LPF to BPF
QN) How can you obtain Band pass filter from a given low pass filter? Explain with suitable figures and examples.
fig: Impulse response of BPF
We have \(s = j \omega\) for low pass and \(s' = j \Omega\) for band pass filter
fig: LPF to BPF
\[
s' = \omega \left(s^2 + 2 B s \right)
\]
Where \(B\) is the bandwidth of the band pass filter:
\[
B = \Omega_2 - \Omega_1
\]
For normalized value of \(\omega\) (\(\omega = 1 \text{ rad/sec}\)):
\[
s' = s^2 + 2 B s
\]
For the imaginary axis, put \(s' = j \omega\) to map \(s = j \Omega\):
\[
j \omega = \frac{-\Omega^2 + \Omega_0^2}{B j \Omega}
\]
Or
\[
\Omega^2 - B \omega \Omega - \Omega_0^2 = 0 \quad \cdots A
\]
For \(\omega = 0\):
\[
\Omega^2 - \Omega_0^2 = 0 \quad \Rightarrow \quad \Omega = \pm \Omega_0
\]
The solution of the quadratic equation is:
\[
\Omega = \frac{B \omega \pm \sqrt{(B \omega)^2 + 4 \Omega_0^2}}{2}
\]
Hence a low pass band is mapped onto two band pass regions:
\[
B = \Omega_2 - \Omega_1
\]
\[
\Omega_1 \Omega_2 = - \Omega_0^2
\]
The upper band edge \(\Omega_2\) and lower band edge \(\Omega_1\) are symmetric about the center frequency 0.
The quality factor \(Q\) is:
\[
Q = \frac{\Omega_0}{B}
\]
For resistor:
No change since resistor is frequency independent.
For Inductor:
\[
X_L = L s
\]
Replace \(s\) by \((s^2 + 2 B s)\):
\[
X_L = L (s^2 + 2 B s) = L s^2 + 2 L B s
\]
Comparing with standard inductor-capacitor relations, we have:
\[
L' = L B \quad \text{and} \quad C' = \frac{1}{L \Omega_2 B}
\]
Inductor is replaced by a series of inductor and capacitor:
\[
L' = L B
\]
\[
C' = \frac{1}{L \Omega_2 B}
\]
fig: Inductor representation interm of Band Pass FIlter
For Capacitor:
\[
X_C = \frac{1}{C s}
\]
Replace \(s\) by \((s^2 + 2 B s)\):
\[
X_C = \frac{1}{C (s^2 + 2 B s)} = \frac{1}{C s^2 + 2 C B s}
\]
Comparing with standard inductor-capacitor relations, we have:
\[
L' = B C \Omega_2 \quad \text{and} \quad C' = C B
\]
The new components are an inductor and capacitor in parallel.
fig: Inductor representation interm of Band Pass FIlter