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Insertion loss, return loss and mismatch loss are three different things

These three get used interchangeably and they are not the same thing. One of them can be large while nothing at all is being dissipated, which is the detail that causes most of the confusion.

The short version

• Return loss is how much of the incident power comes back. It is a property of the mismatch, in dB, and bigger is better.
• Mismatch loss is how much of the incident power never gets in. It is the same mismatch, counted the other way round, and it is not dissipated anywhere. It went home.
• Insertion loss is how much less power reaches the load with the network in place than without it. Depending on who is talking, it may or may not include the mismatch.

A mismatch with nothing to dissipate in it

Here is the case that settles the argument. Take a perfectly lossless 75 ohm transmission line, exactly a quarter wave long, and put a perfect 50 ohm resistor on the end of it. Feed it from a 50 ohm source.

There is nothing in that circuit that can turn power into heat except the load, and the load is the thing you are trying to deliver power to. Yet:

0.695 dB of insertion loss, and not one milliwatt of it became heat. The quarter wave transformed 50 ohms up to 112.5 ohms, the source sees a mismatch, and 0.695 dB of what the source wanted to deliver is reflected back into it. That is mismatch loss wearing an insertion loss badge.

Make the line half a wave instead and it becomes transparent: Zin is 50 ohms again, VSWR 1.0000, insertion loss 0.000 dB. Same lossless line, same load, and the loss went away because the mismatch went away. If the 0.695 dB had been dissipation, doubling the length would have doubled it.

Mismatch loss only depends on the mismatch

It does not care which way the mismatch goes, or what caused it:

25 ohms and 100 ohms are both a 2:1 mismatch and both cost 0.512 dB. Worth keeping in your head: a 2:1 VSWR costs about half a dB, 3:1 costs 1.25 dB, 4:1 costs about 2 dB. Which is why chasing a 1.5:1 down to 1.2:1 is usually a waste of a Saturday, and why a 4:1 is worth fixing.

Loss that really is loss

Now put a lossy component in. Same L-network matching 500 ohms to 50 ohms at 100 MHz, once with an ideal inductor and once with a Q of 30:

0.424 dB of genuine heat. Notice the second effect, which people forget: the lossy inductor also moved the match. Its series resistance is part of the circuit, so Zin went from 50.5 to 55.5 ohms and the VSWR degraded from 1.01 to 1.11. A real component does not just cost you power, it detunes you.

What QuickSmith reports, and why

The Insertion loss figure in the readout is a transducer loss: power delivered to the load, compared with the power that a matched load would have taken from the same source. The source is the characteristic impedance, 50 ohms by default, and the termination is whatever load you set.

That means it counts mismatch as well as dissipation, which is why the lossless quarter wave above reads 0.695 dB. It is the definition that answers the question you usually have, which is "how much less power reaches my load because of this thing", rather than "how lossy is this two-port in isolation".

It also means the test that a lossless matched network reads exactly 0.000 dB is a real test, and the regression suite runs it: build a network that matches its load out of ideal parts and the number has to be zero, not nearly zero.

Doing the arithmetic yourself

All three come from the same reflection coefficient, so if you have one you have all of them.

• Return loss in dB is -20 log |gamma|
• Mismatch loss in dB is -10 log (1 - |gamma|^2)
• VSWR is (1 + |gamma|) / (1 - |gamma|)

A perfect match makes gamma zero, so return loss is infinite. QuickSmith prints an infinity sign there rather than a very large number, which is not a display trick: an early version floored gamma at 1e-36 and could never report a perfect match, and fixing that is one of the things the test suite now holds in place.