Minimized expression in SOP form is $f$ = $X+Y'Z$

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+15 votes

Consider the circuit above. Which one of the following options correctly represents $f\left(x,y,z\right)$

- $x\bar{z}+xy+\bar{y}z$
- $x\bar{z}+xy+\overline{yz}$
- $xz+xy+\overline{yz}$
- $xz+x\bar{y}+\bar{y}z$

+35 votes

Best answer

Result of MUX (first one), is, say $f_1 = x\bar z+ \bar yz$

Result of MUX(second one), $f= f_1\bar y +xy$

$\qquad =(x\bar z+\bar yz)\bar y+xy$

$\qquad = x\bar y\bar z +\bar yz +xy$

$\qquad =x(\bar y\bar z +y) +\bar yz$

$\qquad = x(\bar y+y)(\bar z+y) +\bar yz$

$\qquad =xz'+xy+y'z .$

Option A.

Note:

1. $f =I_0\bar S +I_1S,$ for $2:1$ MUX, where $I_0$ and $I_1$ are inputs, $S$ is the select line

2. Distributive property, $A+BC = (A+B)(A+C)$

3. $A+\bar A =1$

Result of MUX(second one), $f= f_1\bar y +xy$

$\qquad =(x\bar z+\bar yz)\bar y+xy$

$\qquad = x\bar y\bar z +\bar yz +xy$

$\qquad =x(\bar y\bar z +y) +\bar yz$

$\qquad = x(\bar y+y)(\bar z+y) +\bar yz$

$\qquad =xz'+xy+y'z .$

Option A.

Note:

1. $f =I_0\bar S +I_1S,$ for $2:1$ MUX, where $I_0$ and $I_1$ are inputs, $S$ is the select line

2. Distributive property, $A+BC = (A+B)(A+C)$

3. $A+\bar A =1$

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