# 2-to-4 Line Decoder Circuit## DescriptionThe image displays a logic circuit of a 2-to-4 line decoder, and the goal is to create a truth table based on the given circuit. A 2-to-4 line decoder uses two input lines (A and B) and one enable line (E) to generate four outputs (\(D_0\), \(D_1\), \(D_2\), \(D_3\)).### Circuit Breakdown- **Inputs**: - A - B - E (Enable)- **Outputs**: - \(D_0\) - \(D_1\) - \(D_2\) - \(D_3\)### Logic GatesThe circuit consists of inverters (NOT gates) and NAND gates.### ExplanationFor a 2-to-4 decoder:- When the enable input (E) is `0`, all outputs are `0`.- When the enable input (E) is `1`: - \(D_0\) is `1` if A = `0` and B = `0`. - \(D_1\) is `1` if A = `0` and B = `1`. - \(D_2\) is `1` if A = `1` and B = `0`. - \(D_3\) is `1` if A = `1` and B = `1`.### Truth Table FormatThe truth table consists of the possible values of inputs A, B, and E, and the corresponding outputs (\(D_0\), \(D_1\), \(D_2\), \(D_3\)).#### Diagram ExplanationThe given diagram shows inverters connected to inputs A, B, and E. These, along with their original inputs, are fed into four NAND gates to determine the outputs \(D_0\), \(D_1\), \(D_2\), and \(D_3\).## Truth TableBelow is the format for constructing the truth table:| E | A | B | \(D_0\) | \(D_1\) | \(D_2\) | \(D_3\) ||---|---|---|--------|--------|--------|--------|| 0 | 0

To create truth table for the given 2-4 line decoder circuit

Make the truth table for given 2-to-4 line decoder circuit, as shown in sample: A B E Do Do D₁ D₂ D3 Answer (7 characters separated by a space in each row): 000 0000 000 0.000 000 0000 A B Do D1 D2 D3

Make the truth table for given 2-to-4 line decoder circuit, as shown in sample: A B E Do Do D₁ D₂ D3 Answer (7 characters separated by a space in each row): 000 0000 000 0.000 000 0000 A B Do D1 D2 D3

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

# 2-to-4 Line Decoder Circuit

## Description
The image displays a logic circuit of a 2-to-4 line decoder, and the goal is to create a truth table based on the given circuit. A 2-to-4 line decoder uses two input lines (A and B) and one enable line (E) to generate four outputs (\(D_0\), \(D_1\), \(D_2\), \(D_3\)).

### Circuit Breakdown
- **Inputs**:
- A
- B
- E (Enable)

- **Outputs**:
- \(D_0\)
- \(D_1\)
- \(D_2\)
- \(D_3\)

### Logic Gates
The circuit consists of inverters (NOT gates) and NAND gates.

### Explanation
For a 2-to-4 decoder:
- When the enable input (E) is `0`, all outputs are `0`.
- When the enable input (E) is `1`:
- \(D_0\) is `1` if A = `0` and B = `0`.
- \(D_1\) is `1` if A = `0` and B = `1`.
- \(D_2\) is `1` if A = `1` and B = `0`.
- \(D_3\) is `1` if A = `1` and B = `1`.

### Truth Table Format
The truth table consists of the possible values of inputs A, B, and E, and the corresponding outputs (\(D_0\), \(D_1\), \(D_2\), \(D_3\)).

#### Diagram Explanation

The given diagram shows inverters connected to inputs A, B, and E. These, along with their original inputs, are fed into four NAND gates to determine the outputs \(D_0\), \(D_1\), \(D_2\), and \(D_3\).

## Truth Table
Below is the format for constructing the truth table:

| E | A | B | \(D_0\) | \(D_1\) | \(D_2\) | \(D_3\) |
|---|---|---|--------|--------|--------|--------|
| 0 | 0

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