--- title: "FDNR realization of 5th order ladder at 25,000 rad/s" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:fdnr-fifth-order-25000" source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-fifth-order-25000 site: BCT Engineering --- # FDNR realization of 5th order ladder at 25,000 rad/s Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2080 Baisakh, 1+3+5 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-fifth-order-25000 - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What is frequency dependent negative resistor (FDNR)? How can it be realized? Realize the following passive filter using FDNR, having ωo = 25000 rad/s and practical element values in your final circuit. [Figure: Butterworth filter at normalized frequency - source V with 1 Ω series resistor, series 0.618 H, shunt 1.618 F, series 2 H, shunt 1.618 F, series 0.618 H, 1 Ω load] ## Short answer FDNR: Z=1/(s²D). Realized by GIC with capacitors in arms 1,5 and resistors in 2,3,4 (D=C²R). For 5th-order Butterworth at 25krad/s with Ct=10nF: series resistors 2.472kΩ, 8kΩ, 2.472kΩ; shunt FDNRs each C=10nF, RD=6.472kΩ; terminating capacitors 10nF; DC bias resistors 100kΩ, 113kΩ. ## 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. #### Realisation The FDNR is realised with a GIC (Antoniou circuit): ```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 ``` 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: 5th-order Butterworth LPF, $\omega_0$ = 25,000 rad/s Given prototype: $R_s = 1\ \Omega$, $L_1 = 0.618$ H, $C_2 = 1.618$ F, $L_3 = 2$ H, $C_4 = 1.618$ F, $L_5 = 0.618$ H, $R_L = 1\ \Omega$. **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 $\omega_0 = 25{,}000\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF, so $$ k_z = \frac{1}{\omega_0 C_t} = \frac{1}{25000 \times 1 \times 10^{-8}} = 4000\ \Omega $$ $$ \begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned} $$ **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** | | $L_1 = 0.618$ H (series) | $R_1 = 0.618\ \Omega$ | $R_1 = 0.618 \times 4000$ | **2.472 kΩ** | | $C_2 = 1.618$ F (shunt) | FDNR $D_2 = 1.618$ | $D_2 = 1.618/(k_z\omega_0^2) = 6.472 \times 10^{-13}$ | GIC: C = 10 nF, $R_D$ = **6.472 kΩ** | | $L_3 = 2$ H (series) | $R_3 = 2\ \Omega$ | $R_3 = 2 \times 4000$ | **8 kΩ** | | $C_4 = 1.618$ F (shunt) | FDNR $D_4 = 1.618$ | $D_4 = 1.618/(k_z\omega_0^2) = 6.472 \times 10^{-13}$ | GIC: C = 10 nF, $R_D$ = **6.472 kΩ** | | $L_5 = 0.618$ H (series) | $R_5 = 0.618\ \Omega$ | $R_5 = 0.618 \times 4000$ | **2.472 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, $\Sigma R$ = 12.94 kΩ, so $R_b$ = **112.9 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]-+----[R5]-+----o Vo | | | [D2] [D4] [CL||Rb] GND ---------------------------------------------- ``` **Answer:** terminating capacitors 10 nF; series resistors 2.472 kΩ, 8 kΩ, 2.472 kΩ; two FDNRs, each a GIC with C = 10 nF and $R_D$ = 6.472 kΩ; $R_a$ = 100 kΩ, $R_b$ = 112.9 kΩ. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the GIC (Antoniou) realisation of an FDNR, which positions hold capacitors? A. Positions 2 and 4 B. Positions 1 and 5 C. Positions 3 and 5 D. Position 3 only Answer: B. Capacitors go in positions 1 and 5 and resistors in 2, 3, 4, giving Zin = R3/(s^2 C1 C5 R2 R4) = 1/(s^2 D). (Key point: GIC realisation of FDNR) 2. With C1 = C5 = C and R2 = R3 = R4 = R, what is D of the GIC? A. D = C R B. D = C^2 R C. D = C R^2 D. D = C^2 / R Answer: B. D = C1 C5 R2 R4 / R3 = C^2 R when the three resistors are equal. (Key point: FDNR constant D) 3. With kz = 4000 Ω, what resistor replaces the prototype L3 = 2 H? A. 2 kΩ B. 12.94 kΩ C. 8 kΩ D. 6.472 kΩ Answer: C. Each inductor becomes a resistor of L ohms, then is scaled: 2 x 4000 = 8 kΩ. (Key point: Scaled series resistor) 4. Why are Ra and Rb connected across the source and load capacitors? A. To set the cut-off frequency B. To cancel the negative resistance of the FDNR C. To make the circuit an inductor again D. To give the GIC op-amps a DC bias path, since the capacitors block DC Answer: D. The terminating capacitors block DC, so large resistors across them provide the bias path for the op-amps. They must be much larger than kz. 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