Skip to content
Related Articles

Related Articles

Improve Article

GATE | GATE-CS-2002 | Question 27

  • Last Updated : 19 Nov, 2018

Consider the following multiplexor where 10, 11, 12, 13 are four data input lines selected by two address line combinations A1A0 = 00, 01, 10, 11 respectively and f is “the output of the multiplexor. EN is the enable input.

GATECS200227
The function f(x, y, z) implemented by the above circuit is :
(A) xyz’
(B) xy + z
(C) x + z
(D) None of these


Answer: (A)

Explanation: F = (A1’A0’I0 + A1’A0I1 + A1A0’I2 + A1A0I3) * EN

F = (XYZ’ +XYZ + Y’ZY + ZY’)Z’
F = (XYZ’ +XYZ + ZY'(Y + 1))Z’
F = (XYZ’ +XYZ + ZY’ * 1)Z’
F = (XY(Z’ + Z) + ZY’)Z’
F = (XY + ZY’)Z’
F = XYZ’ + 0
F = XYZ’

Thus, option (A) is correct.

Please comment below if you find anything wrong in the above post.

Quiz of this Question

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :