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?

  • \$\begingroup\$ If you use Ls or Cs or stublines, perfect matching only occurs at ONE frequency. \$\endgroup\$ – analogsystemsrf Apr 12 at 20:46
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    \$\begingroup\$ It would probably help to include a diagram showing the kind of matching network you're asking about. What do you mean by "inside the matching network there is a variation of impedance"? \$\endgroup\$ – The Photon Apr 12 at 22:01
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    \$\begingroup\$ Reflection ONLY makes sense when the signal travels some distance. For example, if you have a voltage generator connected to a load impedance by way of a transmission line (such as coaxial cable). If Zg is 50 Ohms, and the transmission line is 50 Ohms, then the best thing to do might be to put your matching network at the load so that the load looks like 50 Ohms. Then there will be no reflections inside the transmission line. \$\endgroup\$ – mkeith Apr 13 at 1:41
  • \$\begingroup\$ If the generator has complex impedance, and the load has complex impedance, connected by a transmission line, and you choose to do your matching network at the generator, you MAY have reflections inside the transmission line unless you also match between the line and the load. Anyway, I guess it all depends on where you put your matching network with respect to the transmission line. Usually real systems are designed at a specific impedance such as 50 Ohms. If the load impedance is different than 50, a network is used at the load. \$\endgroup\$ – mkeith Apr 13 at 1:44

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.


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.

  • \$\begingroup\$ So, if I build a matching network with lumped elements which transforms an imput impedance to another, am I sure that if I cut it, each part has an impedance which is complex conjugate of the other? If yes, why is this property verified? \$\endgroup\$ – Kinka-Byo Apr 25 at 19:22
  • \$\begingroup\$ Yes, if "each part" remains connected to its respective portion of the whole system. This would not work if the matching network was removed from the whole system. The verification, well, if the two portions were not complex conjugate of each other, then there would be something in this circuit that would cause reflection and it would not be an optimally matched circuit/system. \$\endgroup\$ – Tim D Apr 26 at 21:09
  • \$\begingroup\$ Ok, I have a last question. Consider a matching network as this one (upload.wikimedia.org/wikipedia/commons/thumb/5/59/…), with a series inductance and a parallel capacitor. I design it to have that the impedance at the input is the complex conjugate of the output impedance of my signal generator. But, if I cut the matching network in the point between the capacitor and the inductor (with the capacitor on left and the inductor on right, who tells me that the impedances on right and left are complex conjugate ( so there is not reflection). \$\endgroup\$ – Kinka-Byo Apr 28 at 8:32
  • \$\begingroup\$ Again, you need to consider the system as a whole. On your input, you are matching to a generator and this generator would need to be connected to the input of the match network. As for the output of match network, you are matching to something; let us call this the load circuit. This load circuit would also need to be connected. When generator, match network and load circuit are all connected together and properly matched to form a whole system, then yes, if you cut the circuit between inductor and capacitor of match network, you should find these are complex conjugate of each other. \$\endgroup\$ – Tim D Apr 29 at 15:23
  • \$\begingroup\$ Perfect, thank you very much. \$\endgroup\$ – Kinka-Byo Apr 30 at 16:03

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