Band pass (1500 rad/s, 300 rad/s) from 4th order ladder

Use the low-pass to band-pass transformation $s \to \dfrac{s^2+\omega_0^2}{Bs}$ with $\omega_0 = 1500$ rad/s ($\omega_0^2 = 2.25\times10^6$) and $B = 300$ rad/s.

Prototype (table of doubly terminated Butterworth values, $n = 4$, $R_s = R_L = 1\ \Omega$, taken as starting with a series inductor): $L_1 = 0.7654$ H, $C_2 = 1.848$ F, $L_3 = 1.848$ H, $C_4 = 0.7654$ F.

Element rules:

• series $L$ → series $L' = L/B$ and $C' = B/(L\omega_0^2)$;
• shunt $C$ → parallel $C' = C/B$ and $L' = B/(C\omega_0^2)$.

$$\begin{aligned} L_1:&\quad L_1' = \frac{0.7654}{300} = 2.5513\ \text{mH},\quad C_1' = \frac{300}{0.7654\times2.25\times10^6} = 174.20\ \mu\text{F} \\ C_2:&\quad C_2' = \frac{1.848}{300} = 6.16\ \text{mF},\quad L_2' = \frac{300}{1.848\times2.25\times10^6} = 72.150\ \mu\text{H} \\ L_3:&\quad L_3' = \frac{1.848}{300} = 6.16\ \text{mH},\quad C_3' = \frac{300}{1.848\times2.25\times10^6} = 72.150\ \mu\text{F} \\ C_4:&\quad C_4' = \frac{0.7654}{300} = 2.5513\ \text{mF},\quad L_4' = \frac{300}{0.7654\times2.25\times10^6} = 174.20\ \mu\text{H} \end{aligned}$$

  Rs=1  [ Z1 ]        [ Z3 ]
 o-/\/\-[    ]--+-----[    ]--+------+---o
 |              |             |      |   +
 Vs          [ Y2 ]        [ Y4 ]  RL=1  Vo
 |              |             |      |   -
 o--------------+-------------+------+---o

Each LC pair resonates at 1500 rad/s. The result is an 8th-order band-pass filter with 3 dB edges at $\omega_{1,2} = \sqrt{1500^2+150^2} \mp 150 = 1357.5$ and $1657.5$ 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 $k_m$.