Modern Physics 6 Question 6
6. The logic gate equivalent to the given logic circuit is
(Main 2019, 9 April II)
(a) NOR
(b) NAND
(c) OR
(d) AND
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Solution:
- Truth table for given combination of logic gates is
$\boldsymbol{A}$ | $\boldsymbol{B}$ | $\boldsymbol{Y} _{\mathbf{1}}=\overline{\boldsymbol{A}}$ | $\boldsymbol{Y} _{\mathbf{2}}=\overline{\boldsymbol{B}}$ | $\boldsymbol{Y}=\overline{\boldsymbol{Y} _{\mathbf{1}} \cdot \boldsymbol{Y} _{\mathbf{2}}}$ |
---|---|---|---|---|
0 | 0 | 1 | 1 | 0 |
1 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
1 | 1 | 0 | 0 | 1 |
Output $Y$ resembles output of an OR gate. So, given combination acts like an OR gate.
Alternate Solution
The given logic gate circuit can be drawn as shown below
Here,
$$ Y=\overline{Y _1 \cdot Y _2}=\overline{\bar{A} \cdot \bar{B}} $$
Using de-Morgan’s theorem, i.e.
$$ \begin{aligned} \overline{x \cdot y} & =\bar{x}+\bar{y} \\ \therefore \quad Y & =\overline{\bar{A}}+\overline{\bar{B}}=A+B \quad[\because \overline{\bar{x}}=x] \end{aligned} $$
This represents the boolean expression for OR gate.