--- title: "FDNR 4th order Butterworth LPF at 20,000 rad/s" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:fdnr-fourth-order-20000" source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-fourth-order-20000 site: BCT Engineering --- # FDNR 4th order Butterworth LPF at 20,000 rad/s Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2078 Bhadra, 2+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-20000 - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What do you mean by Frequency Dependent Negative Resistor (FDNR)? Design the 4th order Butterworth low pass filter with ωp = 20,000 rad/sec using FDNR. In your final circuit all the elements should be practically realizable. [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: impedance 1/(s^2 D), gives negative frequency-dependent resistance. Bruton transformation: R→C, L→R, C→FDNR, eliminates inductors. For ωp=20,000 rad/s, Ct=10 nF, kz=5kΩ. Terminating caps 10 nF; series R: 3.83k, 9.24k; FDNRs: GIC with C=10 nF, RD=9.24k (D=1.848) and 3.83k (D=0.7654). DC bias: Ra=100k, Rb≈113k. ## 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 is realised with a GIC containing two capacitors. It is used with the **Bruton transformation** (divide all impedances by $s$): inductors become resistors and grounded capacitors become grounded FDNRs, so the ladder has no inductors. #### Design: 4th-order Butterworth LPF, $\omega_p$ = 20,000 rad/s 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 $\omega_0 = 20{,}000\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF, so $$ k_z = \frac{1}{\omega_0 C_t} = \frac{1}{20000 \times 1 \times 10^{-8}} = 5000\ \Omega $$ $$ \begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned} $$ **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) | $R_1 = 0.7654\ \Omega$ | $R_1 = 0.7654 \times 5000$ | **3.827 kΩ** | | $C_2 = 1.848$ F (shunt) | FDNR $D_2 = 1.848$ | $D_2 = 1.848/(k_z\omega_0^2) = 9.24 \times 10^{-13}$ | GIC: C = 10 nF, $R_D$ = **9.24 kΩ** | | $L_3 = 1.848$ H (series) | $R_3 = 1.848\ \Omega$ | $R_3 = 1.848 \times 5000$ | **9.24 kΩ** | | $C_4 = 0.7654$ F (shunt) | FDNR $D_4 = 0.7654$ | $D_4 = 0.7654/(k_z\omega_0^2) = 3.827 \times 10^{-13}$ | GIC: C = 10 nF, $R_D$ = **3.827 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, $\Sigma R$ = 13.07 kΩ, so $R_b$ = **113.1 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 3.827 kΩ and 9.24 kΩ; FDNRs: $D_2$ = GIC (C = 10 nF, $R_D$ = 9.24 kΩ), $D_4$ = GIC (C = 10 nF, $R_D$ = 3.827 kΩ); $R_a$ = 100 kΩ, $R_b$ = 113.1 kΩ. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At s = jw, an FDNR behaves as which kind of element? A. A positive resistance that rises with frequency B. A negative real resistance whose magnitude falls as 1/w^2 C. An inductor with negative value D. A capacitor with value D Answer: B. Z = 1/(s^2 D) = -1/(w^2 D): negative, real, and decreasing as 1/w^2. (Key point: What an FDNR is) 2. For w0 = 20,000 rad/s and terminating capacitors Ct = 10 nF, what is kz? A. 10 kohm B. 2 kohm C. 5 kohm D. 50 kohm Answer: C. kz = 1/(w0 Ct) = 1/(20000 x 1e-8) = 5000 ohm. (Key point: Scaling factor) 3. For an FDNR GIC with C1 = C5 = Ct and R2 = R3 = R4 = RD, how is RD obtained for a given D_new? A. RD = D_new x Ct^2 B. RD = D_new / Ct^2 = kz D C. RD = D_new / Ct D. RD = 1/(w0 Ct) Answer: B. D = C^2 RD, so RD = D_new/Ct^2, which equals kz times the normalised D. For D4 = 0.7654 this gives 3.827 kohm. (Key point: FDNR realisation) 4. Why are Ra and Rb added across the source and load capacitors? A. To set the cutoff frequency B. To make the FDNRs positive C. To remove the op-amp gain D. To give the op-amps a DC bias path, since the capacitors block DC; they must be much larger than kz Answer: D. The terminating capacitors block DC, so large resistors Ra and Rb provide a bias path without disturbing the response near w0. Rb = Ra + sum R keeps the DC gain at 1/2. 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