--- title: "GIC simulation of floating inductor" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:gic-floating-inductor" source: https://www.bctengineer.com/q/semester_5/filter_design/gic-floating-inductor site: BCT Engineering --- # GIC simulation of floating inductor Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2079 Baisakh, 5 marks - Video: [L5 | Generalized Impedance converter, Simulation of Inductors | Integrated Circuits (KEC501) | Hindi](https://www.youtube.com/watch?v=GTQUnLYe7vc) — Dr. Vaibhav Jain - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/gic-floating-inductor - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question What is GIC? How GIC can be used to simulate the floating inductor in the passive filter? Explain. ## Short answer A generalized impedance converter (GIC) is an active two-op-amp, five-impedance circuit with input impedance Zin = (Z1Z3/Z2Z4) Z5. A floating inductor is realized by connecting two identical GICs, each simulating a grounded inductor, back-to-back with a resistor Rx between their output ports; the resulting series impedance is sCRRx, giving a floating inductance L = C R Rx (all GIC resistors R, Z4 a capacitor). ## 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. #### Input impedance of the 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 $V_1 = V_3 = V_5 = V_{in}$. $$ \begin{aligned} \text{Node 5: } & \frac{V_4 - V_{in}}{Z_4} = \frac{V_{in}}{Z_5} \;\Rightarrow\; V_4 = V_{in}\left(1 + \frac{Z_4}{Z_5}\right) \\ \text{Node 3: } & \frac{V_2 - V_{in}}{Z_2} = \frac{V_{in} - V_4}{Z_3} = -\frac{Z_4}{Z_3 Z_5}V_{in} \;\Rightarrow\; V_2 = V_{in}\left(1 - \frac{Z_2 Z_4}{Z_3 Z_5}\right) \\ \text{Node 1: } & I_{in} = \frac{V_{in} - V_2}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3 Z_5}V_{in} \end{aligned} $$ $$ Z_{in} = \frac{V_{in}}{I_{in}} = \frac{Z_1 Z_3 Z_5}{Z_2 Z_4} $$ #### Grounded inductor With $Z_4 = 1/sC$ and all other positions resistors $R$: $Z_{in} = sCR^2$, i.e. $L = CR^2$. This works only when one end of the inductor is grounded. #### Floating inductor using two GICs A series (floating) inductor, as in a lowpass ladder, has neither end grounded. It is simulated by **two identical GICs connected back to back with a resistor $R_x$ between their port-2 terminals**: ```text a o--[GIC 1]--o x --[ Rx ]-- y o--[GIC 2]--o b port1 port2 port2 port1 (GIC 2 is the same circuit, reversed) ``` Treat the GIC with $Z_5$ removed as a two-port (port 2 at node 5). From the analysis above, $V_1 = V_2$ and $I_1 = I_2/k(s)$, with $k(s) = Z_1Z_3/(Z_2Z_4)$ (this follows from $Z_{in} = k(s)Z_5$ and $I_2 = V_2/Z_5$). With $Z_4 = 1/sC$ and equal resistors, $k(s) = sCR$. In transmission (ABCD) form: $$ \text{GIC 1: } \begin{bmatrix} 1 & 0 \\ 0 & 1/k \end{bmatrix}, \quad \text{series } R_x: \begin{bmatrix} 1 & R_x \\ 0 & 1 \end{bmatrix}, \quad \text{GIC 2 reversed: } \begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix} $$ $$ \begin{bmatrix} 1 & 0 \\ 0 & 1/k \end{bmatrix}\begin{bmatrix} 1 & R_x \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 1 & kR_x \\ 0 & 1 \end{bmatrix} $$ This is the matrix of a single **series (floating) impedance** $Z = k(s)R_x = s\,(CRR_x)$, so between a and b we get a floating inductor $$ L = C R R_x $$ Example: $C$ = 10 nF, $R = R_x$ = 10 kΩ gives $L = 10^{-8} \times 10^8$ = 1 H. The two GICs must be well matched; a mismatch leaves a small unwanted grounded element. (The Bruton/FDNR method is the usual alternative for lowpass ladders.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is one GIC not enough to simulate a series inductor in a lowpass ladder? A. A single GIC cannot produce inductance B. The simulated inductor has one end grounded, but a series inductor is floating C. A single GIC gives a capacitor D. The op-amps saturate with floating loads Answer: B. A GIC inductor is grounded; a floating inductor needs a two-GIC arrangement. (Key point: Why two GICs) 2. How is the floating inductor built? A. Two GICs in cascade with Rx to ground B. One GIC with a resistor in series C. Two identical GICs back to back with a resistor Rx between their port-2 terminals D. Two GICs in parallel with a capacitor Answer: C. GIC 1, series Rx and reversed GIC 2 give the transmission matrix of a series impedance. (Key point: Back-to-back structure) 3. The ABCD product of GIC 1, Rx and reversed GIC 2 equals [1 kRx; 0 1]. What does this represent? A. An ideal transformer B. A shunt admittance C. A single series impedance Z = k(s)Rx D. A gyrator Answer: C. The matrix [1 Z; 0 1] is a series impedance, here Z = k(s)Rx. (Key point: Series impedance result) 4. With k(s) = sCR, the floating inductance is: A. L = C R Rx B. L = C R^2 / Rx C. L = C R / Rx D. L = R Rx / C Answer: A. Z = k Rx = s C R Rx, so L = C R Rx. With C = 10 nF and R = Rx = 10 kΩ this is 1 H. (Key point: Inductance formula) ## Related in Filter Design - [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× - [Ideal gyrator and floating inductor simulation](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agyrator-floating-inductor) — Filter Design, chapter 8, asked 3× - [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 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 - [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 - [Gyrator simulation of inductor](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agyrator-inductor-simulation) — Filter Design, chapter 8, asked 3×