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Suppose I have an IIR filter in the $z$-domain in the following form: $$ H\left(z\right)=\frac{1}{1-0.2z^{-1}-0.1z^{-2}} $$

How do I represent this in MATLAB?

I am pretty sure if I just listed the coefficients as H = [1 -0.2 -0.1]; this would be considered wrong. Can somebody help me with this please?

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MATLAB assumes that the transfer function has a form like this:

$$ H(z) = \frac{b_0 + b_1z^{-1} + \ ... \ + b_{M}z^{-M}}{1+a_1z^{-1}+ \ ... \ + a_{N}z^{-N}} $$

So then you can read off the coefficients from your equation as $b_0=1$, $a_0=1$ (always assumed), $a_1 = -0.2$, and $a_2 = -0.1$. In MATLAB world, you define the coefficients as:

b = 1;
a = [1; -0.2; -0.1];

Further, if you wanted to actually use the filter on some signal, you can call the in-built function:

filterOutput = filter(b, a, filterInput);
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    $\begingroup$ That filter command wouldn't give the correct output. It should be filter(b,a,x). $\endgroup$
    – Matt L.
    Dec 2 '20 at 12:38
  • $\begingroup$ But i don't have x in the first place, nor its part of the question $\endgroup$
    – JordenSH
    Dec 2 '20 at 21:33
  • $\begingroup$ @MattL thanks I fixed that $\endgroup$
    – Engineer
    Dec 2 '20 at 21:42
  • $\begingroup$ @Raykh I guess I assumed you wanted to use the filter, but maybe you meant something else. When you say "How do I represent this in MATLAB?", what do you mean? Do you want to plot the transfer function, look at the pole-zero diagram, or maybe something else? $\endgroup$
    – Engineer
    Dec 2 '20 at 21:44
  • $\begingroup$ I think I can take it from here. Thank you very much for your help $\endgroup$
    – JordenSH
    Dec 2 '20 at 22:40
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You have to define numerator and denominator polynomials. In your case you have

b = 1;
a = [1,-0.2, -0.1];
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  • $\begingroup$ But what's to it from here? Do you do H = b/a? $\endgroup$
    – JordenSH
    Dec 2 '20 at 21:32
  • $\begingroup$ @Raykh: The coefficient vectors a and b completely represent your filter. That was your question. You have to tell us what else you want to do. $\endgroup$
    – Matt L.
    Dec 2 '20 at 21:37
  • $\begingroup$ No worries, thank you very much for your help $\endgroup$
    – JordenSH
    Dec 2 '20 at 22:39

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