Band pass (1 krad/s, 100 rad/s) from 4th order ladder

Frequency transformation is the change of the complex-frequency variable $s$ in a normalized low-pass prototype (cut-off $\omega = 1$ rad/s) so that the same circuit or transfer function gives a high-pass, band-pass or band-stop response, or a low-pass response at another frequency. Typical substitutions are $s \to s/\omega_p$ (LP→LP), $s \to \omega_p/s$ (LP→HP) and $s \to \dfrac{s^2+\omega_0^2}{Bs}$ (LP→BP).

Importance

• Only one set of design tables (Butterworth, Chebyshev, Bessel low-pass prototypes) is needed; every other filter type is derived from it.
• The approximation is done once, on the simple low-pass problem; the transformed filter keeps the same ripple and selectivity. Note that a band-pass or band-stop filter has twice the order of the prototype (here a 4th-order low-pass gives an 8th-order band-pass).
• Element values can be converted directly in the circuit (each $L$ and $C$ is replaced by a simple combination), so no new synthesis is needed.
• Normalized values ($\omega = 1$, $1\ \Omega$) are easy to handle and are later scaled to practical values.

Band-pass design

Prototype (Table 2, 4th-order Butterworth, $R_s = R_L = 1\ \Omega$, $\omega_c = 1$ rad/s): $g_1 = 0.7654$, $g_2 = 1.848$, $g_3 = 1.848$, $g_4 = 0.7654$. Taking the ladder that starts with a series inductor: $L_1 = 0.7654$ H (series), $C_2 = 1.848$ F (shunt), $L_3 = 1.848$ H (series), $C_4 = 0.7654$ F (shunt).

Given $\omega_0 = 1000$ rad/s and $B = 100$ rad/s, so $\omega_0^2 = 10^6$.

LP → BP rules (substitute $s \to \dfrac{s^2+\omega_0^2}{Bs}$):

Each new pair resonates at $\omega_0$ since $L'C' = L''C'' = 1/\omega_0^2$.

Calculations

$$\begin{aligned} L_1' &= \frac{0.7654}{100} = 7.654\ \text{mH}, & C_1' &= \frac{100}{0.7654\times10^6} = 130.65\ \mu\text{F} \\ C_2'' &= \frac{1.848}{100} = 18.48\ \text{mF}, & L_2'' &= \frac{100}{1.848\times10^6} = 54.11\ \mu\text{H} \\ L_3' &= \frac{1.848}{100} = 18.48\ \text{mH}, & C_3' &= \frac{100}{1.848\times10^6} = 54.11\ \mu\text{F} \\ C_4'' &= \frac{0.7654}{100} = 7.654\ \text{mF}, & L_4'' &= \frac{100}{0.7654\times10^6} = 130.65\ \mu\text{H} \end{aligned}$$

Terminations stay $R_s = R_L = 1\ \Omega$.

o-[Rs]-[L1 C1]-+-[L3 C3]-+-----+
               |         |     |
          [L2||C2]  [L4||C4]  [RL]
               |         |     |
o--------------+---------+-----+

Check: $L_1'C_1' = 7.654\times10^{-3}\times130.65\times10^{-6} = 10^{-6} = 1/\omega_0^2$. The resulting filter is 8th order, centred at 1000 rad/s with a 3 dB bandwidth of 100 rad/s. For practical values, the circuit may finally be impedance scaled (e.g. to 1 kΩ: multiply each $L$ by 1000, divide each $C$ by 1000).

Answer: series arms 7.654 mH + 130.65 µF and 18.48 mH + 54.11 µF; shunt arms 54.11 µH ∥ 18.48 mF and 130.65 µH ∥ 7.654 mF, with 1 Ω terminations.