--- title: "Leapfrog simulation of 4th order Butterworth LPF" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:leapfrog-fourth-order-table" source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-fourth-order-table site: BCT Engineering --- # Leapfrog simulation of 4th order Butterworth LPF Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2081 Baisakh, 8 marks; 2081 Bhadra, 7 marks (asked 2 times) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-fourth-order-table - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question Design (simulate) the 4th order Butterworth LPF in doubly-terminated (resistively-terminated lossless) network using Leapfrog simulation. The necessary information is listed in the given table: Order (n) = 4 and LPF; R1 = 1; L1 = 0.7654; C2 = 1.848; L3 = 1.848; C4 = 0.7654; R2 = 1 ## Short answer Leapfrog simulation realizes the ladder equations using op-amp integrators. For the 4th-order Butterworth prototype (L1=0.7654, C2=1.848, L3=1.848, C4=0.7654) scaled to ω0=6283.2 rad/s (1 kHz) with R=10 kΩ, capacitors are A1/A5: 12.18 nF, A2/A4: 29.41 nF; two inverters complete the structure. ## 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 prototype (normalised to 1 rad/s, 1 Ω terminations): $R_1 = 1$, $L_1 = 0.7654$ H, $C_2 = 1.848$ F, $L_3 = 1.848$ H, $C_4 = 0.7654$ F, $R_2 = 1\ \Omega$. ```text Rs=R1=1 L1=0.7654 L3=1.848 o----[R1]-----(LLL)-----+-----(LLL)-----+-----+-----o Vo | | | C2=1.848 C4=0.7654 [R2=1] | | | GND --------------------+---------------+-----+-----o ``` 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: $$ \begin{aligned} I_1 &= \frac{V_{in} - V_2}{sL_1 + 1} \\ V_2 &= \frac{I_1 - I_3}{sC_2} \\ I_3 &= \frac{V_2 - V_4}{sL_3} \\ V_4 &= \frac{I_3}{sC_4 + 1} \\ V_o &= V_4 \end{aligned} $$ 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 | $V_{in}$, $-V_2$ | $-I_1$ | 0.7654 F | | A2 | integrator | $-I_1$, $I_3$ | $V_2$ | 1.848 F | | A3 | inverter | $V_2$ | $-V_2$ | R / R | | A4 | integrator | $-V_2$, $V_4$ | $I_3$ | 1.848 F | | A5 | lossy integrator | $I_3$ | $-V_4$ | 0.7654 F | | A6 | inverter | $-V_4$ | $V_4 = V_o$ | R / R | Check of one stage: A2 gives $-\frac{(-I_1) + I_3}{sC_{A2}} = \frac{I_1 - I_3}{s(1.848)}$, which is the ladder equation for $V_2$. **Denormalisation.** Frequency scale by $\omega_0 = 2\pi(1000) = 6283.2\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$: $$ \begin{aligned} C_{A1} &= \frac{0.7654}{6283.19 \times 10^4} = \text{12.18 nF} \\ C_{A2} &= \frac{1.848}{6283.19 \times 10^4} = \text{29.41 nF} \\ C_{A4} &= \frac{1.848}{6283.19 \times 10^4} = \text{29.41 nF} \\ C_{A5} &= \frac{0.7654}{6283.19 \times 10^4} = \text{12.18 nF} \end{aligned} $$ **Final element values:** | Op-amp | Resistors | Capacitor | |---|---|---| | A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 12.18 nF | | A2 | input(s) 10 kΩ | 29.41 nF | | A4 | input(s) 10 kΩ | 29.41 nF | | A5 | input(s) 10 kΩ, feedback 10 kΩ across C | 12.18 nF | | inverters | 10 kΩ in, 10 kΩ feedback | none | ```text Lossy integrator (A1 and A5) +-----[ C ]-----+ | | +-----[ R ]-----+ | | Va --[R]-----+---|-\ | Vb --[R]-----+ | >--------+---- Vo GND-|+/ Lossless integrator (A2 and A4): the same circuit without the feedback resistor (only C from output to (-)). ``` 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. Since no frequency is given, the normalised design (all R = 1 Ω, C = 0.7654, 1.848, 1.848, 0.7654 F) is the answer; the 1 kHz values above show how it is made practical. **Answer (1 kHz example):** R = 10 kΩ everywhere, capacitors 12.18 nF, 29.41 nF, 29.41 nF, 12.18 nF. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are inverters needed in the leapfrog circuit using inverting integrators? A. To supply signals of opposite sign to the integrator inputs B. To add gain C. To block DC D. To realise the terminations Answer: A. The state equations need both signs of V2 and V4, which inverters provide. (Key point: Why inverters) 2. What is the passband gain of the doubly terminated Butterworth leapfrog filter? A. 1 B. 1/2 (-6 dB) C. 2 D. 0.707 Answer: B. Equal source and load resistors form a divider, so DC gain is 1/2. (Key point: Passband gain) 3. For 1 kHz, what w0 is used? A. 1000 rad/s B. 2 pi x 1000 = 6283.2 rad/s C. 500 rad/s D. 2000 rad/s Answer: B. w0 = 2 pi f = 6283.2 rad/s. (Key point: Frequency scaling) 4. With R = 10 kOhm and w0 = 6283.2, what are the capacitors for 0.7654 and 1.848? A. 12.18 nF and 29.41 nF B. 76.54 nF and 184.8 nF C. 1.218 nF and 2.941 nF D. 121.8 nF and 294.1 nF Answer: A. 0.7654/(6283.19 x 1e4) = 12.18 nF; 1.848/(6283.19 x 1e4) = 29.41 nF. (Key point: Capacitor values) ## Related in Filter Design - [FDNR simulation of 4th order Butterworth LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fourth-order-table) — Filter Design, chapter 8 - [Leapfrog 4th order Butterworth LPF at 1000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-fourth-order-1000) — Filter Design, chapter 8, asked 4× - [Leapfrog simulation of 3rd order Butterworth LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-third-order-butterworth) — Filter Design, chapter 8 - [Leapfrog 4th order Butterworth LPF at 10,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-fourth-order-10000) — Filter Design, chapter 8, asked 6× - [Leapfrog 4th order Butterworth LPF at 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-fourth-order-4000) — Filter Design, chapter 8, asked 4× - [4th order Butterworth HPF lossless ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afourth-order-butterworth-hpf-ladder) — Filter Design, chapter 5