Band pass (400 rad/s, 150 rad/s) from 1 dB ladder

Use the low-pass to band-pass transformation $s \to \dfrac{s^2+\omega_0^2}{Bs}$ with $\omega_0 = 400$ rad/s ($\omega_0^2 = 1.6\times10^5$) and $B = 150$ rad/s. The prototype has $\alpha_p = 1$ dB at $\omega_p = 1$ rad/s, so the band-pass filter will have 1 dB ripple between band edges $\omega_{1,2} = \sqrt{400^2+75^2} \mp 75 = 331.97$ and $481.97$ rad/s.

Element rules:

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

Given ladder: series 1.2817 H, shunt 1.9093 F, series 1.4126 H, shunt 1.0495 F, 1 Ω terminations.

$$\begin{aligned} 1.2817\ \text{H}:&\quad L' = \frac{1.2817}{150} = 8.5447\ \text{mH},\quad C' = \frac{150}{1.2817\times1.6\times10^5} = 731.45\ \mu\text{F} \\ 1.9093\ \text{F}:&\quad C' = \frac{1.9093}{150} = 12.729\ \text{mF},\quad L' = \frac{150}{1.9093\times1.6\times10^5} = 491.02\ \mu\text{H} \\ 1.4126\ \text{H}:&\quad L' = \frac{1.4126}{150} = 9.4173\ \text{mH},\quad C' = \frac{150}{1.4126\times1.6\times10^5} = 663.67\ \mu\text{F} \\ 1.0495\ \text{F}:&\quad C' = \frac{1.0495}{150} = 6.9967\ \text{mF},\quad L' = \frac{150}{1.0495\times1.6\times10^5} = 893.28\ \mu\text{H} \end{aligned}$$

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

Each pair resonates at 400 rad/s (e.g. $1/\sqrt{8.5447\text{ mH}\times731.45\ \mu\text{F}} = 400$). 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 $k_m$.