Frequency and magnitude scaling to 10⁵ rad/s

Normalization and denormalization

Normalization is designing the filter for $\omega_o$ = 1 rad/s and 1 Ω terminations, so that standard tables and simple numbers can be used. Denormalization is converting the normalized values to the actual frequency and impedance level using 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: $R_1 = R_2 = 1\ \Omega$, $L_1 = L_2 = 2.024$ H, $C_1 = 0.994$ F at $\omega_o$ = 1 rad/s. Required $\omega_o = 10^5$ rad/s.

Frequency scaling factor: $k_f = 10^5/1 = 10^5$.

Magnitude scaling factor: with $k_m = 1$ the capacitor would be $0.994/10^5 \approx 10\ \mu$F and the resistors 1 Ω, which are not practical. Choose $k_m = 1000$ (1 kΩ terminations).

$$\begin{aligned} R_1' = R_2' &= k_m R = 1000 \times 1 = 1\ \text{k}\Omega \\ L_1' = L_2' &= \frac{k_m L}{k_f} = \frac{1000 \times 2.024}{10^5} = 20.24\ \text{mH} \\ C_1' &= \frac{C}{k_m k_f} = \frac{0.994}{1000 \times 10^5} = 9.94\ \text{nF} \end{aligned}$$

 1k     20.24 mH      20.24 mH
 o-/\/\--UUUU----+----UUUU----+
 V1              |            |
              9.94 nF        1k   V2
                 |            |
 o---------------+------------+

Answer: $R_1 = R_2$ = 1 kΩ, $L_1 = L_2$ = 20.24 mH, $C_1$ = 9.94 nF (≈ 10 nF standard), with $\omega_o = 10^5$ rad/s (≈ 15.9 kHz).

Check: $L'C'$ product scales by $1/k_f^2$: $2.024\times0.994\times10^{-10} = 20.24\times10^{-3}\times9.94\times10^{-9}$, so the response shape is unchanged and only moved to $10^5$ rad/s. Any other $k_m$ (e.g. 10 kΩ) is also correct if it gives practical values.