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

1. $x\bar{z}+xy+\bar{y}z$
2. $x\bar{z}+xy+\overline{yz}$
3. $xz+xy+\overline{yz}$
4. $xz+x\bar{y}+\bar{y}z$
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0
Minimized expression in SOP form is $f$ = $X+Y'Z$

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. $A+\bar A =1$
answered by Veteran (55k points)
edited by
0
You combined

xy¯z¯ & xy

If we combine

xy¯z¯ & y¯z  we get different answer
0
No, you will get the same, steps may increase/decrease in numbers
if you solve this you will get XY' + Y'Z + XY (this can be simplified to X + Y'Z) with min terms as (1,4,5,6,7)

and option A has the same min terms

so option A is equivalent to XY' + Y'Z + XY

Ans (A)
answered by Boss (13.6k points)
0
need correction ...

its xy+xz'+y'z