1-find the minimum cost SOP and POS forms for the function f(x1,x2,x3,x4)= Sm(0,
ID: 3638990 • Letter: 1
Question
1-find the minimum cost SOP and POS forms for the function f(x1,x2,x3,x4)= Sm(0,2,8,9,10,15) +D(1,3,6,7)2- Derive a minimum cost realization of the four varible function that is equal to 1 if exactly two or exactly three of its variables are equal to 1 ; eotherwise it is equal to 0
3- find 5 three- varible function for which the product -of- sums form has lower cost than the sum-of- products function
4- find the minimum cost SOP and POS forms for the function f(x1,x2,x3,x4)= IIm(0,1,2,4,5,7,8,9,10,12,14,15)
Explanation / Answer
1)a) The function can be represented in the form of a Karnaugh map as shown in Figure 4.1a. Note that the location of minterms in the map is as indicated in Figure 4.6 (in the text book). To find the minimum-cost SOP expression, it is necessary to find the prime implicants that cover all 1s in the map. The don’t-cares may be used as desired. Minterm m6 is covered only by the prime implicant x1x2, hence this prime implicant is essential and it must be included in the final expression. Similarly, the prime implicants x1x2 and x3x4 are essential because they are the only ones that cover m10 and m15, respectively. These three prime implicants cover all minterms for which f = 1 except m12. This minterm can be covered in two ways, by choosing either x1x3x4 or x2x3x4. Since both of these prime implicants have the same cost, we can choose either of them. Choosing the former, the desired SOP expression is f = x1x2 + x1x2 + x3x4 + x1x3x4 These prime implicants are encircled in the map. The desired POS expression can be found as indicated in Figure 4.1b. In this case, we have to find the sum terms that cover all 0s in the function. Note that we have written 0s explicitly in the map to emphasize this fact. The term (x1 + x2) is essential to cover the 0s in squares 0 and 2, which correspond to maxterms M0 and M2. The terms (x3 + x4) and (x1 + x2 + x3 + x4) must be used to cover the 0s in squares 13 and 14, respectively. Since these three sum terms cover all 0s in the map, the POS expression is f = (x1 + x2)(x3 + x4)(x1 + x2 + x3 + x4) The chosen sum terms are encircled in the map. Observe the use of don’t-cares in this example. To get a minimum-cost SOP expression we assumed that all four don’t-cares have the value 1. But, the minimum-cost POS expression becomes 1 Page 2 possible only if we assume that don’t-cares 3, 5, and 9 have the value 0 while the don’t-care 7 has the value 1. This means that the resulting SOP and POS expressions are not identical in terms of the functions they represent. They cover identically all valuations for which f is specified as 1 or 0, but they differ in the valuations 3, 5, and 9. Of course, this difference does not matter, because the don’t-care valuations will never be applied as inputs to the implemented circuits
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