---
title: "Bruton transformation; FDNR using GIC"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:bruton-importance-fdnr-gic"
source: https://www.bctengineer.com/q/semester_5/filter_design/bruton-importance-fdnr-gic
site: BCT Engineering
---

# Bruton transformation; FDNR using GIC

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/bruton-importance-fdnr-gic
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

What is the importance of Bruton transformation in filter design? How can you simulate FDNR using generalized impedance converter (GIC)? Explain with example.

## Short answer

Bruton transformation (divide impedances by s) eliminates inductors: series L→resistor, shunt C→grounded FDNR. FDNR is realized with a GIC: Z1=Z5=1/sC, Z2=Z3=Z4=R gives Zin=1/(s²C²R). Example: 3rd-order Butterworth (1,2,1) at 10k rad/s gives R1,R3=10kΩ, FDNR C=10nF, RD=20kΩ.

## Answer

#### Importance of the Bruton transformation

The Bruton transformation divides every impedance of an RLC ladder by $s$: $R \to$ capacitor $1/R$, $L \to$ resistor $L$, $C \to$ FDNR $D = C$. Because a voltage transfer function is a ratio of impedances, it is unchanged. Its importance:

1. **Removes all inductors**, including floating (series) ones in lowpass ladders, which are hard to simulate directly.
2. The only active elements needed are **grounded FDNRs**, each made with one GIC; floating elements are avoided.
3. The design starts from a doubly terminated LC ladder, so the very **low sensitivity** of the ladder to element changes is kept.
4. The final circuit has only resistors, capacitors and op-amps, suitable for integration and for low frequencies where inductors would be huge.
5. Extra care: a DC path must be added (large resistors across the terminating capacitors).

#### FDNR using a GIC

```text
   Iin  node 1
 o-->-----+--------------(+) A1
 Vin      |
         [Z1]
          +--- node 2 ---- output of A1
         [Z2]
          +--- node 3 ---- (-) A1 and (+) A2
         [Z3]
          +--- node 4 ---- output of A2
         [Z4]
          +--- node 5 ---- (-) A2
         [Z5]
          |
         GND
```

Each op-amp has negative feedback (A1 through $Z_2$, A2 through $Z_4$). With ideal op-amps the two input terminals of each op-amp are at the same voltage and draw no current, so $V_1 = V_3 = V_5 = V_{in}$.

$$
\begin{aligned}
\text{Node 5: } & \frac{V_4 - V_{in}}{Z_4} = \frac{V_{in}}{Z_5} \;\Rightarrow\; V_4 = V_{in}\left(1 + \frac{Z_4}{Z_5}\right) \\
\text{Node 3: } & \frac{V_2 - V_{in}}{Z_2} = \frac{V_{in} - V_4}{Z_3} = -\frac{Z_4}{Z_3 Z_5}V_{in} \;\Rightarrow\; V_2 = V_{in}\left(1 - \frac{Z_2 Z_4}{Z_3 Z_5}\right) \\
\text{Node 1: } & I_{in} = \frac{V_{in} - V_2}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3 Z_5}V_{in}
\end{aligned}
$$

$$
Z_{in} = \frac{V_{in}}{I_{in}} = \frac{Z_1 Z_3 Z_5}{Z_2 Z_4}
$$

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).

#### Example

Third-order Butterworth LPF (prototype $R_s = 1$, $L_1 = 1$ H, $C_2 = 2$ F, $L_3 = 1$ H, $R_L = 1$) at $\omega_0$ = 10,000 rad/s.

1. Bruton: $C_s = 1$ F, $R_1 = 1\ \Omega$, $D_2 = 2$, $R_3 = 1\ \Omega$, $C_L = 1$ F.
2. Choose $C_t$ = 10 nF: $k_z = 1/(10^4 \times 10^{-8}) = 10^4\ \Omega$.
3. Resistors: $R_1 = R_3$ = 10 kΩ. Terminating capacitors 10 nF.
4. $D_2 = 2/(10^4 \times 10^8) = 2 \times 10^{-12}$; with GIC capacitors 10 nF, $R_D = D/C^2 = 2 \times 10^{-12}/10^{-16}$ = 20 kΩ.
5. DC path: $R_a$ = 1 MΩ, $R_b = R_a + 20$ kΩ = 1.02 MΩ.

```text
 Vi o--[10nF||Ra]--[10k]--+--[10k]--+---o Vo
                          |         |
                       [GIC D2] [10nF||Rb]
 GND ---------------------+---------+
```

## Check yourself

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

1. Why does dividing all impedances by s leave the voltage transfer function unchanged?
   A. A voltage transfer function is a ratio of impedances
   B. It only holds at DC
   C. The element values are scaled back later
   D. The op-amps compensate
   Answer: A. Common factor 1/s cancels in impedance ratios. (Key point: Transfer function preserved)

2. Which is a main advantage of the Bruton approach?
   A. It works only at DC
   B. It removes the need for capacitors
   C. No inductors, including floating ones, only grounded FDNRs, and low ladder sensitivity is kept
   D. It needs no op-amps
   Answer: C. The result uses only R, C and grounded FDNRs, keeping the doubly terminated ladder's low sensitivity. (Key point: Benefit of the method)

3. For the GIC FDNR with C1 = C5 = C and R2 = R3 = R4 = R, D equals:
   A. C/R
   B. C^2 R
   C. C R^2
   D. R/C^2
   Answer: B. D = C1 C5 R2 R4/R3 = C^2 R. (Key point: FDNR from GIC)

4. In the example, D2 = 2 x 10^-12 with GIC capacitors of 10 nF. R_D is:
   A. 10 kΩ
   B. 20 kΩ
   C. 200 kΩ
   D. 2 kΩ
   Answer: B. R_D = D/C^2 = 2 x 10^-12/10^-16 = 20 kΩ. (Key point: Example numbers)

## Related in Filter Design

- [Bruton transformation and FDNR](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abruton-fdnr-grounded-inductor) — 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×
- [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×
- [GIC current relation and grounded FDNR](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-grounded-fdnr) — Filter Design, chapter 8
- [GIC HPF design at 5 kHz with 10 nF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-hpf-5khz-simulation) — Filter Design, chapter 8
- [GIC HPF design at 4.5 kHz with 0.1 μF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-hpf-4-5khz-simulation) — Filter Design, chapter 8

## Related in other subjects

- [Bilinear transformation method of IIR design](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Abilinear-method-explain) — Digital Signal Analysis and Processing, chapter 6