--- title: "Leapfrog simulation of 3rd order Butterworth LPF" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:leapfrog-third-order-butterworth" source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-third-order-butterworth site: BCT Engineering --- # Leapfrog simulation of 3rd order Butterworth LPF Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2070 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-third-order-butterworth - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question Simulate third order Butterworth low pass filter using Leapfrog simulation. [Table: doubly terminated Butterworth element values normalized to 1 rad/s, 1 Ω terminations: n = 2: 1.414, 1.414; n = 3: 1, 2, 1; n = 4: 0.7654, 1.848, 1.848, 0.7654; n = 5: 0.618, 1.618, 2, 1.618, 0.618] ## Short answer Replace LC ladder elements with op-amp integrators simulating state equations: lossy integrators for the terminations, lossless for the middle. Normalized values: Rs=RL=1 Ω, L1=1 H, C2=2 F, L3=1 H. Denormalized to ω0=10^4 rad/s, R=10 kΩ gives capacitors 10 nF, 20 nF, 10 nF; passband gain 1/2 (-6 dB). ## 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. From the table (n = 3): $R_s = 1\ \Omega$, series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H, $R_L = 1\ \Omega$. ```text 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: | Op-amp | Type | Inputs (each through $R$) | Output | Normalised $C$ (R = 1 Ω) | |---|---|---|---|---| | A1 | lossy integrator | $V_{in}$, $-V_2$ | $-I_1$ | 1 F | | A2 | integrator | $-I_1$, $I_3$ | $V_2$ | 2 F | | A3 | inverter | $V_2$ | $-V_2$ | R / R | | A4 | lossy integrator | $V_2$ | $-I_3$ | 1 F | | A5 | inverter | $-I_3$ | $I_3 = V_o$ | R / R | Check of one stage: A2 gives $-\frac{(-I_1) + I_3}{sC_{A2}} = \frac{I_1 - I_3}{s(2)}$, 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:** | Op-amp | Resistors | Capacitor | |---|---|---| | A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 10 nF | | A2 | input(s) 10 kΩ | 20 nF | | A4 | input(s) 10 kΩ, feedback 10 kΩ across C | 10 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. 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does a leapfrog filter keep the low sensitivity of the passive ladder? A. It uses fewer op-amps B. It simulates the ladder's own equations rather than cascading independent sections C. It uses larger capacitors D. It has no resistors Answer: B. The feedback structure mirrors the ladder, so element variations affect the response like they do in the doubly terminated LC ladder. (Key point: Key idea) 2. Which ladder equation is realised by the middle (lossless) integrator A2? A. V2 = (I1 - I3)/(s C2) with C2 = 2 F B. V2 = I3/(s + 1) C. V2 = (Vin - V2)/(s L1 + 1) D. V2 = I1 s C2 Answer: A. V2 is the capacitor voltage fed by I1 - I3; it is lossless since no resistor is in parallel with C2. (Key point: Equation for V2) 3. What are the normalised integrator capacitors for R = 1 ohm and w0 = 1 rad/s? A. 2 F, 1 F, 2 F B. 1 F, 2 F, 1 F C. 0.5 F, 1 F, 0.5 F D. 1 F, 1 F, 1 F Answer: B. The capacitors equal the prototype values L1 = 1, C2 = 2, L3 = 1. (Key point: Normalised capacitors) 4. Choosing w0 = 10^4 rad/s and R = 10 kohm, what capacitors result? A. 1 nF, 2 nF, 1 nF B. 100 nF, 200 nF, 100 nF C. 10 nF, 20 nF, 10 nF D. 10 uF, 20 uF, 10 uF Answer: C. C = g/(w0 R) = g/10^8, so 10 nF, 20 nF, 10 nF. 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