GATE CSE 2006 | Question: 35

Question

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

• A. $x\bar{z}+xy+\bar{y}z$
• B. $x\bar{z}+xy+\overline{yz}$
• C. $xz+xy+\overline{yz}$
• D. $xz+x\bar{y}+\bar{y}z$

Comments

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

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After getting x+y'z
Draw kmap check with option
This eliminate B,C then again check eliminate D
Option A true

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Detailed Video Solution: https://youtu.be/wEhjAAs8iqk?t=9051&feature=shared 

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Option A is correct

Answer 1

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. $A+\bar A =1$

Comments

Sir, can you please explain how we will get the same one?

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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. :)

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$f(x,y,z) = xy'z' + y'z+xy$

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

Answer 2

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)

Comments

need correction ...

its xy+xz'+y'z

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No, XY' + Y'Z + XY is also correct!

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Yes brother, instead of taking x as common. Why can't we take y' as coomon