Example

We wish to truncate the impulse response of

$$H^+(z) = \frac{B^+(z)}{A^+(z)} = \frac{1}{1-1.9z^{-1} + 0.98 z^{-2} }$$

after $N=300$ samples to obtain a length $301$ FIR filter $H^+_{\rm FIR}(z)$

Steps:

1. Perform synthetic division on $z^{300}B^+(z)$ by $A(z)$ to obtain the remainder
   $$B'^+(z)=-0.162126 z +0.139770$$

2. Form the TIIR filter as
   $$\begin{eqnarray*} H^+_{\rm FIR}(z) &=& \sum_{k=0}^N h^+_k z^{-k}=\frac{B^+(z) - z^{-N} B'^+(z)}{A^+(z)} \\ &=& \frac{1 + 0.162126\, z^{-299} - 0.139770\,z^{-300}}{ 1-1.9z^{-1} + 0.98 z^{-2} } \end{eqnarray*}$$