--- title: "Leapfrog simulation of given ladder at 6 kHz" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:leapfrog-ladder-6khz" source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-ladder-6khz site: BCT Engineering --- # Leapfrog simulation of given ladder at 6 kHz Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2071 Chaitra, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-ladder-6khz - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question Using leapfrog method simulate the LC ladder circuit given below to obtain a low pass filter having passband of 6KHz and suitable element values. [Figure: source V1, 1 Ω source resistor, shunt 2.0237 F, series 0.9941 H, shunt 2.0237 F, 1 Ω load (output V2)] ## Short answer Simulate the π-type ladder using leapfrog: lossy integrator for first shunt capacitor voltage, lossless for series inductor current, lossy for last shunt capacitor voltage, plus two inverters. Scale to passband edge 6 kHz (ω0=37699 rad/s) with 10 kΩ resistors. Capacitors: 5.368 nF, 2.637 nF, 5.368 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 (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: | Op-amp | Type | Inputs (each through $R$) | Output | Normalised $C$ (R = 1 Ω) | |---|---|---|---|---| | A1 | lossy integrator | $V_{in}$, $-I_2$ | $-V_1$ | 2.0237 F | | A2 | integrator | $-V_1$, $V_3$ | $I_2$ | 0.9941 F | | A3 | inverter | $I_2$ | $-I_2$ | R / R | | A4 | lossy integrator | $I_2$ | $-V_3$ | 2.0237 F | | A5 | inverter | $-V_3$ | $V_3 = V_o$ | R / R | 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:** | Op-amp | Resistors | Capacitor | |---|---|---| | A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 5.368 nF | | A2 | input(s) 10 kΩ | 2.637 nF | | A4 | input(s) 10 kΩ, feedback 10 kΩ across C | 5.368 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 5.368 nF, 2.637 nF, 5.368 nF; passband edge 6 kHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For this pi-form ladder (shunt C, series L, shunt C), which are the state variables? A. Inductor voltages only B. Capacitor voltages V1 and V3 and inductor current I2 C. All branch currents D. Only the output voltage Answer: B. The state variables are the energy-storage quantities: capacitor voltages V1, V3 and inductor current I2. (Key point: State variables in pi ladder) 2. Which equation is correct for V1? A. V1 = (Vin - I2)/(s C1 + 1) B. V1 = I2/(s C1) C. V1 = (Vin + I2) s C1 D. V1 = (V1 - V3)/(s L2) Answer: A. With the source converted to a current Vin/Rs, the net current into C1 and Rs is Vin - I2, giving a lossy integrator. (Key point: State equation for V1) 3. What w0 is used for a 6 kHz passband? A. 6000 rad/s B. 12,000 rad/s C. 37,699 rad/s D. 1000 rad/s Answer: C. w0 = 2 pi x 6000 = 37,699 rad/s. (Key point: Frequency scale) 4. With R = 10 kohm, what capacitor is used for the 2.0237 F elements? A. 5.368 nF B. 2.637 nF C. 53.68 nF D. 20.24 nF Answer: A. C = 2.0237/(37699 x 10^4) = 5.368 nF; the 0.9941 F element gives 2.637 nF. 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