Which of the following is NOT true about self-dual function?
The function must contain ‘n’ mutually exclusive terms
In digital logic and Boolean algebra, understanding the characteristics of different types of Boolean functions is crucial. This question focuses on the properties of a self-dual function. A self-dual function possesses unique attributes that distinguish it from other Boolean functions.
| Key Concept | Description |
|---|---|
| Boolean Function | A mathematical function that maps input Boolean values (true/false, 1/0) to a single Boolean output value. |
| Dual Function (\(F^D\)) | Obtained from a Boolean function \(F\) by interchanging logical AND (·) with OR (+), OR (+) with AND (·), 0 with 1, and 1 with 0. Variables are kept as they are. |
| Self-Dual Function | A Boolean function \(F\) is said to be self-dual if it is equal to its dual, i.e., \(F = F^D\). |
Let's analyze each option to determine which statement is NOT true about a self-dual function.
Based on the analysis:
The question asks which of the given statements is NOT true. Option 4 presents a condition that is not universally applicable to all self-dual functions.
Consider the following four variable Boolean function in sum-of-product form
$F(b_3, b_2, b_1, b_0) = \sum(0, 2, 4, 8, 10, 11, 12)$.
where the value of the function is computed by considering $b_3b_2b_1b_0$ as a 4-bit binary number, where $b_3$ denotes the most significant bit and $b_0$ denotes the least significant bit. Note that there are no don't care terms. Which ONE of the following options is the CORRECT minimized Boolean expression for F?
| x | y | z | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
The simplified form of a Boolean equation $(A\overline{B} + A\overline{B}C + AC)(\overline{A}\overline{C} + \overline{B})$ is :
Simplify the following Boolean expression.
E(E + F) + DE + D(E + F)
What is the value of \( \bar{F}\)?
\(F = AB + \bar{C}\bar{D} + \bar{B}D\)