Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

Yeah, Aditya Warnulkar is correct.

That's a short cut to getting your SOP from a POS expression instead of going by the truth table.

How to convert SOP to POS?

In the question, it is said the SOP form. Using K-map we find the position of respective boolean expression.

That is 1, 4, 5, 6, 7 (these are the SOP forms i.e. 1). In SOP we substitute the 1 to A and 0 to bar A (eg. 101 = A (bar B) C.

It is asked to find the POS form from this sop expression. From k-map, the rest positions which are '0' are the POS I. E 0, 2, 3. In POS we substitute the 1 to bar A and 0 to A (eg. 101 (bar A) B (bar C).

As it is asked POS so we have to write in POS form. So 0, 2, 3 in POS form will be like the option B.

****It would be better if it has some option to attach a photo so that a photo of K-map can be attached.

I think the answer should be A + A'B'C.

The sum of products into the product of sum that means we have to replace in the place of * with +.

Can any one explain it clearly?

I don't get how you did these?

Solving using k maps in somewhat hard.

Sop is not compliment of POS or vice versa both are similar.

As @Priyanka said we are taking max term and according to that the answer is B but when I check I found A is answer according to the explanation.