Question
Look at the truth table below. A, B, and C are inputs and Z
is the output. How many cells must its Karnaugh Map
contain?
A B C Z
0 0 1 1 0 0 1
1 0 0 0 0 1 1 1
1
0 1 0 1 0 1 0
1 9. Refer to the problem above. Draw its Karnaugh Map,
populate its contents, and write the logic expression in its
most simpliied SOP form.
is the output. How many cells must its Karnaugh Map
contain?
A B C Z
0 0 1 1 0 0 1
1 0 0 0 0 1 1 1
1
0 1 0 1 0 1 0
1 9. Refer to the problem above. Draw its Karnaugh Map,
populate its contents, and write the logic expression in its
most simpliied SOP form.
Answers
GPT-4o mini
To determine how many cells the Karnaugh map (K-map) should contain, we need to consider the number of inputs. In this case, there are three inputs (A, B, C), which means the K-map will have \(2^3 = 8\) cells.
### Truth Table Rearrangement
Let's correct and arrange the truth table correctly for clarity:
| A | B | C | Z |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
### K-map Setup
The K-map for three variables (A, B, C) is organized as follows, where each cell corresponds to a combination of the variables:
```
BC
00 01 11 10
+----------------
A 0 | 0 1 1 0
1 | 0 1 1 1
```
### Population of the K-map
Now we will fill in the K-map using the mapped values from the truth table:
1. (000) → 0
2. (001) → 1
3. (010) → 0
4. (011) → 1
5. (100) → 0
6. (101) → 1
7. (110) → 1
8. (111) → 1
This means we get:
```
BC
00 01 11 10
+----------------
A 0 | 0 1 1 0
1 | 0 1 1 1
```
### Expression Formation
To find the simplified Sum of Products (SOP) expression from the K-map:
1. Identify the groups of 1s in the K-map:
- There are groups of 1s at (0,1), (0,2), (1,1), (1,2), and (1,3).
2. From these groups, we get the following terms:
- For (0,1) and (0,2): In True when A=0 and B=1, giving us \(B C'\):
- For (1,1): In True when A=1, B=0, and C=1, giving us \(A B'C\).
- For (1,2): In True when A=1 and B=1, C can be anything, giving us \(A B\).
- For (1,3): In True when A=1 and B=1, and C can be anything, giving us \(A B C\).
### Final SOP Expression
Combining these terms in their most simplified form:
\[ Z = A B + A B'C + B C \]
This represents the logic expression for the output Z based on inputs A, B, and C. Thus, the answer includes knowing the K-map has 8 cells, populating it accurately, and deriving the final minterm expression based on grouped 1s.
### Truth Table Rearrangement
Let's correct and arrange the truth table correctly for clarity:
| A | B | C | Z |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
### K-map Setup
The K-map for three variables (A, B, C) is organized as follows, where each cell corresponds to a combination of the variables:
```
BC
00 01 11 10
+----------------
A 0 | 0 1 1 0
1 | 0 1 1 1
```
### Population of the K-map
Now we will fill in the K-map using the mapped values from the truth table:
1. (000) → 0
2. (001) → 1
3. (010) → 0
4. (011) → 1
5. (100) → 0
6. (101) → 1
7. (110) → 1
8. (111) → 1
This means we get:
```
BC
00 01 11 10
+----------------
A 0 | 0 1 1 0
1 | 0 1 1 1
```
### Expression Formation
To find the simplified Sum of Products (SOP) expression from the K-map:
1. Identify the groups of 1s in the K-map:
- There are groups of 1s at (0,1), (0,2), (1,1), (1,2), and (1,3).
2. From these groups, we get the following terms:
- For (0,1) and (0,2): In True when A=0 and B=1, giving us \(B C'\):
- For (1,1): In True when A=1, B=0, and C=1, giving us \(A B'C\).
- For (1,2): In True when A=1 and B=1, C can be anything, giving us \(A B\).
- For (1,3): In True when A=1 and B=1, and C can be anything, giving us \(A B C\).
### Final SOP Expression
Combining these terms in their most simplified form:
\[ Z = A B + A B'C + B C \]
This represents the logic expression for the output Z based on inputs A, B, and C. Thus, the answer includes knowing the K-map has 8 cells, populating it accurately, and deriving the final minterm expression based on grouped 1s.