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Assume I have a FIR, stable and causal system. I want to know the deepness of non-causality on the inverse of my FIR system. It's obvious that the system is non-minimum-phase, since minimum-phase inverse is always stable and causal.

Definition: The system with $h(n)$ having non-zero elements on $$n> -L$$ have deeper non-causality if $L$ is bigger.

Maybe the reference can be the minimum phase inverse. We can slightly make system worther to introduce none causalty in time reverse of $h[n]$. We know even one element on only just $n=-1$ on $h[n]$ causes system to be none-causal. I'm search for a merit to understand what is affects the deepness of none-causality.

Example: Under which condition the none-causal part of this inverse system will decay faster, or shifted or in some manner decreased>? enter image description here

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If the given causal and stable FIR system is non-minimum phase and if it has no zeros on the unit circle, the inverse system is IIR with poles outside and possibly inside the unit circle. The stable realization of that system must be either left-sided or two-sided, i.e., its impulse response extends to minus infinity. That means that shifting the impulse response to the right (i.e., adding delay) will never make the system causal.

In such a case, the best you can do is design an equalizer with impulse response $w[n]$ such that

$$(h\star w)[n]=\delta[n-k],\quad k>0\tag{1}$$

is approximately satisfied, where $h[n]$ is the given FIR system, and $k$ is some delay that needs to be chosen appropriately. The optimal value for the delay will depend on the orders of $h[n]$ and $w[n]$. In practice it's easier if you choose $w[n]$ to be FIR.

As an example I used a second-order non-minimum phase FIR filter with zeros at $z=\frac12$ and $z=3$, and I designed a $15$ tap FIR equalizer $w[n]$ by solving $(1)$ in a least squares sense. I chose a desired delay of $k=5$ samples. Note that the ideal equalizer would be a two-sided (i.e., non-causal) IIR filter.

The figure below shows the result. Of course, the equalization is not perfect but it might be good enough for most practical purposes. Increasing the length of the FIR equalizer and adjusting the delay $k$ will result in equalization of arbitrary precision.

enter image description here

Also take a look at this related question and its answer.

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  • $\begingroup$ But why we need left sided or two sided inverse? How you've got it? $\endgroup$
    – mohammadsdtmnd
    Mar 7 at 5:47
  • $\begingroup$ Does this $k$ causes inverse system to became causal? right handed or will be right handed with shift equal to $k$? Does this shift extrictly causes the system to became causal or it's just an estimate? $\endgroup$
    – mohammadsdtmnd
    Mar 7 at 7:32
  • $\begingroup$ @mohammadsdtmnd: The delay $k$ will not make the exact inverse causal, because the impulse response of the exact inverse extends to minus infinity, as explained in my answer. The value of $k$ in Eq. (1) can be used to minimize the approximation error of the approximate inverse $e[n]$. By definition, $e[n]$ is causal (and usually FIR) but it can only approximately invert the given FIR system. $\endgroup$
    – Matt L.
    Mar 7 at 8:07
  • $\begingroup$ Yes exact inverse have bad property(you say $h$ extends to minus inf) but if you don't want exact inverse by introducing $k$ , we relax causality in the implemented inverse system. Maybe causes none-causal of $h$ to damping faster (though think it won't!). I think in some system $k$ must satisfy exact causality problem, the exception is when original system has frequency component with too high delay. In case of not having this exception case $k$ have to provide exact inverse, now if FIR can approximate, IIR have to reconstrut losslessly, What's your idea? Am I wrong? $\endgroup$
    – mohammadsdtmnd
    Mar 7 at 9:57
  • $\begingroup$ @mohammadsdtmnd: If the system to be equalized is non-minimum phase (i.e., it has finite zeros outside the unit circle), an additional delay will never result in a perfect equalizer that is also causal and stable. The ideal equalizer always extends to minus infinity, that's why a delay doesn't help. But a non-ideal (causal and stable) equalizer is always feasible, as shown in the example I added to my answer. $\endgroup$
    – Matt L.
    Mar 7 at 11:43

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