Lecture 8 (09/20/2002): Supplementary Note on POS with K-Map
Instructor: Yong Kim (Section 1)
During the lecture, I asked you to think about how would you derives a Product of Sum
expression given the K-map.
Product Of Sums Simplification
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Remember, all the minimized Boolean functions derived from the maps in all
previous examples in the lectures were expressed in sum of products (SOP) form.
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With a minor modification, the product of sums form can be obtained.
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The 1s placed in the squares of the map represent the minterms of the function F.
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The 0s represent the minterms of the complement of the function, i.e., F’.
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The complement of F’ is the original function F (=(F’)’) by Involution..
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Because of the generalized DeMorgan’s theorem, the function so obtained is
automatically in the product of sums form.
Step 1: Find F’ in a sum of product form
(simply group and express 0’s in product terms).
Step 2: Find (F’)’ = F by applying De Morgan’s Law, which will automatically convert
the expression in product of sum form.
Or Quick & Dirty ways
Express F’ in minterms, find dual, complement each literals.
Or even faster method:
Express F in maxterm by expressing 0’s in product of maxterms.
Problem: Simplify the Boolean function F(A,B,C,D) = Σ(1, 3, 4, 5, 6, 12,
13, 14) in both sum of products and product of sums form.
F(A,B,C,D) = Σ(1, 3, 4, 5, 6, 12, 13, 14)
AB\CD
A’B’
A’B
AB
AB’
C’D’
0
1
1
0
C’D
1
1
1
0
CD
1
0
0
0
CD’
0
1
1
0
sum of products (what we have been working on most of time)
grouping of 1’s
Make a group of two by grouping m1 and m 3 = A’B’D.
Make a group of four by grouping m4, m5, m12, and m13 = BC’.
Make a group of four by grouping m4, m12, m 6, and m14 = BD’.
read the terms from each group and sum them
F(A,B,C,D) = A’B’D + BC’ + BD’
product of sums
Step 1 – Find F’ by grouping 0’s, expressing 0’s (not 1’
Make a group of four by grouping m0, m2, m8, and m 10 = B’D’.
Make a group of four by grouping m8, m9, m11, and m10 = AB’.
Make a group of two by grouping m7 and m 15 = BCD.
read the terms from each group and sum them
F’(A,B,C,D) = B’D’ + AB’ + BCD
Step 2 – apply DeMorgan’s theorem
(F’)’ = (B’D’ + AB’ + BCD)’
F(A,B,C,D) = (B + D)(A’ + B)(B’ + C’ + D’)
5-variable K-map example: Simplify the Boolean function F(A,B,C,D,E) = Σ(0,
2, 4, 6, 16, 18, 20, 22, 26, 30).
BC\DE
B’C’
B’C
BC
BC’
D’E’
1
1
0
0
D’E
0
0
0
0
DE
0
0
0
0
DE’
1
1
0
0
DE
0
0
0
0
DE’
1
1
1
1
A’
BC\DE
B’C’
B’C
BC
BC’
D’E’
1
1
0
0
D’E
0
0
0
0
A
Step 1 – grouping
Make a group of eight by grouping m0, m4, m 2, m6, m16, m 20, m18, and m22
= B’E’.
Make a group of four by grouping m18, m22, m30, and m26 = ADE’.
Step 2 – read the terms from each group and sum them
F(A,B,C,D,E) = B’E’ + ADE’