--- title: "Bruton transformation and FDNR" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:bruton-fdnr-grounded-inductor" source: https://www.bctengineer.com/q/semester_5/filter_design/bruton-fdnr-grounded-inductor site: BCT Engineering --- # Bruton transformation and FDNR Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2082 Bhadra, 2+4 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/bruton-fdnr-grounded-inductor - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What is Bruton Transformation? How can you simulate grounded inductor using FDNR? Explain with necessary figures and derivations. ## Short answer Bruton transformation divides all impedances by s, replacing R→1/sC, L→R, C→1/(s²D) (FDNR). A grounded inductor L becomes a grounded resistor R=L. The method is used in lowpass ladders: series inductors become resistors, shunt capacitors become grounded FDNRs realized with GICs. So no actual inductor remains, and the grounded inductor is simulated indirectly as a resistor after transformation. ## Answer #### Bruton transformation **Bruton transformation**: divide every impedance of the RLC ladder by $s$ (more generally by $ks$). A voltage transfer function is a ratio of impedances, so it does not change: | Element | Impedance | After dividing by $s$ | New element | |---|---|---|---| | Resistor $R$ | $R$ | $R/s$ | capacitor $C = 1/R$ | | Inductor $L$ | $sL$ | $L$ | resistor $R = L$ | | Capacitor $C$ | $1/sC$ | $1/(s^2C)$ | FDNR $D = C$ | In a lowpass ladder the inductors are in the series arms (often floating) and the capacitors are shunt (grounded). After the transformation the inductors become plain resistors and the grounded capacitors become **grounded FDNRs**, which one GIC can realise. So no inductor is left. #### Simulating the inductor problem with FDNR A lowpass LC ladder has inductors in the series arms. Inductors are bulky, lossy and cannot be integrated. After the Bruton transformation (all impedances divided by $s$): - each inductor $L$ becomes a resistor $R = L$ (no inductor left), - each grounded capacitor $C$ becomes a grounded FDNR $D = C$, - each terminating resistor $R$ becomes a capacitor $1/R$. So the inductor is "simulated" indirectly: the whole network is converted to R, C and grounded FDNRs with the same voltage transfer function. **Grounded inductors specifically.** A grounded inductor $L$ (for example in a highpass ladder) becomes a grounded resistor $R = L$. The series capacitors of such a ladder would become floating FDNRs, which are hard to build, so the Bruton method is used mainly for lowpass ladders, whose shunt capacitors become grounded FDNRs. ```text Original: o-[Rs]-(L1)-+-(L3)-+---o C2 C4 [RL] Bruton: o-[Cs]-[R1]-+-[R3]-+---o D2 D4 [CL] ``` #### Realising the FDNR with 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). #### Practical points - Because the terminations become capacitors, there is no DC path; resistors $R_a$ (across the source capacitor) and $R_b$ (across the load capacitor) are added, much larger than the capacitor impedances in the passband, with $R_b = R_a + \Sigma R$ to keep the DC gain at 1/2. - Only grounded FDNRs are needed, so one GIC per shunt capacitor is enough, and the low sensitivity of the ladder is kept. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the Bruton transformation remove the inductor problem in a lowpass ladder? A. Each inductor becomes a resistor and each grounded capacitor a grounded FDNR B. Inductors are moved to ground C. Each inductor is replaced by a gyrator D. Each inductor becomes a larger inductor Answer: A. Dividing by s turns sL into L ohms, and shunt capacitors into FDNRs, so no inductor remains. (Key point: Why Bruton helps) 2. The GIC gives Zin = 1/(s^2 D) with D = C1 C5 R2 R4/R3. Which positions hold capacitors? A. 2 and 4 B. 4 only C. 1 and 5 D. 3 and 5 Answer: C. Capacitors at Z1 and Z5, resistors at Z2, Z3, Z4. (Key point: FDNR with GIC) 3. What happens to the terminating resistors, and what must be added? A. They become inductors B. They stay resistors, no change needed C. They become FDNRs D. They become capacitors, so large resistors Ra and Rb are added for a DC path Answer: D. R becomes a capacitor 1/R which blocks DC, so Ra and Rb provide the DC bias path. (Key point: Terminations) 4. To keep the prototype DC gain of 1/2, Rb should be set to: A. ΣR B. Ra C. Ra - ΣR D. Ra + ΣR Answer: D. At DC the FDNRs are open, so gain is Rb/(Ra + ΣR + Rb); equal to 1/2 needs Rb = Ra + ΣR. (Key point: DC gain) ## Related in Filter Design - [Bruton transformation; FDNR using GIC](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abruton-importance-fdnr-gic) — 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 simulation of a grounded inductor](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-grounded-inductor-simulation) — Filter Design, chapter 8, asked 7× - [GIC current relation and grounded FDNR](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-grounded-fdnr) — Filter Design, chapter 8 - [Why simulate resistors with switched capacitors in MOS](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asc-resistor-simulation-mos) — Filter Design, chapter 9