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).
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)
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