Low pass to band pass transformation

Frequency transformation

Frequency transformation is a change of the complex frequency variable that converts a normalized low-pass prototype filter (cutoff 1 rad/s) into a high-pass, band-pass or band-stop filter (or a low-pass with a different cutoff). The magnitude response of the prototype is mapped onto the new frequency axis, so the same attenuation values appear at the corresponding new frequencies.

Low-pass to band-pass transformation

Transformation: replace $s$ in the low-pass prototype by

$$s \;\rightarrow\; \frac{s^2 + \omega_0^2}{B\,s}, \qquad \omega_0 = \sqrt{\omega_1\omega_2},\quad B = \omega_2 - \omega_1$$

On the $j\omega$ axis this gives $\Omega_{LP} = \dfrac{\omega^2 - \omega_0^2}{B\omega}$. So:

• $\omega = \omega_0 \Rightarrow \Omega_{LP} = 0$ (centre of passband maps to DC of the LP);
• $\omega = \omega_1, \omega_2 \Rightarrow \Omega_{LP} = \mp1$, since $\omega_1\omega_2 = \omega_0^2$ and $\omega_2 - \omega_1 = B$;
• $\omega = 0$ and $\omega = \infty \Rightarrow \Omega_{LP} = \mp\infty$ (stopbands).

The band-pass response is geometrically symmetric about $\omega_0$, and its order is twice that of the prototype.

Element transformation:

Series inductor $L$:

$$Z = L\,\frac{s^2+\omega_0^2}{Bs} = \frac{L}{B}s + \frac{1}{\left(\dfrac{B}{L\omega_0^2}\right)s}$$

becomes a series LC: $L' = \dfrac{L}{B}$, $C' = \dfrac{B}{L\omega_0^2}$.

Shunt capacitor $C$:

$$Y = C\,\frac{s^2+\omega_0^2}{Bs} = \frac{C}{B}s + \frac{1}{\left(\dfrac{B}{C\omega_0^2}\right)s}$$

becomes a parallel LC to ground: $C' = \dfrac{C}{B}$, $L' = \dfrac{B}{C\omega_0^2}$.

Resistor $R$: unchanged (frequency independent). Every new LC pair resonates at $\omega_0$ ($L'C' = 1/\omega_0^2$).

 LP element       BP element
 series L   -->   series L'--C'
 shunt C    -->   shunt  L' || C'
 R          -->   R

Example: third-order Butterworth prototype ($R_s = R_L = 1\ \Omega$, series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H) converted to a band-pass filter with $\omega_0 = 2000$ rad/s and $B = 400$ rad/s ($\omega_0^2 = 4\times10^6$).

$$\begin{aligned} L_1 = 1:&\quad L_1' = \frac{1}{400} = 2.5\ \text{mH},\quad C_1' = \frac{400}{1\times4\times10^6} = 100\ \mu\text{F} \\ C_2 = 2:&\quad C_2' = \frac{2}{400} = 5\ \text{mF},\quad L_2' = \frac{400}{2\times4\times10^6} = 50\ \mu\text{H} \\ L_3 = 1:&\quad L_3' = 2.5\ \text{mH},\quad C_3' = 100\ \mu\text{F} \end{aligned}$$

  Rs=1   [ Z1 ]          [ Z3 ]
 o-/\/\--[    ]---+------[    ]---+-----o
 |                |               |     +
 Vs            [ Y2 ]           RL=1    Vo
 |                |               |     -
 o----------------+---------------+-----o

Check: $1/\sqrt{2.5\text{ mH}\times100\ \mu\text{F}} = 2000$ rad/s. The 3 dB band edges are $\omega_{1,2} = \sqrt{\omega_0^2+(B/2)^2} \mp B/2 = 1810.0$ and $2210.0$ rad/s, so $\omega_2 - \omega_1 = 400$ rad/s. The result is a 6th-order band-pass filter.

These values are for $1\ \Omega$ terminations. For practical values, magnitude-scale by $k_m$ (e.g. $k_m = 1000$ for 1 kΩ terminations): multiply every L by $k_m$ and divide every C by $k_m$.