Simplify the following function using K-map: F(A, B, C, D) = Σ (1, 3, 5, 7, 8, 12, 14, 15). Draw the circuit of the simplified function using NAND gates

Sure, let’s simplify the given function ( F(A, B, C, D) ) using Karnaugh Map (K-map):

[
\begin{array}{|c|c|c|c|c|}
\hline
\text{AB} & \text{00} & \text{01} & \text{11} & \text{10} \
\hline
\text{CD} & & & & \
\hline
\text{00} & 1 & 3 & 7 & 5 \
\hline
\text{01} & 8 & & & \
\hline
\text{11} & 12 & & 14 & 15 \
\hline
\text{10} & & & & \
\hline
\end{array}
]

Now, we group the adjacent 1’s:

[
\begin{align} \text{Group 1:} & \quad 1, 3, 7, 5 \ \text{Group 2:} & \quad 8 \ \text{Group 3:} & \quad 12, 14, 15 \ \end{align}
]

[
\begin{array}{|c|c|c|c|c|}
\hline
\text{AB} & \text{00} & \text{01} & \text{11} & \text{10} \
\hline
\text{CD} & & & & \
\hline
\text{00} & X & X & X & X \
\hline
\text{01} & X & & & X \
\hline
\text{11} & X & & X & X \
\hline
\text{10} & & & & \
\hline
\end{array}
]

From the K-map, we get the simplified expression:

[ F(A, B, C, D) = \bar{A}CD + A\bar{B}D + AB ]

Now, let’s draw the circuit using NAND gates:

1. Let’s define intermediate terms:

• ( X = \bar{A}CD )
• ( Y = A\bar{B}D )
• ( Z = AB )

1. Implementing ( X ):

• ( X = NAND(NAND(\bar{A}, C), D) )

1. Implementing ( Y ):

• ( Y = NAND(NAND(A, \bar{B}), D) )

1. Implementing ( Z ):

• ( Z = NAND(A, B) )

1. Finally, ( F ):

• ( F = NAND(NAND(X, Y), Z) )

The circuit will have NAND gates arranged in such a way that they implement the logic described above.