Minimum number of 2-input NAND gates required to implement the function F = (X+Y)(Z’+W’) is ___.

Steps to Solve

1. Identify the Logical Expression Components

The given expression is ( F = (X + Y)(Z' + W') ). We need to implement this using NAND gates.

2. Rewrite Using NAND Gates

Recall that:

• ( A + B ) can be implemented as ( (A \text{ NAND } A) \text{ NAND } (B \text{ NAND } B) )
• ( A' ) can be implemented as ( A \text{ NAND } A )

3. Break Down the Expression

• For ( (X + Y) ):

  • Convert ( X + Y ): $$ (X + Y) = (X \text{ NAND } X) \text{ NAND } (Y \text{ NAND } Y) $$ This requires 2 NAND gates for ( X ) and ( Y ) respectively, plus one more for the NAND operation.

• For ( (Z' + W') ):

  • Convert ( Z' ) and ( W' ): $$ Z' = Z \text{ NAND } Z $$ $$ W' = W \text{ NAND } W $$ This requires 2 NAND gates for ( Z' ) and ( W' ) respectively, plus one more for ( Z' + W' ).

4. Combine the Results

Now combine the results:

• ( (X + Y)(Z' + W') ) can be computed as: $$ F = (X + Y) \text{ NAND } (Z' + W') $$ You have already implemented ( (X + Y) ) and ( (Z' + W') ) each using 3 NAND gates.

5. Total Number of NAND Gates

Now, sum up all the gates used:

• 3 gates for ( (X + Y) )
• 3 gates for ( (Z' + W') )
• 1 gate for the final NAND operation, making a total of: $$ 3 + 3 + 1 = 7 $$

However, there is an incorrect assumption here; we only utilize NAND gates that were reused. Let's simplify again considering shared gates.

1. Final calculations indicate ( = 4 ).