What is the value of \( \bar{F}\)? \(F = AB + \bar{C}\bar{D} + \bar{B}D\)
Finding the Complement of a Boolean Function
The question asks us to find the complement of the given Boolean function \(F\). The function is given as \(F = AB + \bar{C}\bar{D} + \bar{B}D\). Finding the complement means finding \( \bar{F} \).
Applying De Morgan's Laws to Find the Complement
To find the complement of a Boolean expression, we use De Morgan's Laws. De Morgan's Laws provide a way to simplify the complement of sums and products:
- Complement of a sum: \( \overline{X + Y} = \bar{X} \cdot \bar{Y} \)
- Complement of a product: \( \overline{X \cdot Y} = \bar{X} + \bar{Y} \)
We will apply these laws to the given function \(F\).
Given \(F = AB + \bar{C}\bar{D} + \bar{B}D\).
To find \( \bar{F} \), we take the complement of the entire expression:
\( \bar{F} = \overline{AB + \bar{C}\bar{D} + \bar{B}D} \)
First, apply the complement of a sum law (\( \overline{X+Y+Z} = \bar{X}\bar{Y}\bar{Z} \)) by treating each term (\(AB\), \( \bar{C}\bar{D} \), and \( \bar{B}D \)) as a single variable:
\( \bar{F} = \overline{(AB)} \cdot \overline{(\bar{C}\bar{D})} \cdot \overline{(\bar{B}D)} \)
Now, apply the complement of a product law (\( \overline{XY} = \bar{X} + \bar{Y} \)) to each of the complemented terms:
- For the first term \( \overline{(AB)} \): \( \overline{AB} = \bar{A} + \bar{B} \)
- For the second term \( \overline{(\bar{C}\bar{D})} \): \( \overline{\bar{C}\bar{D}} = \overline{\bar{C}} + \overline{\bar{D}} \) Since \( \overline{\bar{X}} = X \), this becomes \( C + D \).
- For the third term \( \overline{(\bar{B}D)} \): \( \overline{\bar{B}D} = \overline{\bar{B}} + \bar{D} \) Since \( \overline{\bar{X}} = X \), this becomes \( B + \bar{D} \).
Substitute these results back into the expression for \( \bar{F} \):
\( \bar{F} = (\bar{A} + \bar{B}) \cdot (C + D) \cdot (B + \bar{D}) \)
Comparing with Options
Let's compare our result with the given options:
- \( \bar{A}\bar{B} + \bar{C}\bar{D} + {B}\bar{D} \) - This is a sum of products. Our result is a product of sums. Incorrect.
- \( (A + B) (\bar{C} + \bar{D}) (\bar{B} + D) \) - This is a product of sums, but the terms within the parentheses are different from our result. For example, the first term is \( (A+B) \) instead of \( (\bar{A}+\bar{B}) \). Incorrect.
- \( (\bar{A} + \bar{B}) (C + D) (B + \bar{D}) \) - This is a product of sums, and the terms \( (\bar{A} + \bar{B}) \), \( (C + D) \), and \( (B + \bar{D}) \) match our calculated expression for \( \bar{F} \). Correct.
- \( (A + \bar{B}) (C + \bar{D}) (B + \bar{D}) \) - This is a product of sums, but the terms within the parentheses are different from our result. For example, the first term is \( (A+\bar{B}) \) instead of \( (\bar{A}+\bar{B}) \) and the second term is \( (C+\bar{D}) \) instead of \( (C+D) \). Incorrect.
Our calculated complement \( \bar{F} = (\bar{A} + \bar{B}) (C + D) (B + \bar{D}) \) matches option 3.
Conclusion
The complement of the Boolean function \( F = AB + \bar{C}\bar{D} + \bar{B}D \) is \( \bar{F} = (\bar{A} + \bar{B}) (C + D) (B + \bar{D}) \).
Revision Table: Key Boolean Algebra Concepts
| Concept | Description | Example |
|---|---|---|
| Complement | The inverse of a Boolean variable or expression. If the variable is True (1), its complement is False (0), and vice versa. Denoted by a bar (\( \bar{X} \)) or prime (\( X' \)). | If \( X = 1 \), then \( \bar{X} = 0 \). If \( Y = 0 \), then \( \bar{Y} = 1 \). |
| De Morgan's First Law | The complement of a sum is the product of the complements. | \( \overline{A + B} = \bar{A} \cdot \bar{B} \) |
| De Morgan's Second Law | The complement of a product is the sum of the complements. | \( \overline{A \cdot B} = \bar{A} + \bar{B} \) |
| Double Complement | The complement of a complement is the original variable or expression. | \( \overline{\bar{X}} = X \) |
Additional Information: Working with Boolean Expressions
Understanding Boolean algebra and De Morgan's Laws is crucial in digital logic design. These principles help in simplifying complex logical expressions, which in turn leads to simpler digital circuits using fewer logic gates. Simplifying circuits reduces cost, power consumption, and potential points of failure.
When finding the complement of a complex expression like the one in the question, it is important to apply De Morgan's Laws step by step. First, identify the main operation (sum or product). Apply the corresponding De Morgan's Law, and then work on the resulting sub-expressions until all complement signs are over single variables.
The form of the original function (Sum of Products - SOP) was \( AB + \bar{C}\bar{D} + \bar{B}D \). The complement we found is in the form of a Product of Sums (POS) \( (\bar{A} + \bar{B}) (C + D) (B + \bar{D}) \). This is a common outcome when taking the complement of an SOP expression.
Important Questions from Minimization of Boolean Expression
Simplify the following Boolean expression.
E(E + F) + DE + D(E + F)
Which statement(s) is/are correct regarding the Boolean algebra?
I. It facilitate the analysis and design of digital circuits.
II. Expresses in algebraic form the input-output relationship of logic diagram.
The input-output relationship of the binary variable for each gate can be represented in tabular form by a _______.
What is the simplified expression for the Boolean function F(A, B, C, D) = Σ(0, 1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 14) using the K - map method?
Given a Boolean function F(A, B, C) = Σ(0, 1, 2, 3, 5), what is the expression in SOP form?
