I want to understand a technique for calculating the inverse of a digital filter.
Let A be a function from $R$ to $R$ such that $$ A(0) = 1;\quad A(1) = a_1;\quad A(2) = a_2; \quad\text{, otherwise } A(t) = 0. $$ The $\mathcal{Z}$ transform of $A$ is $$ \mathcal{Z}(A)(z) = 1 + a_1z^{-1} + a_2z^{-2}. $$ The function $A$ defines a digital filter.
The inverse digital filter is a function B from $R$ to $R$ such that $A*B=\delta$, where "$*$" denotes convolution and $\delta$ is the Kronecker delta function.
I know that I can find $B$ by using partial fraction expansion on $$ \mathcal{Z}(B)(z)= \frac{1}{\mathcal{Z}(A)(z)} = \frac{1}{1 + a_1z^{-1} + a_2z^{-2}} $$ to express it in terms that have a inverse $\mathcal{Z}$ transform available in a table of well-known $\mathcal{Z}$ transformations.
The technique I am interested in knowing more about works as follows.
If $r$ is complex number such that $|r|<1$, then the geometric sum is $$ \sum_{k=0}^{\infty} r^k = \frac{1}{1-r}. $$ Let us find $B(1)$. Using the geometric sum we have that \begin{align} \frac{1}{\mathcal{Z}(A)(z)} &= \frac{1}{1 + a_1z^{-1} + a_2z^{-2}} \\ &= \sum_{k=0}^{\infty} (-a_1z^{-1} - a_2z^{-2})^k \\ &= 1 -a_1z^{-1} - a_2z^{-2} + \sum_{k=1}^{\infty} (-a_1z^{-1} - a_2z^{-2})^k. \\ \end{align} By inspection we find that the coefficient of the term $z^{-1}$ is $\;-a_1$. Hence, $B(1) = -a_1$.
To find $B(n)$ we continue to evaluate the sum and identify terms.
My questions: Does this work for any function $A$? Does the condition $|r|<1$ matter for this technique to work?
Yes, it works for any $A$ in the form $$ A(0) = 1;\quad A(1) = a_1;\quad A(2) = a_2; \quad\text{, otherwise } A(t) = 0. $$ where $a_1 , a_2\in R$, provided that $|r|<1$.
$|r|<1$ is necessary for convergence.