Leapfrog simulation of given ladder at 6 kHz

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 ladder (π form, a 1 dB-ripple Chebyshev prototype with passband edge 1 rad/s): $R_s = 1\ \Omega$, shunt $C_1 = 2.0237$ F, series $L_2 = 0.9941$ H, shunt $C_3 = 2.0237$ F, $R_L = 1\ \Omega$. Here the state variables are the capacitor voltages $V_1$, $V_3$ and the inductor current $I_2$ (the source is converted to a current $V_{in}/R_s$).

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:

$$\begin{aligned} V_1 &= \frac{V_{in} - I_2}{sC_1 + 1} \\ I_2 &= \frac{V_1 - V_3}{sL_2} \\ V_3 &= \frac{I_2}{sC_3 + 1} \\ V_o &= V_3 \end{aligned}$$

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 $-\frac{(-V_1) + V_3}{sC_{A2}} = \frac{V_1 - V_3}{s(0.9941)}$, which is the ladder equation for $I_2$.

Denormalisation. Frequency scale by $\omega_0 = 2\pi(6000) = 37{,}699\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$\begin{aligned} C_{A1} &= \frac{2.0237}{37699.1 \times 10^4} = \text{5.368 nF} \\ C_{A2} &= \frac{0.9941}{37699.1 \times 10^4} = \text{2.637 nF} \\ C_{A4} &= \frac{2.0237}{37699.1 \times 10^4} = \text{5.368 nF} \end{aligned}$$

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.

Answer: three integrators and two inverters, all resistors 10 kΩ; integrator capacitors 5.368 nF, 2.637 nF, 5.368 nF; passband edge 6 kHz.