Laurent expansion

Given an unknown filter B(Z), to understand its inverse, we need to factor B(Z) into two parts: B(Z) = B_out(Z) B_in(Z), where B_out contains all the roots outside the unit circle and B_in contains all the roots inside. Then the inverse of B_out is expressed as a Taylor series about the origin, and the inverse of B_in is expressed as a Taylor series about infinity. The final expression for 1/B(Z) is called a ``Laurent expansion'' for 1/B(Z), and it converges on a ring including the unit circle. Cases with zeros exactly on the unit circle present special problems. For example, the differentiation filter (1-Z) is the inverse of integration, but the converse is not true, because of the additive constant of integration.

EXERCISES:

1. Find the filter that is inverse to (2 - 5Z + 2Z²). You may just drop higher-order powers of Z, but an exact expression for the coefficients of any power of Z is preferable. (Partial fractions is a useful, though not a necessary, technique.) Sketch the impulse response.
2. Describe a general method for determining A(Z) and B(Z) from a Taylor series of $B(Z)/A(Z) = C_0 + C_{1}Z + C_{2}Z^2 + \cdots + C_{\infty}Z^\infty$,where B(Z) and A(Z) are polynomials of unknown degree n and m, respectively. Work out the case $C(Z) = {1 \over 2} - {3 \over 4}Z - {3 \over 8}Z^2 - {3 \over 16}Z^3 -$${3 \over 32}Z^4 - \cdots$.Do not try this problem unless you are familiar with determinants. (HINT: identify coefficients of B(Z) = A(Z) C(Z).)