Numerical I

QN) The given circuit is a fourth order Butterworth low pass filter having ω = 2000 rad/sec. Obtain Band pass filter having B = 400 rad/sec

fig: LPF

We have:  

Center Frequency: \(\Omega_0 = 2000 \, \text{rad/sec}\)  

Bandwidth: \(B = 400 \, \text{rad/sec} = \Omega_2 - \Omega_1\)  

For low pass to Band pass, replace  

\[
s \to B s + \frac{s^2}{2}
\]

For Resistor:
No change since resistor is frequency independent.  

For Inductor:
Inductor is replaced by inductor and capacitor in parallel:

fig: Inductor interms of BSF

\[
L' = L \cdot B^2 \quad \text{and} \quad C' = \frac{1}{L \cdot B}
\]

For \(L_1 = 1.848 \, \text{H}\):  

\[
L' = L \cdot B^2 = 1.848 \times 400 \times (2000)^2 = 1.848 \times 10^{-4} \, \text{H} = 0.184 \, \text{mH}
\]

\[
C' = \frac{1}{L \cdot B} = \frac{1}{1.848 \times 4000} = 1.35 \times 10^{-3} \, \text{F} = 1.35 \, \text{mF}
\]

For Capacitor:
Capacitor is replaced by inductor and capacitor in series:

fig: Capactior interms of BSF

\[
L' = \frac{1}{C \cdot B} \quad \text{and} \quad C' = C \cdot B^2
\]

For \(C_1 = 0.765 \, \text{F}\):  

\[
L' = \frac{1}{C \cdot B} = \frac{1}{0.765 \times 400} = 3.27 \times 10^{-3} \, \text{H} = 3.27 \, \text{mH}
\]

\[
C' = C \cdot B^2 = 0.765 \times 400 \times (2000)^2 = 7.65 \times 10^{-5} \, \text{F} = 76.5 \, \mu\text{F}
\]

For \(C_2 = 1.848 \, \text{F}\):  

\[
L' = \frac{1}{C \cdot B} = \frac{1}{1.848 \times 400} = 1.35 \times 10^{-3} \, \text{H} = 1.35 \, \text{mH}
\]

\[
C' = C \cdot B^2 = 1.848 \times 400 \times (2000)^2 = 1.848 \times 10^{-5} \, \text{F} = 184.8 \, \mu\text{F}
\]

The new circuit becomes the combination of these transformed inductors and capacitors in series and parallel as per the above calculations.

fig: BSF