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0 votes

- III only
- IV only
- I, II, III only
- I, II, IV only

Clearly IV is wrong and III is right. I feel I and II need not be true. Anyone please clarify.

+1 vote

1,2,3 are correct

SOP is when you take minterms together and add them up(+)

POS is when you take maxterms and perform a (.) operation on them (a+b).(a’+b’)

SOP uses NAND NAND or you can say AND-OR realization while POS is OR-AND or NOR-NOR realisation.

Thats it.

SOP is when you take minterms together and add them up(+)

POS is when you take maxterms and perform a (.) operation on them (a+b).(a’+b’)

SOP uses NAND NAND or you can say AND-OR realization while POS is OR-AND or NOR-NOR realisation.

Thats it.

0

But generally when we take OR of min terms, we will get Standard SOP, to get Canonical SOP we need to solve the obtained one again using some Boolean algebra rules. But in question they specified "taking OR of min terms". If at all they specified like "solving OR of min terms" then it would have been correct, isn't it ? (same thing for POS)

0 votes

Correct version of (I):- Taking OR of minterms ** for which f takes the value 1** ,canonical SOP can be obtained.

Correct version of(II):- Taking AND of maxterms ** for which f takes the value 0**, canonical POS can be obtained.

(III) is correct.

So option a) III only , should be correct here.

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SOURCE: http://www.cs.ucr.edu/~ehwang/courses/cs120a/minterms.pdf

Definition: Any Boolean function that is expressed as a sum of minterms or as a product of maxterms is said to be in its canonical form.

0

@Ashutosh, see in your pdf it is written that “ANY BOOLEAN FUNCTION that is expressed as a sum of minterms (or maxterms) is called canonical SOP”. To express the function in sum of products form you have to take those minterms for which it is 1. When expressing a function in SOP form, do you include those terms for which it is zero?

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Canonical SOP of F is m3+m5+m6+m7

I dont think m3+m2 is also canonical SOP of F because m3+m2 doesnt even represent F. If what you say is write then m3+m2 should also be canonical SOP of F.

0

they are talking just about the representation. they are just mentioning the procedure about how to convert a boolean equation to canonical form.

suppose you are given a func with minterms m1 and m5(already implies it is true for them)

then they are saying that we just OR them and get the canonical form of the function

suppose you are given a func with minterms m1 and m5(already implies it is true for them)

then they are saying that we just OR them and get the canonical form of the function

0

So what you are saying is the minterms they are talking about ,they are implicitly treating f to be 1 for those minterms? For a function f with n variables there are always exactly 2$^{n}$ minterms. f need not be true for all of them. When talking about random minterm of f i cannot assume that f is true for this minterm so i cannot say ORing it will give canonical SOPof f.

What you are saying maybe they wanted to say that, but anyway i have written more proper version of the statement in my answer.

I dont know what they were trying to say.

What you are saying maybe they wanted to say that, but anyway i have written more proper version of the statement in my answer.

I dont know what they were trying to say.

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