--- title: "Leapfrog simulation of 3rd order Chebyshev LPF" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:leapfrog-third-order-chebyshev" source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-third-order-chebyshev site: BCT Engineering --- # Leapfrog simulation of 3rd order Chebyshev LPF Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2073 Shrawan, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-third-order-chebyshev - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question The following circuit is a third-order Chebyshev lowpass filter. Simulate it using the leap-frog method. The final design should have ω0 = 4000 rad/s and practically realizable element values. [Figure: source V1, 1 Ω source resistor, series 2.026 H, shunt 0.9941 F, series 2.026 H, 1 Ω load] ## Short answer Simulate the given third-order Chebyshev ladder using leapfrog: lossy integrators for the inductor currents, lossless for the shunt capacitor voltage, plus two inverters. Denormalize to 4000 rad/s with 10 kΩ resistors. Capacitors: 50.65 nF, 24.85 nF, 50.65 nF. ## Answer **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 (3rd-order Chebyshev, normalised so the passband edge is 1 rad/s): $R_s = 1\ \Omega$, series $L_1 = 2.026$ H, shunt $C_2 = 0.9941$ F, series $L_3 = 2.026$ H, $R_L = 1\ \Omega$. 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: | Op-amp | Type | Inputs (each through $R$) | Output | Normalised $C$ (R = 1 Ω) | |---|---|---|---|---| | A1 | lossy integrator | $V_{in}$, $-V_2$ | $-I_1$ | 2.026 F | | A2 | integrator | $-I_1$, $I_3$ | $V_2$ | 0.9941 F | | A3 | inverter | $V_2$ | $-V_2$ | R / R | | A4 | lossy integrator | $V_2$ | $-I_3$ | 2.026 F | | A5 | inverter | $-I_3$ | $I_3 = V_o$ | R / R | Check of one stage: A2 gives $-\frac{(-I_1) + I_3}{sC_{A2}} = \frac{I_1 - I_3}{s(0.9941)}$, which is the ladder equation for $V_2$. **Denormalisation.** Frequency scale by $\omega_0 = 4000\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$: $$ \begin{aligned} C_{A1} &= \frac{2.026}{4000 \times 10^4} = \text{50.65 nF} \\ C_{A2} &= \frac{0.9941}{4000 \times 10^4} = \text{24.85 nF} \\ C_{A4} &= \frac{2.026}{4000 \times 10^4} = \text{50.65 nF} \end{aligned} $$ **Final element values:** | Op-amp | Resistors | Capacitor | |---|---|---| | A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 50.65 nF | | A2 | input(s) 10 kΩ | 24.85 nF | | A4 | input(s) 10 kΩ, feedback 10 kΩ across C | 50.65 nF | | inverters | 10 kΩ in, 10 kΩ feedback | none | ```text 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 50.65 nF, 24.85 nF, 50.65 nF; passband edge 4000 rad/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the series-first 3rd-order ladder, which is the equation for the last stage? A. I3 = V2/(s L3 + 1), Vo = I3 RL B. I3 = V2 s L3 C. V3 = I3/(s C3) D. I3 = (V2 - Vin)/(s L3) Answer: A. The last inductor is in series with RL, so I3 = V2/(sL3 + RL) with RL = 1, and Vo = I3 RL. This is a lossy integrator. (Key point: State equations) 2. How many integrators and inverters does this design use? A. Five integrators, no inverters B. Two integrators and three inverters C. Three integrators and two inverters D. Three integrators and no inverters Answer: C. A1, A2, A4 are integrators (L1, C2, L3) and A3, A5 are inverters for sign correction. (Key point: Number of op-amps) 3. With R = 10 kohm and w0 = 4000 rad/s, what capacitor corresponds to L1 = 2.026? A. 5.065 nF B. 20.26 nF C. 50.65 nF D. 506.5 nF Answer: C. C = 2.026/(4000 x 10^4) = 50.65 nF. (Key point: Capacitor calculation) 4. A2 has inputs -I1 and I3 into an inverting integrator with C = 0.9941. What is its output? A. -(I1 - I3)/(0.9941 s) B. (I1 - I3)/(0.9941 s), which is V2 C. (I1 + I3)/(0.9941 s) D. (I3 - I1)/s Answer: B. -((-I1)+I3)/(s C) = (I1 - I3)/(s 0.9941), matching the ladder equation for V2. 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