LPF to BSP

QN) How can you obtain Band stop filter from a given low pass filter? Explain with suitable figures and examples.

Similarly, for Band Stop Filter:  

Replace 

\[
s' = \omega \left( \frac{\Omega_2 - \Omega_1}{s} + \frac{s}{2} \right) = \omega \left( B \frac{1}{s} + \frac{s}{2} \right)
\]

Where \(B\) is the bandwidth of the band stop filter  

\[
B = \Omega_2 - \Omega_1
\]

and 

\[
\Omega_1 \Omega_2 = \Omega_0^2
\]

For normalized value of \(\omega\) (\(\omega = 1 \text{ rad/sec}\)):

\[
s' = B \frac{1}{s} + \frac{s}{2}
\]

fig: LPF to BSF

For low pass: cut-off frequency = \(\omega\) for \(s =\) cut-off frequency \(\omega\)  

For new band stop: cut-off frequency = \(\Omega\)  

Along the \(j\)-axis, put \(s = j \omega\) in the relations  

In band stop, the usual concern is what happens in the stop band.  

For low pass filter with \(\omega = 1 \text{ rad/sec}\), the result is similar to the band pass filter. The point \(\omega = 1\) is mapped to \(j \Omega\) axis by solving the above quadratic equations similarly to the band pass filter.  

fig: Impulse Response for BSF

For resistor:  
No change since resistor is frequency independent.  

For Inductor:  

\[
X_L = L s
\]

Replace \(s'\) by \(B \frac{1}{s} + \frac{s}{2}\)

\[
X_L = L \left( B \frac{1}{s} + \frac{s}{2} \right) = \frac{B L}{s} + \frac{L s}{2}
\]

Comparing with standard inductor-capacitor relations, inductor is replaced by inductor and capacitor in parallel:

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

fig: Inductor interms of BSF

For Capacitor:  

\[
X_C = \frac{1}{C s}
\]

Replace \(s\) by \(B \frac{1}{s} + \frac{s}{2}\)

\[
X_C = \frac{1}{C \left( B \frac{1}{s} + \frac{s}{2} \right)} = \frac{s}{C B + \frac{C s^2}{2}} = (C B) s + \frac{C B^2}{s}
\]

Comparing with standard inductor-capacitor relations, capacitor is replaced by inductor and capacitor in series:

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

fig: Capacitance interms of BSF