Q.
Figure gives a system of logic gates. From the study of truth table it can be found that to produce a high output
(1) at $R$, we must have
Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
The truth table can be written as
$X$
$Y$
$\bar{X}$
$\bar{Y}$
$P= \bar{X} +Y$
$Q=\overline{X \bar{Y}}$
$R=\overline{P+Q}$
0
1
1
0
1
1
0
1
1
0
0
1
1
0
1
0
0
1
0
0
1
0
0
1
1
1
1
0
Hence $X=1,\, Y=0$ gives output $R=1$
| $X$ | $Y$ | $\bar{X}$ | $\bar{Y}$ | $P= \bar{X} +Y$ | $Q=\overline{X \bar{Y}}$ | $R=\overline{P+Q}$ |
|---|---|---|---|---|---|---|
| 0 | 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 1 | 1 | 0 |