Numerical 1

 

 

 

QN) The given circuit is a third order Butterworth low pass filter having ω = 10,000 rad/sec. Obtain Band pass filter having B = 1000 rad/sec

fig: LPF

 

We have:  

Center Frequency (\(\Omega_0\)) = 10,000 rad/sec  

Bandwidth (\(B\)) = 1,000 rad/sec \(= \Omega_2 - \Omega_1\)  

For low pass to band pass, we replace \(s\) by \((s^2 + 2 B s)\)  

For resistor:  
No change since resistor is frequency independent.  

For Inductor:  
Inductor is replaced by a series of inductor and capacitor:

\[
L' = L B \quad \text{and} \quad C' = \frac{1}{L \Omega_2 B}
\]

For \(L = 2\,\text{H}\):

\[
L' = L B = 2 \times 10^{-3} \, \text{H} = 2\,\text{mH}
\]

\[
C' = \frac{1}{L \Omega_2 B} = \frac{1}{2 \cdot (10000)^2 \cdot 1000} = 5 \times 10^{-6}\, \text{F} = 5\,\mu\text{F}
\]

fig: Inductor representation for Bandpass Filter

Similarly, for Capacitor:  
It is replaced by an inductor and capacitor in parallel:

fig: Capacitor representation for Bandpass Filter

\[
L' = B C \Omega_2 \quad \text{and} \quad C' = C B
\]

\[
L' = B C \Omega_2 = 1000 \cdot 1 \cdot (10000)^2 = 1 \times 10^{-5}\, \text{H} = 0.1\,\mu\text{H}
\]

\[
C' = C B = 1 \cdot 1000 = 10^{-3}\, \text{F} = 1\,\text{mF}
\]

Then the new circuit becomes:

fig: BPF