Electronic – Impedance of waveguides shorter than 1/10 of wavelength

impedanceimpedance-matchingwaveguide

I read already several times that it is a common rule of thumb, that matching the impedance of a waveguide to the external network impedance is only necessary for waveguides longer than 1/10 of the wavelength of the incident signal. Is that true? And why is that so? Can someone give me a physical science basis for why that rule of thumb is true? Maybe someone knows literature where this rule is confirmed?

Thanks a lot in advance!

Best Answer

A rule of thumb cannot really be true or false. It does not belong to science, but to to engineering or practice. So it can be useful or not useful. The basic idea, at an intuitive level is that a short waveguide will not effect the signal much, because the reflection gets back to the source before the source phase has changed much (recall that for a sine wave, phase advances linearly at a constant rate with time). The big problems caused by reflections have to do with creating minima and maxima in space on the waveguide and self cancellation associated with that.

In digitial circuits, it is similar, but it is the rise and fall time that matter, not the frequency. If the rise/fall time is long compared to the waveguide flight time, then impedance matching is not needed. Because the reflection will get back to the driving source while the edge is still changing. So in other words, the source feels the load, even though there is a mismatch at the transmission line.

Another way to look at it is this. (This is highly non-technical.) When a source puts energy into a waveguide, it does not know what the load looks like. It only knows what the waveguide looks like. It does not receive any feedback from the load until energy reaches the load, reflects, and comes back to the source. If the waveguide is long, and the mismatch is large, then the feedback, when it arrives, may be far out of phase with the source. This can cause big problems. But if the waveguide is short, the feedback will not be far out of phase, and all will be well, even if the mismatch is larger. And if the load is perfectly matched to the waveguide, then there will be no reflection, no feedback, and all will be well.

One last thing. This has to do with RF propagation in the presence of obstacles. In order for an RF wave in space to be reflected (or absorbed, for that matter), you need an object of a certain size with respect to the wavelength. Longer wavelengths tend to refract around small objects. This happens with waveguides also. If the waveguide is "small" the energy sort of refracts around it.