---
title: "Leapfrog 4th order Butterworth LPF at 10,000 rad/s"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:leapfrog-fourth-order-10000"
source: https://www.bctengineer.com/q/semester_5/filter_design/leapfrog-fourth-order-10000
site: BCT Engineering
---

# Leapfrog 4th order Butterworth LPF at 10,000 rad/s

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

- Asked in: 2074 Chaitra, 7 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-10000
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

Design the fourth order Butterworth low pass filter using leapfrog simulation. In your final design the half power frequency should be 10000 rad/s 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

Simulate the doubly terminated fourth-order Butterworth LC ladder equations with op-amp integrators; first and last stages are lossy, middle ones lossless, and inverters correct sign. Scale to half-power frequency 10000 rad/s using 10 kΩ resistors. Capacitors: 7.654 nF, 18.48 nF, 18.48 nF, 7.654 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 = 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 $\omega_0 = 10^4\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$
\begin{aligned}
C_{A1} &= \frac{0.7654}{10000 \times 10^4} = \text{7.654 nF} \\
C_{A2} &= \frac{1.848}{10000 \times 10^4} = \text{18.48 nF} \\
C_{A4} &= \frac{1.848}{10000 \times 10^4} = \text{18.48 nF} \\
C_{A5} &= \frac{0.7654}{10000 \times 10^4} = \text{7.654 nF}
\end{aligned}
$$

**Final element values:**

| Op-amp | Resistors | Capacitor |
|---|---|---|
| A1 | input(s) 10 kΩ, feedback 10 kΩ across C | 7.654 nF |
| A2 | input(s) 10 kΩ | 18.48 nF |
| A4 | input(s) 10 kΩ | 18.48 nF |
| A5 | input(s) 10 kΩ, feedback 10 kΩ across C | 7.654 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 7.654 nF, 18.48 nF, 18.48 nF, 7.654 nF; half-power frequency 10,000 rad/s.

## Check yourself

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

1. Which equation describes the shunt capacitor C2 voltage in the ladder?
   A. V2 = (I1 - I3)/(s C2)
   B. V2 = (Vin - I1) s C2
   C. V2 = I1/(s C2 + 1)
   D. V2 = (I1 + I3) s C2
   Answer: A. The current into C2 is I1 - I3, and the capacitor voltage is that current integrated over C2. (Key point: State equations)

2. How is a lossy inverting integrator formed and what is its output?
   A. A resistor in series with C, output -(sum Vi)/(sRC)
   B. A resistor R across C, output -(sum Vi/R)/(sC + 1/R)
   C. A capacitor at the input only
   D. Two op-amps in series
   Answer: B. Adding R across the feedback capacitor gives the +1/R term in the denominator that represents the termination. (Key point: Lossy integrator)

3. With R = 10 kohm and w0 = 10^4 rad/s, which set of capacitors results for g = 0.7654, 1.848, 1.848, 0.7654?
   A. 76.54 nF, 184.8 nF, 184.8 nF, 76.54 nF
   B. 7.654 nF, 18.48 nF, 18.48 nF, 7.654 nF
   C. 0.7654 nF, 1.848 nF, 1.848 nF, 0.7654 nF
   D. 19.13 nF, 46.2 nF, 46.2 nF, 19.13 nF
   Answer: B. C = g/(w0 R) = g/10^8, giving 7.654 nF and 18.48 nF. (Key point: Denormalised capacitors)

4. What is the passband gain of the leapfrog filter?
   A. 1 (0 dB)
   B. 2 (6 dB)
   C. 1/2 (-6 dB), same as the doubly terminated prototype
   D. 0.707
   Answer: C. Equal source and load terminations give a divider of 1/2 at DC, which the leapfrog structure reproduces. (Key point: Passband gain)

## Close variants set in other exams

- [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×
- [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 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×
- [Bruton transformation; FDNR 4th order LPF at 20,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abruton-fdnr-fourth-order-20000) — Filter Design, chapter 8, asked 5×

## Related in Filter Design

- [Sallen-Key 4th order Butterworth LPF at 10,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-fourth-order-10000) — Filter Design, chapter 8
- [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 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 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×
- [FDNR 5th order Butterworth LPF at 10,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fifth-order-10000) — Filter Design, chapter 8
- [Bruton transformation; FDNR 4th order LPF at 2000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abruton-fdnr-fourth-order-2000) — Filter Design, chapter 8, asked 3×