Impedance matching and reflection

Question

I have a simple question about reflection and impedance matching.

Consider a voltage generator Vg with its output impedance Zg, and suppose it drives a load Zl. In order not to have reflection, we should have Zg = Zl. Since this is not automatically true, it is important to put an input matching network which transforms zg into an impedance equal to zl.

But this transformation will ensure that the reflection coefficient at the connection between the matching network and the load is 0. But inside the matching network there is a variation of impedance, so there may be reflection inside it. Why is this not a problem?

Answer 1

Your question is a bit confusing because you are talking about reflection, but you also seem to be talking about discrete matching elements, and you make no mention of any transmission line.

Typically, the way this topic is taught is that two cases are presented.

1. Case 1 is when the distance between source and load is less than 10% or 15% of a wavelength. In this case, you don't need to give any consideration to reflections or transmission line theory.
2. Case 2 is when there is a traveling electromagnetic wave, and it covers a distance more than 10% or 15% of its own wavelength. In this case, you use transmission line theory and consider reflections rather than just complex impedance calculations.

For example, if you are building an FM radio receiver (88-108 MHz) on a small circuit board, you probably don't need to think of your PCB traces as a transmission line, because the wavelength at 100 MHz is around 3 meters. But if you connect a 5 meter antenna cable to the PCB, then you need to consider the impedance of the cable and antenna and PCB and make sure the reflections (which will be in the transmission line) are acceptable.

Commercial radio gear is almost always designed to have 50 Ohm impedance. If you build an antenna and it has an impedance much different from that, you will probably try to match the antenna right at the antenna feed point so that it presents a 50 Ohm load to the cable. Likewise, the transmitter designer will also probably match the transmitter source impedance to 50 Ohms right at the transmitter output. So you end up with a 50 Ohm source, 50 Ohm transmission line and 50 Ohm antenna, and no major problem with reflections.

In theory, you could have an antenna with a complex impedance. You could connect it to a 50 Ohm cable, and put a matching network at the source end of the cable to make sure the transmitter sees a 50 Ohm load. But this will lead to reflections in the cable, and it could be a problem in some cases, because standing waves will build up in the cable, and the voltage may be much higher than the transmit voltage, for example.

You may wish to google voltage standing wave ratio (VSWR) for further reading.

Answer 2

In mathematical terms, you can "cut" your network, let us suppose cutting your complete network into two parts, at any arbitrary point and measure the impedance looking into each part. If these two impedances are complex conjugates of each other, then this is a "matched" network which will transfer all signal power without reflection. This mathematical concept applies in practice to, for example, a coaxial line where impedances are uniformly distributed throughout the geometry of the line. If you build a matching network from components such as capacitors and inductors, now you have "lumped" impedances rather than distributed impedances. The same mathematics still applies even though it may not be practical to cut a lumped network at some arbitrary point. If you could find a way to cut through those capacitors and inductors of your properly designed matching network (without altering their values nonlinearly and avoiding many other difficulties with such an experiment) and could make those impedance measurement on both sides, you would surely find the two impedances you measure are complex conjugates of each other regardless of where you decide to make the cut. I suppose a practical experiment might be to choose cutting through the matching network only at points of simple conductors, i.e. the wires in the circuit. I hope this helps with your understanding.