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:

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?

## Best Answer

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 DomainLet'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 DomainA \$\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.

SummaryAs 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

willbe reflections because the linesdohave 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'.