---
title: "Leapfrog 4th order Butterworth LPF at 1000 rad/s"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:leapfrog-fourth-order-1000"
source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-fourth-order-1000
site: BCT Engineering
---

# Leapfrog 4th order Butterworth LPF at 1000 rad/s

Filter Design · Chapter 8: Design of High-Order Active Filters

- Asked in: 2082 Bhadra, 6 marks; 2082 Baisakh, 6 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-1000
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

Design the fourth order Butterworth low-pass filter in resistively terminated lossless realization using a leapfrog simulation. Your final design should accommodate half power frequency of 1000 rad/sec and should have practically realizable elements in it. [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 LC ladder elements with op-amp integrators, preserving low sensitivity. For 4th-order Butterworth at 1000 rad/s using prototype L1=0.7654, C2=1.848, L3=1.848, C4=0.7654 and R=10kΩ, integrator capacitors are A1/A5: 76.54 nF, A2/A4: 184.8 nF; two inverters complete the circuit.

## 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, $R_L = 1\ \Omega$ (T-ladder: 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 $\omega_0 = 1000\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$
\begin{aligned}
C_{A1} &= \frac{0.7654}{1000 \times 10^4} = \text{76.54 nF} \\
C_{A2} &= \frac{1.848}{1000 \times 10^4} = \text{184.8 nF} \\
C_{A4} &= \frac{1.848}{1000 \times 10^4} = \text{184.8 nF} \\
C_{A5} &= \frac{0.7654}{1000 \times 10^4} = \text{76.54 nF}
\end{aligned}
$$

**Final element values:**

| Op-amp | Resistors | Capacitor |
|---|---|---|
| A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 76.54 nF |
| A2 | input(s) 10 kΩ | 184.8 nF |
| A4 | input(s) 10 kΩ | 184.8 nF |
| A5 | input(s) 10 kΩ, feedback 10 kΩ across C | 76.54 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.

**Answer:** six op-amps (four integrators, two inverters), all resistors 10 kΩ, integrator capacitors 76.54 nF, 184.8 nF, 184.8 nF and 76.54 nF; half-power frequency 1000 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?
   A. Replaces each inductor by a GIC
   B. Simulates the ladder's state equations using coupled integrators
   C. Uses gyrators for every element
   D. Replaces capacitors with FDNRs
   Answer: B. Each reactive element becomes an op-amp integrator coupled as the ladder voltages and currents are. (Key point: Leapfrog idea)

2. Which integrators are lossy, and why?
   A. The middle ones, because of the capacitors
   B. The first and last, because the terminations add the +1 term
   C. All of them
   D. None, they are ideal
   Answer: B. I1 = (Vin - V2)/(sL1 + 1) and V4 = I3/(sC4 + 1): the terminations give the resistor across C. (Key point: Lossy integrators)

3. How is each integrator capacitor found after frequency scaling to w0 with resistor R?
   A. C = g/(w0 R)
   B. C = g w0 R
   C. C = g R/w0
   D. C = w0/(g R)
   Answer: A. For w0 = 1000 and R = 10 kOhm, C = g/(1e7). (Key point: Capacitor formula)

4. What are the capacitors for 1000 rad/s and R = 10 kOhm?
   A. 76.54 nF, 184.8 nF, 184.8 nF, 76.54 nF
   B. 765.4 nF, 1.848 uF, 1.848 uF, 765.4 nF
   C. 12.18 nF, 29.41 nF, 29.41 nF, 12.18 nF
   D. 7.654 nF, 18.48 nF, 18.48 nF, 7.654 nF
   Answer: A. 0.7654/1e7 = 76.54 nF and 1.848/1e7 = 184.8 nF. (Key point: Capacitor values)

## Close variants set in other exams

- [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 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×

## Related in Filter Design

- [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×
- [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
- [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 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
- [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×