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+23 votes

Consider the following multiplexer 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.

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

- $xyz'$
- $xy + z$
- $x + y$
- None of the above

+35 votes

Best answer

As $X$ connected to $I0$ & $I1$, $Y$ connected to $I2,$ $Y'$ connected to $I3$ & $A1,$ $Z$ connected to $A0$ and $Z'$ connected to ENABLE (EN),

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

$\implies F = (XYZ' +XYZ + Y'Z'Y + ZY')Z'$

$\qquad= (XYZ' +XYZ + ZY')Z' =XYZ'$

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

$\implies F = (XYZ' +XYZ + Y'Z'Y + ZY')Z'$

$\qquad= (XYZ' +XYZ + ZY')Z' =XYZ'$

+1

how to simply this:

F = (XYZ' +XYZ + Y'ZY + ZY')Z'

= (XYZ' +XYZ + ZY')Z' //this step to final step XYZ'

=XYZ'

+1

yes it will multiply

When EN = 1, the**multiplexer** performs its operation depending on the selection line

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