HW2_Solutions

# HW2_Solutions - ECE 2030 E,F Homework#2 Solutions(1...

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ECE 2030 E,F Homework #2 - Solutions (1) Textbook #2-1 – Use Truth Tables to prove the validity of the following identities: (a) DeMorgan’s Theorem for three variables: (xyz)’=x’+y’+z’ x y z (xyz) (xyz)’ x’ y’ z’ (x’+y’+z’) 0 0 0 0 1 1 1 1 1 0 0 1 0 1 1 1 0 1 0 1 0 0 1 1 0 1 1 0 1 1 0 1 1 0 0 1 1 0 0 0 1 0 1 1 1 1 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 (b) The 2 nd Distributive Law: x+yz = (x+y)(x+z) x y z (yz) (x+yz) (x+y) (x+z) (x+y)(x+z) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 (c) x’y+y’z+xz’ = xy’+yz’+x’z x y z x’y y’z xz’ x’y+y’z+xz’ xy’ yz’ x’z xy’+yz’+x’z 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 1 1 1 1 0 0 0 0 0 0 0 0

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(2) Textbook #2-6 – Simplify to a minimum number of literals: (a) A’C’+A’BC+B’C = A’C’ + A’BC + (A’B’C+ B’C) [X+XY=X] = A’C’ + (A’BC + A’B’C)+ B’C [(X+Y)+Z=X+(Y+Z)] = A’C’ + A’(B+B’)C + B’C [XY+XZ=X(Y+Z)] = A’C’ + A’(1)C + B’C [X+X’=1] = A’C’ + A’C +B’C [X(1)=X] = A’(C’ + C)+ B’C [XY+XZ=X(Y+Z)] = A’+ B’C [X+X’=1] (b) (A+B+C)’(ABC)’ = A’B’C’(A’+B’+C’) DeMorgan = A’A’B’C’ + A’B’B’C’ + A’B’C’C’ [X(Y+Z)=XY+XZ] = A’B’C’ + A’B’C’ + A’B’C’ [XX=X] = A’B’C’ [X+X=X] (c) ABC’ + AC = A(BC’ +C) [XY+XZ=X(Y+Z)] = A(B+C) [X+X’Y=X+Y] OR : = ABC’ + (ABC + AC) [X+XY=X] = (ABC’ + ABC) + AC [(X+Y)+Z=X+(Y+Z)] = AB(C’+C) + AC [XY+XZ=X(Y+Z)] = AB(1)+AC [X+X’=1] = AB + AC [X(1)=X]
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HW2_Solutions - ECE 2030 E,F Homework#2 Solutions(1...

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