---
title: "GIC to avoid shunt inductors in LC ladder"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:gic-avoid-shunt-inductors"
source: https://www.bctengineer.com/q/semester_5/filter_design/gic-avoid-shunt-inductors
site: BCT Engineering
---

# GIC to avoid shunt inductors in LC ladder

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

- Asked in: 2081 Bhadra, 5 marks
- Video: [Inductance Simulator Circuit Using GIC](https://www.youtube.com/watch?v=ZanoUNK7jjk) — A 2 Z Engineering Classes
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/gic-avoid-shunt-inductors
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

What is GIC? How can it be used to avoid shunt inductors in LC ladder circuit?

## Short answer

A GIC is an active circuit (two op-amps, five impedances) with input impedance Zin = Z1 Z3 Z5/(Z2 Z4). By placing a capacitor in Z4 and resistors elsewhere, Zin becomes an inductance L = C R1 R3 R5/R2. With equal resistors R, L = C R^2. This simulates a grounded inductor, eliminating shunt inductors in LC ladders.

## 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.

#### GIC circuit and input impedance

```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 $V_1 = V_3 = V_5 = V_{in}$.

$$
\begin{aligned}
\text{Node 5: } & \frac{V_4 - V_{in}}{Z_4} = \frac{V_{in}}{Z_5} \;\Rightarrow\; V_4 = V_{in}\left(1 + \frac{Z_4}{Z_5}\right) \\
\text{Node 3: } & \frac{V_2 - V_{in}}{Z_2} = \frac{V_{in} - V_4}{Z_3} = -\frac{Z_4}{Z_3 Z_5}V_{in} \;\Rightarrow\; V_2 = V_{in}\left(1 - \frac{Z_2 Z_4}{Z_3 Z_5}\right) \\
\text{Node 1: } & I_{in} = \frac{V_{in} - V_2}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3 Z_5}V_{in}
\end{aligned}
$$

$$
Z_{in} = \frac{V_{in}}{I_{in}} = \frac{Z_1 Z_3 Z_5}{Z_2 Z_4}
$$

#### Avoiding shunt inductors

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**.

In an LC ladder the **shunt inductors** (for example in a highpass ladder, where every inductor goes from a node to ground) are exactly such grounded inductors. The procedure is:

1. Obtain the passive ladder (prototype values, then frequency and impedance scaling).
2. Keep the resistors and capacitors as they are.
3. Replace each shunt inductor $L$ by a GIC: choose $C$ (e.g. 10 nF) and $R = \sqrt{L/C}$.

```text
 o--[Rs]--||--+--||--+--||--o Vo
              |      |      |
            [GIC]  [GIC]   [RL]
              |      |      |
 GND ---------+------+------+
```

**Example:** a shunt inductor of 0.1 H with $C$ = 10 nF needs $R = \sqrt{0.1/10^{-8}}$ = 3.162 kΩ in positions 1, 2, 3 and 5.

Advantages: no bulky, lossy inductors; the ladder's low sensitivity is kept; elements are easy to integrate. Floating (series) inductors cannot be replaced by one grounded GIC; for those, use two GICs or the Bruton/FDNR method.

## Check yourself

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

1. In the Antoniou GIC, what is the input impedance in terms of Z1..Z5?
   A. Z1 Z3 Z5 / (Z2 Z4)
   B. Z2 Z4 Z5 / (Z1 Z3)
   C. Z1 Z5 / (Z2 Z3 Z4)
   D. Z1 Z2 Z3 / (Z4 Z5)
   Answer: A. Analysis with ideal op-amps gives Zin = Z1 Z3 Z5/(Z2 Z4), i.e. the load Z5 times k(s) = Z1Z3/(Z2Z4). (Key point: GIC input impedance)

2. Which arrangement makes the GIC behave as a grounded inductor?
   A. Capacitors in positions 1 and 5
   B. Capacitor in position 5 only
   C. Capacitor in position 4, resistors elsewhere
   D. Capacitor in position 1, resistors elsewhere
   Answer: C. Z4 = 1/sC sits in the denominator, so Zin = (R1R3R5/R2)sC = sL. (Key point: Where the capacitor goes)

3. With C in position 4 and all resistors equal to R, the simulated inductance is:
   A. L = R/C
   B. L = C R^2
   C. L = C^2 R
   D. L = C/R
   Answer: B. Zin = sC R^2, so L = C R^2. (Key point: Inductor value)

4. Why can one GIC replace the shunt inductors of a ladder but not the series ones?
   A. Series inductors carry DC current
   B. The GIC inductor has one end grounded, and shunt inductors go to ground
   C. Series inductors need larger values than a GIC can give
   D. The GIC only works at high frequency
   Answer: B. The simulated inductor is grounded at one end, so it only replaces grounded (shunt) inductors; floating ones need two GICs or the Bruton method. (Key point: Which inductors can be replaced)

## Related in Filter Design

- [FDNR to avoid inductors; design from Figure 1 at 10⁴ rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-figure1-ladder-1e4) — Filter Design, chapter 8
- [FDNR to avoid bulky inductors](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-avoid-bulky-inductors) — Filter Design, chapter 8
- [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×
- [Derive reflection coefficient for terminated LC ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Areflection-coefficient-derivation) — Filter Design, chapter 5, asked 2×
- [Synthesize two-port LC ladder from z21 and z22](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atwo-port-lc-ladder-z-parameters) — Filter Design, chapter 4
- [Define coefficients; synthesize t(s) as LC ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atransmission-reflection-synthesize-ladder) — Filter Design, chapter 5

## 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
- [Voltage ratio transfer function of LC ladder](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Avoltage-ratio-tf-lc-ladder-1h) — Electrical Circuit Theory, chapter 5
- [LC circuit analogy, equal maximum energies](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Alc-max-energies-equal) — Engineering Physics, chapter 1, asked 5×
- [Maximum current in LC circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Alc-maximum-current-numerical) — Engineering Physics, chapter 1