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Extension to Higher-Order TIIR Sequences

We can extend this idea from the one-pole case to any rational filter $ H(z) = B(z)/A(z)$ . The general procedure is to find the ``tail filter'' $ H^\prime_{\rm IIR}(z)$ and subtract it off:

$\displaystyle H_{\rm FIR}(z) = H_{\rm IIR}(z) - H^\prime_{\rm IIR}(z)
$

Multiply $ H_{\rm IIR}(z)$ by $ z^{N}$ to obtain

\begin{eqnarray*}
z^{N} H_{\rm IIR}(z)&=& h_0 z^{N} + \cdots +h_{N-1} z + h_N \\
&&\quad {}+h_{N+1} z^{-1} + h_{N+2} z^{-2} + \cdots \\
&\mathrel{\stackrel{\mathrm{\Delta}}{=}}& C(z) + H'_{\rm IIR}(z) \\
&=& \frac{z^{N} B(z)}{A(z)} \mathrel{\stackrel{\mathrm{\Delta}}{=}}C(z)+ \frac{B'(z)}{A(z)}
\end{eqnarray*}


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``Horn Modeling'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2019-02-05 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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