With regards to the quarter wavelength impedance matcher, I understand it is used to correct a mismatch and looks as follows where \$ Z_T \$ is the matching line:

Quarter Wavelength Matcher

I understand that we have set up the problem such that the input impedance of the combined matching line and the load is the same as \$ Z_0 \$ meaning that we have no reflections here at all. We find that \$ Z_T \$ must be the geometric mean of \$ Z_0 \$ and \$ Z_L \$ through this condition.

However, surely the matching line and the load itself aren't matched and hence we should have some reflections at this point itself?


As in any situation like this, it can be analysed in the time domain, or the frequency domain. Both will agree, but one or the other might be easier to comprehend. Flipping between one and the other has to be done carefully, as things that are simple in one domain are not in the other, which is why either might be helpful. In the time domain, a single step that's wideband in frequency is good, in the frequency domain, a single sinewave is easier to handle.

Time Domain

Let's launch a step along the line, and stay with it as it negotiates the junctions.

It gets to the Z0/ZT junction, and some is reflected back. Of the energy that carries on, some is absorbed in the load, and the rest is reflected back. Some of that reflection passes the ZT/Z0 junction, and some is reflected back towards the load. So you can see that some energy, decreasing every bounce, is trapped on the ZT line, and that there have been two steps reflected back, with more to come. This creates a succession of steps, separated by 2*ZT's length.

As out quarter-wave transformer is only supposed to work at a single frequency, we need to concentrate on the effect at that frequency, so we need to put our frequency domain hat on now.

The successsion of steps separated by t has energy at zero frequency, no energy at 1/2t, energy at 1/t, no energy at 3/2t and so on.

A \$\frac{\lambda}{4}\$ 'transformer' only works for signals with wavelength \$\lambda\$. And at that wavelength, we can see that there's no energy in that train of reflected steps.

If instead of a wideband step, we send in a single frequency wave of the right frequency, each reflected pulse will be replaced by a sinewave. The time-shifted sinewaves, with a half-period spacing defined by the length of the ZT line, will add up to give you no nett reflection.

So what's that about no energy at 3/2t? Yes, a \$\frac{3\lambda}{4}\$, and in fact every odd multiple, works as well.

Frequency Domain

A \$\frac{\lambda}{4}\$ transformer makes its load 'appear' as if it has an impedance of \$\frac{{Z_T}^2}{Z_L}\$, when you do the line input impedance sums properly, which accounts for taking the ZT/ZL reflection and phase shifting it to the start of the line. This reflection has therefore already been accounted for, and must not be double counted.

As the ZT line now appears to have an input impedance of Z0, there is no reflection at the Z0/ZL junction either.


As a \$\frac{\lambda}{4}\$ transformer works only for certain frequencies, it's most appropriate to analyse it in the frequency domain, where steps and other wideband signals don't exist. Which means if you do try a time domain approach, ie 'there will be reflections because the lines do have different impedances', then you have to follow the analysis all the way through to analysing the results in the frequency domain again, where the 'transformer' 'works'.

  • \$\begingroup\$ @NightStrider It's also important to bear in mind that quarter-wave matching can mathematically only be done for one single frequency (although in practice the match is acceptable over a small range). The reflections only cancel out at the matched frequency. \$\endgroup\$ – Tom Carpenter Apr 19 '17 at 21:57
  • \$\begingroup\$ Thank you for your response. I thought that the entire purpose of the impedance matching was to eliminate reflections at the first junction because the impedance there matches that of the line perfectly. Is this not the case? \$\endgroup\$ – DreamsOfHummus Apr 19 '17 at 22:16
  • \$\begingroup\$ @NightStrider, yes. It does that by arranging that the reflections from the second junction interfere destructively with the reflections from the first junction. \$\endgroup\$ – The Photon Apr 19 '17 at 22:19
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    \$\begingroup\$ That's a different way to say the same thing. It "sees Z_0" because the reflections from the second junction are interfering with the immediate reflections at the first junction to make it that way. \$\endgroup\$ – The Photon Apr 19 '17 at 22:31
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    \$\begingroup\$ What third junction? Can you edit your diagram and label what you mean by "first junction" and "2nd junction". I thought you meant the junction between the Z0 line and the ZT line when you said "first junction", and the junction between the ZT line and the load when you said "second junction". I don't see any third junction in your diagram. \$\endgroup\$ – The Photon Apr 19 '17 at 23:05

If you look at the situation from within the time/reflections framework, then of course there are reflections. Infinite reflections, in fact.

However, the widely used (and more convenient) frequency/impedance framework allows us to "ignore" those multiple reflections. How can this be?

Well, because it's just a steady-state simplification: those reflections are actually there, but they're only relevant for a time/transient analysis.

From Chapter 2.5 in Pozar, D., Microwave Engineering:

Multiple reflections

[...] the matching property of the quarter-wave transformer comes about by properly selecting the characteristic impedance and length of the matching section so that the superposition of all of the partial reflections adds to zero. Under steady-state conditions, an infinite sum of waves traveling in the same direction with the same phase velocity can be combined into a single traveling wave. Thus, the infinite set of waves traveling in the forward and reverse directions on the matching section can be reduced to two waves traveling in opposite directions.

(emphasis added)


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