---
title: "Leapfrog 3rd order Butterworth LPF at 4 kHz"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:leapfrog-third-order-4khz"
source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-third-order-4khz
site: BCT Engineering
---

# Leapfrog 3rd order Butterworth LPF at 4 kHz

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

- Asked in: 2072 Kartik, 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-4khz
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

Design third order Butterworth low pass filter using Leapfrog simulation. Your final design should have half power frequency of 4KHz and practically realizable elements. [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

Use a third-order Butterworth prototype (L1=1, C2=2, L3=1) and simulate with leapfrog integrators: lossy first and last, lossless middle, with two inverters for sign. Scale to half-power frequency 4 kHz (ω0=25133 rad/s) with 10 kΩ resistors. Capacitors: 3.979 nF, 7.958 nF, 3.979 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.

From the table (n = 3): $R_s = 1\ \Omega$, series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H, $R_L = 1\ \Omega$.

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 $\omega_0 = 2\pi(4000) = 25{,}133\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$
\begin{aligned}
C_{A1} &= \frac{1}{25132.7 \times 10^4} = \text{3.979 nF} \\
C_{A2} &= \frac{2}{25132.7 \times 10^4} = \text{7.958 nF} \\
C_{A4} &= \frac{1}{25132.7 \times 10^4} = \text{3.979 nF}
\end{aligned}
$$

**Final element values:**

| Op-amp | Resistors | Capacitor |
|---|---|---|
| A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 3.979 nF |
| A2 | input(s) 10 kΩ | 7.958 nF |
| A4 | input(s) 10 kΩ, feedback 10 kΩ across C | 3.979 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 3.979 nF, 7.958 nF, 3.979 nF; half-power frequency 4 kHz.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. From the table for n = 3, what are the ladder elements?
   A. L1 = 1 H, C2 = 2 F, L3 = 1 H
   B. L1 = 2 H, C2 = 1 F, L3 = 2 H
   C. L1 = 1.414 H, C2 = 1.414 F
   D. L1 = 0.618 H, C2 = 1.618 F, L3 = 2 H
   Answer: A. The n = 3 row is 1, 2, 1: series L1 = 1, shunt C2 = 2, series L3 = 1. (Key point: Prototype values)

2. What w0 is used for a half-power frequency of 4 kHz?
   A. 4000 rad/s
   B. 2 x pi x 4000 = 25,133 rad/s
   C. 8000 rad/s
   D. pi x 4000 rad/s
   Answer: B. The frequency is given in Hz, so w0 = 2 pi f = 25,133 rad/s. (Key point: Frequency scale)

3. What capacitor is used in A2 (g = 2) with R = 10 kohm?
   A. 3.979 nF
   B. 7.958 nF
   C. 79.58 nF
   D. 2 nF
   Answer: B. C = 2/(25133 x 10^4) = 7.958 nF. (Key point: Capacitor for C2)

4. Which integrators have a 10 kohm resistor across the capacitor?
   A. A2 only
   B. A1 and A4, because the terminations add the +1
   C. All three
   D. None
   Answer: B. A1 and A4 realise I1 and I3, whose denominators sL + 1 include the source and load resistances. A2 is lossless. (Key point: Lossy integrators)

## 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 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×

## Related in Filter Design

- [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×
- [Leapfrog simulation of 3rd order Chebyshev LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-third-order-chebyshev) — Filter Design, chapter 8
- [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×
- [Sallen-Key 2nd order Butterworth LPF at 10 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-butterworth-10khz) — Filter Design, chapter 6
- [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