Design a circuit that takes three bits, X2, X1, X0 as input and produces one output, F. F is 0 if and only if 4<=X<=6 when X = (X2, X1, X0) is read as an unsigned integer.
X2 |
X1 |
Xo |
F |
0 |
0 |
0 |
|
0 |
0 |
1 |
|
0 |
1 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
0 |
|
1 |
0 |
1 |
|
1 |
1 |
0 |
|
1 |
1 |
1 |
|
truth table, equation, and map
Solution:
Given,
=>Input bits are X2, X1 and X0 and output is F.
=>F is 0 if an only if 4 <= X <= 6 where X = (X2, X1, X0)
Explanation:
Creating truth table:
=>When X does not belong to the range 4 <= X <= 6 the value of F = 1 otherwise 0.
X2 | X1 | X0 | F |
0 | 0 | 0 | 1 |
0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 |
Creating equation:
=>We can write F = X2'X1'X0' + X2'X1'X0 + X2'X1X0' + X2'X1X0 + X2X1X0 using minterms(where F = 1)
K-map and circuit diagram:
I have explained each and every part with the help of statements as well image attached to it.
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