Numerical 3: Scaling

The Given Circuit is a Butterworth Low pass filter having half power frequency 1 rad/sec. Obtain a low pass filter having half power frequency of 10 Kilo hertz and practically reliable elements with value of load resister 1 KΩ.

fig: Butterworth LPF

\[
\textbf{Given:} \quad W = 1 \ \text{rad/sec}
\]

\[
f_c = 10 \ \text{kHz} = 10 \times 10^3 \ \text{Hz}
\]

\[
\omega(C) = 2\pi f_c = 2\pi \times 10^4 \ \text{rad/sec}
\]

\[
K_f = \frac{\omega(C)}{W} = \frac{2\pi \times 10^4}{1} = 2\pi \times 10^4
\]

---

\[
\textbf{Given:} \quad R_{\text{new}} = 10 \ \text{k}\Omega
\]

\[
R_{\text{new}} = K_m \cdot R_{\text{old}} \quad \Rightarrow \quad 10^4 = K_m \cdot 1 \ \Omega \quad \Rightarrow \quad K_m = 10^4
\]

---

### Inductor:

\[
L_{\text{new}} = L_{\text{old}} \cdot \frac{K_m}{K_f}
\]

\[
L_{\text{new}} = 1 \ \text{H} \cdot \frac{10^4}{2\pi \cdot 10^4} = \frac{1}{2\pi} \ \text{H} \approx 0.159 \ \text{H}
\]

---

### Capacitor:

\[
C_{\text{new}} = C_{\text{old}} \cdot \frac{1}{K_f \cdot K_m}
\]

\[
C_{\text{new}} = \frac{2 \ \text{F}}{2\pi \cdot 10^4 \cdot 10^4} = \frac{2}{2\pi \cdot 10^8} \ \text{F} \approx 0.1 \ \mu\text{F}
\]

The new circuit is given by

fig: Scaled Butterworth LPF