To simplify the Boolean function $F(W, X, Y, Z) = \Sigma m(1, 3, 4, 6, 9, 11, 12, 14)$, we'll use a Karnaugh map (K-map) to minimize it. The K-map will help us visualize the minterms and group them for simplification.

Step 1: Set up the Karnaugh map

We create a 4-variable K-map for variables $W, X, Y, Z$. Each minterm corresponds to a specific cell in the K-map.

WX\YZ	00	01	11	10
00	m(0)	m(1)	m(3)	m(2)
01	m(4)	m(5)	m(7)	m(6)
11	m(12)	m(13)	m(15)	m(14)
10	m(8)	m(9)	m(11)	m(10)

Now, place 1's in the K-map for the minterms given in the problem: $1, 3, 4, 6, 9, 11, 12, 14$.

WX\YZ	00	01	11	10
00	0	1	1	0
01	1	0	0	1
11	1	0	0	1
10	0	1	1	0

Step 2: Grouping the 1's

Now, let's form the largest possible groups of 1's to minimize the function.

• Group 1: The four corners of the map, corresponding to minterms 1, 3, 9, and 11. This group gives the simplified term $\overline{W} \overline{Y}$.
• Group 2: The four cells (4, 6, 12, 14) in a 2x2 block. This group gives the term $\overline{X} Z$.

Step 3: Write the simplified expression

The minimal Boolean expression is the OR of the simplified terms: $F(W, X, Y, Z) = \overline{W} \overline{Y} + \overline{X} Z$

Final Answer:

$F(W, X, Y, Z) = \overline{W} \overline{Y} + \overline{X} Z$

Would you like a deeper explanation or any further details on Karnaugh maps or Boolean algebra?

5 Related Questions:

1. How do Karnaugh maps work for more than four variables?
2. What is the difference between minterms and maxterms?
3. How can you implement this Boolean function using logic gates?
4. What are some common mistakes when minimizing Boolean functions?
5. Can Karnaugh maps be used for don't-care conditions, and how?

Tip:

When working with Karnaugh maps, always aim to form the largest groups of 1's possible, as this leads to the simplest Boolean expression.