POS FORM
1. K-map of 3 variables
K-map 3 variable POS form
F(A,B,C)=?(0,3,6,7)
From red group we find terms
A B
Taking complement of these two
A' B'
Now sum up them
(A' + B')
From brown group we find terms
B C
Taking complement of these two terms
B’ C’
Now sum up them
(B’+C’)
From yellow group we find terms
A' B' C’
Taking complement of these two
A B C
Now sum up them
(A + B + C)
We will take product of these three terms : Final expression –
(A' + B’) (B’ + C’) (A + B + C)
2. K-map of 4 variables
K-map 4 variable POS form
F(A,B,C,D)=?(3,5,7,8,10,11,12,13)
From green group we find terms
C’ D B
Taking their complement and summing them
(C+D’+B’)
From red group we find terms
C D A’
Taking their complement and summing them
(C’+D’+A)
From blue group we find terms
A C’ D’
Taking their complement and summing them
(A’+C+D)
From brown group we find terms
A B’ C
Taking their complement and summing them
(A’+B+C’)
Finally we express these as product –
(C+D’+B’).(C’+D’+A).(A’+C+D).(A’+B+C’)
PITFALL– *Always remember POS ? (SOP)’
*The correct form is (POS of F)=(SOP of F’)’
Introduction of K-Map (Karnaugh Map)
In many digital circuits and practical problems, we need to find expressions with minimum variables. We can minimize Boolean expressions of 3, 4 variables very easily using K-map without using any Boolean algebra theorems.
K-map can take two forms:
- Sum of product (SOP)
- Product of Sum (POS)
According to the need of problem. K-map is a table-like representation, but it gives more information than the TABLE. We fill a grid of the K-map with 0’s and 1’s then solve it by making groups.