--- title: "FDNR simulation of 4th order Butterworth LPF" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:fdnr-fourth-order-table" source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-fourth-order-table site: BCT Engineering --- # FDNR simulation of 4th order Butterworth LPF Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2075 Asoj, 6 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-table - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question Simulate the Butterworth 4th order low pass filter in resistively-terminated lossless network 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 After Bruton transformation the normalized 4th-order Butterworth ladder becomes: source capacitor1 F, series resistors0.7654 Ω and1.848 Ω, shunt FD NRs with D=1.848 and D=0.7654, load capacitor1 F. Each FDNR is realized by a GIC with two equal capacitors C in positions1,5 and three equal resistors R in2,3,4 yielding D=C²R. ## Answer The inductors of the resistively terminated LC ladder are removed with the **Bruton transformation** (divide every impedance by $s$), and the grounded capacitors become **FDNRs** ($Z = 1/s^2D$), each realised by a GIC. 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). **Normalised FDNR network** ($\omega_0 = 1$ rad/s): source capacitor 1 F, $R_1 = 0.7654\ \Omega$, FDNR $D_2 = 1.848$, $R_3 = 1.848\ \Omega$, FDNR $D_4 = 0.7654$, load capacitor 1 F. No frequency is given, so the design is shown denormalised to a practical example of $\omega_0$ = 10,000 rad/s: **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 $\omega_0 = 10^4\ \text{rad/s}$. 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** | | $L_1 = 0.7654$ H (series) | $R_1 = 0.7654\ \Omega$ | $R_1 = 0.7654 \times 10000$ | **7.654 kΩ** | | $C_2 = 1.848$ F (shunt) | FDNR $D_2 = 1.848$ | $D_2 = 1.848/(k_z\omega_0^2) = 1.848 \times 10^{-12}$ | GIC: C = 10 nF, $R_D$ = **18.48 kΩ** | | $L_3 = 1.848$ H (series) | $R_3 = 1.848\ \Omega$ | $R_3 = 1.848 \times 10000$ | **18.48 kΩ** | | $C_4 = 0.7654$ F (shunt) | FDNR $D_4 = 0.7654$ | $D_4 = 0.7654/(k_z\omega_0^2) = 7.654 \times 10^{-13}$ | GIC: C = 10 nF, $R_D$ = **7.654 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, $\Sigma R$ = 26.13 kΩ, so $R_b$ = **1.026 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]-+-------+----o Vo | | | [D2] [D4] [CL||Rb] GND -------------------------------------------- ``` **Answer (for $\omega_0$ = 10⁴ rad/s):** terminating capacitors 10 nF; series resistors 7.654 kΩ and 18.48 kΩ; FDNR GICs with C = 10 nF and $R_D$ = 18.48 kΩ ($D_2$) and 7.654 kΩ ($D_4$); $R_a$ = 1 MΩ, $R_b$ = 1.026 MΩ. For any other $\omega_0$, use $k_z = 1/(\omega_0 C_t)$ and the same steps. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the normalised FDNR network (ω0 = 1 rad/s), what are the source and load elements? A. 1 Ω resistors B. 1 H inductors C. 1 F capacitors D. FDNRs with D = 1 Answer: C. Under Bruton, each 1 Ω termination becomes a capacitor of 1/R = 1 F. (Key point: Normalised FDNR network) 2. With kz = 10 kΩ, what is the series resistor from L1 = 0.7654 H? A. 7.654 kΩ B. 76.54 kΩ C. 0.7654 kΩ D. 18.48 kΩ Answer: A. R1 = 0.7654 x 10000 = 7.654 kΩ. (Key point: Scaled resistors) 3. What is RD of the GIC (C = 10 nF) for D2 = 1.848? A. 7.654 kΩ B. 1.848 kΩ C. 10 kΩ D. 18.48 kΩ Answer: D. RD = kz x D = 10000 x 1.848 = 18.48 kΩ. 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