Leapfrog 4th order Butterworth LPF at 1000 rad/s

Leapfrog (operational) simulation realises the doubly terminated LC ladder by simulating its equations rather than its elements: each reactive element is replaced by an op-amp integrator, and the integrators are coupled exactly as the ladder voltages and currents are coupled. The result keeps the low sensitivity of the passive ladder.

From the table (n = 4), the doubly terminated Butterworth prototype (1 Ω terminations, $\omega_0 = 1$ rad/s) is $R_s = 1$, $L_1 = 0.7654$ H, $C_2 = 1.848$ F, $L_3 = 1.848$ H, $C_4 = 0.7654$ F, $R_L = 1\ \Omega$ (T-ladder: series L first, shunt C to ground).

Write the state equations of the doubly terminated ladder (normalised, $R_s = R_L = 1\ \Omega$). Every current is turned into a voltage by multiplying by a 1 Ω resistance, so each equation becomes an integrator:

$$\begin{aligned} I_1 &= \frac{V_{in} - V_2}{sL_1 + 1} \\ V_2 &= \frac{I_1 - I_3}{sC_2} \\ I_3 &= \frac{V_2 - V_4}{sL_3} \\ V_4 &= \frac{I_3}{sC_4 + 1} \\ V_o &= V_4 \end{aligned}$$

The first and last equations are lossy integrators (the terminations give the "+1"); the middle ones are lossless integrators. Neighbouring integrators feed each other in alternate directions, which gives the "leapfrog" structure.

Using inverting op-amp integrators, $V_o = -\frac{1}{sRC}\sum V_i$ (lossless) or $V_o = -\frac{\sum V_i/R}{sC + 1/R}$ (lossy, resistor $R$ across $C$), the signs are arranged as follows. Inverters supply the signals of opposite sign:

Check of one stage: A2 gives $-\frac{(-I_1) + I_3}{sC_{A2}} = \frac{I_1 - I_3}{s(1.848)}$, which is the ladder equation for $V_2$.

Denormalisation. Frequency scale by $\omega_0 = 1000\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$\begin{aligned} C_{A1} &= \frac{0.7654}{1000 \times 10^4} = \text{76.54 nF} \\ C_{A2} &= \frac{1.848}{1000 \times 10^4} = \text{184.8 nF} \\ C_{A4} &= \frac{1.848}{1000 \times 10^4} = \text{184.8 nF} \\ C_{A5} &= \frac{0.7654}{1000 \times 10^4} = \text{76.54 nF} \end{aligned}$$

Final element values:

 Lossy integrator (A1 and A5)
              +-----[ C ]-----+
              |               |
              +-----[ R ]-----+
              |               |
 Va --[R]-----+---|-\         |
 Vb --[R]-----+   |  >--------+---- Vo
              GND-|+/
 Lossless integrator (A2 and A4): the same circuit
 without the feedback resistor (only C from output to (-)).

The passband gain is 1/2 (−6 dB), the same as the doubly terminated prototype; the response is that of the passive ladder but with no inductors, and the low sensitivity of the ladder is kept.

Answer: six op-amps (four integrators, two inverters), all resistors 10 kΩ, integrator capacitors 76.54 nF, 184.8 nF, 184.8 nF and 76.54 nF; half-power frequency 1000 rad/s.