Determine the Boolean expression for the logic circuit. Simplify the Boolean expression using Boolean laws and De Morgan’s theorem. Redraw the logic circuit using the simplified Bo... Determine the Boolean expression for the logic circuit. Simplify the Boolean expression using Boolean laws and De Morgan’s theorem. Redraw the logic circuit using the simplified Boolean expression.
Understand the Problem
The question requires determining the Boolean expression for a given logic circuit, simplifying it using Boolean algebra (including De Morgan's theorem), and then redrawing the circuit based on the simplified expression. This involves understanding logic gates (AND, OR, NOT) and their corresponding Boolean operations.
Answer
$X = (\overline{A} \cdot B) + (\overline{C} \cdot D)$
Answer for screen readers
The simplified Boolean expression is: $X = (\overline{A} \cdot B) + (\overline{C} \cdot D)$
The equivalent logic circuit consists of:
- A NOT gate for input A, producing $\overline{A}$.
- An AND gate with inputs $\overline{A}$ and B, producing $\overline{A} \cdot B$.
- A NOT gate for input C, producing $\overline{C}$.
- An AND gate with inputs $\overline{C}$ and D, producing $\overline{C} \cdot D$.
- An OR gate with inputs $\overline{A} \cdot B$ and $\overline{C} \cdot D$, producing $(\overline{A} \cdot B) + (\overline{C} \cdot D)$.
Steps to Solve
- Write the Boolean expression for the original circuit
Starting from the left, the top branch has input A that is inverted by a NOT gate, giving $\overline{A}$. The bottom branch has input B. These two are inputs to an AND gate, giving $\overline{A} \cdot B$. The third branch has input C that is inverted by a NOT gate, giving $\overline{C}$. The fourth branch has input D. These two are inputs to an AND gate, giving $\overline{C} \cdot D$. The outputs of the two AND gates are inputs to an OR gate. Therefore, the Boolean expression for the original circuit is:
$$X = (\overline{A} \cdot B) + (\overline{C} \cdot D)$$
- Simplify the Boolean expression using Boolean algebra
The expression $(\overline{A} \cdot B) + (\overline{C} \cdot D)$ is already in its simplest form. There are no common terms to factor, and De Morgan's laws or other Boolean algebra identities do not directly lead to further simplification.
- Draw the simplified logic circuit
Since the original expression is already simplified, we can redraw the equivalent logic circuit using the expression $X = (\overline{A} \cdot B) + (\overline{C} \cdot D)$. This involves two NOT gates, two AND gates, and one OR gate.
The simplified Boolean expression is: $X = (\overline{A} \cdot B) + (\overline{C} \cdot D)$
The equivalent logic circuit consists of:
- A NOT gate for input A, producing $\overline{A}$.
- An AND gate with inputs $\overline{A}$ and B, producing $\overline{A} \cdot B$.
- A NOT gate for input C, producing $\overline{C}$.
- An AND gate with inputs $\overline{C}$ and D, producing $\overline{C} \cdot D$.
- An OR gate with inputs $\overline{A} \cdot B$ and $\overline{C} \cdot D$, producing $(\overline{A} \cdot B) + (\overline{C} \cdot D)$.
More Information
The De Morgan's Law states: $\overline{A + B} = \overline{A} \cdot \overline{B}$ $\overline{A \cdot B} = \overline{A} + \overline{B}$
Tips
A common mistake is incorrectly applying De Morgan's theorem or other Boolean identities, leading to an incorrect simplification. Another mistake is misinterpreting the logic gates and their corresponding Boolean operations, resulting in an incorrect initial Boolean expression.
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