--- title: "GIC simulation of HPF at ωo = 4000 rad/s" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:gic-hpf-4000-simulation" source: https://www.bctengineer.com/q/semester_5/filter_design/gic-hpf-4000-simulation site: BCT Engineering --- # GIC simulation of HPF at ωo = 4000 rad/s Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2074 Asoj, 4+4 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/gic-hpf-4000-simulation - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What is generalized impedance converter (GIC)? Explain how inductors can be simulated using GIC? Simulate the following highpass filter by active simulation of grounded inductors such that ω0 is 4000 rad/s and practically realizable elements. [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 is an active impedance converter; a grounded inductor L is simulated by a GIC with Z4 = 1/(sC) and all other positions resistors R, yielding L = C R². For the 5th-order Butterworth high-pass filter scaled to ω0 = 4000 rad/s with impedance factor 10⁴, Rs = RL = 10 kΩ; series capacitors are 40.45 nF, 12.5 nF, 40.45 nF; each 1.545 H shunt inductor is replaced by a GIC with C = 10 nF and R = 12.43 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 given ladder is a 5th-order Butterworth highpass prototype ($\omega_0 = 1$ rad/s, 1 Ω terminations): series 1.618 F, shunt 0.618 H, series 0.5 F, shunt 0.618 H, series 1.618 F. Both inductors are grounded, so each is replaced by a GIC. Frequency scale to $\omega_0 = 4000\ \text{rad/s}$ and impedance scale by $k_z$ (chosen as $10^4$ so that capacitors are in the nF range): $$ R_{new} = k_z R, \qquad L_{new} = \frac{k_z L}{\omega_0}, \qquad C_{new} = \frac{C}{k_z\,\omega_0} $$ $k_z = 10000\ \Omega$, so the source and load resistors become $R_s = R_L$ = 10 kΩ. | Prototype | Scaled | Final | |---|---|---| | $C_1$ = 1.618 F (series) | $C = 1.618/(k_z\omega_0)$ | **40.45 nF** | | $L_2$ = 0.618 H (shunt) | $k_z(0.618)/\omega_0$ = 1.545 H | GIC: C = 10 nF, $R = \sqrt{L/C}$ = **12.43 kΩ** | | $C_3$ = 0.5 F (series) | $C = 0.5/(k_z\omega_0)$ | **12.5 nF** | | $L_4$ = 0.618 H (shunt) | $k_z(0.618)/\omega_0$ = 1.545 H | GIC: C = 10 nF, $R = \sqrt{L/C}$ = **12.43 kΩ** | | $C_5$ = 1.618 F (series) | $C = 1.618/(k_z\omega_0)$ | **40.45 nF** | | $R_s, R_L$ = 1 Ω | $k_z \times 1$ | **10 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:** $R_s = R_L$ = 10 kΩ; series capacitors 40.45 nF, 12.5 nF, 40.45 nF; each simulated inductor 1.545 H = GIC with C = 10 nF and four resistors of 12.43 kΩ; $\omega_0$ = 4000 rad/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With a capacitor in position 4 and equal resistors R, a GIC input looks like: A. A floating inductor B. An FDNR C. A grounded inductor L = C R^2 D. A grounded capacitor Answer: C. Zin = sCR^2, a grounded inductor. (Key point: What the GIC does) 2. In the 5th-order highpass ladder, why can each inductor be replaced by a GIC? A. They are lossless B. They are shunt inductors with one end at ground C. They are in series with the source D. They are the smallest elements Answer: B. Shunt inductors are grounded, which is what a GIC simulates. (Key point: Which inductors are simulated) 3. With kz = 10^4 and ω0 = 4000 rad/s, the 0.618 H inductor becomes: A. 24.7 mH B. 1.545 H C. 0.618 H D. 6.18 H Answer: B. L_new = kz L/ω0 = 10^4 x 0.618/4000 = 1.545 H. (Key point: Scaled inductor value) 4. With C = 10 nF, what R realises 1.545 H? A. About 3.9 kΩ B. About 12.43 kΩ C. About 1.24 kΩ D. About 154 kΩ Answer: B. R = sqrt(L/C) = sqrt(1.545/10^-8) = 12.43 kΩ. (Key point: GIC resistor) ## Related in Filter Design - [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 4th order Butterworth HPF at 16,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-fourth-order-hpf-16000) — Filter Design, chapter 8 - [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 HPF design at 5 kHz with 10 nF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-hpf-5khz-simulation) — Filter Design, chapter 8 - [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× - [FDNR to avoid inductors; 3rd order LPF at 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-third-order-4000) — Filter Design, chapter 8, asked 2×