Analog IIR Filter
Analog IIR filters: Introduction

One of the fruitful way to design digital infinite impulse response (IIR) filter is to convert analog filters by impulse invariant method or bilinear transformation method. Analog filter plays important tool in the design of digital filters with the given requirements.

The transform function of analog filters $H_{a}(s)$ is expressed as:

$$H_{a}(s)=\frac{B(s)}{A(s)}=\frac{\sum\limits_{k= 0 }^{M} {b_{k}s^{k} }}{1+\sum\limits_{k= 1 }^{N} {a_{k}s^{k} }}=\frac{H_{0}\prod\limits_{k=1}^{M}(s-z_{k})}{\prod\limits_{k=1}^{N} (s-p_{k})}$$

For the stability of the filter the poles must lie on the left-half of the s - plane, i.e for poles $p_{k}=\sigma_{k}$ + j $\Omega_{k}$ the real part {$\sigma_{k}$} must be less than the zero . Also to realize analog filters, it is important to have an order of zeros is less than or equal to order of poles and the coefficients $b_k$ and $a_k$ are real   It simplifies the analysis of analog filter by using $|H_{a}(j\Omega)|^2$ instead of ${H}_{a}(s)$  . So the magnitude form of above equation can be represented as :

$$|H_{a}(j\Omega)|^2 =H_{a}(s) H_{a}(-s)_{|s=j\Omega}= H_{a}(j\Omega) H_{a}(-j\Omega)$$ $$=\frac{B(j\Omega) B(-j\Omega)}{A(j\Omega) A(-j\Omega)}=\frac{\sum\limits_{k= 0 }^{M}e_{k}(\Omega^2)^k}{1+\sum\limits_{k= 1 }^{N}f_{k}(\Omega^2)^k}=\frac{E_{M}(\Omega^2)}{F_{N}(\Omega^2)}$$ where $$E_{M}(\Omega^2)=\sum\limits_{k= 0 }^{M}e_{k}(\Omega^2)^k,$$ and $$F_{N}(\Omega^2)=1+\sum\limits_{k= 1 }^{N}f_{k}(\Omega^2)^k$$
• Note that for analog filters, the imaginary axis $s=j\Omega$ (analog frequency domain) plays the same role as the unit circle $z=e^{j\omega}$ ( digital frequency domain) for digital filters.