Question

A 4 \(\times\) 1 multiplexer with two selector lines is used to realize a Boolean function \( F \) having four Boolean variables X, Y, Z and W as shown below. \( S_0 \) and \( S_1 \) denote the least significant bit (LSB) and most significant bit (MSB) of the selector lines of the multiplexer respectively. \( I_0, I_1, I_2, I_3 \) are the input lines of the multiplexer. 

The canonical sum of product representation of F is 
 

• A. \( F(X, Y, Z, W) = \sum m(0, 1, 3, 14, 15) \)
• B. \( F(X, Y, Z, W) = \sum m(0, 1, 3, 11, 14) \)
• C. \( F(X, Y, Z, W) = \sum m(2, 5, 9, 11, 14) \)
• D. \( F(X, Y, Z, W) = \sum m(1, 3, 7, 9, 15) \)

Hint:

For 4:1 multiplexers, use the selector lines to identify the corresponding minterms for the canonical sum of products form.

Answer 1

The Correct Option is B

We are given a 4:1 multiplexer and the task is to find the canonical sum of products form for the Boolean function \( F \). The inputs of the multiplexer are based on the values of the selector lines \( S_0 \) and \( S_1 \) which are the Boolean variables \( X \) and \( Y \), respectively. We then analyze the inputs \( I_0, I_1, I_2, I_3 \) based on the given conditions for \( Z \) and \( W \). The corresponding minterms where the output is 1 are \( m(0, 1, 3, 11, 14) \). Therefore, the correct canonical sum of products is: \[ F(X, Y, Z, W) = \sum m(0, 1, 3, 11, 14) \] Hence, the correct answer is option (B).

Step 1: The multiplexer configuration corresponds to the sum of minterms where the output is 1.

Step 2: After checking the input combinations, we conclude that the canonical sum of products representation is \( F(X, Y, Z, W) = \sum m(0, 1, 3, 11, 14) \).

Final Answer: (B) \( F(X, Y, Z, W) = \sum m(0, 1, 3, 11, 14) \)