Implementing the Target Binary Equation using NAND Gates

This solution determines the minimum number of NAND gates required to implement the given binary equation $Y = (A+B)(\overline{C \cdot D})$. NAND gates are universal logic gates, meaning any boolean function can be constructed solely using them. The process involves converting each part of the expression into its equivalent NAND gate implementation.

Implementing the OR Term $(A+B)$

The logical OR operation $(A+B)$ is implemented using three NAND gates as follows:

1. Gate 1: Produces $\overline{A}$ using $NAND(A, A)$.
2. Gate 2: Produces $\overline{B}$ using $NAND(B, B)$.
3. Gate 3: Produces $A+B$ using $NAND(\overline{A}, \overline{B})$. This step utilizes the identity $A+B = \overline{(\overline{A} \cdot \overline{B})}$.

This segment requires 3 NAND gates.

Implementing the Term $(\overline{C \cdot D})$

The term $\overline{C \cdot D}$ represents a logical NAND operation directly.

• Gate 4: Produces $\overline{C \cdot D}$ using $NAND(C, D)$.

This segment requires 1 NAND gate.

Implementing the Final AND Combination

The complete equation requires combining the results of the two terms $(A+B)$ and $(\overline{C \cdot D})$ using a logical AND operation: $Y = (A+B) \cdot (\overline{C \cdot D})$. A standard two-input AND operation $(X \cdot Z)$ can be implemented using two NAND gates:

• Gate 5: Computes the NAND of the two terms: $NAND(\text{Output of Gate 3}, \text{Output of Gate 4}) = \overline{(A+B)(\overline{C \cdot D})}$.
• Gate 6: Inverts the output of Gate 5 to achieve the AND result: $NAND(\text{Output of Gate 5}, \text{Output of Gate 5}) = Y$.

This final combination requires 2 NAND gates.

Calculating the Total Number of NAND Gates

The total minimum number of NAND gates is the sum of gates required for each part:

Total Gates = (Gates for $A+B$) + (Gates for $\overline{C \cdot D}$) + (Gates for final AND)

Total Gates = 3 + 1 + 2 = 6.

Therefore, the minimum number of NAND gates required is 6.