--- title: "FDNR 4th order Butterworth LPF at 16,000 rad/s" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:fdnr-fourth-order-16000" source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-fourth-order-16000 site: BCT Engineering --- # FDNR 4th order Butterworth LPF at 16,000 rad/s Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2079 Baisakh, 5 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-16000 - 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 16000 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 Bruton transform the 4th-order Butterworth prototype (Rs=1, L1=0.7654, C2=1.848, L3=1.848, C4=0.7654, RL=1) and scale to 16000 rad/s with Ct=10 nF (kz=6250 Ω). Series resistors: 4.784 kΩ and 11.55 kΩ. FDNRs: D2=1.848 (GIC: C=10 nF, RD=11.55 kΩ), D4=0.7654 (GIC: C=10 nF, RD=4.784 kΩ). DC bias: Ra=100 kΩ, Rb=116.3 kΩ. ## Answer The inductors of the doubly terminated ladder are removed with the **Bruton transformation**, and the shunt capacitors become grounded FDNRs ($Z = 1/s^2D$) made with GICs. 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 $\omega_0 = 16{,}000\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF, so $$ k_z = \frac{1}{\omega_0 C_t} = \frac{1}{16000 \times 1 \times 10^{-8}} = 6250\ \Omega $$ $$ \begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned} $$ **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 | $1/(k_z\omega_0)$ | **10 nF** | | $L_1 = 0.7654$ H (series) | $R_1 = 0.7654\ \Omega$ | $R_1 = 0.7654 \times 6250$ | **4.784 kΩ** | | $C_2 = 1.848$ F (shunt) | FDNR $D_2 = 1.848$ | $D_2 = 1.848/(k_z\omega_0^2) = 1.155 \times 10^{-12}$ | GIC: C = 10 nF, $R_D$ = **11.55 kΩ** | | $L_3 = 1.848$ H (series) | $R_3 = 1.848\ \Omega$ | $R_3 = 1.848 \times 6250$ | **11.55 kΩ** | | $C_4 = 0.7654$ F (shunt) | FDNR $D_4 = 0.7654$ | $D_4 = 0.7654/(k_z\omega_0^2) = 4.784 \times 10^{-13}$ | GIC: C = 10 nF, $R_D$ = **4.784 kΩ** | | $R_L = 1\ \Omega$ | $C_L = 1$ F | $1/(k_z\omega_0)$ | **10 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, $\Sigma R$ = 16.33 kΩ, so $R_b$ = **116.3 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 10 nF; series resistors 4.784 kΩ and 11.55 kΩ; FDNRs: $D_2$ = GIC (C = 10 nF, $R_D$ = 11.55 kΩ), $D_4$ = GIC (C = 10 nF, $R_D$ = 4.784 kΩ); $R_a$ = 100 kΩ, $R_b$ = 116.3 kΩ. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After the Bruton transformation of the 4th-order ladder, which elements become grounded FDNRs? A. The series inductors B. The source and load resistors C. The shunt capacitors D. All the capacitors and inductors Answer: C. Shunt (grounded) capacitors become grounded FDNRs with D = C; series inductors become ordinary resistors. (Key point: Which elements become FDNRs) 2. For ω0 = 16000 rad/s and Ct = 10 nF, what is kz? A. 6250 Ω B. 625 Ω C. 62.5 kΩ D. 1600 Ω Answer: A. kz = 1/(16000 x 1e-8) = 6250 Ω. (Key point: Impedance scaling factor) 3. For C4 = 0.7654 F (D4 = 0.7654), what is RD of the GIC with C = 10 nF? A. 11.55 kΩ B. 7.654 kΩ C. 0.7654 kΩ D. 4.784 kΩ Answer: D. RD = kz x D = 6250 x 0.7654 = 4.784 kΩ. (Key point: GIC resistor value) 4. With Ra = 100 kΩ and series resistors 4.784 kΩ and 11.55 kΩ, what Rb keeps the prototype DC gain of 1/2? A. 100 kΩ B. 116.3 kΩ C. 16.33 kΩ D. 11.63 kΩ Answer: B. Set Rb = Ra + ΣR = 100 + 16.33 = 116.3 kΩ; then the DC gain Rb/(Ra + ΣR + Rb) = 1/2. (Key point: Bias resistor Rb) ## Close variants set in other exams - [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 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× ## 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 - [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× - [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× - [GIC 4th order Butterworth HPF at 16,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-fourth-order-hpf-16000) — Filter Design, chapter 8 - [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× - [FDNR simulation of 4th order Butterworth LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fourth-order-table) — Filter Design, chapter 8