It is well known that there are two usual methods in impedance matching: if we have a network consisting of a source with impedance \$Z_S\$ and load with impedance \$Z_L\$ then we have two options to match the the source to the load:

  1. If we want to minimize the power loss, then we do maximum power transfer. This requires the condition \$Z_S= Z^*_L\$

  2. If we want to minimize the wave reflection from the load we should realize the condition \$Z_S= Z_L\$

Question: I'm confused since my intuition leaves me here in the stick. Doesn't the reflected waves contribute to the amount of power loss? What is precisely the difference between two effects? Can the difference between them be demonstrated on a prefereably as simple as possible analogy?

Seemingly here fails the standard analogy for absorbed and reflected signals/waves given by sea waves and sandy beach. There, the energy transfered by the waves can be only absorbed by the sandy ground or reflected - we obtain reflected waves. There are no other ways the system can lose the energy. So in this simple analogy the non absorbed energy is exactly the reflected energy.

This analogy seemingy can be used to explain the case when we are dealing with network having only resistant components. Then there is to reactance that condition 1 is equivalent to condition 2). Lost power= power reflected by the load

What about the case when our network has nonzero reactance components. Obviously 1) and 2) are not the same as the mathematical conditions tell us. I would like to understand this difference more intuitively. Is there a way to visualize the difference of 1) and 2) on an understandable analogy?

  • 1
    \$\begingroup\$ MPT doesn't minimise the loss, it maximises the power delivered to the load. Not the same thing. (Consider, at DC it enormously increases losses, but increases power drawn from the source, delivering exactly half of it to the load; the remainder is pure loss. Same applies in transmission lines) \$\endgroup\$
    – user16324
    Nov 14, 2021 at 11:26

3 Answers 3


Here's a good reference that would be of help in understanding the difference between the two.

enter image description here

Courtesy: Rutgers, The State University of New Jersey



The problem you have is that you are applying all the results of traveling waves on transmission lines with real characteristic impedance to a problem where the base impedance is complex. The normal reflection coefficient when \$Z_0\$ is real:

\$\Gamma_L = \frac{b}{a} = \frac{Z_L-Z_0}{Z_L+Z_0}\$ eqn(1)

represents the ratio of the reflected to forward waves and \$|\Gamma|^2\$ is their power ratio.

When \$Z_0\$ is complex, this is no longer true. Read, for example:

S. Llorente-Romano, A. Garca-Lampérez, T. K. Sarkar and M. Salazar-Palma, "An Exposition on the Choice of the Proper S Parameters in Characterizing Devices Including Transmission Lines with Complex Reference Impedances and a General Methodology for Computing Them," in IEEE Antennas and Propagation Magazine, vol. 55, no. 4, pp. 94-112, Aug. 2013, doi: 10.1109/MAP.2013.6645145.

It's complicated, but it's easy to see the problem, as at conjugate match with the normal definition of reflection coefficient,

enter image description here

which is not zero if \$X_0\$ is not zero. You need to work through the maths to see that when the base impedance is complex, the reflection coefficient defined in eqn(1) does NOT represent the ratio of reflected power to forward power. In the paper above, then define something they call power wave scattering parameters, and these are much the same except that the reflection coefficient is defined as:

\$\Gamma = \frac{Z_L-Z_0^*}{Z_L+Z_0}\$

and they show that this has similar characteristics as the one we know well when \$Z_0\$ is real. Clearly when the load is conjugately matched then \$\Gamma=0\$ and it behaves as we expect.

Read the paper, it's too involved to reproduce here.

I guess I should apologise - it's not intuitive - when the impedance is complex I think you need to follow the maths.


Question: I'm confused since my intuition leaves me here in the stick. Doesn't the reflected waves contribute to the amount of power loss? What is precisely the difference between two effects? Can the difference between them be demonstrated on a prefereably as simple as possible analogy?

IMO, The OP wants to understand the nature of it, close to the intuitive level, not relearning the calculation. @CyrilMechkov will explain it better, I am sure.

I expect responses of "this is an absurd". If so, please post your answer. It is easier to judge others than self.

See what "Reflection coefficient" means here. Honestly, I did not read that lengthy text, but relying on staled memory.

That Wikipedia text explains how to draw "reflection coeff" = (ZL - Z0)/(ZL + Z0). But, still wonder why that happens in first place: Gamma = V_reflected/VI_incident. Wikipedia has a good illustration, too. A curious mind still would ask why it happens.

I cannot describe the temporal side in general, and it is explained well at many other places. So, let's look at the moment of an "incident" signal just arrived at the load,and the load just have been fully influenced. Then the incident side and the load side see each other, face to face, while standing on the signal they shared. And the amount of share is proportional to the resistance (we start with pure resistive load first). The resulting effect would look like below. Thus, the load effectively drives the source with "VL - Vi". When the matching impedance of two sides results in sharing the equal amount of the voltage, there will be no reflection, 0 = VL - Vi. Thus the energy is fully consumed on the load.


simulate this circuit – Schematic created using CircuitLab

Now, we can rush to the rest (and I need to do some other works), "why conjugate" part. The reasoning above is based on the pure resistive load, where complex form of signal does not occur. And, we observed how the maximum (no waste, no return of excess) consumption of the delivered signal was achieved. So, that can happen, too, with "reactive" circuit, if the reactive parts get canceled, thus the remaining pure resistive part (real part of complex), and that pure impedance matches.

I hope Cyrill Mechkov drops by, and explain it in the way that makes more sense.


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