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Problem 4-A: Simplify the following expression to sum of three products. F = (a + b’ + c)(c’ + d + b)(a + b + c)(a + d + b) Solution: - Using algebraic manipulation: Using the (dual of) consensus law, that is: (x+y)(x’+z)(y+z)=(x+y)(x’+z) we can combine the second, third and forth terms to get: F = (a + b’ + c)(a + b +c)(b + c’ + d) F = (a + c)(b + c’ + d) = ab + ac’ + ad + bc + cd Hint: (x + y)(x + y’) = x Again by twice using the consensus law (bc + ac’ + ab = ac’ + bc) in the above equation: F = ac’ + ad + bc + cd = ac’ + bc + cd - Using K map: We have: F’ = a’bc’ + b’cd’ + a’b’c’ + a’b’d’ Using these data we can draw the following Kmap (pay attention that in the above equation we have the inverse of F). Using this map: F = ac’ + bc + cd Problem 4-B: Find two different minimum sum of products forms for the function f(a,b,c,d) = m(0,1,2,5,6,13,14,15) + d(3,11) whose prime implicant chart is given below. 0 (1,5) 1 2 X (2,6) 5 6 13 14 X X (5,13) X X (6,14) X X X (11,15) X (13,15) X (14,15) EPI (0,1,2,3) 15 X X X X X X Solution: In order to find the essential prime implicants, we look at each column to see which one has only one X. The related PI is an essential prime implicant. So, from the above table there is only one essential PI which is a’b’ (0,1,2,3). To specify the minimum sum of products besides the above essential PI we should include some other PIs so that they cover all the minterms, that is: f = a’b’ + bc’d + bcd’ + (abd or abc)