Question

Truth table for the following digital circuit will be :

• A.

  x	y	z
  0	0	0
  0	1	0
  1	0	0
  1	1	1

• B.

  x	y	z
  0	0	1
  0	1	1
  1	0	1
  1	1	0

• C.

  x	y	z
  0	0	1
  0	1	1
  1	0	1
  1	1	1

• D.

  x	y	z
  0	0	0
  0	1	1
  1	0	1
  1	1	1

Hint:

Find the outputs of the two AND gates (one uses x and y, the other uses NOT x and y) and feed them to the NAND gate; the two AND outputs can never be 1 together.

Answer 1

The Correct Option is C

Top AND gate output: \(A = x\cdot y\).
Bottom AND gate inputs are the NOT of x and y, so its output: \(B = \bar{x}\cdot y\).
The NAND gate gives \(z = \overline{A\cdot B} = \overline{(x y)(\bar{x} y)}\).
Now \((xy)(\bar{x}y) = (x\bar{x})\,y = 0\cdot y = 0\), since \(x\cdot\bar{x}=0\).
Hence \(z = \bar{0} = 1\) for every combination of x and y.
So the truth table has z = 1 in all four rows: (0,0) gives 1, (0,1) gives 1, (1,0) gives 1, (1,1) gives 1.
Answer: (3) the table with z = 1, 1, 1, 1