I need a closed form equation for the return loss \$ ~S_{11}~\$ of an n-th order Butterworth low-pass prototype filter.

I am designing a high power RF (lumped element) n-th order Butterworth bandpass filter for the transmitter side, so will be transforming the return loss equation for the low-pass prototype to a bandpass filter. I have the equations for the transfer function and for the insertion loss, but not the return loss.

Also, if someone can provide me with a reference text or handbook for this type of question, I'd much appreciate it.

I did check the first 7 suggested posts, but this question has not been answered before.

\$ Z_{source} = Z_{load} \$, in this case, 50 Ω

  • \$\begingroup\$ What is the input impedance of the circuit and what, ideally would you want it to be. From those parameters you can calculate return loss. \$\endgroup\$
    – Andy aka
    Commented Jul 25, 2015 at 11:28
  • \$\begingroup\$ I thought there might be some closed form equation that I could use, rather than performing circuit analysis on the circuit. \$\endgroup\$ Commented Jul 25, 2015 at 11:46
  • 1
    \$\begingroup\$ Well, you're going to have to perform circuit simulation on it to get any sort of accuracy in your filter response, especially if you take into account the component parasitics. Depending on the frequency and powers you're looking at, you may have to take those into account to get an accurate simulation, so you might as well start out with a simulation \$\endgroup\$
    – rfdave
    Commented Jul 25, 2015 at 15:25

1 Answer 1


There is no single answer to this question. There are numerous circuits that can implement any particular filter design, as we discussed in a recent question The return loss (\$S_{11}\$) depends on the topology you choose to implement the filter.

If you use an active topology, for example, then only the first stage will affect \$Z_{in}\$, and thus only the first stage will affect \$S_{11}\$.

If you choose a passive topology it depends if you construct the filter from pi sections or T-sections or some other topology.

For example, if you use pi sections (with LC parallel elements in the shunt members), then \$S_{11}\$ will go to -1 in the stop-bands. If you use T sections (with LC series elements in the through members) it will go to +1.

If you use microstrip elements, the behavior will likely have some complex periodic behavior in the stop bands.

Whichever one you choose, \$S_{11}\$ should be near zero in the pass band. Whether you achieve -40 or -50 or -60 dB probably depends more on choosing very tight-tolerance parts or trimming the circuit carefully, rather than on the nominal design. Although some design choices might be more or less sensitive to component variation. So a closed form solution for reflections in the nominal design won't help as much as doing a Monte Carlo simulation accounting for likely component variations.

If \$n\$ is more than 2, I'd suggest to just simulate the design rather than try to find a closed-form solution, because the equations will get rather tedious to deal with very quickly.

  • \$\begingroup\$ I'm a bit puzzled. I stated the filter is for high power RF (commonly meaning 20+ dBm and below 1GHz) and lumped-element, so microstrip and active topology is not applicable, and that I am after the filter prototype equation, which I thought means a Cauer ladder topology by definition. I agree \$ S_{11} \$ should be near zero, so for practical purposes I'm aiming for -30dB, -40dB, -60dB if I can get it. I'm still examining the recent question you mentioned. Thanks for your help so far. \$\endgroup\$ Commented Jul 25, 2015 at 17:52
  • \$\begingroup\$ Even if you restrict it to Cauer ladder topology, you can still choose whether the initial section is pi or T, and if n > 2 you can still choose different ways to arrange the sections to implement the poles and zeros. Also, if you really want high power handling, you should consider microstrip implementations rather than lumped elements. \$\endgroup\$
    – The Photon
    Commented Jul 25, 2015 at 18:03
  • \$\begingroup\$ Also, "high power" does not imply less than 1 GHz. There are lots of high power applications above 1 GHz. Most of the define radar bands (L, K, X, ...), for example, are above 1 GHz. "High power" doesn't even really say anything about power. If you want us to know what band you're operating in and how much power you're using, you need to tell us what band and how much power. \$\endgroup\$
    – The Photon
    Commented Jul 25, 2015 at 18:08
  • \$\begingroup\$ Apologies. I meant 50+ dBm and below 1 GHz (VHF\UHF). \$\endgroup\$ Commented Jul 25, 2015 at 18:09
  • \$\begingroup\$ Just checking a reference that might answer this question and my subsequent one on Chebyshev type 2: Matthei, Young and Jones, 1963. I didn't realise I've had it all along, but was reluctant to wade through the 1000 pages or so of it. It will be interesting to see what they say about parasitics. Back in a day or two. \$\endgroup\$ Commented Jul 25, 2015 at 18:34

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