--- title: "GIC HPF design at 5 kHz with 10 nF" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:gic-hpf-5khz-simulation" source: https://www.bctengineer.com/q/semester_5/filter_design/gic-hpf-5khz-simulation site: BCT Engineering --- # GIC HPF design at 5 kHz with 10 nF Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2081 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/gic-hpf-5khz-simulation - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What is a generalized impedance converter (GIC)? How can you simulate the grounded inductor using GIC? From the LC ladder given in figure below, design a highpass filter with a half power frequency of 5 kHz and the largest capacitance of 10nF using inductor simulation. [Figure: source V1, 1 Ω source resistor, series 1.618 F, shunt 0.618 H, series 0.5 F, shunt 0.618 H, series 1.618 F, 1 Ω load (output V2)] ## Short answer A GIC converts impedance; a grounded inductor L is realized by a GIC with Z4 = 1/(sC) and all other positions R, giving L = C R². For the 5th-order Butterworth high-pass filter at 5 kHz with 10 nF maximum capacitance, Rs = RL = 5.15 kΩ; series capacitors are 10 nF, 3.09 nF, 10 nF; each 101.3 mH shunt inductor is simulated by a GIC with C = 10 nF and R = 3.183 kΩ. ## Answer A **generalized impedance converter (GIC)** is an active two-port, usually Antoniou's two-op-amp, five-impedance circuit. Its input impedance is the load impedance $Z_5$ multiplied by a conversion function $k(s) = Z_1Z_3/(Z_2Z_4)$. By choosing which positions hold capacitors, a GIC can make an inductor or an FDNR from only R, C and op-amps. #### Simulating a grounded inductor with a GIC ```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 ``` Each op-amp has negative feedback (A1 through $Z_2$, A2 through $Z_4$). With ideal op-amps the two input terminals of each op-amp are at the same voltage and draw no current, so $V_1 = V_3 = V_5 = V_{in}$. $$ \begin{aligned} \text{Node 5: } & \frac{V_4 - V_{in}}{Z_4} = \frac{V_{in}}{Z_5} \;\Rightarrow\; V_4 = V_{in}\left(1 + \frac{Z_4}{Z_5}\right) \\ \text{Node 3: } & \frac{V_2 - V_{in}}{Z_2} = \frac{V_{in} - V_4}{Z_3} = -\frac{Z_4}{Z_3 Z_5}V_{in} \;\Rightarrow\; V_2 = V_{in}\left(1 - \frac{Z_2 Z_4}{Z_3 Z_5}\right) \\ \text{Node 1: } & I_{in} = \frac{V_{in} - V_2}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3 Z_5}V_{in} \end{aligned} $$ $$ Z_{in} = \frac{V_{in}}{I_{in}} = \frac{Z_1 Z_3 Z_5}{Z_2 Z_4} $$ Put a capacitor in position 4 ($Z_4 = 1/sC$) and resistors elsewhere: $$ Z_{in} = \frac{R_1 R_3 R_5}{R_2}\, sC = sL, \qquad L = \frac{C R_1 R_3 R_5}{R_2} $$ With all resistors equal to $R$: $L = CR^2$. One end of this inductor is ground, so it replaces a **grounded (shunt) inductor**. #### Design The ladder is the 5th-order Butterworth highpass prototype (1 rad/s, 1 Ω): series 1.618 F, shunt 0.618 H, series 0.5 F, shunt 0.618 H, series 1.618 F. The shunt inductors are grounded and are simulated by GICs. Frequency scale to $\omega_0 = 2\pi(5000) = 31{,}416\ \text{rad/s}$ and impedance scale by $k_z$ (chosen so that the largest capacitor, 1.618 F, becomes 10 nF: $k_z = 1.618/(\omega_0 \times 10^{-8})$): $$ R_{new} = k_z R, \qquad L_{new} = \frac{k_z L}{\omega_0}, \qquad C_{new} = \frac{C}{k_z\,\omega_0} $$ $k_z = 5150.3\ \Omega$, so the source and load resistors become $R_s = R_L$ = 5.15 kΩ. | Prototype | Scaled | Final | |---|---|---| | $C_1$ = 1.618 F (series) | $C = 1.618/(k_z\omega_0)$ | **10 nF** | | $L_2$ = 0.618 H (shunt) | $k_z(0.618)/\omega_0$ = 101.3 mH | GIC: C = 10 nF, $R = \sqrt{L/C}$ = **3.183 kΩ** | | $C_3$ = 0.5 F (series) | $C = 0.5/(k_z\omega_0)$ | **3.09 nF** | | $L_4$ = 0.618 H (shunt) | $k_z(0.618)/\omega_0$ = 101.3 mH | GIC: C = 10 nF, $R = \sqrt{L/C}$ = **3.183 kΩ** | | $C_5$ = 1.618 F (series) | $C = 1.618/(k_z\omega_0)$ | **10 nF** | | $R_s, R_L$ = 1 Ω | $k_z \times 1$ | **5.15 kΩ** | Each grounded inductor is replaced by a GIC with $Z_4 = C$ and $R_1 = R_2 = R_3 = R_5 = R$, so $L = CR^2$ and $R = \sqrt{L/C}$. **Final circuit** (each GIC realises the grounded inductor in that position): ```text Vi o--[Rs]--[C1]-+----[C3]-+----[C5]-+----o Vo | | | [GIC-L2] [GIC-L4] [RL] GND ------------------------------------------ ``` **Answer:** $R_s = R_L$ = 5.15 kΩ; capacitors 10 nF, 3.09 nF, 10 nF; each 101.3 mH inductor = GIC with C = 10 nF and R = 3.183 kΩ; half-power frequency 5 kHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the GIC with Z4 = 1/sC and the rest equal resistors R, which expression gives R for a wanted L? A. R = sqrt(LC) B. R = sqrt(L/C) C. R = LC D. R = L/C Answer: B. L = C R^2 so R = sqrt(L/C). (Key point: Inductance of GIC) 2. How is the impedance scale kz fixed in this design? A. So that inductors become 1 H B. So that the largest capacitor 1.618 F becomes 10 nF C. So that the source resistance becomes 1 kΩ D. So that ω0 equals 1 rad/s Answer: B. kz = 1.618/(ω0 x 10^-8) = 5150 Ω makes C1 = 10 nF. (Key point: Choosing kz) 3. What ω0 is used for a 5 kHz half-power frequency? A. 1 rad/s B. 5000 rad/s C. 31,416 rad/s D. 15,708 rad/s Answer: C. ω0 = 2π x 5000 = 31,416 rad/s. (Key point: Frequency scale) 4. The 0.5 F series capacitor scales to about: A. 3.09 nF B. 31 nF C. 10 nF D. 1.0 nF Answer: A. C = 0.5/(kz ω0) = 0.5/(5150 x 31416) = 3.09 nF. (Key point: Capacitor result) ## Related in Filter Design - [GIC HPF design at 4.5 kHz with 0.1 μF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-hpf-4-5khz-simulation) — Filter Design, chapter 8 - [GIC simulation of HPF with Kf = 2000](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-hpf-kf-2000-simulation) — Filter Design, chapter 8, asked 2× - [GIC simulation of a grounded inductor](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-grounded-inductor-simulation) — Filter Design, chapter 8, asked 7× - [GIC simulation of HPF at ωo = 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-hpf-4000-simulation) — Filter Design, chapter 8 - [Sallen-Key LPF at 10 kHz, largest C 0.01 μF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-10khz-largest-capacitance) — Filter Design, chapter 6 - [GIC to avoid shunt inductors in LC ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-avoid-shunt-inductors) — Filter Design, chapter 8 ## Related in other subjects - [Insertion loss design of given LC ladder filter](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Afilter-ladder-insertion-loss-microstrip) — RF and Microwave Engineering, chapter 6