Solutions-HW2

# Solutions-HW2 - Homework 2 Solution Problem 1 (2 marks)...

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Homework 2 Solution Problem 1 (2 marks) Find the complement of the following function: x + Y(Z + (X + Z)') Change every variable to its complement and exchange ands and ors. F' = X' (Y' + (Z' (X + Z))) = X' (Y' + (Z'X +Z'Z)) = (Y' + XZ') Or applying De-Morgan theorem multiple times. F' = X'(Y(Z + (X + Z)'))' = X' + (Z + (X + Z)')') = + (Z' (X + Z)")) = + (Z' (X + = 01'' + (Z' (X + Z))) = (Y' + (Z'X +Z'Z)) = + XZ') 2 (5 For the Boolean Function F, as given in the following truth table: [XlYlZlFI (a) List its minterms F = Zrn(O,l, 3'5'7) (b) List its maxterms F = lIM(2,4,6) (c) List the minterms of F = Zm(2,4,6) (d) Express F in sum-of-minterms algebraic form F = X'Y'Z' + X'Y'Z + X'YZ + XY'Z + XYZ (e) Express F in product-of-maxterms algebraic form F=(X+Y9+Z).(X'+Y+Z).(X'+Y'+Z)

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Problem 3 ( 2 w.4~~) Draw the logic diagram for the following Boolean expression. The diagram should correspond exactly to the equation. Assume the complements of the inputs are not available. B(A'C7 + AC) + D'(A + B'C)
Problem 4 (6 marks) Optimize the following Boolean functions by means of Karnaugh map: (a) F(X, Y, Z) = Cm(O,2,3,4,6) F = Z' + X'Y (b) F(W, X, Y, Z) = Cm(O,2,5,6,8, 10, 13, 14, 15) Or F = X'Z' + YZ' + XY'Z + WXY d' (c) F(W, X, Y, Z) = Cm(2,4,9, 12, 15), with the don't-care conditions d(W, X, Y, Z) = Cm(3, 5,6, 13) F = XY' + WY'Z + WXZ + W'X'Y F = XY' + WY'Z + WXZ + W'YZ' J

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## This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.

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Solutions-HW2 - Homework 2 Solution Problem 1 (2 marks)...

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