--- title: "FDNR to avoid inductors; 3rd order LPF at 4000 rad/s" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:fdnr-third-order-4000" source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-third-order-4000 site: BCT Engineering --- # FDNR to avoid inductors; 3rd order LPF at 4000 rad/s Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2080 Bhadra, 4+6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-third-order-4000 - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What is FDNR? How can you use FDNR to avoid the inductor in filter design? Explain. Design third order Butterworth low pass filter having half power frequency 4000 rad/s using 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 FDNR has impedance Z=1/(s²D). Bruton transformation converts inductors to resistors, grounded capacitors to FDNRs, removing inductors. For 3rd-order Butterworth at ω0=4000 rad/s, set terminating capacitors 100 nF: series resistors 2.5 kΩ; FDNR D=5×10⁻¹¹ F²Ω realized with GIC: C=100 nF, RD=5 kΩ. Add large DC bias resistors. ## Answer #### FDNR A **frequency dependent negative resistor (FDNR)** is a one-port with impedance $Z(s) = \dfrac{1}{s^2D}$. At $s = j\omega$, $Z = -\dfrac{1}{\omega^2 D}$: a real, negative resistance whose size falls as $1/\omega^2$. $D$ (unit F²Ω) is its constant. It cannot be made from passive elements; a GIC with two capacitors realises it. #### Using FDNR to avoid inductors **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. The FDNR itself is 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). #### Design: 3rd-order Butterworth LPF, $\omega_0$ = 4000 rad/s From the table (n = 3): $R_s = 1\ \Omega$, $L_1 = 1$ H (series), $C_2 = 2$ F (shunt), $L_3 = 1$ H (series), $R_L = 1\ \Omega$. **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** | | $L_1 = 1$ H (series) | $R_1 = 1\ \Omega$ | $R_1 = 1 \times 2500$ | **2.5 kΩ** | | $C_2 = 2$ F (shunt) | FDNR $D_2 = 2$ | $D_2 = 2/(k_z\omega_0^2) = 5 \times 10^{-11}$ | GIC: C = 100 nF, $R_D$ = **5 kΩ** | | $L_3 = 1$ H (series) | $R_3 = 1\ \Omega$ | $R_3 = 1 \times 2500$ | **2.5 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, $\Sigma R$ = 5 kΩ, so $R_b$ = **105 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] [CL||Rb] GND ------------------------------------ ``` **Answer:** terminating capacitors 0.1 μF; $R_1 = R_3$ = 2.5 kΩ; FDNR $D_2$ = GIC with C = 0.1 μF and $R_D$ = 5 kΩ; $R_a$ = 100 kΩ, $R_b$ = 105 kΩ. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the impedance of an FDNR? A. Z = 1/(sD) B. Z = sD C. Z = s^2 D D. Z = 1/(s^2 D) Answer: D. An FDNR has Z(s) = 1/(s^2 D); at s = jω this is -1/(ω^2 D), a negative real resistance that falls as 1/ω^2. (Key point: FDNR impedance) 2. When every impedance of the LC ladder is divided by s, what does a series inductor L become? A. A resistor of value L B. A capacitor of value L C. An FDNR with D = L D. An open circuit Answer: A. sL divided by s gives L, a plain resistor, so the inductors disappear from the lowpass ladder. (Key point: Bruton transformation) 3. With ω0 = 4000 rad/s and terminating capacitors of 100 nF, what is kz = 1/(ω0 Ct)? A. 400 Ω B. 25 kΩ C. 2500 Ω D. 250 Ω Answer: C. kz = 1/(4000 x 1e-7) = 2500 Ω. (Key point: Impedance scaling factor) 4. The shunt capacitor C2 = 2 F gives D = 2. With C = 100 nF in the GIC and RD = kz x D, what is RD? A. 2.5 kΩ B. 5 kΩ C. 10 kΩ D. 1.25 kΩ Answer: B. RD = 2500 x 2 = 5 kΩ, so the FDNR is a GIC with C = 100 nF and R = 5 kΩ. (Key point: GIC resistor for the FDNR) ## 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× ## Related in Filter Design - [FDNR 4th order Butterworth LPF at 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fourth-order-4000) — Filter Design, chapter 8, asked 3× - [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× - [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; design from Figure 1 at 10⁴ rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-figure1-ladder-1e4) — Filter Design, chapter 8 - [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×