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.[1] 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 [2] [1] It simplifies the analysis of analog filter by using $|H_{a}(j\Omega)|^2$ instead of ${H}_{a}(s)$ [2] . 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.[1]