PhysicsSemiconductors and Electronic Devices

Which of the following truth tables is true for the following circuit?

$Figure for the question$

Answer & explanation

Correct answer: option 1

The correct answer is Option (1)

$Figure for option 1$

The circuit is two NOT gates (the input lines cross, so $ Y_1 $ is the inverse of B and $ Y_2 $ the inverse of A), two NAND gates and a final NOR gate:

    $ Y_1 = \overline{B},\quad Y_2 = \overline{A} $
    $ Y_3 = \overline{A\cdot Y_1},\quad Y_4 = \overline{Y_2\cdot B} $
    $ Y = \overline{Y_3 + Y_4} $

Taking each input pair in turn (A, B):

    (0, 0): $ Y_1 = 1,\ Y_2 = 1,\ Y_3 = 1,\ Y_4 = 1,\ Y = 0 $
    (1, 0): $ Y_1 = 1,\ Y_2 = 0,\ Y_3 = 0,\ Y_4 = 1,\ Y = 0 $
    (0, 1): $ Y_1 = 0,\ Y_2 = 1,\ Y_3 = 1,\ Y_4 = 0,\ Y = 0 $
    (1, 1): $ Y_1 = 0,\ Y_2 = 0,\ Y_3 = 1,\ Y_4 = 1,\ Y = 0 $

The output is 0 for every input, which is the truth table of Option (1).

Shortcut: Since $ Y = \overline{Y_3}\cdot\overline{Y_4} $, we get $ Y = (A\cdot\overline{B})(\overline{A}\cdot B) $. This product contains both A and $ \overline{A} $, so it is always 0, and the all-zero table follows without any rows.