1k views

Which functions does NOT implement the Karnaugh map given below?

1. $(w + x) y$
2. $xy + yw$
3. $(w + x) (\bar{w} + y) (\bar{x} + y)$
4. None of the above
edited | 1k views
0
Please derive the expression which is in option C

Solving $K$ map gives $xy +wy$

edited
+1
Why answer is D why not B,  xy+wy = xy + yw Why considering them different?
+1
^^ There is NOT in question

a, b and c are same.
0
I am sorry, I didn't see NOT in the question.
+1
how to get the 2nd option..??
0
Please derive the expression which is in option c
0

@Praveen Saini Sir  @Anu007  xy+wy ,   xy + yw These two are different terms..?

0
No, in binary logic
0

Answer is D.See we can simplify each equation given in the option and get that all of them gives xy +wy. But let think in other way.

1st option is written in POS form, as we can check we get the same if we consider the following impicants.

which is (w+x)y

for second one

Which gives wy+xy

now for 3rd one we can verify like this

which is (w+x)(ˉw+y)(ˉx+y)

So as we can verify each equation in a given KMap, so answer is D

After simplification we get WY+XY
So Ans is Option D

+1 vote

Answer is D.See we can simplify each equation given in the option and get that all of them gives xy +wy. But let think in other way.

1st option is written in POS form, as we can check we get the same if we consider the following impicants.

which is (w+x)y

for second one

Which gives wy+xy

now for 3rd one we can verify like this

which is (w+x)(ˉw+y)(ˉx+y)

So as we can verify each equation in a given KMap, so answer is D

0
–1 vote
Ans is C