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@jeffrey-xA7lUP • Oct 31, 2012
@pikachu1994-UcQIIJ • Oct 31, 2012
@rahul69-97fAOs • Oct 31, 2012
@pikachu1994-UcQIIJ • Oct 31, 2012
Thank youjeffrey samuelWell in logic expression say
A'B+AB we group like this B(A'+A) so that the Logical law can be applied and the resultant can be verified as B
Like wise in any of the laws direct simplification is done only when a pair of variable and its complement exist in an equation
That is why we take them as pairs of 2^n
Thank yourahul69Well the concept is simple, as there are only Two States , 1 and 0 (eg: A and A'), so for covering all the combinations, it is constructed in the form of 2^n
@jeffrey-xA7lUP • Oct 31, 2012
@pikachu1994-UcQIIJ • Oct 31, 2012
Thanksjeffrey samuelWe need even no of common entries to cancel them out
At any time we need a variable and its complement
Just check this out
ABC+ABC'+AB'C+A'B'C
When simplifying this equation we do this
AB(C+C') + B'C(A+A')
Here in this part A and C are variable and A' and C' are their complements when we add the two we get 1
ie A+A' = 1 (by Boolean law)
We do the same in a Kmap we root out terms which exist in both as a variable and its complement to simplify the equation That is why pairs always are in 2 ^ n terms
Take a simple eg simplify this f=sum of (m0,m1,m2)
This needs a two variable Kmap and eqn is rewritten as f=sigma(0,1,2)
A\B 0 1
0 1 1
1 1
We pair both vertically to get an expression B'
And horizontally to get the expression A'
So F=A'+B'
Now solve the same problem as minterms
F=A'B'+A'B+AB'
F=B'(A+A')+A'B
F=B'+A'B AS (A+A'=1)
F=B'+A' AS{X+YZ=(X+Y).(X+Z)}
Here X=B' ::Y=A':: Z=B
and so (B'+B)(B'+A')=B'+A'
This is a long and tedious process by which we can clearly see why we pair in even no of terms
In the case of higher pair rates like 4 or 8 we do the pairing in steps till we get the simpler expressions
@simplycoder-NsBEdD • Oct 31, 2012
