LPF to LPF Transformation

fig: LPF to LPF Transform

 

For old low-pass: \(\text{cut-off frequency} = \omega\) for \(s = \text{cut-off frequency } \omega\)  

For new low-pass: \(\text{cut-off frequency} = \Omega\)  

Similarly, for \(s'\) (Response) = cut-off frequency = \(\Omega\)  

Then we have:  
For new response, the ratio can be \(\omega s\)  
Replace \(s\) by \(\omega s\)  
For normalized value for \(\omega\) (\(\omega = 1 \text{ rad/sec}\)), then  
\[
s' = \omega s
\]  

The frequency response of the new LPF remains same whereas the cut-off of new low-pass filter changes.  

Eg:  
\[
T_{LP}(s) = \frac{1}{s+1}
\]  

New transfer function is given by:  
Replace \(s\) with \(s'\), then  
\[
T_{\text{NewLP}}(s) = \frac{1}{s'+1} = \frac{1}{s + \Omega}
\]  

The new component is given as:  

For Resistor:  
No change since resistor is frequency independent.  

For Inductor:  
\[
X_L = L \cdot s
\]  
Replace \(s\) by \(s'\),  
\[
X'_L = L \cdot s' = L \cdot s
\]  
\[
L_{\text{new}} = L_{\text{old}}
\]  

For Capacitor:  
\[
X_C = \frac{1}{C \cdot s}
\]  
Replace \(s\) by \(s'\),  
\[
X'_C = \frac{1}{C \cdot s'} = \frac{1}{C \cdot s}
\]  
\[
C_{\text{new}} = C_{\text{old}}
\]