Logic gateMedium
Question
For the given logic gate circuit, which of the following is the correct truth table?
Options
A.
B.
C.
D.
Solution
Gate 1: At bottom there is an OR gate with inputs n & m
out put $= n + m$
Gate 2 : A NAND gate, its Input are direct n & the output of OR gate ( $n + m$ )
out put $z = \overline{n \cdot (n + m)}$
since $n.(n + m) = (n.n) + (n.m)$
$$= n + n.m = n(1 + m) = n $$∴ output $z = \overline{n \cdot (n + m)} = \overline{n}$
| n | m | $$z = \overline{n}$$ |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 1 | 0 |
| 1 | 0 | 0 |
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