---
title: "GIC simulation of a grounded inductor"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:gic-grounded-inductor-simulation"
source: https://www.bctengineer.com/q/semester_5/filter_design/gic-grounded-inductor-simulation
site: BCT Engineering
---

# GIC simulation of a grounded inductor

Filter Design · Chapter 8: Design of High-Order Active Filters

- Asked in: 2083 Baisakh, 2+4 marks; 2075 Asoj, 5 marks; 2073 Shrawan, 5 marks; 2071 Shrawan, 6 marks; 2070 Chaitra, 5 marks; 2081 Bhadra, 4 marks; 2080 Bhadra, 5 marks (asked 7 times)
- 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-grounded-inductor-simulation
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

What is GIC (Antoniou's GIC)? How can a GIC be used to simulate a grounded inductor (in the passive filter)? Explain with necessary figures and derivations.

## Short answer

Antoniou's GIC is a two-op-amp circuit with five impedances giving Z_in = Z1 Z3 Z5 / (Z2 Z4). To simulate a grounded inductor, set Z4 = 1/(sC4) and the others as resistors, yielding Z_in = sL with L = C4 R1 R3 R5 / R2 (or Z2 capacitive with L = C2 R1 R3 R5 / R4).

## Answer

#### GIC (Antoniou's GIC)
The **GIC (Antoniou's generalized impedance converter)** is a two-op-amp circuit with five impedances $Z_1 \dots Z_5$ in a chain. Its input impedance is
$$
Z_{in} = \frac{Z_1Z_3Z_5}{Z_2Z_4}
$$
By choosing which $Z$'s are resistors and which are capacitors, it can simulate a grounded inductor or an FDNR.

```text
 Iin -->
 V1 o-----+---------- A1 input
          |
          Z1
          |
          +---------- A1 output (Va)
          Z2
          |
     (3)  +---------- A1 and A2 inputs
          Z3
          |
          +---------- A2 output (Vb)
          Z4
          |
     (5)  +---------- A2 input
          Z5
          |
         GND
```

Op-amp A1 has its inputs at nodes 1 and 3; A2 has its inputs at nodes 3 and 5.

**Derivation** (ideal op-amps: virtual short gives $V_1 = V_3 = V_5 = V$; no input current):
$$
\begin{aligned}
I_{Z5} &= \frac{V}{Z_5} \quad\text{(flows through } Z_4 \text{ too)} \\
V_b &= V + Z_4\frac{V}{Z_5} \\
I_{Z3} &= \frac{V - V_b}{Z_3} = -\frac{Z_4}{Z_3Z_5}V \quad\text{(flows through } Z_2 \text{ too)} \\
V_a &= V + Z_2 I_{Z3} = V - \frac{Z_2Z_4}{Z_3Z_5}V \\
I_{in} &= \frac{V - V_a}{Z_1} = \frac{Z_2Z_4}{Z_1Z_3Z_5}V \\
Z_{in} &= \frac{V}{I_{in}} = \frac{Z_1Z_3Z_5}{Z_2Z_4}
\end{aligned}
$$

#### Simulating a grounded inductor
**Grounded inductor simulation:** make $Z_4 = \dfrac{1}{sC_4}$ and $Z_1, Z_2, Z_3, Z_5$ resistors:
$$
Z_{in} = \frac{R_1R_3R_5}{R_2}\,sC_4 = sL, \qquad L = \frac{C_4R_1R_3R_5}{R_2}
$$
(Equally, $Z_2$ can be the capacitor, giving $L = C_2R_1R_3R_5/R_4$.) With all resistors equal to $R$, $L = R^2C$. One end of the simulated inductor is ground, so the GIC replaces the **shunt (grounded) inductors** of an LC ladder, e.g. in a high-pass ladder.

**Example:** $L = 0.5$ H with $C_4 = 0.1\ \mu$F and $R_1 = R_2 = R_3 = R_5 = R$: $R = \sqrt{L/C} = \sqrt{0.5/10^{-7}} = 2.236\ \text{k}\Omega$.

**Advantages:** the realised inductor has a good Q over a useful frequency range (limited by the op-amp gain-bandwidth); it uses only R, C and op-amps, so the low-sensitivity property of the passive LC ladder is kept.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the input impedance of Antoniou's GIC?
   A. Z1 Z3 Z5 /(Z2 Z4)
   B. Z2 Z4 /(Z1 Z3 Z5)
   C. Z1 Z2 Z3 /(Z4 Z5)
   D. Z1 + Z3 + Z5
   Answer: A. The derivation gives Zin = V/Iin = Z1Z3Z5/(Z2Z4). (Key point: GIC input impedance)

2. What property of ideal op-amps lets the derivation set V1 = V3 = V5 = V?
   A. Infinite output resistance
   B. Virtual short between inputs, with no input current
   C. Unity gain
   D. Zero bandwidth
   Answer: B. The virtual short makes the node voltages equal and zero input current lets the same current flow through adjacent impedances. (Key point: Derivation basis)

3. Which choice makes the GIC simulate a grounded inductor?
   A. Z1 and Z3 capacitors, rest resistors
   B. Z4 a capacitor, Z1, Z2, Z3, Z5 resistors
   C. All five capacitors
   D. Z5 an inductor
   Answer: B. Zin = R1R3R5 sC4 / R2 = sL. (Key point: Making an inductor)

4. With Z4 = C4, what is L?
   A. C4 R2 R4 / R1
   B. C4 R1 R3 R5 / R2
   C. C4 R2 /(R1 R3 R5)
   D. C4 (R1 + R3 + R5)
   Answer: B. L = C4 R1 R3 R5 / R2; with equal R this is R^2 C. (Key point: Value of L)

## Related in Filter Design

- [GIC simulation of floating inductor](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-floating-inductor) — 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 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
- [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 current relation and grounded FDNR](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-grounded-fdnr) — 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