--- title: "Leapfrog 4th order Butterworth LPF at 4000 rad/s" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:leapfrog-fourth-order-4000" source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-fourth-order-4000 site: BCT Engineering --- # Leapfrog 4th order Butterworth LPF at 4000 rad/s Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2075 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-fourth-order-4000 - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question Design a fourth order Butterworth low pass filter having half power frequency of 4000 rad/s 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 Leapfrog simulation replaces ladder reactive elements with op-amp integrators, keeping low sensitivity. For 4th-order Butterworth at ωo=4000 rad/s, use all resistors R=10kΩ. Lossy integrators (first and last) have C=19.14 nF; lossless integrators (middle two) have C=46.2 nF each; two inverters. Passband gain is –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 = 4), the doubly terminated Butterworth prototype (1 Ω terminations, $\omega_0 = 1$ rad/s) is $R_s = 1$, $L_1 = 0.7654$ H, $C_2 = 1.848$ F, $L_3 = 1.848$ H, $C_4 = 0.7654$ F, $R_L = 1\ \Omega$ (series L first, shunt C to ground). 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 $\omega_0 = 4000\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$: $$ \begin{aligned} C_{A1} &= \frac{0.7654}{4000 \times 10^4} = \text{19.14 nF} \\ C_{A2} &= \frac{1.848}{4000 \times 10^4} = \text{46.2 nF} \\ C_{A4} &= \frac{1.848}{4000 \times 10^4} = \text{46.2 nF} \\ C_{A5} &= \frac{0.7654}{4000 \times 10^4} = \text{19.14 nF} \end{aligned} $$ **Final element values:** | Op-amp | Resistors | Capacitor | |---|---|---| | A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 19.14 nF | | A2 | input(s) 10 kΩ | 46.2 nF | | A4 | input(s) 10 kΩ | 46.2 nF | | A5 | input(s) 10 kΩ, feedback 10 kΩ across C | 19.14 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:** four integrators and two inverters, all resistors 10 kΩ; integrator capacitors 19.14 nF, 46.2 nF, 46.2 nF, 19.14 nF; half-power frequency 4000 rad/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does leapfrog simulation do to a doubly terminated LC ladder? A. Replaces each inductor by a gyrator B. Simulates the ladder's state equations with op-amp integrators coupled like the ladder voltages and currents C. Cascades biquads of different Q D. Replaces capacitors by FDNRs Answer: B. Each reactive element becomes an op-amp integrator wired according to the ladder equations, keeping the ladder's low sensitivity. (Key point: Leapfrog idea) 2. Which integrators in the 4th-order design are lossy (resistor across C), and why? A. The middle two, because of the inductor B. All four C. The first and last, because the terminations give the +1 term D. None; only inverters are lossy Answer: C. I1 = (Vin - V2)/(sL1 + 1) and V4 = I3/(sC4 + 1) include the source and load resistances, so those integrators have a resistor across C. (Key point: Lossy vs lossless integrators) 3. With R = 10 kohm and w0 = 4000 rad/s, what is the capacitor for g = 1.848? A. 46.2 nF B. 4.62 nF C. 19.13 nF D. 462 nF Answer: A. C = g/(w0 R) = 1.848/(4000 x 10^4) = 46.2 nF. (Key point: Capacitor values) 4. Why are inverters A3 and A6 needed? A. To amplify the signal B. To supply signals of opposite sign, since inverting integrators flip the sign C. To set the cutoff D. To provide DC bias Answer: B. Each inverting integrator negates its output, so inverters give the opposite-sign signals needed for the alternating feed structure. (Key point: Role of inverters) ## Close variants set in other exams - [FDNR 4th order Butterworth LPF at 16,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fourth-order-16000) — Filter Design, chapter 8, asked 5× - [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× - [FDNR 4th order Butterworth LPF at 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fourth-order-4000) — Filter Design, chapter 8, asked 3× ## Related in Filter Design - [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 simulation of 4th order Butterworth LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-fourth-order-table) — Filter Design, chapter 8, asked 2× - [FDNR 4th order Butterworth LPF at 20,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fourth-order-20000) — Filter Design, chapter 8 - [Leapfrog 3rd order Butterworth LPF at 4 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-third-order-4khz) — Filter Design, chapter 8, asked 4× - [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