34 votes 34 votes Consider the circuit above. Which one of the following options correctly represents $f\left(x,y,z\right)$ $x\bar{z}+xy+\bar{y}z$ $x\bar{z}+xy+\overline{yz}$ $xz+xy+\overline{yz}$ $xz+x\bar{y}+\bar{y}z$ Digital Logic gatecse-2006 digital-logic circuit-output normal + – Rucha Shelke asked Sep 22, 2014 edited May 17, 2019 by Arjun Rucha Shelke 9.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Chhotu commented Dec 27, 2017 reply Follow Share Minimized expression in SOP form is $f$ = $X+Y'Z$ 4 votes 4 votes Ram Swaroop commented Jan 8, 2020 reply Follow Share After getting x+y'z Draw kmap check with option This eliminate B,C then again check eliminate D Option A true 1 votes 1 votes Please log in or register to add a comment.
Best answer 63 votes 63 votes 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: $f =I_0\bar S +I_1S,$ for $2:1$ MUX, where $I_0$ and $I_1$ are inputs, $S$ is the select line Distributive property, $A+BC = (A+B)(A+C)$ $A+\bar A =1$ Praveen Saini answered Aug 11, 2015 edited Jun 28, 2018 by Arjun Praveen Saini comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments vishalshrm539 commented Oct 30, 2018 reply Follow Share Sir, can you please explain how we will get the same one? 0 votes 0 votes tusharp commented Dec 10, 2018 reply Follow Share See applying these properties may increase your adrenaline in exam if you could not convert your SOP to desired answer. Better minimize the SOP obtained from circuit and then minimize each min term from options. It will be accurate. :) 1 votes 1 votes Lakshman Bhaiya commented Oct 14, 2019 i edited by Lakshman Bhaiya Oct 14, 2019 reply Follow Share $f(x,y,z) = xy'z' + y'z+xy$ If we take $y'$ is common,then we get $f(x,y,z) = y'(xz' + z)+xy$ Apply the distributive property $x+yz = (x+y)\cdot(x+z)$ $f(x,y,z) = y'\big[(x + z)\cdot(z'+z)\big]+xy$ $f(x,y,z) = y'(x + z)+xy$ $\boxed{f(x,y,z) = xy' + y'z + xy}$ $f(x,y,z) = x(y'+y) + y'z$ $\boxed{f(x,y,z) = x + y'z}$ This is minimal SOP. $$(OR)$$ $f(x,y,z) = xy'z' + y'z+xy$ If we take $x$ is common,then we get $f(x,y,z) = x(y'z' + y) + y'z$ Apply the distributive property $x+yz = (x+y)\cdot(x+z)$ $f(x,y,z) = x\big[(y' + y)\cdot(z'+y)\big] + y'z$ $f(x,y,z) = x(z'+y) + y'z$ $\boxed{f(x,y,z) = xz' + xy + y'z}$ $f(x,y,z)= xz' + (xy+y')\cdot(xy + z)$ $f(x,y,z)= xz' + [(x+y')\cdot(y+y')]\cdot(xy + z)$ $f(x,y,z)= xz' + (x+y')\cdot(xy + z)$ $f(x,y,z)= xz' + xyz + xz + y'z$ $f(x,y,z)= xz' + xz(y + 1) + y'z$ $f(x,y,z)= xz' + xz + y'z$ $f(x,y,z)= x(z' + z) + y'z$ $\boxed{f(x,y,z)= x+ y'z}$ This is minimal SOP. So, both are same. $$(OR)$$ $f(x,y,z) = xy'z' + y'z+xy$ We can expand this expression and get the canonical SOP. $f(x,y,z) = \underbrace{xy'z'}_{100} + \underbrace{x'y'z}_{001} + \underbrace{xy'z}_{101}+\underbrace{xyz'}_{110}+ \underbrace{xyz}_{111}$ $f(x,y,z) = \sum(1,4,5,6,7)$ $\boxed{f(x,y,z)= x+ y'z}$ see this for reference: minimization 8 votes 8 votes Please log in or register to add a comment.
16 votes 16 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) Vikrant Singh answered Dec 25, 2014 Vikrant Singh comment Share Follow See all 2 Comments See all 2 2 Comments reply sid1221 commented Dec 11, 2017 reply Follow Share need correction ... its xy+xz'+y'z 2 votes 2 votes vishalshrm539 commented Oct 30, 2018 reply Follow Share No, XY' + Y'Z + XY is also correct! 2 votes 2 votes Please log in or register to add a comment.