Filter definition and denormalization from ωold to ωnew

Filter

A filter is a frequency-selective two-port network that passes signals in a chosen band of frequencies (passband) with little attenuation and attenuates signals at other frequencies (stopband).

Significance 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.

New element values for changing $\omega_{old}$ to $\omega_{new}$

Let $k_f = \omega_{new}/\omega_{old}$. For the same response at the new frequency, each element's reactance at $\omega_{new}$ must equal its old reactance at $\omega_{old}$:

$$\begin{aligned} \omega_{new} L_{new} = \omega_{old} L_{old} &\Rightarrow L_{new} = \frac{L_{old}}{k_f} \\ \frac{1}{\omega_{new} C_{new}} = \frac{1}{\omega_{old} C_{old}} &\Rightarrow C_{new} = \frac{C_{old}}{k_f} \\ R_{new} &= R_{old} \end{aligned}$$

The transfer function becomes $T_{new}(s) = T_{old}(s/k_f)$. Combined with impedance scaling by $k_m$: $R' = k_m R$, $L' = k_m L/k_f$, $C' = C/(k_m k_f)$.

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.