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If we insert a tanh function (or any other activation function) between the feedback summation and the unit delays, how will such an IIR filter behave for values of $|a| < $1 and $a > ±1$ ?

IIR Structure

The filter becomes stable for all values of $a$ but what happens to the frequency response of the filter?

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  • $\begingroup$ Ok, so we're talking about nonlinear filters, I guess! Can you draw an example IIR structure, just to make it clear what you're considering? I don't think a general answer can be given, but using a bit of $z$-transform magic might help make specific statements (not sure, though). If you could come up with a equation describing your filter, that would be perfect. $\endgroup$
    – Marcus Müller
    Apr 14, 2017 at 12:14
  • $\begingroup$ dropbox.com/s/n4kp23vfo780j88/IIR2.png?dl=0 $\endgroup$
    – Kosmas Giannoutakis
    Apr 14, 2017 at 12:32
  • $\begingroup$ Made this in LTspice. Since the .AC analysis linearizes the circuit, the frequency responses for two Butterworth IIRs (normal and modified like you showed) looked the same, but in .TRAN they looked different: i.stack.imgur.com/TLACJ.png . Black is modified IIR, red is normal, and blue is the tanh of the normal IIR. The last two come close, but not quite; the FFT shows similar odd harmonics. Overall, I don't know where this setup might be favourable, or usable, since the output looks unpredictable. If tanh limiting is wanted, it might be better applied at output. $\endgroup$
    – a concerned citizen
    Aug 13, 2017 at 6:46

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A transfer function is only defined for linear time variant systems. Since your proposed structure is non-linear, the concept of a transfer function doesn't apply any more. You could make approximations for certain special cases and assumptions, but overall the answer to "what happens to the frequency response" is "undefined".

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    $\begingroup$ okay, strictly speaking Hilmar is correct, but if your signal is small, then it behaves much like the IIR filter would behave if the $\tanh(\cdot)$ function was a straight line with the same slope. the derivative $\tfrac{d}{dx}\tanh(x)$ is equal to 1 when $x$=0. so that's a wire and $\tanh(x)\approx x$. the frequency response should be mostly unchanged. when the signal gets large enough that your soft-clipping starts to kick in, then the $\tanh(x)$ deviates from $x$ sufficiently that the behavior of the system might get a little funky. could even get a limit cycle. $\endgroup$
    – robert bristow-johnson
    Apr 15, 2017 at 7:17
  • $\begingroup$ Thanks for your answers! So, which would be the best approach for understanding the behavior in the extremes? Would it be possible to track this down mathematically or with iterative methods? $\endgroup$
    – Kosmas Giannoutakis
    Apr 15, 2017 at 10:45
  • $\begingroup$ i think all you can do is simulate the system. try it out for different signals. it is possible that you could have a stable system with an input of one level and it become unstable at an input much larger. it's also possible that you could excite the system with a non-zero input, cut the input to zero, and watch the system continue on ringing forever. that might be a limit cycle or it might just be an unstable system. $\endgroup$
    – robert bristow-johnson
    Apr 15, 2017 at 20:14
  • $\begingroup$ The claim that the system is stable for all values of "a" is misleading. It's true that the output will be bounded by the tanh function, but clearly the system will oscillate for certain values of a1 and a2, and behave like an oscillator that is stabilized by a saturation non-linearity. $\endgroup$
    – Bob
    Oct 12, 2017 at 11:46

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