--- title: "FDNR 5th order Butterworth LPF at 10,000 rad/s" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:fdnr-fifth-order-10000" source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-fifth-order-10000 site: BCT Engineering --- # FDNR 5th order Butterworth LPF at 10,000 rad/s Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2080 Baisakh, 2+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-10000 - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What is FDNR? How can FDNR be used to avoid the inductor in passive filter? Simulate the Butterworth 5th order lowpass filter using FDNR referring table below. Your final design should have ωo = 10,000 rad/s and practically realizable elements. [Table: Order (n) = 5; R1 = R2 = 1; L1 = 0.6180; C2 = 1.618; L3 = 2.000; C4 = 1.618; L5 = 0.6180] ## Short answer FDNR has impedance 1/(s^2 D), a frequency-dependent negative resistor. Bruton transformation divides impedances by s: L becomes R, grounded C becomes FDNR, removing inductors. For ωo=10,000 rad/s, choose Ct=10 nF, kz=10kΩ. Terminating caps 10 nF; series R: 6.18k, 20k, 6.18k; two FDNRs (D=1.618) use GICs with C=10 nF, RD=16.18kΩ. Add 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. #### Avoiding the inductor **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. In the 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: 5th-order Butterworth LPF, $\omega_0$ = 10,000 rad/s Prototype from the table: source resistor $R_s = R_1 = 1\ \Omega$, $L_1 = 0.618$ H, $C_2 = 1.618$ F, $L_3 = 2$ H, $C_4 = 1.618$ F, $L_5 = 0.618$ H, load resistor $R_L = R_2 = 1\ \Omega$. **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 $\omega_0 = 10^4\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF, so $$ k_z = \frac{1}{\omega_0 C_t} = \frac{1}{10000 \times 1 \times 10^{-8}} = 10000\ \Omega $$ $$ \begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned} $$ **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) | $R_1 = 0.618\ \Omega$ | $R_1 = 0.618 \times 10000$ | **6.18 kΩ** | | $C_2 = 1.618$ F (shunt) | FDNR $D_2 = 1.618$ | $D_2 = 1.618/(k_z\omega_0^2) = 1.618 \times 10^{-12}$ | GIC: C = 10 nF, $R_D$ = **16.18 kΩ** | | $L_3 = 2$ H (series) | $R_3 = 2\ \Omega$ | $R_3 = 2 \times 10000$ | **20 kΩ** | | $C_4 = 1.618$ F (shunt) | FDNR $D_4 = 1.618$ | $D_4 = 1.618/(k_z\omega_0^2) = 1.618 \times 10^{-12}$ | GIC: C = 10 nF, $R_D$ = **16.18 kΩ** | | $L_5 = 0.618$ H (series) | $R_5 = 0.618\ \Omega$ | $R_5 = 0.618 \times 10000$ | **6.18 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 $R_b = R_a + \Sigma R$: $R_a$ = 1 MΩ, $\Sigma R$ = 32.36 kΩ, so $R_b$ = **1.032 MΩ**. **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 6.18 kΩ, 20 kΩ, 6.18 kΩ; two FDNRs, each a GIC with C = 10 nF and $R_D$ = 16.18 kΩ; $R_a$ = 1 MΩ, $R_b$ = 1.032 MΩ. ## 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 = s D B. Z = 1/(s^2 D), which is -1/(w^2 D) at s = jw C. Z = 1/(s D), a capacitor-like reactance D. Z = -D, a constant negative resistance Answer: B. An FDNR has Z(s) = 1/(s^2 D); at s = jw this is a real negative resistance -1/(w^2 D) that falls as 1/w^2. (Key point: FDNR impedance) 2. Under the Bruton transformation (all impedances divided by s), what do an inductor L and a shunt capacitor C become? A. L becomes a capacitor, C becomes an inductor B. L becomes a resistor of L ohms, C becomes an FDNR with D = C C. L becomes an FDNR, C becomes a resistor D. Both become resistors Answer: B. sL/s = L (a resistor) and 1/(sC)/s = 1/(s^2 C), an FDNR with D = C. The voltage ratio is unchanged. (Key point: Bruton transformation) 3. Why does the transformed ladder need no inductors and can use grounded GICs? A. The inductors become resistors and the grounded shunt capacitors become grounded FDNRs B. The inductors are replaced by transformers C. The source resistor is removed D. All elements become capacitors Answer: A. Series inductors turn into plain resistors and grounded capacitors turn into grounded FDNRs, which one GIC can realise. (Key point: Why inductors are avoided) 4. With Ct = 10 nF and w0 = 10^4 rad/s, what is kz and what resistor replaces L1 = 0.618 H? A. kz = 1000 ohm, R = 618 ohm B. kz = 10 kohm, R = 6.18 kohm C. kz = 10 kohm, R = 618 kohm D. kz = 100 kohm, R = 61.8 kohm Answer: B. kz = 1/(w0 Ct) = 10 kohm, and R = kz x 0.618 = 6.18 kohm. 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