Method

As can be seen from the code listing, this implementation of residuez simply calls residue, which was written to carry out the partial fraction expansions of $s$ -plane (continuous-time) transfer functions $H(s)$ :

$$\begin{eqnarray*} H(s) &=& \frac{B(s)}{A(s)}\\ &=& \frac{b_0 s^M + b_1 s^{M-1} + \cdots + b_{M-1}s + b_M} {a_0 s^N + a_1 s^{N-1} + \cdots + a_{N-1}s + a_N}\\ &=& F(s) + R(s) \end{eqnarray*}$$

where $F(s)$ is the ``quotient'' and $R(s)$ is the ``remainder'' in the PFE:

$$F(s) \isdef f_0 s^L + f_1 s^{L-1} + \cdots + f_{L-1} s + f_L$$

$$R(s) \isdef \frac{r_1}{(s-p_1)^m_1} + \cdots + \frac{r_N}{(s-p_N)^m_N} \protect$$ (J.1)

where $L=M-N$ is the order of the quotient polynomial in $s$ , and $m_i$ is the multiplicity of the $i$ th pole. (When all poles are distinct, we have $m_i=1$ for all $i$ .) For $M<N$ , we define $F(s)=0$ .

In the discrete-time case, we have the $z$ -plane transfer function

$$H(z) = \frac{B(z)}{A(z)}$$

$$= \frac{b_0 + b_1 z^{-1}+ \cdots + b_{M-1}z^{-(M-1)} + b_M z^{-M}} {a_0 + a_1 z^{-1}+ \cdots + a_{N-1}z^{-(N-1)} + a_N z^{-N}}. \protect$$ (J.2)

For compatibility with Matlab's residuez, we need a PFE of the form $H(z)=F(z)+R(z)$ such that

$$\begin{eqnarray*} F(z) &\isdef & f_0 + f_1 z^{-1}+ \cdots + f_{L-1} z^{-(L-1)} + f_L z^{-L}\\ R(z) &\isdef & \frac{r_1}{(1-p_1z^{-1})^m_1} + \cdots \frac{r_N}{(1-p_Nz^{-1})^m_N} \end{eqnarray*}$$

where $L=M-N$ .

We see that the $s$ -plane case formally does what we desire if we treat $z$ -plane polynomials as polynomials in $z^{-1}$ instead of $z$ . From Eq.(J.2), we see that this requires reversing the coefficient-order of B and A in the call to residue. In the returned result, we obtain terms such as

$$\frac{\rho_i}{(s-\pi_i)^{m_i}} \isdef \frac{\rho_i}{(z^{-1}-\pi_i)^{m_i}} =\frac{(-1)^{m_i}\rho_i \pi_i^{-m_i}}{(1-\pi_i^{-1}z^{-1})^{m_i}}$$

where the second form is simply the desired canonical form for $z$ -plane PFE terms. Thus, the $i$ th pole is

$$p_i = \pi_i^{-1}$$

and the $i$ th residue is

$$r_i = \frac{\rho_i}{(-\pi_i)^{m_i}}.$$

Finally, the returned quotient polynomial must be flipped for the same reason that the input polynomials needed to be flipped (to convert from left-to-right descending powers of $z^{-1}$ [$s$ ] in the returned result to ascending powers of $z^{-1}$ ).