Leapfrog simulation of 4th order Butterworth LPF

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.

Given prototype (normalised to 1 rad/s, 1 Ω terminations): $R_1 = 1$, $L_1 = 0.7654$ H, $C_2 = 1.848$ F, $L_3 = 1.848$ H, $C_4 = 0.7654$ F, $R_2 = 1\ \Omega$.

 Rs=R1=1      L1=0.7654        L3=1.848
 o----[R1]-----(LLL)-----+-----(LLL)-----+-----+-----o Vo
                         |               |     |
                     C2=1.848     C4=0.7654  [R2=1]
                         |               |     |
 GND --------------------+---------------+-----+-----o

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 $\omega_0 = 2\pi(1000) = 6283.2\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$\begin{aligned} C_{A1} &= \frac{0.7654}{6283.19 \times 10^4} = \text{12.18 nF} \\ C_{A2} &= \frac{1.848}{6283.19 \times 10^4} = \text{29.41 nF} \\ C_{A4} &= \frac{1.848}{6283.19 \times 10^4} = \text{29.41 nF} \\ C_{A5} &= \frac{0.7654}{6283.19 \times 10^4} = \text{12.18 nF} \end{aligned}$$

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.

Since no frequency is given, the normalised design (all R = 1 Ω, C = 0.7654, 1.848, 1.848, 0.7654 F) is the answer; the 1 kHz values above show how it is made practical.

Answer (1 kHz example): R = 10 kΩ everywhere, capacitors 12.18 nF, 29.41 nF, 29.41 nF, 12.18 nF.