Element values for changing half-power frequency

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.

Changing the half-power frequency from $\omega_0$ to $\omega_n$

Let the frequency scaling factor be

$$k_f = \frac{\omega_n}{\omega_0}$$

The new filter must show at $\omega_n = k_f\omega_0$ the same response the old one had at $\omega_0$. So each element's impedance at the new frequency must equal the old element's impedance at the old frequency.

Inductor:

$$j\omega_n L_n = j\omega_0 L_0 \;\Rightarrow\; L_n = \frac{\omega_0}{\omega_n}L_0 = \frac{L_0}{k_f}$$

Capacitor:

$$\frac{1}{j\omega_n C_n} = \frac{1}{j\omega_0 C_0} \;\Rightarrow\; C_n = \frac{\omega_0}{\omega_n}C_0 = \frac{C_0}{k_f}$$

Resistor: the impedance of a resistor does not depend on frequency, so $R_n = R_0$.

Thus $T_n(s) = T_0(s/k_f)$ and the half-power point moves from $\omega_0$ to $\omega_n$ with the same shape of response.

If the impedance level is also changed by $k_m$ (to get practical values):

$$R_n = k_m R_0, \qquad L_n = \frac{k_m L_0}{k_f}, \qquad C_n = \frac{C_0}{k_m k_f}$$

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.