Leapfrog simulation of 3rd 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.

From the table (n = 3): $R_s = 1\ \Omega$, series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H, $R_L = 1\ \Omega$.

 Vin  Rs=1   L1=1H        L3=1H
 o---[ R ]--(LLL)--+--(LLL)--+----o Vo
                   |         |
                 C2=2F     [RL=1]
                   |         |
 GND --------------+---------+

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:

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

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 $-\frac{(-I_1) + I_3}{sC_{A2}} = \frac{I_1 - I_3}{s(2)}$, which is the ladder equation for $V_2$.

Denormalisation. Frequency scale by $\omega_0 = 10^4\ \text{rad/s}$ (an example, since no frequency is given) and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$\begin{aligned} C_{A1} &= \frac{1}{10000 \times 10^4} = \text{10 nF} \\ C_{A2} &= \frac{2}{10000 \times 10^4} = \text{20 nF} \\ C_{A4} &= \frac{1}{10000 \times 10^4} = \text{10 nF} \end{aligned}$$

Final element values:

 Lossy integrator (the first and last stages)

                +-----[ Rf = R ]-----+
                |                    |
                +-------| C |--------+
                |                    |
 Va --[ R ]-----+                    |
 Vb --[ R ]-----+-----(-)            |
                          A ---------+---- Vo
          GND ----------(+)

 Lossless integrator: the same circuit without Rf
 (only C from the output to the (-) input).

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.

The normalised simulation ($\omega_0$ = 1 rad/s) has all resistors 1 Ω and capacitors 1 F, 2 F, 1 F.

Answer (example, $\omega_0$ = 10⁴ rad/s): all resistors 10 kΩ; integrator capacitors 10 nF, 20 nF, 10 nF.