Page 66 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS

Unit 4: Minimization of Boolean Algebra

Figure 4.10: Three Sum-terms in our POS Notes

We have three groups, so we expect to have three sum-terms in our POS result. The group
of 4-cells yields a 2-variable sum-term. The two groups of 2-cells give us two 3-variable
sum-terms. Details are shown for how we arrived at the sum-terms above. For a group,
write the binary group input address, then complement it, converting that to the Boolean
sum-term. The final result is product of the three sums.

Simplify the Product-Of-Sums Boolean expression below, providing a result in SOP form:
Out = (A + B + C + D ) (A + B + C + D ) (A + B + C + D ) (A + B + C + D )
+
(A + B + C + D ) (A + B + C + D ) (A + B + C + D)
Solution:
This looks like a repeat of the last problem. It is except that we ask for a Sum-Of-Products
Solution instead of the Product-Of-Sums which we just finished. Map the maxterm 0s
from the Product-Of-Sums given as in the previous problem, below left:
Figure 4.11: Sum-of-Products

Then fill in the implied 1s in the remaining cells of the map right.
Figure 4.12: Remaining Cells of the Map

Form groups of 1s to cover all 1s.

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