Solve Digital Logic Homework: Simplify w/ K-map, Sum of Prods & NAND Gates

[TimTimBoBim]

Homework Statement

Using k-map simplify the boolean function:

f(w, x, y, z) = $$\prod$$(1, 3, 4, 5, 7, 11, 12, 13, 15)

into sum of products form

product of sums form

implement using nand gates

Homework Equations

The Attempt at a Solution

I have the k-map, I understand that part. What I don't understand is how to extract the boolean function from the k-map. The NAND gates part I'm clueless on.

Notes
I'm studying for a test and this is a random example problem. I would thoroughly appreciate a walk-through.

Also, for a different question, how do I extract a boolean function from a circuit drawing? Meaning, there are 4 XOR's, leading to 2 AND's, leading to 1 AND.How do I create a boolean function by looking at that drawing?

Thank you.

 

[KataKoniK]

Hi there,

I would be happy to walk you through the process of simplifying a boolean function using a k-map and implementing it with NAND gates. Let's start with the given boolean function:

f(w, x, y, z) = \prod(1, 3, 4, 5, 7, 11, 12, 13, 15)

To simplify this function, we will use a k-map. First, we need to arrange the minterms in the k-map according to their binary representation. Since we have 4 variables (w, x, y, z), we will need a 4-variable k-map. The k-map will look like this:

| 00 | 01 | 11 | 10 |
|----|----|----|----|
| 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 |

Next, we will group the minterms based on their adjacent 1's in the k-map. The groups should be as large as possible and should be in the shape of squares or rectangles. In this case, we have two groups: one with 4 minterms and one with 2 minterms. The groups are shown in the k-map below:

| 00 | 01 | 11 | 10 |
|----|----|----|----|
| 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 |

Next, we will write the simplified boolean function by combining the variables that remain constant within each group. For the first group, the variable y remains constant at 0, so we can write it as y' (y prime). For the second group, the variable x remains constant at 0, so we can write it as x'. The final simplified boolean function is:

f(w, x, y, z) = y'z' + x'z

To implement this function with

 

[Mene]

To solve this digital logic homework, we will first need to understand how to use a k-map to simplify a boolean function. The k-map is a graphical method that helps us to identify patterns in the truth table of a boolean function and simplify it using boolean algebra. In this case, we have a boolean function with four variables, w, x, y, and z. The k-map will have 16 cells, each representing a possible combination of the four variables.

To extract the boolean function from the k-map, we will first group the cells that have a value of 1 together. These groups should be as large as possible and can be in the form of squares, rectangles, or adjacent cells. Each group will represent a term in the boolean function. For example, in the k-map given, we can group cells 1, 3, 4, 5, 7, 11, 12, 13, and 15 together to form a rectangle. This group represents the term w'x'yz' in the boolean function.

After identifying all the groups, we can write the simplified boolean function in sum of products form. This means that we will take the complement of the variables in each group and multiply them together. In this case, our simplified boolean function would be:

f(w, x, y, z) = w'x'yz' + w'x'yz + w'x'y'z' + w'x'y'z + wxy'z' + wxy'z + wxyz' + wxyz + wx'yz'

To implement this function using NAND gates, we can use DeMorgan's theorem to convert the sum of products form into a product of sums form. This means that we will take the complement of the entire simplified boolean function and then use NAND gates to implement it. The simplified boolean function in product of sums form would be:

f(w, x, y, z) = (w + x + y' + z)(w + x + y + z')(w + x' + y' + z')(w + x' + y' + z)(w + x' + y + z')(w + x' + y + z)(w + x + y' + z')(w + x + y' + z)(w + x + y + z)

To implement this using NAND gates, we will first