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The "maximum power transfer theorem" states that "maximum power transfer occurs when the load impedance is equal to the complex conjugate of the source impedance", that "for two impedances to be complex conjugates their resistances must be equal, and their reactances must be equal in magnitude but of opposite signs. " I understand reflections occur when impedance is not matched (short and open being the extremes). But, in "maximum power transfer theorem", why does reactance have to be matched and resistance have to be matched, rather than just impedance (why cannot increase or decrease in resistive impedance compensate for more or less reactive impedance on either side, and vice versa)?

Edit: To clarify a bit, my understanding of reflections is that impedance has to be matched, X Ohm on one side and X Ohm on the other. I have not understood it as that reactance has to be neutralized and then resistance matched. To my understanding, 10 Ohm resistance only vs 10 Ohm reactance only, should be "impedance matched" (from literal sense of term "impedance"). I might miss something about that "impedance matching" treats resistance and reactance differently, or, maybe only the "maximum power transfer theorem" does.

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But, in "maximum power transfer theorem", why does reactance have to be matched and resistance have to be matched, rather than just impedance

It doesn't say that.

For maximum power transfer, the reactive impedance magnitudes need to be matched but, they need to have opposite signs to each other. This means series resonance and, of course maximum power transfer.

I mention series resonance because, an inductor in series with a capacitor forms a short circuit when their impedances are equal in magnitude. And, of course, the impedance polarity of an inductor is the negative of a capacitor. Thus, their impedances cancel to zero.

If you just matched the impedances, it would not be a case of maximum power transfer because, the two reactive components would be exactly equal and their net impedance would double (rather than cancel to zero when conjugate impedances are used).


From a comment by the OP it becomes clearer where the misconception arises: -

My understanding of reflections is that impedance has to be matched, X Ohm on one side and X Ohm on the other. I have not understood it as that reactance has to be neutralized and then resistance matched.

Strictly speaking, if a cable has a characteristic impedance that is complex then, to avoid reflections, it needs to be terminated in that same complex impedance. However, cables don't have a complex characteristic impedance at RF; they have a purely resistive impedance and, need to be terminated in a pure resistance.

So, the main reflection-causing component is the connecting cable. This is because cables create significant signal delay over their length and, if significantly mismatched, will create problematic RF reflections.

But, I'll also mention antennas. Some antennas (such as an ideal \$\frac{\lambda}{4}\$monopole) present an impedance at their electrical terminals of 36.5 + j21.25 Ω i.e. it's got reactance already "built in". However, for best operation of the antenna, we want to neutralize that reactance with a conjugate reactance. And, we get the added benefit of a bit more tuning selectivity due to series resonance.

Going back to cables; at pretty much all frequencies from 100 kHz to over 1 GHz we can reasonably assume that the characteristic impedance is \$\sqrt{L/C}\$ and, when you do the dimensional analysis on it, it is purely resistive. So, we want to match a cable (at one end or the other) with a purely resistive termination in order to prevent or dissipate reflections.

As I said earlier, if a cable has a complex characteristic impedance then, we'd have to terminate the cable in that identical impedance but, at RF, it doesn't; it's resistive.

Then, if you consider the chip/IC that drives a length of cable; the chip might have a complex output impedance but, the chip itself won't care if you cancel out it's complex reactance with a conjugate impedance. And anyway, you have to do this to ensure you produce a purely resistive impedance at the cable interface.

So, it's all down to the cable and the antenna. We want the cable to be terminated by resistance at the "chip end" and, we want the antenna to look like resistance at the other end. Hence, we cancel reactances at the cable interfaces and, avoid reflections.

As mentioned earlier, if a cable has a complex characteristic impedance then, we'd have to drive the cable with that identical impedance but, it doesn't; it's resistive at RF and therefore, the chip impedance driving it has to be be "manipulated" to obtain a purely resistive drive impedance.

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  • \$\begingroup\$ I know they should be opposite, cancel out, and in resonance. So, matched. And, all impedance is then in the resistance. What I do not see is that total impedance is minimized that way, compared to matching just total impedance. R = 10 Ohm, Z = 0 Ohm, R=0 Ohm, Z = 10 Ohm for example. The total Ohm becomes 20 if just impedance is matched, compared to both reactance and resistance. \$\endgroup\$
    – BipedalJoe
    May 3, 2023 at 15:37
  • \$\begingroup\$ I'm not following the argument you put forward in that comment @BipedalJoe <-- maybe you are considering not the power transfer but, the maximum power dissipation in the source impedance. Remember we are trying to send power to a load from a source that has some resistance and, we are not interested in maximizing the power dissipated by the source. \$\endgroup\$
    – Andy aka
    May 3, 2023 at 15:43
  • \$\begingroup\$ Well you don't have to. But, it is what I ask about. Answers from anyone is welcome. \$\endgroup\$
    – BipedalJoe
    May 3, 2023 at 15:46
  • \$\begingroup\$ Please read my full comment. \$\endgroup\$
    – Andy aka
    May 3, 2023 at 15:47
  • \$\begingroup\$ My understanding of reflections is that impedance has to be matched, X Ohm on one side and X Ohm on the other. I have not understood it as that reactance has to be neutralized and then resistance matched. To my understanding, 10 Ohm resistance only vs 10 Ohm reactance only, should be "impedance matched". I might miss something about that "impedance matching" treats resistance and reactance differently, or, only the "maximum power transfer theorem" does. \$\endgroup\$
    – BipedalJoe
    May 3, 2023 at 15:53
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why does reactance have to be matched and resistance have to be matched, rather than just impedance

To my understanding, 10 Ohm resistance only vs 10 Ohm reactance only, should be "impedance matched" (from literal sense of term "impedance").

The literal sense of impedance ' in electronics" has a phase angle that cannot be ignored.

Impedance is the sum of resistance and reactance. I think that what you are calling impedance is what is actually called is magnitude.

Impedance is $$Z=R +jX=|Z|e^{j\theta}$$ and the magnitude of impedance is $$|Z|=\sqrt{R^2+X^2}$$

I have not understood it as that reactance has to be neutralized and then resistance matched.

The reactance in a source stores and releases energy to the circuit on a cyclical basis. The load reactance also stores and releases energy to the circuit on a cyclical basis.

If the load and source reactances are not complex conjugates, then the cyclic transfer of energy between them is unbalanced. When balanced this transfer of energy between the load and source is constant, but does not contribute the the load energy.

If unbalanced, the excess energy is radiated as electromagnetic energy. This appears as a resistance in parallel with the load thus requiring more power from the source. This is noticeable as a standing wave on transmission lines.

So maximum power transfer for complex systems includes the reduction of the radiated energy.

Circuits with no transmission line will demonstrate higher electronic emissions when the load is not "matched" with complex conjugate.

So the resistance matching is compensating to distribute \$I_2R\$ losses. The reactive matching is reducing the "other" radiative load resistance that we just can't see.

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  • \$\begingroup\$ Thanks. To clear up any misunderstanding I have: the same thing applies to normal reflection theory too, and reflection coefficient, such that (ZL-ZS)/(ZL+ZS), is not true if just the Ohm value is taken? If ZL is 100 Ohm purely resistive, and ZS is 100 Ohm purely reactive, then reflection coefficient is not 0? \$\endgroup\$
    – BipedalJoe
    May 3, 2023 at 17:34
  • \$\begingroup\$ The “Ohm value “ is just the units. Resistive, reactive, or impedance, the Ohm value has the same meaning. The reflection coefficient is not zero. \$\endgroup\$
    – RussellH
    May 3, 2023 at 17:58
  • \$\begingroup\$ Impedance is complex. Trying to sidestep will get you off track. \$\endgroup\$
    – RussellH
    May 3, 2023 at 17:59
  • \$\begingroup\$ I'm not trying to side-step it. I know it is complex. Assuming source & load resistance behaves as series resistance, then R_load = 1/sqrt(2), R_source = sqrt(2)-1, Z_load = j, Z_source = 0j, matches the Ohm for Z_source & Z_load. Assuming source & load resistance can be treated independently, R_source = sqrt(2), R_load=1, Z_load =j, Z_source = 0j, matches impedance, sqrt(2) Ohm on both source and load. Is there any way to pin point to exactly why reflections would still be a problem, if the impedance is matched in Ohm, but not necessarily matching the reatance and resistance separately? \$\endgroup\$
    – BipedalJoe
    May 4, 2023 at 6:07
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On the question about reflections: I mentioned in an answer to the other post (link), I think it is that ability for reactive impedance to "resist or conduct current" is time varying. So while it may be equal to a resistive impedance over time, it is not equal at the same time. So, a signal reaching the end of a transmission line and a load, will still not be matched in "Ohm" at that exact moment. Then, for "maximum power transfer" (as question was about that as well), the reactance of the source happens to be removable, while the resistance of source cannot be removed, and lowering series resistance through circuit increases power (and, as voltage has to be dropped over source, maximum power is achieved when exactly half is dropped, since more means less power into load, and less means more resistance in series. ) So in that case, the reactance is removed simply because it is removable, and the decrease of "phase-based reflections" is probably secondary.

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    \$\begingroup\$ This honestly just seems wrong. The maximum per transfer theorem is as stated in your question but describing magnitudes, and your answer is not compatible with it. The words you string together here so not really reflect how linear high frequency circuits work. \$\endgroup\$ May 8, 2023 at 23:38
  • \$\begingroup\$ It's a good answer to the question. I asked the question because I'd learnt about reflections prior to maximum power transfer theorem. In reflections, my impression was magnitude had to be matched, not phase and magnitude. But, since reactive impedance is time-varying, I can see why it is also important to match phase. I mention both, since what motivated the question, assumed the prior (reflections), and if anyone else had a similar question they would possibly come from similar place. \$\endgroup\$
    – BipedalJoe
    May 8, 2023 at 23:41
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    \$\begingroup\$ It is not a good answer. It is plain wrong. No, reactive impedance is not time-varying. \$\endgroup\$ May 8, 2023 at 23:42
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    \$\begingroup\$ You are plain wrong. Saying "I'm not looking for an argument" in that situation just means you don't want to face that you're misunderstanding what you're talking about. Misunderstanding is no shame - it takes a lot of work to learn these things - but insisting ones poorly constructed view of the world is equally valid as you know, physics, that's a shame. \$\endgroup\$ May 8, 2023 at 23:50
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    \$\begingroup\$ The use of the term reactive impedance is for when considering steady-state sinusoidal signals. It is a meaningless quantity when talking about transient signals of voltage and current. Your use of the term reactive impedance is incorrect; you are wrong to say that reactive impedance is time-varying. Just plain wrong. \$\endgroup\$
    – Andy aka
    May 10, 2023 at 9:35

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