Realization of LPF

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

fig: Passive LPF

Note: Applying Kirchoff’s law,

\[
V_1 = I \cdot R + \frac{1}{C} \int i \, dt, \quad V_2 = \frac{1}{C} \int i \, dt
\]

\[
\text{After Laplace Transform:}
\quad V_1(s) = I(s) \cdot R + \frac{1}{C s} \cdot I(s), \quad V_2(s) = \frac{1}{C s} \cdot I(s)
\]

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

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

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

\[
\text{Let } \omega_0 = \frac{1}{R C}, \quad \Rightarrow T(s) = \frac{\omega_0}{s + \omega_0}
\]

For Magnitude Plot

\[
T(s) = \frac{\omega_0}{s + \omega_0}
\quad \text{Let } s = j\omega
\]

\[
T(j\omega) = \frac{\omega_0}{j\omega + \omega_0}
\]

\[
|T(j\omega)| = \left| \frac{\omega_0}{j\omega + \omega_0} \right|
= \frac{\omega_0}{\sqrt{\omega^2 + \omega_0^2}}
\]

\[
\textbf{Case 1: } \omega = 0
\quad \Rightarrow |T(j0)| = \frac{\omega_0}{\sqrt{0 + \omega_0^2}} = \frac{\omega_0}{\omega_0} = 1
\]

\[
\textbf{Case 2: } \omega \to \infty
\quad \Rightarrow |T(j\omega)| = \frac{\omega_0}{\sqrt{\omega^2 + \omega_0^2}} \to 0
\]

\[
\textbf{Case 3: } \omega = \omega_0
\quad \Rightarrow |T(j\omega_0)| = \frac{\omega_0}{\sqrt{\omega_0^2 + \omega_0^2}} = \frac{\omega_0}{\sqrt{2}\omega_0} = \frac{1}{\sqrt{2}}
\]

fig: Magnitude Plot

 

For Phase plot(k=0)

\[
T(j\omega) = \frac{\omega_0}{j\omega + \omega_0}
\quad \Rightarrow \angle T(j\omega) = \angle \omega_0 - \angle (j\omega + \omega_0)
\]

\[
\text{Since } \omega_0 \text{ is real and positive, } \angle \omega_0 = 0^\circ
\]

\[
\therefore \angle T(j\omega) = -\tan^{-1}\left( \frac{\omega}{\omega_0} \right)
\]

Phase at specific frequencies:

\[
\omega = 0 \Rightarrow \angle T(j0) = -\tan^{-1}(0) = 0^\circ
\]

\[
\omega = \omega_0 \Rightarrow \angle T(j\omega_0) = -\tan^{-1}(1) = -45^\circ
\]

\[
\omega \to \infty \Rightarrow \angle T(j\omega) = -\tan^{-1}(\infty) = -90^\circ
\]

fig: Phase Plot