---
title: "FDNR 4th order Butterworth LPF at 4000 rad/s"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:fdnr-fourth-order-4000"
source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-fourth-order-4000
site: BCT Engineering
---

# FDNR 4th order Butterworth LPF at 4000 rad/s

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

- Asked in: 2072 Chaitra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-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 Frequency dependent negative resistor (FDNR). [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

Bruton transform the4th-order Butterworth prototype (Rs=1, L1=0.7654, C2=1.848, L3=1.848, C4=0.7654, RL=1) and scale to4000 rad/s with Ct=100 nF (kz=2500 Ω). Series resistors1.914 kΩ and4.62 kΩ. FDNRs: D2=1.848 (GIC: C=100 nF, RD=4.62 kΩ), D4=0.7654 (GIC: C=100 nF, RD=1.914 kΩ). DC bias: Ra=100 kΩ, Rb=106.5 kΩ.

## Answer

The design uses the **Bruton transformation** to remove the inductors of the doubly terminated Butterworth ladder; the shunt capacitors become grounded FDNRs ($Z = 1/s^2D$), each realised with a GIC.

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).

**FDNR from a GIC.** Each FDNR is made with an Antoniou GIC (two op-amps, five impedances $Z_1$ to $Z_5$ in a chain with $Z_5$ grounded, so that $Z_{in} = Z_1Z_3Z_5/(Z_2Z_4)$). Put capacitors in positions 1 and 5 ($Z_1 = 1/sC_1$, $Z_5 = 1/sC_5$) and resistors in 2, 3, 4:

$$
Z_{in} = \frac{R_3}{s^2 C_1 C_5 R_2 R_4} = \frac{1}{s^2 D}, \qquad D = \frac{C_1 C_5 R_2 R_4}{R_3}
$$

With $C_1 = C_5 = C$ and $R_2 = R_3 = R_4 = R$: $D = C^2 R$. At $s = j\omega$, $Z_{in} = -1/(\omega^2 D)$, a negative resistance whose value depends on frequency: an **FDNR** (also called a D-element or supercapacitor).

**Step 1: Bruton transformation** (divide all impedances by $s$): source and load resistors (1 Ω) become 1 F capacitors, each inductor $L$ becomes a resistor of $L$ Ω, each shunt capacitor $C$ becomes a grounded FDNR with $D = C$.

**Step 2: Scaling.** Frequency scaling factor $\omega_0 = 4000\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 100 nF, so

$$
k_z = \frac{1}{\omega_0 C_t} = \frac{1}{4000 \times 1 \times 10^{-7}} = 2500\ \Omega
$$

$$
\begin{aligned}
R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2}
\end{aligned}
$$

**Step 3: Element values.** Each FDNR is a GIC with $C_1 = C_5 = C$ and $R_2 = R_3 = R_4 = R_D$, so $D = C^2R_D$. Taking $C = C_t$ gives $R_D = D_{new}/C_t^2 = k_z D$.

| Prototype | After Bruton | Scaled | Final element |
|---|---|---|---|
| $R_s = 1\ \Omega$ | $C_s = 1$ F | $1/(k_z\omega_0)$ | **100 nF** |
| $L_1 = 0.7654$ H (series) | $R_1 = 0.7654\ \Omega$ | $R_1 = 0.7654 \times 2500$ | **1.914 kΩ** |
| $C_2 = 1.848$ F (shunt) | FDNR $D_2 = 1.848$ | $D_2 = 1.848/(k_z\omega_0^2) = 4.62 \times 10^{-11}$ | GIC: C = 100 nF, $R_D$ = **4.62 kΩ** |
| $L_3 = 1.848$ H (series) | $R_3 = 1.848\ \Omega$ | $R_3 = 1.848 \times 2500$ | **4.62 kΩ** |
| $C_4 = 0.7654$ F (shunt) | FDNR $D_4 = 0.7654$ | $D_4 = 0.7654/(k_z\omega_0^2) = 1.914 \times 10^{-11}$ | GIC: C = 100 nF, $R_D$ = **1.914 kΩ** |
| $R_L = 1\ \Omega$ | $C_L = 1$ F | $1/(k_z\omega_0)$ | **100 nF** |

**Step 4: DC path.** The source and load capacitors block DC, and the op-amps in the GICs need a DC bias path. Put $R_a$ across the source capacitor and $R_b$ across the load capacitor. They must be much larger than $k_z$ (the capacitor impedance at $\omega_0$). At DC the FDNRs are open, so the DC gain is $R_b/(R_a + \Sigma R + R_b)$. To keep the prototype DC gain of 1/2, set $R_b = R_a + \Sigma R$:

$R_a$ = 100 kΩ, $\Sigma R$ = 6.534 kΩ, so $R_b$ = **106.5 kΩ**.

**Final circuit** ($C_s \| R_a$ in series from the source, $C_L \| R_b$ as the load, each $D_k$ a GIC FDNR to ground):

```text
Vi o--[Cs||Ra]--[R1]-+----[R3]-+-------+----o Vo
                     |         |       |
                    [D2]      [D4]  [CL||Rb]
GND --------------------------------------------
```

**Answer:** terminating capacitors 0.1 μF; series resistors 1.914 kΩ and 4.62 kΩ; FDNRs: $D_2$ = GIC (C = 0.1 μF, $R_D$ = 4.62 kΩ), $D_4$ = GIC (C = 0.1 μF, $R_D$ = 1.914 kΩ); $R_a$ = 100 kΩ, $R_b$ = 106.5 kΩ.

## Check yourself

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

1. Why is the Bruton transformation used on the Butterworth ladder?
   A. To change the cut-off frequency
   B. To remove the inductors without changing the voltage transfer function
   C. To convert it to a highpass filter
   D. To raise the filter order
   Answer: B. Dividing all impedances by s leaves the voltage ratio unchanged but turns inductors into resistors and capacitors into FDNRs. (Key point: Purpose of Bruton transformation)

2. For ω0 = 4000 rad/s and Ct = 100 nF, kz is:
   A. 250 Ω
   B. 25 kΩ
   C. 4000 Ω
   D. 2500 Ω
   Answer: D. kz = 1/(4000 x 1e-7) = 2500 Ω. (Key point: Impedance scaling factor)

3. Which pair gives the series resistors from L1 = 0.7654 H and L3 = 1.848 H?
   A. 1.913 kΩ and 4.62 kΩ
   B. 4.62 kΩ and 1.913 kΩ
   C. 0.7654 kΩ and 1.848 kΩ
   D. 19.1 kΩ and 46.2 kΩ
   Answer: A. Multiply by 2500: 0.7654 x 2500 = 1.913 kΩ and 1.848 x 2500 = 4.62 kΩ. (Key point: Series resistors)

4. What RD values (C = 100 nF) realise D2 = 1.848 and D4 = 0.7654?
   A. 1.913 kΩ and 4.62 kΩ
   B. 4.62 kΩ and 1.913 kΩ
   C. 2.5 kΩ each
   D. 4.62 kΩ and 4.62 kΩ
   Answer: B. RD = kz x D: 2500 x 1.848 = 4.62 kΩ for D2 and 2500 x 0.7654 = 1.913 kΩ for D4. (Key point: GIC resistors for the FDNRs)

## 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 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-fourth-order-4000) — Filter Design, chapter 8, asked 4×

## Related in Filter Design

- [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 to avoid inductors; 3rd order LPF at 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-third-order-4000) — Filter Design, chapter 8, asked 2×
- [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×
- [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×
- [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×