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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. $xz+x\bar{y}+\bar{y}z$
in Digital Logic by Active (3.3k points)
edited by | 2.9k views
+1
Minimized expression in SOP form is $f$ = $X+Y'Z$
0
After getting x+y'z
Draw kmap check with option
This eliminate B,C then again check eliminate D
Option A true

2 Answers

+45 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. $A+\bar A =1$
by Veteran (57k 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
0
Sir, can you please explain how we will get the same one?
+1
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. :)
0

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

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

+13 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)
by Boss (13.6k points)
0
need correction ...

its xy+xz'+y'z
+1

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

Answer:

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