A Boolean digital circuit is composed using two 4-input multiplexers (M1 and M2) and one 2-input multiplexer (M3) as shown in the figure. X0–X7 are the inputs of M1 and M2 and can be set to 0 or 1. The select lines of M1 and M2 are $(A,C)$, and the select line of M3 is $B$. The output of M3 is the final circuit output.
Which one of the following sets of values of $(X0,X1,X2,X3,X4,X5,X6,X7)$ will realise the Boolean function \[ F(A,B,C) = \overline{A} + \overline{A}C + ABC \; ? \]
A Boolean digital circuit is composed using two 4-input multiplexers (M1 and M2) and one 2-input multiplexer (M3) as shown in the figure. X0–X7 are the inputs of M1 and M2 and can be set to 0 or 1. The select lines of M1 and M2 are $(A,C)$, and the select line of M3 is $B$. The output of M3 is the final circuit output.
Which one of the following sets of values of $(X0,X1,X2,X3,X4,X5,X6,X7)$ will realise the Boolean function \[ F(A,B,C) = \overline{A} + \overline{A}C + ABC \; ? \]
Show Hint
- (1, 1, 0, 0, 1, 1, 1, 0)
- (1, 1, 0, 0, 1, 1, 0, 1)
- (1, 1, 0, 1, 1, 1, 0, 0)
- (0, 0, 1, 1, 0, 1, 1, 1)
The Correct Option is C
Solution and Explanation
To solve this problem, we need to determine which combination of values for \((X0, X1, X2, X3, X4, X5, X6, X7)\) will realize the Boolean function \(F(A, B, C) = \overline{A} + \overline{A}C + ABC\) using the given multiplexers.
The circuit comprises two 4-input multiplexers (M1 and M2) and one 2-input multiplexer (M3).
Step 1: Writing Truth Table of the Function
The given Boolean function is \(F(A, B, C) = \overline{A} + \overline{A}C + ABC\).
| A | B | C | F(A, B, C) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Step 2: Configure the Multiplexers
M1 and M2 have select lines \((A, C)\), meaning their outputs are dependent on these inputs. M1 provides the output for \(B=0\) and M2 for \(B=1\).
- When \(B = 0\):
- M1 outputs based on values \((X0, X1, X2, X3)\) selected per \((A, C)\) combination.
- When \(B = 1\):
- M2 outputs based on values \((X4, X5, X6, X7)\) selected per \((A, C)\) combination.
Step 3: Map to the Truth Table
We map the truth table to the inputs of M1 and M2:
- M1 outputs match the truth table for \(B=0\), covered by \((X0, X1, X2, X3)\).
- M2 outputs match the truth table for \(B=1\), covered by \((X4, X5, X6, X7)\).
The set of values that satisfies this condition is (1, 1, 0, 1, 1, 1, 0, 0):
- \(X0 = 1\), \(X1 = 1\), \(X2 = 0\), \(X3 = 1\) for \(B=0\).
- \(X4 = 1\), \(X5 = 1\), \(X6 = 0\), \(X7 = 0\) for \(B=1\).
The correct option is indeed (1, 1, 0, 1, 1, 1, 0, 0).
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