# IIR Adaptive Filter in MATLAB

Suppose I have a IIR filter represented by $$G_0\left(z\right)=\frac{1}{1-0.2z^{-1}-0.1z^{-2}}$$

I would like to use the LMS algorithm to model an FIR filter $$G\left(z\right)$$ of order $$N = 15$$ such that it would adaptively reach $$G_0\left(z\right)$$ coefficients values

So we know that $$G_0$$ can be represented by:

num = 1;
den = [1; -0.2; -0.1];


my issue is that I don't know how to initiate $$G(z)$$ given its order is $$N=15$$.

Is it valid to say:

G(z) = zeros(N,1)??


because my confusion is that if $$G_0(z)$$ has only 3 coefficients, how would $$G(z)$$ converge to 3 coefficients given that it is 15 taps long?

Please give me some clear insights into this. Thank you in advance.

• The title says "IIR adaptive filter", but isn't it the FIR filter that's adaptive? And how is the FIR filter supposed to converge to the coefficients of the IIR filter if it is to model (i.e., approximate) the IIR filter? Dec 3, 2020 at 11:58

An adaptive FIR filter is a FIR filter, that uses some kind of an adaptive algorithm to change the filter weights and reach a desired state. In case of using an LMS algorithm the general update equation is the following:

$$\mathbf{w}(n+1) = \mathbf{w}(n) + \mu\cdot e(n) \cdot \mathbf{x}(n),$$

where $$\mathbf{w}(n)$$, $$e(n)$$ and $$\mathbf{x}(n)$$ are respectively filter weights, an error and tap values at $$n^{th}$$ sample and $$\mu$$ is an adaptation step size.

That means, that the result of adaptation of a $$N$$-tap filter will be $$N$$ new values of its weights.

In your case, you can make a FIR filter $$G(z)$$ with $$N=15$$ try to adapt, such that its impulse response will be close to the impulse response of the IIR filter $$G_0(z)$$. And it should be able to do so starting from zero weights. However, after the convergence it will certainly still have an error.

Useful references on the LMS algorithms are Chapter 9 "Adaptive Filtering" in Statistical Digital Signal Processing and Modeling by M.H. Hayes and a Least_mean_squares_filter entry on Wikipedia.

• Thank you very much Dec 5, 2020 at 6:03