Normalization, denormalization and element scaling equations

Importance of normalization and denormalization

Normalization means designing the filter with convenient reference values, usually a cut-off (or half-power) frequency of $\omega_o = 1$ rad/s and a termination of $R = 1\ \Omega$. Denormalization (scaling) converts the normalized element values to the actual frequency and impedance level required.

Importance:

• Filter tables (Butterworth, Chebyshev, Bessel poles and ladder element values) are published only in normalized form; one table serves every frequency and impedance level.
• Calculations use simple numbers like 1, 1.414, 2 instead of values like $10^{-9}$ and $10^{4}$, so errors are fewer.
• Designs can be compared and reused: a 1 rad/s prototype becomes a 1 kHz or 1 MHz filter by simple scaling.
• Scaling lets the designer choose practical element values (e.g. capacitors in nF–μF, resistors in kΩ) without changing the shape of the response.

Magnitude (impedance) scaling

Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change.

$$\begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned}$$

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Combined scaling

Applying both:

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

Example: the normalized third-order Butterworth ladder (1 Ω, 1 H, 2 F, 1 H, 1 Ω, $\omega_o$ = 1 rad/s) scaled to $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and 1 kΩ terminations ($k_m = 1000$) gives $R$ = 1 kΩ, $L = 1\times1000/10^4 = 0.1$ H and $C = 2/(1000\times10^4) = 0.2\ \mu$F.