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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. $xz+x\bar{y}+\bar{y}z$
asked in Digital Logic by Active (3.7k points) | 1.4k views
0
Minimized expression in SOP form is $f$ = $X+Y'Z$

2 Answers

+31 votes
Best answer

Result of MUX (first one), is,say f1, = xz'+ y'z
Result of MUX(second one , f= f1y' +xy

=(xz'+y'z)y'+xy  = xy'z' +y'z +xy =x(y'z' +y) +y'z  = x(y'+y)(z'+y) +y'z   =xz'+xy+y'z .

Option A.

Note:

1. f =I0S' +I1S ,for 2:1 MUX , where I0 and I1 are inputs , S is select line

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

3. A+A' =1

 

answered by Veteran (54.5k points)
selected by
+12 votes
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 (13.5k points)
0
need correction ...

its xy+xz'+y'z
Answer:


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