the load-line wouldn't change. That impedance is some distance away. For the source to even know that the impedance is changing, some sort of feedback must occur from the changing impedance back to the load-line. Exactly what is the nature of the feedback from the changing impedance located some distance away from the load-line? Given that feedback is necessarily photonic in nature, i.e. limited to the speed of light, what could it possibly be except reflected voltage/current? Are you introducing a brand new phenomenon in the field of physics? If so, why?

    Tim, you and others are going to laugh at yourselves when this very simple concept finally soaks in. Let me say it again. In my schematic example, the V/I ratio (impedance) at point 'Y' changes one nanosecond before point 'X' can possibly detect that V/I change in impedance. What is the nature of the feedback from the new V/I at point 'Y' back to point 'X' that takes 1ns and allows (causes?) V/I to change at point 'X'? Please show your equations.

    At t0 at point 'Y', Vy/Iy changes from Vold/Iold to Vnew/Inew

    At t0 at point 'X', Vx/Ix equals Vold/Iold

    At t0+1ns at point 'X', Vx/Ix changes from Vold/Iold to Vnew/Inew

    It obviously took 1ns for the feedback involving the V/I change at point 'Y' to reach point 'X'. What is the nature of the feedback that travels at the speed of light from Point 'Y' back to Point 'X' in 1 ns?
     
    Last edited: Sep 22, 2010
  3. AB0WR

    AB0WR Ham Member QRZ Page

    Cecil,

    1. You've changed the argument. Move Point Y to be at the same place you have marked as the "Load". The discussion has been as to whether the reflection from the load traverses the tank circuit. You are now trying to argue as to whether there is a reflection at Point "Y" from the driving source signal. That's a totally different argument.

    2. You didn't answer my question at all.

    If a reflected wave that is incident at the point you have labeled as the "Load" traverses the tank circuit to Point "X" and gets changed by the intervening tank circuit then how does the impedance at Point "Load" ever become representative of the actual impedance at the far end of the transmission line? How does the SWR on the transmission line ever even approach what theory says it should? How does a bounce diagram for the transmission line ever begin to give the right answers?

    I am beginning to suspect that I will never get an answer from you about this question. Trying to reconcile actual SWR values on the transmission line with your claim that the reflection from the load travels all the way to the tube *DOES* seem to be an impossible task.

    I think Walt has explained how this all works, maybe in 2004? If there is a complex conjugate match at Point "Load" then there has to be total re-reflection of the wave returning from the mismatched load or otherwise nothing works right. This is a totally separate issue from what happens between the tube and the tank.
     
  4. AB0WR

    AB0WR Ham Member QRZ Page

    "OK, given the location of that impedance, how is the magnitude/phase of that impedance communicated over the physical distance back to the tube at no faster than the speed of light so that the tube can adjust its load-line? Please be specific."

    Cecil,

    You keep trying to force this False Dilemma as if it *means* something.

    The load line is based on impedance seen by the tube from the matching network. If the impedance presented by the network changes because a change is needed to match a different impedance presented by the transmission line then the tube sees that different impedance.

    That will affect the conditions seen between the tube and the tank but it doesn't mean that the reflected waves from the changed load are traveling all the way to the tube.

    tim ab0wr
     
  5. AB0WR

    AB0WR Ham Member QRZ Page

    "You misunderstand what I have said. I didn't say all of the reflected energy reaches the source under mismatched conditions, just that enough reflected energy reaches the source under mismatched conditions to cause the load-line to change. Your above statement is simply false. The SWR on the line is 100% dictated by the Z0 of the line and the Z of the load. Whether reflected energy is incident upon the source or not has ZERO effect on the SWR on the transmission line."

    Oh, give me a break.

    The reflection coefficient can also be calculated from the impedances and the reflection coefficient can be used to calculate the VSWR.

    The VSWR is certainly dependent upon the reflected energy -- by definition. That means that the SWR is *also* dependent on the reflected energy.
     
  6. W5DXP

    W5DXP Ham Member QRZ Page

    I'm working on an article for my web page which will contain the following information.

    We hams and other RF types are not free to invent rules for amateur radio that violate the known laws of physics. If we have shortcuts and models that violate the known laws of physics, we must acknowledge that fact (or bury our heads in the sand). We have seen such responses as: "It doesn't matter" and "It's a waste of time". Consider the following generalized case:

    Source-------some distance 'L'----------Load

    The generalized load doesn't have to be an antenna - the above load seen by the source could be the tank circuit. Any location at which an impedance discontinuity exists can be considered to be the "Load". The following discussion will be about how RF energy is transferred from one place to another.

    RF energy, which travels at the speed of light, is known to consist of photons. This applies to RF fields and waves. There is no other method of transferring RF energy from one place to another at the speed of light except through photonic fields and waves. The free electron carriers move much too slowly to transfer energy at the speed of light so the photons win by process of elimination.

    Since the source and load cannot occupy the same space, they must necessarily be located some distance, L, apart. With photons traveling at the speed of light (in the medium) that associates a delay time, delta-t, with the time it takes for a photon to travel from the source to the load. The magnitude of the separation distance and delay time doesn't matter. This is a general discussion about the concepts involved in RF energy transfer.

    Note that L/delta-t always equals the speed of light in the medium and that neither L nor delta-t can ever be zero. (If L and delta-t were ever zero, the term L/delta-t would be undefined).

    Ideally, the source initiates a forward traveling field/wave that is incident upon the load delta-t later. If the load accepts all of the RF energy, no reflections occur and there is zero RF energy flowing back toward the source. The source "assumes" that the load is accepting all of the RF energy because there are no reflections.

    If the load does not accept all of the forward RF energy, some energy will be reflected back toward the source. However, the source will not "know" that the load is rejecting energy until 2 delta-t times have passed (the time it takes the forward wave to make a round trip to the load and back). The reflected energy incident upon the source is in the form of a reflected RF field/wave again consisting of photons. The reflected voltage/current superposes with the load-line in the source to create a new load-line.

    For purposes of discussion, the original load-line can be considered to be the default matched load-line while the new load-line can be considered to be the mismatched load-line. In any case, any deviation from the default (matched) load-line is caused by reflected energy no matter what is the length L because we know that L cannot equal zero.

    There is no other known mechanism in the field of physics that can cause a change in the default load-line besides reflected energy feedback from the load.

    It is obvious that some folks on this newsgroup believe that RF energy can travel faster than the speed of light because that's one of the presuppositions of the lumped-circuit model but that presupposition is known to be false and a violation of the speed of light limit. There is always a delay between the source's initial forward wavefront and the feedback information from the load. The feedback information must necessarily involve reflected energy because there is no other method known to physics for obtaining that feedback information, especially at faster than light speeds as required by the lumped-circuit model. The source simply cannot "know" that the load has changed at faster than light speeds.
     
  7. W5DXP

    W5DXP Ham Member QRZ Page

    When one detunes the network, one destroys the conjugate match and allows some reflected energy, that was previously 100% re-reflected back toward the load, to become incident upon the source. The load-line cannot tell the difference because there is no difference and exactly the same incident reflected phasors are incident upon the source in both cases.

    If it's not the reflected phasors that are incident upon the source, exactly what mechanism of physics does the detuned network use to tell the source to change its load-line? What kind of energy exists besides the forward fields/waves and the reflected fields/waves?
     
  8. KD8GFC

    KD8GFC Ham Member QRZ Page

    Why dont you guys agree to disagree this is going NOWHERE !!!
     
  9. W8JI

    W8JI Ham Member QRZ Page

    Just don't read it.
     
  10. KD8GFC

    KD8GFC Ham Member QRZ Page

    Well OK....... :eek:
     
Thread Status:
Not open for further replies.

Share This Page