Scaling a Butterworth ladder to 100 Hz

Normalization and de-normalization

Normalization is designing a filter for a reference frequency of 1 rad/s and a reference impedance of 1 Ω, so standard tables and simple numbers can be used. De-normalization is converting the normalized element values to the actual cut-off frequency and impedance level by frequency scaling ($k_f$) and magnitude scaling ($k_m$):

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

Numerical

Given: $R_s = R_L = 1\ \Omega$, $L_1 = L_3 = 1$ H, $C_2 = 2$ F, $\omega_o$ = 1 rad/s. Required: $f_o$ = 100 Hz, capacitor = 0.01 μF.

Frequency scaling factor:

$$k_f = \frac{\omega_{new}}{\omega_{old}} = \frac{2\pi\times100}{1} = 628.32$$

Magnitude scaling factor (chosen so that the 2 F capacitor becomes 0.01 μF):

$$\begin{aligned} C' = \frac{C}{k_m k_f} &\Rightarrow k_m = \frac{C}{k_f C'} = \frac{2}{628.32 \times 0.01\times10^{-6}} \\ &= 3.183\times10^{5} \end{aligned}$$

New element values:

$$\begin{aligned} R_s' = R_L' &= k_m \times 1 = 318.3\ \text{k}\Omega \\ L_1' = L_3' &= \frac{k_m L}{k_f} = \frac{3.183\times10^5 \times 1}{628.32} = 506.6\ \text{H} \\ C_2' &= 0.01\ \mu\text{F} \end{aligned}$$

 318.3k   506.6 H      506.6 H
 o-/\/\/--UUUU----+----UUUU----+
 Vs               |            |
               0.01uF       318.3k  Vo
                  |            |
 o----------------+------------+

Answer: $R_s = R_L$ = 318.3 kΩ, $L_1 = L_3$ = 506.6 H, $C_2$ = 0.01 μF; half-power frequency 100 Hz (628.3 rad/s).

Note: the inductors are very large because the capacitor is small and the frequency is low. In practice such a filter would be built as an active RC filter or with inductors simulated by a GIC.