Realization of Biquad Transfer Function

Let us now see how the transfer function and its various special cases can be realized with passive elements.

fig: Biquad Filter

From Figure

$Y_1 = \frac{1}{R_1} + Cs$

$Y_1 = \frac{1 + R_1 C s}{R_1}$

The admittance can be given as 

$Z_1 = \frac{R_1}{1 + R_1 C s}$

The Transfer function can be represented as

$
\begin{aligned}
T(s) &= \frac{R_2}{R_2 + Z_1} = \frac{R_2}{R_2 + \frac{R_1}{1 + R_1 C_1 s}} \\
&= \frac{R_2 (1 + R_1 C_1 s)}{R_2 (1 + R_1 C_1 s) + R_1} \\
&= \frac{R_2 R_1 C_1 s + R_2}{R_2 R_1 C_1 s + R_2 + R_1} \\
&= \frac{s + \frac{1}{R_1 C_1}}{s + \frac{1}{R_1 C_1} + \frac{1}{R_2 C_1}} \\
&= \frac{s + z}{s + p} \quad \text{where } z = \frac{1}{R_1 C_1},\; p = z + \frac{1}{R_2 C_1}
\end{aligned}
$

Magnitude Response 

For s = jw

$
\begin{aligned}
T(j\omega) &= \frac{j\omega + z}{j\omega + p} \\
|T(j\omega)| &= \left| \frac{j\omega + z}{j\omega + p} \right| = \frac{\sqrt{\omega^2 + z^2}}{\sqrt{\omega^2 + p^2}} \\[10pt]
\text{At } \omega = 0: \quad |T(j0)| &= \frac{z}{p} = \frac{\frac{1}{R_1 C_1}}{\frac{1}{R_1 C_1} + \frac{1}{R_2 C_1}} = \frac{1}{1 + \frac{R_1}{R_2}} = \frac{R_2}{R_1 + R_2} \\[10pt]
\text{As } \omega \to \infty: \quad |T(j\omega)| &\to \frac{\omega}{\omega} = 1
\end{aligned}
$

fig: Magnitude Plot 

For Phase Plot 

$
\begin{aligned}
\theta(j\omega) &= \tan^{-1}\left(\frac{\omega}{z}\right) - \tan^{-1}\left(\frac{\omega}{p}\right) \\[10pt]
\text{For } \omega = 0: && \theta(j0) = \tan^{-1}(0) - \tan^{-1}(0) = 0 \\[10pt]
\text{For } \omega \to \infty: && \theta(j\infty) = \frac{\pi}{2} - \frac{\pi}{2} = 0 \\[10pt]
\text{For } \omega = z = \frac{1}{R_1 C_1}: && \theta(jz) = \tan^{-1}(1) - \tan^{-1}\left(\frac{z}{p}\right) = 45^\circ - \tan^{-1}\left(\frac{z}{p}\right) \\[10pt]
\text{For } \omega = p = z + \frac{1}{R_2 C_1}: && \theta(jp) = \tan^{-1}\left(\frac{p}{z}\right) - \tan^{-1}(1) = \tan^{-1}\left(\frac{p}{z}\right) - 45^\circ
\end{aligned}
$

fig: Phase Plot