Question: Ex 4 (15 points) For each three-variable K-map, find the optimized expression. (c) (a) (b) A W 1 0 1 YZ 00 0 1
Ex 4 (15 points) For each three-variable K-map, find the optimized expression. (c) (a) (b) A W 1 0 1 YZ 00 0 1 0 BC 00 1 1 1 1 1 01 1 UV 00 01 11 11 10 01 11 1 1 1 1 1 1 1 10 1 10 1 Ex 5 (14 points) Optimize the following Boolean expressions using a K-map (a) F = XYZ + ZY + XZ + XYZ (b) G=AXCW + AXC + CAX+XCW+WACX + AWC + AXWC Ex 1 (20 points) Using algebraic manipulation, show that the left and right sides of each of the following Boolean expressions are equivalent: (a) XY+XY+XY=X+Y (b) BC + BC + ABC= B (c) XY + ZY+XZ + XY+Y2 = XY + XZ+Y2 (d) ABCD + BCD + ACD + BCD + BCD + ABCD + ACD=CD+BD + BD Ex 2 (18 points) Reduce the following Boolean expressions to the minimum number of literals possible using algebraic reduction. (a) AC + ABC + BC (b) (A+B+C). ABC (c) ABD + ACD + BD Ex 3 (18 points) Find the complement of the following Boolean expressions: (a) AB+ AB(C+ D) (b) WXY (CD +Z) + WX( + Z)(Y+ Z) (c) (A+C+B)(AB+ C)
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Ex 4 a For the given 3variable Kmap rz x 00 01 11 10 0 1 1 1 1 1 The optimized expression is rz x Explanation The Kmap shows that the output is 1 for all combinations of the input variables r z and x ... View full answer
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