Logic and Computer Design Fundamentals lab course.ppt

* * * * * * * * F = A’ B’ C + A (B’ C’ + B C’ + B’ C + B C) = A’ B’ C + A (B’ + B) (C’ + C) = A’ B’ C + A.1.1 = A’ B’ C + A = B’C + A * * * * * * * * Dual G = ((X+Y) · (W · Z)) = ((X+Y) ·(W + Z) Dual H = (A + B)(A + C)(B + C). Using the Boolean identities, = (A +BC) (B+C) = AB + AC + BC. So H is self-dual. * * Justification 1: 1 . X = X Justification 2: X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X * = X’ Y’ Z + X Y’ (A + B)’ = A’ . B’ (DeMorgan’s Law) = Y’ X’ Z + Y’ X A . B = B . A (Commutative Law) = Y’ (X’ Z + X) A(B + C) = AB + AC (Distributive Law) = Y’ (X’ + X)(Z + X) A + BC = (A + B)(A + C) (Distributive Law) = Y’ . 1 . (Z + X) A + A’ = 1 = Y’ (X + Z) 1 . A = A, A + B = B + A (Commutative Law) * * x . y + x’ . y = (x + x’) . y X(Y + Z) = XY + XZ (Distributive Law) = 1 . y X + X’ = 1 = y X . 1 = X * * * * G’ = (a (b’ + c’) + d) e’ * * * * * * * * * m6 = X Y Z’ M6 = (X’ + Y’ + Z) * M1 = a + b + c + d’ m3 = a’ b’ c d m7 = a’ b c d M 13 = a’ + b’ + c + d’ * * * * * F(A,B,C,D,E) = A’B’C’DE’ + A’BC’D’E + AB’C’D’E + AB’CDE * * F = (A + B + C’ + D’) (A’ + B + C + D) (A’ + B + C’ + D’) (A’ + B’ + C’ + D) * * F = A(B + B’)(C + C’) + (A + A’) B’ C = ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C = ABC + ABC’ + AB’C + AB’C’ + A’B’C = m7 + m6 + m5 + m4 + m1 = m1 + m4 + m5 + m6 + m7 Chapter 2 - Part 1 */64 Shorthand SOM Form From the previous example, we started with: We ended up with: F = m1+m4+m5+m6+m7 This can be denoted in the formal shorthand: Note that we explicitly show the standard variables in order and drop