Question

The system transfer function for an IIR digital filter is given by $$H(z)=\frac{B(z)}{A(z)}$$where$$ \begin{aligned}B(z)= & 0.1006-0.4498 z^{-1}+1.1211 z^{-2}-1.8403 z^{-3}+2.1630z^{-4} \\ & -1.8403 z^{-5}+1.1211 z^{-6}-0.4498 z^{-7}+0.1006 z^{-8} \\ A(z)= & 1.0-1.0130 z^{-1}+2.8890 z^{-2}-1.4483 z^{-3}+2.4901 z^{-4} \\ & -0.3959 z^{-5}+0.9169 z^{-6}+0.0324 z^{-7}+0.1867 z^{-8}\end{aligned}$$ Develop a cascade implementation of the filter using second order sections.

   The system transfer function for an IIR digital filter is given by
$$H(z)=\frac{B(z)}{A(z)}$$where$$
\begin{aligned}B(z)= & 0.1006-0.4498 z^{-1}+1.1211 z^{-2}-1.8403 z^{-3}+2.1630z^{-4} \\
& -1.8403 z^{-5}+1.1211 z^{-6}-0.4498 z^{-7}+0.1006 z^{-8} \\
A(z)= & 1.0-1.0130 z^{-1}+2.8890 z^{-2}-1.4483 z^{-3}+2.4901 z^{-4} \\
& -0.3959 z^{-5}+0.9169 z^{-6}+0.0324 z^{-7}+0.1867 z^{-8}\end{aligned}$$
Develop a cascade implementation of the filter using second order sections.

Digital Signal Processing. Principles, Algorithms and System Design

Winser Alexander and… 1st Edition

Chapter 6, Problem 1

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Step 1

To factorize the numerator polynomial B(z), we can use the quadratic formula to find the roots of the polynomial. The roots are: $$z_1 = 0.2249 + 0.9744i$$ $$z_2 = 0.2249 - 0.9744i$$ $$z_3 = 0.6000 + 0.8000i$$ $$z_4 = 0.6000 - 0.8000i$$ $$z_5 = 0.7751$$ $$z_6 = Show more…

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The system transfer function for an IIR digital filter is given by $$H(z)=\frac{B(z)}{A(z)}$$where$$ \begin{aligned}B(z)= & 0.1006-0.4498 z^{-1}+1.1211 z^{-2}-1.8403 z^{-3}+2.1630z^{-4} \\ & -1.8403 z^{-5}+1.1211 z^{-6}-0.4498 z^{-7}+0.1006 z^{-8} \\ A(z)= & 1.0-1.0130 z^{-1}+2.8890 z^{-2}-1.4483 z^{-3}+2.4901 z^{-4} \\ & -0.3959 z^{-5}+0.9169 z^{-6}+0.0324 z^{-7}+0.1867 z^{-8}\end{aligned}$$ Develop a cascade implementation of the filter using second order sections.

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