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Unit 2 Solutions

17

(A + B + C + D) (A + B + C + E) (A + B + C + F)

= A + B + C + DEF

Apply second Distributive Law twice

See FLD p. 731 for solution.

2.1

Unit 2 Problem Solutions

Answer is in FLD p. 731

2.3

AB + C'D' = (AB + C') (AB + D')

= (A + C') (B + C') (A + D') (B + D')

2.6 (a) WX + WY'X + ZYX = X(W + WY' + ZY)

= X(W + ZY) {By Absorption}

= X(W +Z) (W + Y)

2.6 (b)

X

X Y

X

2.7 (a) WXYZ + VXYZ + UXYZ = XYZ (W + V + U)

By first Distributive Law

2.7 (b)

Y

X

Y Z

X

Z

X

Unit 2 Solutions

18

(A' + B' + C)(A' + B' + C)' = 0 By

Complementarity Law

AB + (C' + D)(AB)' = AB + C' + D

By Elimination Theorem

[AB' + (C + D)' +E'F](C + D)

= AB'(C + D) + E'F(C + D) Distributive Law

F1 = A'A + B + (B + B) = 0 + B + B = B

F3 = [(AB + C)'D][(AB + C) + D]

= (AB + C)'D (AB + C) + (AB + C)' D

= (AB + C)' D By Absorption

AB(C' + D) + B(C' + D) = B(C' + D) By

Absorption

(A'BF + CD')(A'BF + CEG) = A'BF + CD'EG

By Distributive Law

A'(B + C)(D'E + F)' + (D'E + F)

= A'(B + C) + D'E + F By Elimination

2.11 (a)

2.13 (a) F2 = A'A' + AB' = A' + AB' = A' + B'

Z = [(A + B)C]' + (A + B)CD = [(A + B)C]' + D

By Elimination with X = [(A + B) C]'

= A'B' + C' + D'

2.11 (c)

2.11 (e)

2.11 (b)

2.11 (d)

2.11 (f)

2.13 (c)

2.13 (b)

2.13 (d)

X

Y

X

2.10 (a) 2.10 (b)

X

YX Y

Unit 2 Solutions

19

2.18 (b)

2.18 (d)

sum-of-products

sum term, sum-of-products, product-of-sums

2.18 (c)

2.18 (e)

none apply

product-of-sums

F = D[(A' + B' )C + AC' ]

= D(A' + B' + AC' )(C + AC' )

2.20 (c)

A B C H F G

0 0 0 0 0 0

2.21

A'B' + A'CD + A'DE'

= A'(B' + CD + DE')

= A'[B' + D(C + E')]

= A'(B' + D)(B' + C + E')

2.22 (a)

A'B' + (CD' + E) = A'B' + (C + E)(D' + E)

= (A'B' + C + E)(A'B' + D' + E)

= (A' + C + E)(B' + C + E)

(A' + D' + E)(B' + D' + E)

2.22 (d)

2.23 (a) 2.23 (b)

W + U'YV = (W + U')(W + Y)(W + V)TW + UY' + V

= (T+U+Z)(T+Y'+V)(W+U+V)(W+Y'+V)

Unit 2 Solutions

H = [W'X'(Y' + Z')]' = W + X + YZ

2.27

2.28 (a)

F = (V + X + W) (V + X + Y) (V + Z)

= (V + X + WY)(V + Z) = V + Z (X + WY)

By Distributive Law with X = V

F = ABC + A'BC + AB'C + ABC'

Beginning with the answer to (a):

2.28 (b)

2.26 (c)

+

+

W

YXZVF

[(XY')' + (X' + Y)'Z] = X' + Y + (X' + Y)'Z

= X' + Y'+ Z By Elimination Theorem with X

= (X' + Y)

2.24 (a) (X + (Y'(Z + W)')')' = X'Y'(Z + W)' = X'Y'Z'W'

2.24 (b)

X Y Z X+Y X'+Z (X+Y)

(X'+Z)

XZ X'Y XZ+X'Y

0 0 0 0 1 0 0 0 0

0 0 1 0 1 0 0 0 0

2.29 (b)2.29 (a) X Y Z X+Y Y+Z X'+Z (X+Y)

(Y+Z)

(X'+Z)

(X+Y)

(X'+Z)

0 0 0 0 0 1 0 0

Unit 2 Solutions

21

2.29 (d)

2-29 (c)

W X Y Z W'XY WZ W'XY+WZ W'+Z W+XY (W'+Z)(W+XY)

0 0 0 0 0 0 0 1 0 0

0 0 0 1 0 0 0 1 0 0

0 0 1 0 0 0 0 1 0 0

0 0 1 1 0 0 0 1 0 0

2.29 (e)

X Y Z XY YZ X'Z XY+YZ+X'Z XY+X'Z

0 0 0 0 0 0 0 0

0 0 1 0 0 1 1 1

0 1 0 0 0 0 0 0

A B C A+C AB+C' (A+C)

(AB+C')

AB AC' AB

+AC'

0 0 0 0 1 0 0 0 0

0 0 1 1 0 0 0 0 0

2.30 F = (X+Y')Z + X'YZ' (from the circuit)

= (X+Y'+ X'YZ')( Z+X'YZ') (Distributive Law)

= (X+Y'+X')(X+Y'+Y)(X+Y'+Z')(Z+X')(Z+Y)(Z+Z') (Distributive Law)

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