Realization of Biquad Transfer Function III

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

fig: Biquad

The Transfer function can be represented as

\[
T(s) = \frac{s + \frac{1}{C_2 R_2}}{\left( s + \frac{1}{C_1 R_1} \right) + \frac{1}{C_2 R_2}}
\]

\[
\text{Let } z = \frac{1}{C_2 R_2}
\Rightarrow T(s) = \frac{s + z}{s + z + \frac{1}{C_1 R_1}}
\]

\[
\text{Let } p = z + \frac{1}{C_1 R_1}
\Rightarrow T(s) = \frac{s + z}{s + p}
\]

Magnitude Response

S=jω

\[
T(j\omega) = \frac{j\omega + z}{j\omega + p}
\]

\[
|T(j\omega)| = \frac{\sqrt{\omega^2 + z^2}}{\sqrt{\omega^2 + p^2}}
\]

ω = 0:
\[
|T(j\omega)| = \frac{\sqrt{z^2}}{\sqrt{p^2}} = \frac{z}{p}
\]

\[
\text{Recall:} \quad z = \frac{1}{R_2 C_2}, \quad p = z + \frac{1}{R_1 C_1}
= \frac{1}{R_2 C_2} + \frac{1}{R_1 C_1}
\]

\[
\Rightarrow \frac{z}{p} = \frac{\frac{1}{R_2 C_2}}{\frac{1}{R_2 C_2} + \frac{1}{R_1 C_1}}
= \frac{1}{1 + \frac{R_2 C_2}{R_1 C_1}}
= \frac{R_1 C_1}{R_1 C_1 + R_2 C_2}
\]

Alternative form:
\[
\frac{z}{p} = \frac{1/(R_2 C_2)}{1/(R_2 C_2) + 1/(R_1 C_1)}
= \frac{R_1 C_1}{R_1 C_1 + R_2 C_2}
\]

ω=∞
\[
|T(j\omega)| = \frac{\sqrt{\omega^2 + z^2}}{\sqrt{\omega^2 + p^2}} \approx \frac{\omega}{\omega} = 1
\]

fig: Magnitude Plot

Phase Response

\[
\theta(j\omega) = \arg\left( \frac{j\omega + z}{j\omega + p} \right) 
= \tan^{-1}\left( \frac{\omega}{z} \right) - \tan^{-1}\left( \frac{\omega}{p} \right)
\]

ω = 0:
\[
\theta(j\omega) = \tan^{-1}(0) - \tan^{-1}(0) = 0^\circ
\]

ω=∞
\[
\theta(j\omega) = \tan^{-1}(\infty) - \tan^{-1}(\infty) = 90^\circ - 90^\circ = 0^\circ
\]

\[
\omega = \frac{1}{R_1 C_1}:
\quad
\theta(j\omega) = \tan^{-1} \left( \frac{1}{z R_1 C_1} \right) - \tan^{-1}(1)
= \tan^{-1} \left( \frac{R_2 C_2}{R_1 C_1} \right) - 45^\circ
\]

\[
\theta(j\omega) = 45^\circ - \tan^{-1}\left( \frac{R_2 C_2}{R_1 + R_2} \right)
\]

\[
\omega = \frac{1}{R_1 C_1} + \frac{1}{R_2 C_2}:
\quad
\theta(j\omega) = \tan^{-1}\left( \frac{R_1 + R_2}{R_2 C_2} \right) - 45^\circ
\]

fig: Phase Plot