Simplification of POS Expression

Simplication of POS Expression  AU May-09,15, Dec.-09,12,14,19

• In practice, the designer should examine both the sum of products and product of sums reductions to ascertain which is more simplified.

• Once the expression is plotted on the K-map instead of making the groups of ones, we have to make groups of zeros.

• Each group of zero results a sum term and it is nothing but the prime implicate. The technique for using maps for POS reductions is a simple step by step process and it is similar to the one used earlier.

1. Plot the K-map and place Os in those cells corresponding to the Os in the truth table or maxterms in the product of sums expression.

2. Check the K-map for adjacent Os and encircle those Os which are not adjacent to any other Os. These are called isolated Os.

3. Check for those Os which are adjacent to only one other 0 and encircle such pairs.

4. Check for quads and octets of adjacent Os even if it contains some Os that have already been encircled. While doing this make sure that there are minimum number of groups.

5. Combine any pairs necessary to include any Os that have not yet been grouped.

6. Form the simplified POS expression for F by taking product of sum terms of all the groups.

To get familiar with these steps we will solve some examples.

Examples for Understanding

Step 1: Fig. 1.4.1 (a) shows the K-map for four variable and it is plotted according to given maxterms.

Step 2: There are no isolated Os.

Step 3: 0 in the cell 0 is adjacent only to 0 in the cell 8. This pair is combined and referred to as group 1.

Step 4: There are two quads. Cells 12, 13, 14 and 15 forms a quad 1 and cells 6, 7, 14, 15 forms a quad 2. These two quads are referred to as group 2 and group 3, respectively.

Step 5:All Os have already been grouped.

Step 6:In group 1, variable Ais eliminated. In group 2, variableC and D areeliminated and ingroup 3 variables A and D are eliminated. Thereforewe getsimplified POS expression as,

Example 1.4.2 Reduce the following function using K-map technique

f(A, B, C, D) = II M (0, 2, 3, 8, 9, 12, 13, 15)

Solution:

Step 1: Fig. 1.4.2 (a) shows the K-map for four variables and it is plotted according to givenmaxterms.

Step 2:There are no isolated Os.

Step 3: The 0 in the cell 15 is adjacent only to 0 in the cell 13 and 0 in the cell 3 is adjacent only to 0 in the cell 2. These two pairs are combined and referred to as group 1 and group 2, respectively.

Step 4: The cells 8, 9, 12 and 13 form a quad which is referred to as group 3.

Step 5: The remaining 0 in the cell 0 is combined with the 0 in the cell 2 to form a pair, which is referred to as group 4.

Step 6: In group 1 and in group 4 variable C is eliminated. In group 2 variable D is eliminated and in group 3 variables B and D are eliminated. Therefore, we get simplified expression in POS form as,

Example 1.4.3 Reduce the following function using K-map technique.

f(A, B, C, D) = л (0, 3, 4, 7, 8, 10, 12, 14) + d (2, 6)  AU: Dec.-09, Marks 10

Example 1.4.4 Simplify F (A, B, C, D) = ∑(0, 1, 2, 5, 8, 9, 10) in sum of products and product of sums using K-map.  AU: Dec.-12,14,19, Marks 12.

Solution :

Example 1.4.5 Simplify the follwing Boolean expression in

i) Sum-of-product

ii) Product-of-sum using Karnaugh map

AC' + B'D + A'CD + ABCD.  AU May-15, Marks 16

Example 1.4.6 Expresss the following function in sum of min-terms and product of max-terms

F(x, y, z) = x + yz.AU May-15, Marks 8

Solution :

Step 1:Represent the expression using k-map

Examples for Practice

Review Question

1. Give the steps for simplification of POS expression.AU: May-08, Dec.-11