Question

Write the Boolean function according to the following Verilog source code. Implement it by minimum number of gates. Please sketch the schematic of your circuit.

module problem_A_2_2(input a, b, c,

output y, z);

assign y = a & b & cla& b&cla & -b & c;

assign z=a&-b-a & b;

endmodule

Answer 1

To implement the given Verilog code using the minimum number of gates, we can break down the boolean function into smaller parts and use AND, OR, and NOT gates to implement each part. The y output can be implemented using two AND gates, two NOT gates, and one OR gate, while the z output can be implemented using two AND gates and one OR gate.

Step-by-step explanation:

The given Verilog code represents a Boolean function with three input variables (a, b, and c) and two output variables (y and z). The goal is to implement this function using the minimum number of gates. Let's break down the code and analyze each output assignment:

y = a & b & cla& b&cla & -b & c;

The expression above can be simplified as follows: y = a & b & cla & -b & c; This can be implemented using two AND gates, two NOT gates, and one OR gate. The first AND gate is used to compute a & b & cla. The second AND gate is used to compute -b & c. Finally, the OR gate combines the results of the two AND gates to get the final y value.

z=a&-b-a & b;

The expression above can be simplified as follows: z = (a & -b) | (a & b); This can be implemented using two AND gates and one OR gate. The first AND gate computes a & -b, while the second AND gate computes a & b. The OR gate combines the results of the two AND gates to get the final z value.