Q.
Truth table for the following digital circuit will be :
Solution:
The given circuit is
x
y
$\overline{ x }$
$x \cdot y$
$\bar{x} \cdot y$
$(x \cdot y) \cdot(\bar{x} y)$
$z=\overline{(x \cdot y) \cdot(\bar{x} y)}$
0
0
1
0
0
0
1
0
1
1
0
1
0
1
1
0
0
0
0
0
1
1
1
0
1
0
0
1
x
y
z
0
0
1
0
1
1
1
0
1
1
1
1
| x | y | $\overline{ x }$ | $x \cdot y$ | $\bar{x} \cdot y$ | $(x \cdot y) \cdot(\bar{x} y)$ | $z=\overline{(x \cdot y) \cdot(\bar{x} y)}$ |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 | 0 | 0 | 1 |
| x | y | z |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |