Transcription of EEC180A Homework 1 Solution 2.5(a) …
1 EEC180A Homework 1 Solution Chapter 2 (a) (A+B)(B+C)(B+D )(ACD +E) = (B+CA)(B+D )(ACD +E) = (B+ACD )(ACD +E) = ACD + BE (b) (A +B+C )(A +C +D)(B +D ) = (A +C +BD) (B +D ) = A B +A D +B C +C D (a) (A+B+C+D)(A+B+C+E)(A+B+C+F) = (A+B+C+DEF) (b) WXYZ+VXYZ+UXYZ =XYZ(W+V+U) (a)F1 = B (b)F2 = A +AB =A +B (c)F3 = (AB+C) D [(AB+C)+D] = (AB+C) D (d)F4 = (A+BC) +(A+BC)D = (A+BC) + D (a) W+X YZ = (W+X )(W+Y)(W+Z) (b)VW+XY +Z = (X+V+Z)(X+W+Z)(V+Y +Z)(W+Y +Z) (c) A B C+B CD +B E = B (A C+CD +E ) = B [C(A +D )+E ] = B (C+E )(A +D +E ) (d) ABC+ADE +ABF = A(BC+DE +BF) =A[B(C+F )+DE ]=A(DE +B)(DE +C+F )=A(B+D)(B+E )(C+D+F )(C+E +F ) F=ABC+A BC+AB C+ABC =BC+AB C+ABC = BC+AC+ABC = AB+BC+AC Chapter 3 (A +D)(AB) +(A +D) (AB) = (A +D)(A +B )+AB(AD ) = A +B D+ABD = (A+B +CE+DE )(B +C +D +E ) = B + (A+CE+D E )(C +D +E ) = B + AC +AE +D C+D E A CD E+A B D +ABCE+ABD = A CD E+A B D +ABCE+ABD + BCD E = A B D +ABD + BCD E (a) (A+D)(A+E)(B +C+D +E) (b) AD+AE+B CD E (a) = (X +Y )(XZ+X Z ) +(X+Y)(XZ +X Z) = X Z +XY Z+X Y Z +XZ +XYZ +X YZ = X Z +XY Z+X Y Z +XZ +XYZ +X YZ = Z +XY Z+X Y Z +XYZ +X YZ = Z +XY Z+X YZ = Z +Z(XY +X Y) = Z +(XY +X Y) (b) = (W +X+Y )(W+X +Y)(W+Y +Z) = (W +X+Y )[W+(X +Y)(Y +Z)] = (W +X+Y )(W+X Y +YZ)
2 = W X Y + X YZ + XW + XYZ + Y W + X Y = W X Y + XW + XYZ + X Y (c) = ABC+A C D +A BD +ACD = (A+B+C)(A +C +D )(A +B+D )(A+C+D) = (A+C+BD)(A +D +BC ) = A C+A BD+AD +ABC +BC D = A C+AD +BC D (a) True (b) False When A = 0, LHS must be equal to RHS regardless of B+C=D (c) True (d) False When C = 1, LHS must be equal to RHS regardless of A+B=D
