GATE CSE 2002 | Question: 2.2

Question

Consider the following multiplexer where $I0, I1, I2, I3$ 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.

The function $f(x,y,z)$ implemented by the above circuit is 

• A. $xyz'$
• B. $xy + z$
• C. $x + y$
• D. None of the above

Comments

It could be done via Table and K-map but that will be more time consuming.

Answer 1

As $x$ is connected to $I0$  & $I1$, $y$ connected to $I2,$ $y'$ connected to $I3$ & $A1,$ $z$ connected to $A0$ and $z'$ connected to ENABLE $(EN),$

$f= ( {\overline{A1}}.{\overline{A0}}.I0+ {\overline{A1}}.A0.I1 +A1.{\overline{A0}}.I2 + A1.A0.I3 ).EN$

$\implies f = (xyz' +xyz + y'z'y + zy')z'$
$\qquad= (xyz' +xyz + zy')z' =xyz'$

Correct Answer: $A$

Comments

how come ZY' for A1.A0.I3  why not Y'ZY'

---

ZY' equals to ZY'Y' @sandeep verma

---

@Prasad patil EN is basically for enabling and disabling the multiplexer 

Answer 2

Option A