Conversion II

QN) The given circuit is a third order Butterworth low pass filter having ω = 1 rad/sec. Obtain High pass filter having =1 rad/sec.

fig: LPF

For low pass Cut off frequency = ω = 1 rad/sec  
For High pass Cut off frequency = Ω = 1 rad/sec  

For resistor  
No change since Resistor is frequency independent.  

For Inductor  
Inductor in Low pass filter is replaced by capacitance in High pass filter  

\[
C' = \frac{1}{L \Omega}
\]  

For Ω = 1 rad/sec  

\[
C' = \frac{1}{L}
\]  

For L = 2H  

\[
C' = \frac{1}{2} = 0.5\,\text{F}
\]  

For L = 1H  

\[
C' = \frac{1}{1} = 1\,\text{F}
\]  

For Capacitor  
Capacitor in Low pass filter is replaced by Inductor in High pass filter  

\[
L' = \frac{1}{C}
\]  

For C = 1 F  

\[
L' = \frac{1}{1} = 1\,\text{H}
\]

Then the new circuit for high pass filter becomes

fig: HPF