Derive the truth tables, K-maps and the Logic circuits for the following
f(x,y,z)- (0,3,4,7)
f(x,y,z)=(0,3,4,7) (6 Marks)
To derive the truth table for the given function f(x,y,z) = (0, 3, 4, 7), we need to determine the output for all possible combinations of the input variables x, y, and z.
We can represent these combinations in a truth table as follows:
x | y | z | f(x,y,z)
---------------------------
0 | 0 | 0 | 0
0 | 0 | 1 | ?
0 | 1 | 0 | ?
0 | 1 | 1 | ?
1 | 0 | 0 | ?
1 | 0 | 1 | ?
1 | 1 | 0 | ?
1 | 1 | 1 | ?
To determine the values for the remaining rows, we need to look at the given function values.
f(x=0, y=0, z=1) = 3
f(x=0, y=1, z=0) = 4
f(x=0, y=1, z=1) = 7
f(x=1, y=0, z=0) = 0
f(x=1, y=0, z=1) = 0
f(x=1, y=1, z=0) = 0
f(x=1, y=1, z=1) = 0
Now we can fill in the remaining rows of the truth table:
x | y | z | f(x,y,z)
---------------------------
0 | 0 | 0 | 0
0 | 0 | 1 | 3
0 | 1 | 0 | 4
0 | 1 | 1 | 7
1 | 0 | 0 | 0
1 | 0 | 1 | 0
1 | 1 | 0 | 0
1 | 1 | 1 | 0
Next, we can use the truth table to create a Karnaugh map (K-map). The K-map will help us simplify the function to its minimal Boolean expression.
The K-map for f(x,y,z) is as follows:
\begin{matrix}
& z=0 & z=0 & z=1 & z=1 \\
y=0 & 0 & 0 & 0 & 0 \\
y=0 & 0 & 0 & 0 & 3 \\
y=1 & 0 & 0 & 4 & 7 \\
y=1 & 0 & 0 & 0 & 0
\end{matrix}
From the K-map, we can see that f(x,y,z) has four groups of 1's. These groups can be used to construct the logical expression:
f(x,y,z) = x'y'z' + x'y'z + x'yz + xyz
Finally, we can represent this logical expression as a logic circuit using AND, OR, and NOT gates:
_______
x ----| ___| |
| | | OR |---- f(x,y,z)
y ----| AND --|___|_______|
_____
z ----| NOT |
|_____|
This logic circuit accurately represents the function f(x,y,z) = (0, 3, 4, 7).
To derive the truth table, K-map, and logic circuit for the given function f(x, y, z) = (0, 3, 4, 7), follow the steps below:
1. Derive the Truth Table:
The truth table lists all possible combinations of input values (x, y, z) and their corresponding output values (f). Since there are three input variables (x, y, z), there are 2^3 = 8 possible input combinations.
Here is the truth table for the given function:
| x | y | z | f |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | |
| 0 | 1 | 0 | |
| 0 | 1 | 1 | 3 |
| 1 | 0 | 0 | 4 |
| 1 | 0 | 1 | |
| 1 | 1 | 0 | |
| 1 | 1 | 1 | 7 |
2. Derive K-Maps:
To draw the K-maps, group the output values (f) based on their corresponding input combinations. Since there are three input variables (x, y, z), you will have three K-maps, each representing one input variable. The number of cells in each K-map will be 2^(number of input variables - 1).
K-Map for variable x:
```
yz
00 01 11 10
x __________________
0 | | | | |
|---|---|---|---|
1 | | | | |
|---|---|---|---|
```
K-Map for variable y:
```
xz
00 01 11 10
y __________________
0 | | | | |
|---|---|---|---|
1 | | | | |
|---|---|---|---|
```
K-Map for variable z:
```
xy
00 01 11 10
z __________________
0 | | | | |
|---|---|---|---|
1 | | | | |
|---|---|---|---|
```
3. Fill in the K-Maps:
Using the truth table, fill in the K-maps with the corresponding output values (f) for each input combination.
K-Map for variable x:
```
yz
00 01 11 10
x __________________
0 | 0 | | | 4 |
|---|---|---|---|
1 | | | | |
|---|---|---|---|
```
K-Map for variable y:
```
xz
00 01 11 10
y __________________
0 | 0 | | | |
|---|---|---|---|
1 | | 3 | 7 | |
|---|---|---|---|
```
K-Map for variable z:
```
xy
00 01 11 10
z __________________
0 | 0 | | | |
|---|---|---|---|
1 | 3 | | | 7 |
|---|---|---|---|
```
4. Simplify each K-Map:
Simplify each K-map by grouping adjacent cells that have the same output value (f). Groupings should be in powers of 2 (1, 2, 4, 8, etc.), and they must be square or rectangular. Each cell can only belong to one grouping.
For the x K-Map, there is only one grouping:
```
yz
00 01 11 10
x __________________
0 | 0 | - | - | 4 |
|---|---|---|---|
1 | - | - | - | - |
|---|---|---|---|
```
For the y K-Map, there are two groupings:
```
xz
00 01 11 10
y __________________
0 | 0 | - | - | - |
|---|---|---|---|
1 | - | 3 | 7 | - |
|---|---|---|---|
```
For the z K-Map, there are two groupings:
```
xy
00 01 11 10
z __________________
0 | 0 | - | - | - |
|---|---|---|---|
1 | 3 | - | - | 7 |
|---|---|---|---|
```
5. Derive the Logic Circuit:
Convert the groupings from each K-Map into Boolean expressions, and then simplify those expressions. From these simplified expressions, you can design the logic circuit using logic gates.
Based on the groupings from the K-maps, the Boolean expression for f(x, y, z) is:
f(x, y, z) = x'yz + xy'z' + xy'z + xyz
The logic circuit can be designed by using AND, OR, and NOT gates to implement the above Boolean expression.
Note: The simplified Boolean expression may differ depending on the grouping of cells in the K-maps.