Digital Electronics - Boolean Algebra and Logic Simplification - Discussion

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.

From the given sop we can find min terms they are:

Min terms = 1, 4, 5, 6, 7.

From the above, we can know that the sop & pos are compliments.

So Max terms = 0, 2, 3.

Then according to pos.

The answer is B.

There is no need of drawing k-map here because its take some extra time for that compared to this procedure.

F(A,B,C) = m7+m4+m5+m6+m1
= A'B'C' + AB'C' + AB'C + ABC' + A'B'C
F'(A,B,C)= m0 + m2 + m3

If we complement F', we get F;
F = (m0 + m2 + m3)'
By DeMorgan's Theorem;

F = (A'B'C')'(A'BC')'(A'BC)'
F = (A''+B''+C'')(A''+B'+C'')(A''+B'+C')
F = (A+B+C)(A+B'+C)(A+B'+C')

So answer is letter B.

SHORT CUT :NO.OF COMBINATION POSSIBLE WITH 3 VARIABLE IS(01234567)
min terms of the abv eq is: 7,4,5,6,1
IN POS IS TAKING MAX TERMS :REMAINING NUMBERS IS =0,2,3
MIN TERM IS :1 IF U GET 0 IS COMPLIMENT USING METHOD IS :SOP
MAX TERM IS :0 IF U GET 1 IS COMPLEMENT USING METHOD IS :POS

When you find the maxterms corresponding to the question you will obtain ( 0,1,2,3,6) and when you do same for the possible answers.
A. (7,5,4)
B. (0,2,3)
C. (0,2,2)
D. (0,5,2)
The answer is B because it's the only one that satisfies the SOP values of the question.

In the addition expression of our original equation (Given one) there is no term of A bar / Not A.

So, check in the options by only A bar / Not A analysis means no tern can have A bar / Not A in it. This is only possible in option B.

Draw a three variable k-map and plot the given SOP forms. The remaining boxes which are not filled will give the answer for POS form. But you have to inverse each variable for example A means Abar.

This answer is not correct. If maxterms are
0 =not A + not B + not C
2=not A + B + not C
3=A + not B + not C

Maybe it cud be A is not correct.So no answer.

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.

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.