Boolean Function AB + CD Analysis

The problem asks us to determine the minimum number of 2-input NAND gates required to implement the Boolean function Y = AB + CD.

Realizing Logic Functions with NAND Gates

NAND gates are universal gates, meaning any logic function can be built using only NAND gates. To find the minimum number of gates, we need to efficiently convert the given expression into a form directly implementable using NAND gates, often by applying De Morgan's laws.

Step-by-Step NAND Gate Implementation

The target function is Y = AB + CD. This expression represents the OR operation between two product terms, AB and CD. Let X = AB and Z = CD. The function is then Y = X + Z.

We can rewrite the OR operation using NAND logic. A key identity derived from De Morgan's laws is: X + Z = $\overline{\overline{X} \cdot \overline{Z}}$.

Substituting our terms back, we get:

Y = AB + CD = $\overline{\overline{AB} \cdot \overline{CD}}$

This expression indicates that the function can be realized by first obtaining the inverted product terms $\overline{AB}$ and $\overline{CD}$, and then performing a NAND operation on these results.

Implementing $\overline{AB}$

A single 2-input NAND gate can directly produce the output $\overline{AB}$ when its inputs are A and B.

• Gate 1: Inputs A, B. Output: $\overline{AB}$.

Implementing $\overline{CD}$

Similarly, another 2-input NAND gate can produce the output $\overline{CD}$ when its inputs are C and D.

• Gate 2: Inputs C, D. Output: $\overline{CD}$.

Combining Terms for Final Output

The final step is to combine the outputs of Gate 1 ($\overline{AB}$) and Gate 2 ($\overline{CD}$) using a third 2-input NAND gate. The output of this gate is $\overline{(\overline{AB}) \cdot (\overline{CD})}$.

As per De Morgan's law, $\overline{(\overline{AB}) \cdot (\overline{CD})} = \overline{\overline{AB}} + \overline{\overline{CD}} = AB + CD. This successfully implements the required function Y.

• Gate 3: Inputs (Output of Gate 1), (Output of Gate 2). Output: Y = AB + CD.

Minimum Gate Count

The implementation requires a total of three 2-input NAND gates:

• One gate to compute $\overline{AB}$.
• One gate to compute $\overline{CD}$.
• One gate to perform the final NAND operation combining the previous two results.

Thus, the minimum number of 2-input NAND gates needed is 3.