ad: ProAudio-1

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: Subscribe
ad: L-MFJ
ad: abrind-2
ad: Left-3
ad: Left-2
  1. W5DXP

    W5DXP Ham Member QRZ Page

    Walt, I agree with your first sentence above - so you appear not to understand my position. Here it is again for the Nth time.

    1. Matched systems - If the source load-line is operating at the designed-for slope, the system is matched and ZERO reflected energy is reaching/invading the source, i.e. reflected energy is blocked from reaching the source. This statement includes all systems whose matching networks have been properly adjusted for a Zg-match, a Z0-match, or a conjugate match. Example: A source designed to drive a 50 ohm load driving 1/2WL of 50 ohm coax terminated with a 50 ohm resistor.

    2. Mismatched systems - If the load-line is operating relatively far away from the designed-for slope (away from the matched condition), the system is mismatched - reflected voltage/current is reaching the source and superposing with the source voltage and current. The load-line only changes when feedback to the source is present in the form of reflected energy incident upon the source. Reflected voltage/current is the only thing that can change the load line. Example: A source designed to drive a 50 ohm load driving 1/2WL of 50 ohm coax terminated with a 600 ohm resistor.

    In a matched system, reflected energy is blocked from being incident upon the source. That is the purpose of matching. The source load-line assumes the designed-for slope (or relatively close to it). This is the most likely case for amateur radio systems.

    In a mismatched system, reflected energy is incident upon the source. The subsequent superposition of source signal and reflected signal changes the source load-line relatively far away from the designed-for slope.

    My main point: Reflected energy incident upon the source is the only thing that can change the load-line away from the designed-for slope. There is no other way for the source to sense a mismatch except for reflected energy feedback to occur. The feedback from the load to the source cannot travel faster than the speed of light, i.e. the feedback cannot take zero time.

    Three things can happen when reflected energy is incident upon the source.
    1. The reflected signal can be re-reflected at the source back toward the load.
    2. The reflected signal can undergo destructive interference at the source and be re-distributed back toward the load.
    3. The reflected signal can be attenuated (dissipated).

    What happens to the reflected energy is dictated by the magnitude of the above three phenomena all of which must add up to 100% of the reflected energy. By adjusting the circuit design and magnitude/phase of the reflections, it is possible to maximize or minimize any one or two of the above three possibilities.

    A special case of interference is when the reflected wave arrives 90 degrees out of phase with the source wave. Since cos(90) = 0, there is no interference. If there are also no reflections (Z0=Zsource) then 100% of the reflected power is dissipated in the source resistor (Source configuration supplied by W7EL). Such a special case is discussed in the following paper:

    http://www.w5dxp.com/nointfr.htm
     
  2. W5DXP

    W5DXP Ham Member QRZ Page

    Tim, I don't have time to educate you on Fourier Analysis (through which you apparently slept) so I will recommend one of my college textbooks as a good reference:

    Higher Mathematics for Engineers and Physicists, by Sokolnikoff.

    Simply put, the partial cycle breaks up into the fundamental frequency which superposes with the rest of the fundamental frequency energy and most likely becomes unmeasurable. It might cause a short transient blip if it is large enough and makes it to the source. The harmonic frequencies are attenuated by the low-pass filtering in the output network and probably never make it to the source.

    No, and your irrelevant attempts to divert the main issue are noted. It's the lumped-element circuit model that requires magical thinking. Again quoting Dr. Corum:

    "The failure of any lumped element circuit model to describe the real world lies at its core inherent presupposition: the speed of light is assumed infinite in the wave equation (all regions of the universe can be communicated with instantaneously)."

    The fact that the load-line changes under steady-state mismatched conditions is proof that reflected voltage/current has made it to the source. There is no other way for the source to detect a mismatch.
     
  3. AB0WR

    AB0WR Ham Member QRZ Page

    Nice ad hominem.

    BTW, the Fourier Transform is between the time domain and the frequency domain. So even you are reduced to "lumped circuit" theory in order to provide an answer. You just throw out the distance part of the wave equation because it is convenient.

    Nice.

    "Unmeasurable"? That's magic right there!! Why would it become unmeasurable? Wouldn't that depend on the phase? It could double or cancel based on superposition or anything in between!

    First you want us to believe that the voltage across a capacitor can never reach the source voltage because e^(-t) requires infinite time to reach zero and then you try to use the vague term "unmeasurable"?

    And what in Pete's name is a short "blip"? Who doesn't understand Fourier Analysis? A short blip *has* to have higher order components in order to *be* a blip! So how does the "blip" get through the filtering?

    You are still waving your hands and saying "magic" with all this. It's nothing more than mumbo-jumbo using technical terms.

    Cecil, exactly what is the characteristic impedance of a coaxial transmission line when it is too short to exhibit the capacity required for a line of its characteristic impedance, i.e. (D/d)? Or, alternatively, what is the characteristic impedance of a coaxial transmission line when it is too short to have the inductance required for a line of its characteristic impedance, i.e. (D/d)?

    Is it still a transmission line? Or just a lumped component with a certain C and a certain L?

    You speak of unmeasurable and then try to use the fog of "infinite speed"?

    The wave equation is a function of time *and* space. I.e. it has terms dx *and* dt in it. When dV/dx becomes small enough that it approaches zero closely enough that the incremental difference can no longer be measured then *exactly* what do you have except an equation in terms of time with distance not involved?

    If you can't measure the difference then does it exist? You want to say that it *does* exist when it suits your argument and then turn around and say that it *does NOT* exist when it suits your argument!

    If the distance between the input capacitor of the tank and the tube is so short that you simply cannot measure any voltage difference based on the distance then how are you supposed to analyze the circuit except using just the time component?

    And then you have the hutzpah to speak of "steady-state"? In steady-state there *is no* information to communicate! So how does that have anything to do with anything?

    And you *still* haven't answered what happens to that reflected wave when it hits the tank circuit. All you've done is wave your hand and say abracadabra!
     
  4. W5DXP

    W5DXP Ham Member QRZ Page

    Of course it depends on the sophistication of the measurement equipment. Do you really think 100 ns of reflected power would show up on a Bird wattmeter? Good luck on that one.

    All that has to happen for reflected energy to reach the source during a transient condition is:

    (Vfor1)(rho1) <> (Vref2)(tau2), where '<>' means "not equal".

    When those two terms differ in amplitude or phase for any reason, total destructive interference cannot occur and reflected voltage gets through to the source. If you understand s-parameter equations, it is when

    s11(a1) <> s12(a2)

    If you don't understand that the presumption of faster than light speeds by the lumped-circuit model is magical thinking, please review the laws of physics.

    Sometime ago, a physicist worked out the length of coax required to establish the Z0 environment. It was less than one inch. It is a rare amp that has less than one inch of wire between the source and the output connector. A Bird wattmeter's truline is long enough to establish a 50 ohm environment.

    Since it is impossible for the source and load to occupy the same space, dv/dx cannot go to zero in the real world. The best we can do in the real world is delta-v/delta-x where x cannot go to zero and the speed of light is inviolate. That means no matter what the separation between the source and load is, it is never zero. What is the length of the path from the amp to the output connector in the average amplifier? Hint: no matter how much magic you try to introduce into the real world, it doesn't work.

    Correction: there is no change in information to communicate but there is continuous communication of information. No change is information. In a mismatched system, the forward wave is continuously communicating with the load and the reflected wave is continuously communicating with the source. A lack of communication would obviously be a change in information.

    Each component in the tank causes reflections so the situation gets too complicated to quantize. That's why I left the tank circuit out of my examples. What we can say is that if the tank circuit is not providing a match, reflected energy reaches the source. I believe that is what Walt said.
     
    Last edited: Sep 21, 2010
  5. AB0WR

    AB0WR Ham Member QRZ Page

    "Of course it depends on the sophistication of the measurement equipment. Do you really think 100 ns of reflected power would show up on a Bird wattmeter? Good luck on that one."

    "Since it is impossible for the source and load to occupy the same space, dv/dx cannot go to zero in the real world."

    I didn't say it went to zero. I said it got so close to zero that you couldn't measure it.

    You are still trying to argue that it is a measurement issue when it benefits you and that it isn't a measurement issue when it doesn't benefit you.

    Can *you* measure the voltage difference between two ends of a one-inch long piece of coax due to the *distance* part of the wave equation and not the time part of the equation?

    There isn't any time-domain reflectometer capable of differentiating such a small measurement.

    "Each component in the tank causes reflections so the situation gets too complicated to quantize."

    And *I* am the one invoking magic?

    You can't analyze the situation but we are supposed to believe *your* handwaving because it is *YOU* doing the hand waving?

    ROFL!!!

    tim ab0wr
     
  6. W5DXP

    W5DXP Ham Member QRZ Page

    Strange question considering that if time is taken into account, the lumped-circuit model falls apart.

    Strange that you attack me for exactly the same argument concerning reflected power measurements. Please make up your mind one way or another. You cannot be allowed to have it both ways.

    Even if the small measurement cannot be measured, it can be calculated. The phase shift in one inch of 0.66 VF coax at 4 MHz is ~0.185 degrees. I suppose that one million dollars of measurement equipment could measure such a change in phase. Do you want to pony up or should I? :)

    I didn't say it was too complicated to be conceptualized. If one wanted to spend an exorbitant amount of time and money proving a foolish notion to be wrong, it could be done. I just don't choose to waste my valuable time/money doing such. All of those foolish notions of yours are obviously false to the most casual observer.

    We are getting bogged down in minute unimportant details. The question is: Does reflected energy reach the source when the source is unmatched? The answer is yes (no matter what magical thinking you attempt to introduce while defending your favorite guru).
     
    Last edited: Sep 21, 2010
  7. KD8GFC

    KD8GFC Ham Member QRZ Page

    Im so tired of seeing this thread. WHO CARES ??? Lock it.
     
  8. W5DXP

    W5DXP Ham Member QRZ Page

    Consider that the source won't change its load-line unless it gets some sort of feedback yet we know that it does change its load-line in response to a mismatch. So we must ask: What is the nature of the feedback that causes the source's load-line to change because of a mismatch? Please consider that question and provide an answer that doesn't involve reflections.
     
    Last edited: Sep 22, 2010
  9. AB0WR

    AB0WR Ham Member QRZ Page

    W2DU,

    If the reflected wave was NOT reflected at the tank circuit then the impedance on the line would never represent the actual load impedance.

    The impedance on the line is developed from a *series* of reflections from the load end of the transmission line and re-reflections at the source end of transmission line. This is what the "bounce diagram" is meant to represent.

    If you never have this series of reflections and re-reflections then the actual load impedance never gets fully represented on the transmission line.

    It is this *impedance* of the transmission line caused by the reflections and re-reflections that has an effect on the tube load line.

    tim ab0wr
     
  10. AB0WR

    AB0WR Ham Member QRZ Page

    Cecil,

    "We are getting bogged down in minute unimportant details. The question is: Does reflected energy reach the source when the source is unmatched? The answer is yes (no matter what magical thinking you attempt to introduce while defending your favorite guru)."

    The answer is NO, it doesn't.

    See my latest post.

    If the reflection goes through the tank circuit and get modified in any way, as even you claimed that it would, then there is no way for the load impedance to affect the SWR on the transmission line as theory suggests.

    That SWR is the result of a never-ending series of reflections and re-reflections. If you interrupt that series of reflections and re-reflections then you will substantially change the SWR shown on the line from what it should be.

    The fact that the SWR *does* change as theory suggests shows that the series of reflections and re-reflections *does*, in fact, occur.

    That means that the re-reflections *must* occur at the tank circuit.

    If you could measure the tube load line quickly enough you would see it making discrete changes as the reflections and re-reflections change the impedance seen at the end of the transmission line.

    Think about it and then tell us how the SWR on a transmission line can be calculated based on the load if the reflections on the transmission line are changed by traversing the tank circuit and being changed by the traverse even if they are fully reflected at the tube instead of being dissapated?

    Are you suggesting that a wave traveling through a tank circuit is not impacted symmetrically? That it gets changed in one direction and restored in the other direction?

    Or is it just more "magic"?
     
Thread Status:
Not open for further replies.

Share This Page

ad: Amateur-1