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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 

  1. $xyz'$
  2. $xy + z$
  3. $x + y$
  4. None of the above
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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: $A$
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Kathleen asked Sep 15, 2014
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Let $f(A,B) = A'+B$. Simplified expression for function $f(f(x+y, y), z)$ is$x' + z$$xyz$$xy' + z$None of the above