Band pass (10⁴ rad/s, 9.9×10⁴ rad/s) from given ladder

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. For band-pass: $s \to \dfrac{s^2+\omega_0^2}{Bs}$.

Band-pass filter from the given low-pass filter

Given: $\omega_0 = 10^4$ rad/s ($\omega_0^2 = 10^8$), $B = 9.9\times10^4$ rad/s (values used as given; they correspond to band edges of about $\omega_1 \approx 1000$ and $\omega_2 \approx 10^5$ rad/s, since $\omega_1\omega_2 = 10^8$ and $\omega_2 - \omega_1 = 9.9\times10^4$).

Prototype (Figure 1): $R_1 = 1\ \Omega$, series $L_1 = 2.024$ H, shunt $C_1 = 0.994$ F, series $L_2 = 2.024$ H, $R_2 = 1\ \Omega$.

Element rules: series $L$ → series $L/B$ and $B/(L\omega_0^2)$; shunt $C$ → parallel $C/B$ and $B/(C\omega_0^2)$.

$$\begin{aligned} L_1 = 2.024:&\quad L_1' = \frac{2.024}{9.9\times10^4} = 20.444\ \mu\text{H},\quad C_1' = \frac{9.9\times10^4}{2.024\times10^8} = 489.13\ \mu\text{F} \\ C_1 = 0.994:&\quad C' = \frac{0.994}{9.9\times10^4} = 10.040\ \mu\text{F},\quad L' = \frac{9.9\times10^4}{0.994\times10^8} = 995.98\ \mu\text{H} \\ L_2 = 2.024:&\quad L_2' = 20.444\ \mu\text{H},\quad C_2' = 489.13\ \mu\text{F} \end{aligned}$$

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

Check: $1/\sqrt{20.444\ \mu\text{H}\times489.13\ \mu\text{F}} = 10^4$ rad/s. 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 $k_m$.