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I have the following equalization problem as shown in the figure below:

Now I can compute the coefficients for my adaptive FIR filter c (dim(c) = N) the following:
$\mathbf{c_{opt}} = (\mathbf{H}^T\mathbf{H})^{-1}\mathbf{H}~\mathbf{h_{ideal}}$
where $\mathbf{H}$ is a convolution matrix with shifted vectors of $\mathbf{h}$ and $\mathbf{h_{ideal}}$ is chosen such that $x[n]=d[n]$ (delay-free equalizer).

The channel impulse response is given as
$\mathbf{h} = [1, 0.5]^T$
$\Rightarrow H(z) = 1+0.5 z^{-1}$ so the inverse of the system would be IIR: $1/H(z) = \frac{z}{z+0.5}$

Now the question is the following: What is the difference between the LS-solution with an adaptive filter and direct inversion of the system? Is it just that one filter is FIR and the other one IIR? Therefore with the FIR-filter we cannot reach full equalization and a residual error stays?

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  • $\begingroup$ Yes: the main issue is that the inverse of a FIR is IIR. $\endgroup$
    – MBaz
    Feb 15, 2020 at 16:57

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Inverting a channel can only be done when the channel is a minimum phase system (trailing echos only). A minimum phase system is characterized as having all zeros in the left half plane (for the s plane, or equivalently in a sampled system and the z plane all zeros inside the unit circle). Inverting such a channel results in poles where every zero exists, and a causal system that has any poles in the right half plane (outside the unit circle) is not stable. So a minimum phase system has a stable causal inverse, while a mixed phase or maximum phase system does not.

leading and trailing echos

recursive vs feedforward

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  • $\begingroup$ Finally youve mentioned: IIR not used since impulse reponses are mixed. But it seems IIR can model the mixed better than FIR, since inverses of all FIR filters have poles and then it seems we need IIR doesn't it? FIR have only zeros but IIR have both zeros and poles and it seems like the mixed case doesn't it? $\endgroup$ Feb 20, 2023 at 7:13
  • $\begingroup$ @mohammadsdtmnd Great observation! As long as the IIR impulse response is decaying over the longer time duration, eventually to insignificance, we can create any IIR with an FIR given the coefficients of the FIR are the samples of that impulse response (it just takes more coefficients than we could otherwise do with an IIR). Does that make sense? $\endgroup$ Feb 20, 2023 at 12:25
  • $\begingroup$ THX. Yes. But you've told typically not use IIR because we have mixed system. I think this is not right. Am I right? $\endgroup$ Feb 20, 2023 at 13:38
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    $\begingroup$ @mohammadsdtmnd Do NOT use IIR because you have a mixed phase system; the IIR will be unstable. Yes an FIR filter can create the same response as an IIR filter as long as the impulse response of the IIR filter is decaying with time and eventually insignificant. It won't have the same poles and zeros but it will have the same response. Please post as another question if that doesn't make sense to you as long dialogue in the comments is discouraged. Also when commenting use @ with my name otherwise I am not alerted. $\endgroup$ Feb 23, 2023 at 3:37
  • $\begingroup$ Then I think saying don't use it, since you have mixed system, is not correct. The corrected one: In mixed system IIR is better option theoretically but they have potential to be unstable, then typycally many guys opting IIR estimation using FIR. It neglects my @, O_O! Maybe because your are the answer composer. $\endgroup$ Feb 23, 2023 at 11:08

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