# Adaptive equalization vs inverse of transfer function

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?

• Yes: the main issue is that the inverse of a FIR is IIR. – MBaz Feb 15 '20 at 16:57

## 1 Answer

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.  