Numerical 1: Realization of Bilinear LPF

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

fig: LPF

The figure can be modified as

fig: Modified LPF

 

Applying Kirchoff’s law,

At node 2 and 3

\[
V_{\text{node 2}} = V_1 \cdot \frac{R_1}{R_1 + R_1} \quad \text{and} \quad
V_{\text{node 3}} = I R \cdot \frac{I R + \frac{1}{C} \int i \, dt}{V_1}
\]

\[
V_{\text{node 2}} = V_{12} \quad \text{and} \quad
V_{\text{node 3}} = I R \cdot \frac{I R + \frac{1}{C} \int i \, dt}{V_1}
\]

Applying Laplace transform,

\[
V_{\text{II}} = V_1(s) \cdot 2
\]

\[
V_{\text{III}} = I(s) R(s) \cdot \frac{I(s) R(s) + \frac{1}{C(s)} I(s)}{V_1(s)} = \frac{R(s)}{R(s) + \frac{1}{C(s)}} \cdot V_1(s)
\]

Applying Kirchoff’s law,

\[
V_2(s) = V_{\text{II}} - V_{\text{III}}
\]

\[
V_2(s) = V_1(s) \cdot 2 - \frac{R}{R + \frac{1}{C(s)}} \cdot V_1(s)
\]

\[
V_2(s) = V_1(s) \cdot \left( 2 - \frac{R}{R + \frac{1}{C(s)}} \right)
\]

\[
\frac{V_2(s)}{V_1(s)} = 2 - \frac{R C(s)}{C(s) R + 1}
\]

\[
T(s) = \frac{R C s - 2 R C s + 1}{C s R + 1} = \frac{-R C s + 1}{C s R + 1} = \frac{-R C (s - \frac{1}{R C})}{2 R C (s - \frac{1}{R C})}
\]

\[
T(s) = -\frac{1}{2} \frac{(s - \frac{1}{R C})}{(s - \frac{1}{R C})}
\]

Let

\[
\omega_0 = \frac{1}{R C}
\]

For Magnitude Plot

For s = jw

\[
T_{jw} = -\frac{1}{2} (j\omega - \omega_0)(j\omega + 0)
\]

\[
|T_j| = \left| -\frac{1}{2} \right| \cdot \sqrt{\omega^2 + \omega_0^2} \cdot \sqrt{\omega^2 + 0^2}
\]

\[
|T_j| = \frac{1}{2} \sqrt{\omega^2 + \omega_0^2} \cdot \omega
\]

fig: Magnitude Plot

For Phase Plot

\[
T_{jw} = -\frac{1}{2} (j\omega - \omega_0)(j\omega + 0)
\]

\[
T_{jw} = -\frac{1}{2} \cdot 0 \cdot 0 \cdot (j\omega_0 - 1)(j\omega_0 + 1)
\]

\[
T_{jw} = -\frac{1}{2} (j\omega_0 - 1)(j\omega_0 + 1)
\]

\[
\theta(j\omega) = 180^\circ + \tan^{-1}(0) - \tan^{-1}(\omega_0)
\]

\[
\theta(j\omega) = 180^\circ + \tan^{-1}(-\omega_0) - \tan^{-1}(\omega_0)
\]

\[
\theta(j\omega) = 180^\circ - \tan^{-1}(0) - \tan^{-1}(\omega_0)
\]

\[
\theta(j\omega) = 180^\circ - 2 \tan^{-1}(\omega_0)
\]

\[
\text{When } \omega = 0, \quad \theta(j\omega) = 180^\circ - 2 \tan^{-1}(0) = 180^\circ
\]

\[
\text{When } \omega = 1, \quad \theta(j\omega) = 180^\circ - 2 \tan^{-1}(1) = 180^\circ - 90^\circ = 90^\circ
\]

\[
\text{When } \omega \to \infty, \quad \theta(j\omega) = 180^\circ - 2 \tan^{-1}(\infty) = 0^\circ
\]

fig: Phase Plot