Question: Use the cubical representation and the method discussed in Section 8.4.2 to find a minimum cost SOP realization of the function f (x 1 ,
Use the cubical representation and the method discussed in Section 8.4.2 to find a minimum cost SOP realization of the function f (x1, . . . , x4) defined by the ON-set ON = {00x0, 100x, x010, 1111} and the don’t-care set DC = {00x1, 011x}.
Section 8.4.2
Assume that the initial specification of a function f is given in terms of implicants that are not necessarily either minterms or prime implicants. Then it is convenient to define an operation that will generate other implicants that are not given explicitly in the initial specification, but which will eventually lead to the prime implicants of f . One such possibility is known as the ∗-product operation, which is usually pronounced the “star-product” operation. We will refer to it simply as the ∗-operation.


*-Operation The *-operation provides a simple way of deriving a new cube by combining two cubes that differ in the value of only one variable. Let A = AAA and B = B B B be two cubes that are implicants of an n-variable function. Thus each coordinate A; and B; is specified as having the value 0, 1, or x. There are two distinct steps in the *-operation. First, the *-operation is evaluated for each pair A; and B;, in coordinates i = 1, 2, ..., n, according to the table in Figure 8.30. Then based on the results of using the table, a set of rules is applied to determine the overall result of the *-operation. The table in Figure 8.30 defines the coordinate *-operation, A; * B. It specifies the result of A; * B; for each possible combination of values of A; and B. This result is the intersection (i.e., the common part) of A and B in this coordinate. Note that when A; and B; have the opposite values (0 and 1, or vice versa), the result of the coordinate *-operation is indicated by the symbol . We say that the intersection of A; and B; is empty. Using the table, the complete *-operation for A and B is defined as follows: C = A * B, such that
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