---
title: "Ideal gyrator and floating inductor simulation"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:gyrator-floating-inductor"
source: https://www.bctengineer.com/q/semester_5/filter_design/gyrator-floating-inductor
site: BCT Engineering
---

# Ideal gyrator and floating inductor simulation

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

- Asked in: 2082 Baisakh, 1+4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/gyrator-floating-inductor
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

What is ideal gyrator? How can you simulate floating inductor using gyrator? Explain with necessary figure and derivations.

## Short answer

An ideal gyrator is a lossless two-port with Zin = r^2/ZL. A floating inductor is realized by cascading two identical gyrators with a grounded capacitor between them, yielding L = r^2 C.

## Answer

#### Ideal gyrator

An **ideal gyrator** is a lossless two-port that inverts impedance: a load $Z_L$ on port 2 appears at port 1 as $Z_{in} = r^2/Z_L$, where $r$ is the gyration resistance. Its defining equations are

$$
V_1 = -r I_2, \qquad V_2 = r I_1 \quad\Longrightarrow\quad Z_{in} = \frac{V_1}{I_1} = \frac{r^2}{Z_L}
$$

So a capacitor $C$ on port 2 gives $Z_{in} = r^2 sC = sL$ with $L = r^2C$. A gyrator is built with op-amps (e.g. two voltage-controlled current sources of opposite sign, $I_1 = gV_2$, $I_2 = -gV_1$, $r = 1/g$).

#### Floating inductor using gyrators

One gyrator with a grounded capacitor gives only a **grounded** inductor. A **floating** inductor is made by **cascading two identical gyrators with a grounded capacitor between them**:

```text
        +-----------+          +-----------+
 a o----| Gyrator 1 |----+-----| Gyrator 2 |----o b
        |    (r)    |    |     |    (r)    |
        +-----------+   === C  +-----------+
                         |
 GND ---------------------------------------------- GND
```

Each gyrator and the capacitor share the common ground. The transmission (ABCD) matrix of an ideal gyrator ($V_1 = rI_2$, $I_1 = V_2/r$, with $I_2$ leaving port 2) and of a shunt capacitor are

$$
G = \begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix}, \qquad Y_C = \begin{bmatrix} 1 & 0 \\ sC & 1 \end{bmatrix}
$$

For the cascade:

$$
\begin{aligned}
G\,Y_C\,G &= \begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ sC & 1 \end{bmatrix}\begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix} \\
&= \begin{bmatrix} rsC & r \\ 1/r & 0 \end{bmatrix}\begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix} = \begin{bmatrix} 1 & sCr^2 \\ 0 & 1 \end{bmatrix}
\end{aligned}
$$

The matrix $\begin{bmatrix} 1 & Z \\ 0 & 1 \end{bmatrix}$ is that of a single **series impedance** $Z$ between the two terminals. Here $Z = s\,r^2C$, so the circuit between a and b is a **floating inductor**

$$
L = r^2 C
$$

Example: $r$ = 10 kΩ and $C$ = 10 nF give $L = 10^8 \times 10^{-8}$ = 1 H. The two gyrators must have equal gyration resistance; any mismatch leaves an unwanted grounded element. The capacitor is grounded, which suits IC realisation.

## Check yourself

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

1. A gyrator with gyration resistance r loaded by ZL has input impedance:
   A. r^2/ZL
   B. r ZL
   C. r + ZL
   D. ZL/r^2
   Answer: A. V1 = -rI2 and V2 = rI1 give Zin = r^2/ZL. (Key point: Gyrator property)

2. A capacitor C on port 2 of a gyrator looks at port 1 like:
   A. An inductor L = C/r^2
   B. An inductor L = r^2 C
   C. A capacitor C/r^2
   D. A resistor r
   Answer: B. Zin = r^2 sC = sL with L = r^2 C. (Key point: Grounded inductor)

3. How is a floating inductor made?
   A. A gyrator with a resistor in series
   B. Two identical gyrators cascaded with a grounded capacitor between them
   C. One gyrator with a floating capacitor
   D. Two gyrators in parallel
   Answer: B. G Y_C G gives a series impedance, so the grounded capacitor appears as a floating inductor. (Key point: Floating structure)

4. The product G Y_C G equals:
   A. [1 0; sC 1], a shunt capacitor
   B. [0 r; 1/r 0], a gyrator
   C. [1 sCr^2; 0 1], a series impedance
   D. [0 1; 1 0]
   Answer: C. The matrix is [1 Z; 0 1] with Z = s r^2 C, so L = r^2 C. (Key point: Cascade result)

## Close variants set in other exams

- [Gyrator simulation of inductor](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agyrator-inductor-simulation) — Filter Design, chapter 8, asked 3×

## 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 a grounded inductor](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-grounded-inductor-simulation) — Filter Design, chapter 8, asked 7×
- [Why simulate resistors with switched capacitors in MOS](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asc-resistor-simulation-mos) — Filter Design, chapter 9