Scaling a ladder to 1000 rad/s and 75 Ω load

Significance of normalization and de-normalization

Filter tables and prototypes are given for $\omega_p$ = 1 rad/s and 1 Ω. Normalization lets one table serve every filter and keeps calculations simple; de-normalization (scaling) converts the prototype to the actual frequency and impedance level with practical element values, without changing the response shape:

$$R' = k_m R, \qquad L' = \frac{k_m L}{k_f}, \qquad C' = \frac{C}{k_m k_f}$$

Numerical

Given prototype ($\omega_p$ = 1 rad/s): shunt $C_1$ = 1.3 F, series $L_2$ = 1.5 H, shunt $C_3$ = 1.2 F, series $L_4$ = 0.5 H, load $R_L$ = 1 Ω.

Required: $\omega_p$ = 1000 rad/s and $R_L$ = 75 Ω.

$$k_f = \frac{1000}{1} = 1000, \qquad k_m = \frac{75}{1} = 75$$ $$\begin{aligned} C_1' &= \frac{1.3}{75 \times 1000} = 17.33\ \mu\text{F} \\ L_2' &= \frac{75 \times 1.5}{1000} = 0.1125\ \text{H} = 112.5\ \text{mH} \\ C_3' &= \frac{1.2}{75 \times 1000} = 16\ \mu\text{F} \\ L_4' &= \frac{75 \times 0.5}{1000} = 0.0375\ \text{H} = 37.5\ \text{mH} \\ R_L' &= 75 \times 1 = 75\ \Omega \end{aligned}$$

        112.5 mH      37.5 mH
 o---+---UUUU----+----UUUU----+
 V1  |           |            |
  17.33uF      16 uF        75R   V2
     |           |            |
 o---+-----------+------------+

Answer: $C_1$ = 17.33 μF, $L_2$ = 112.5 mH, $C_3$ = 16 μF, $L_4$ = 37.5 mH, $R_L$ = 75 Ω; passband edge 1000 rad/s. (If the source has a 1 Ω resistance it also becomes 75 Ω.)