Solution Manual For Fundamentals of Logic Design, 7th Edition

Fundamentals of Logic Design 7th Edition Roth Solutions Manual Full Download: http://testbanklive.com/download/fundamentals-of-logic-design-7th-edition-roth-solutions-manual/ Unit 2 Solutions Unit 2 Problem Solutions 2.1 See FLD p. 731 for solution. 2.2 (a) In both cases, if X = 0, the transmission is 0, and if X = 1, the transmission is 1. X X 2.2 (b) In both cases, if X = 0, the transmission is YZ, and if X = 1, the transmission is 1. X X Y Y Z X X Y Z 2.3 Answer is in FLD p. 731 2.4 (a) F = [(A·1) + (A·1)] + E + BCD = A + E + BCD 2.4 (b) Y = (AB’ + (AB + B)) B + A = (AB’ + B) B + A = (A + B) B + A = AB + B + A = A + B 2.5 (a) (A + B) (C + B) (D’ + B) (ACD’ + E) = (AC + B) (D’ + B) (ACD’ + E) By Dist. Law = (ACD’ + B) (ACD’ + E) By Dist. Law = ACD’ + BE By Dist. Law 2.5 (b) (A’ + B + C’) (A’ + C’ + D) (B’ + D’) = (A’ + C’ + BD) (B’ + D’) {By Distributive Law with X = A’ + C’} = A’B’ + B’C’ + B’BD + A’D’ + C’D’ + BDD’ = A’B’ + A’D’ + C’B’ + C’D’ 2.6 (a) AB + C’D’ = (AB + C’) (AB + D’) = (A + C’) (B + C’) (A + D’) (B + D’) 2.6 (b) WX + WY’X + ZYX = X(W + WY’ + ZY) = X(W + ZY) {By Absorption} = X(W +Z) (W + Y) 2.6 (c) A’BC + EF + DEF’ = A’BC + E(F +DF’) = A’BC + E(F +D) = (A’BC + E) (A’BC + F + D) = (A’ + E) (B + E) (C + E) (A’ + F + D) (B + F + D) (C + F + D) 2.6 (d) XYZ + W’Z + XQ’Z = Z(XY + W’ + XQ’) = Z[W’ + X(Y + Q’)] = Z(W’ + X) (W’ + Y + Q’) By Distributive Law 2.6 (e) ACD’ + C’D’ + A’C = D’ (AC + C’) + A’C = D’ (A + C’) + A’C By Elimination Theorem = (D’ + A’C) (A + C’ + A’C) = (D’ + A’) (D’ + C) (A + C’ + A’) By Distributive Law and Elimination Theorem = (A’ + D’) (C + D’) (A + B + C + D) (A + B + C + E) (A + B + C + F) = A + B + C + DEF Apply second Distributive Law twice 2.6 (f) A + BC + DE = (A + BC + D)( A + BC + E) = (A + B + D)(A + C + D)(A + B + E)(A + C + E) 2.7 (b) WXYZ + VXYZ + UXYZ = XYZ (W + V + U) By first Distributive Law 2.7 (a) D E F 2.8 (a) 2.8 (c) 2.9 (a) A B U V W C X Y Z [(AB)’ + C’D]’ = AB(C’D)’ = AB(C + D’) = ABC + ABD’ ((A + B’) C)’ (A + B) (C + A)’ = (A’B + C’) (A + B)C’A’ = (A’B + C’)A’BC’ = A’BC’ 2.8 (b) [A + B (C’ + D)]’ = A'(B(C’ + D))’ = A'(B’ + (C’ + D)’) = A'(B’ + CD’) = A’B’ + A’CD’ F = [(A + B)’ + (A + (A + B)’)’] (A + (A + B)’)’ = (A + (A + B)’)’ By Elimination Theorem with X=(A+(A+B)’)’ = A'(A + B) = A’B 2.9 (b) G = {[(R + S + T)’ PT(R + S)’]’ T}’ = (R + S + T)’ PT(R + S)’ + T’ = T’ + (R’S’T’) P(R’S’)T = T’ + PR’S’T’T = T’ 17 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com Unit 2 Solutions 2.10 (a) X Y 2.10 (c) X X Y’ X’ 2.10 (e) X X Y X Z X Y X 2.10 (d) Y’ A Y 2.10 (f) B X X Y C A B A Z X Y Z Y 2.11 (a) (A’ + B’ + C)(A’ + B’ + C)’ = 0 By Complementarity Law 2.11 (c) AB + (C’ + D)(AB)’ = AB + C’ + D By Elimination Theorem 2.10 (b) C’ X B X Y 2.11 (b) AB(C’ + D) + B(C’ + D) = B(C’ + D) By Absorption 2.11 (d) (A’BF + CD’)(A’BF + CEG) = A’BF + CD’EG By Distributive Law Z 2.11 (e) [AB’ + (C + D)’ +E’F](C + D) = AB'(C + D) + E’F(C + D) Distributive Law 2.11 (f) 2.12 (a) (X + Y’Z) + (X + Y’Z)’ = 1 By Complementarity Law 2.12 (b) [W + X'(Y +Z)][W’ + X’ (Y + Z)] = X'(Y + Z) By Uniting Theorem 2.12 (c) (V’W + UX)’ (UX + Y + Z + V’W) = (V’W + UX)’ (Y + Z) By Elimination Theorem 2.12 (d) (UV’ + W’X)(UV’ + W’X + Y’Z) = UV’ + W’X By Absorption Theorem 2.12 (e) (W’ + X)(Y + Z’) + (W’ + X)'(Y + Z’) = (Y + Z’) By Uniting Theorem 2.12 (f) (V’ + U + W)[(W + X) + Y + UZ’] + [(W + X) + UZ’ + Y] = (W + X) + UZ’ + Y By Absorption 2.13 (a) F1 = A’A + B + (B + B) = 0 + B + B = B 2.13 (b) F2 = A’A’ + AB’ = A’ + AB’ = A’ + B’ 2.13 (c) F3 = [(AB + C)’D][(AB + C) + D] = (AB + C)’D (AB + C) + (AB + C)’ D = (AB + C)’ D By Absorption 2.13 (d) Z = [(A + B)C]’ + (A + B)CD = [(A + B)C]’ + D By Elimination with X = [(A + B) C]’ = A’B’ + C’ + D’ 2.14 (a) ACF(B + E + D) 2.14 (b) W + Y + Z + VUX 2.15 (a) f ‘ = {[A + (BCD)’][(AD)’ + B(C’ + A)]}’ = [A + (BCD)’]’ + [(AD)’ + B(C’ + A)]’ = A'(BCD)” + (AD)”[B(C’ + A)]’ = A’BCD + AD[B’ + (C’ + A)’] = A’BCD + AD[B’ + C”A’] = A’BCD + AD[B’ + CA’] 2.15(b) 2.16 (a) f D = [A + (BCD)’][(AD)’ + B(C’ + A)]D = [A (B + C + D)’] + [(A + D)'(B + C’A)] 2.16 (b) f D = [AB’C + (A’ + B + D)(ABD’ + B’)]D = (A + B’ + C)[A’BD + (A + B + D’ )B’) 2.17 (a) f = [(A’ + B)C] + [A(B + C’)] = A’C + B’C + AB + AC’ = A’C + B’C + AB + AC’ + BC = A’C + C + AB + AC’ = C + AB + A = C + A 2.17 (c) f = (A’ + B’ + A)(A + C)(A’ + B’ + C’ + B) (B + C + C’) = (A + C) 2.17 (b) f = A’C + B’C + AB + AC’ = A + C 18 A'(B + C)(D’E + F)’ + (D’E + F) = A'(B + C) + D’E + F By Elimination f ‘ = [AB’C + (A’ + B + D)(ABD’ + B’)]’ = (AB’C)'[(A’ + B + D)(ABD’ + B’]’ = (A’ + B” + C’)[(A’ + B + D)’ + (ABD’)’B”] = (A’ + B + C’)[A”B’D’ + (A’ + B’ + D”)B] = (A’ + B + C’)[AB’D’ + (A’ + B’ + D)B] 2.18 (a) product term, sum-of-products, product-of-sums) © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Unit 2 Solutions 2.18 (b) sum-of-products 2.18 (c) none apply 2.18 (d) sum term, sum-of-products, product-of-sums 2.18 (e) product-of-sums 2.19 2.20 (a) F = D[(A’ + B’ )C + AC’ ] W Z Z + X 2.20 (b) F = D[(A’ + B’ )C + AC’ ] = A’ CD + B’ CD +AC’ D + X Y F W + Y 2.20 (c) F = D[(A’ + B’ )C + AC’ ] = D(A’ + B’ + AC’ )(C + AC’ ) = D(A’ + B’ + C’ )(C + A) A’ D B’ C’ 2.21 C A A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 A’ C D B’ C D A C’ D C 0 1 0 1 0 1 0 1 H 0 1 1 1 0 1 0 1 F 0 1 0 1 0 0 0 1 G 0 x 1 x 0 1 0 x 2.22 (a) A’B’ + A’CD + A’DE’ = A'(B’ + CD + DE’) = A'[B’ + D(C + E’)] = A'(B’ + D)(B’ + C + E’) 2.22 (d) 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 (b) H’I’ + JK = (H’I’ + J)(H’I’ + K) = (H’ + J)(I’ + J)(H’ + K)(I’ + K) 2.22 (e) A’B’C + B’CD’ + EF’ = A’B’C + B’CD’ + EF’ = B’C (A’ + D’) + EF’ = (B’C + EF’)(A’ + D’ + EF’) = (B’ + E)(B’ + F’)(C + E)(C + F’ ) (A’ + D’ + E)(A’ + D’ + F’) 2.22 (c) A’BC + AB’C + CD’ = C(A’B + AB’ + D’) = C[(A + B)(A’ + B’) + D’] = C(A + B + D’)(A’ + B’ + D’) 2.22 (f) WX’Y + W’X’ + W’Y’ = X'(WY + W’) + W’Y’ = X'(W’ + Y) + W’Y’ = (X’ + W’)(X’ + Y’)(W’ + Y + W’)(W’ + Y + Y’) = (X’ + W’)(X’ + Y’)(W’ + Y) 2.23 (a) W + U’YV = (W + U’)(W + Y)(W + V) 2.23 (b) TW + UY’ + V = (T+U+Z)(T+Y’+V)(W+U+V)(W+Y’+V) 2.23 (c) A’B’C + B’CD’ + B’E’ = B'(A’C + CD’ + E’) = B'[E’ + C(A’ + D’)] = B'(E’ + C)(E’ + A’ + D’) 2.23 (d) ABC + ADE’ + ABF’ = A(BC + DE’ + BF’) = A[DE’ + B(C + F’)] = A(DE’ + B)(DE’ + C + F’) = A(B + D)(B + E’)(C + F’ + D)(C + F’ + E’) 19 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Unit 2 Solutions 2.24 (a) [(XY’)’ + (X’ + Y)’Z] = X’ + Y + (X’ + Y)’Z = X’ + Y’+ Z By Elimination Theorem with X = (X’ + Y) 2.24 (c) [(A’ + B’)’ + (A’B’C)’ + C’D]’ = (A’ + B’)A’B’C(C + D’) = A’B’C 2.25 (a) F(P, Q, R, S)’ = [(R’ + PQ)S]’ = R(P’ + Q’) + S’ = RP’ + RQ’ + S’ 2.25 (c) F(A, B, C, D)’ = [A’ + B’ + ACD]’ = [A’ + B’ + CD]’ = AB(C’ + D’) 2.26 (a) F = [(A’ + B)’B]’C + B = [A’ + B + B’]C + B =C+B 2.26 (c) H = [W’X'(Y’ + Z’)]’ = W + X + YZ 2.24 (b) (X + (Y'(Z + W)’)’)’ = X’Y'(Z + W)’ = X’Y’Z’W’ 2.24 (d) (A + B)CD + (A + B)’ = CD + (A + B)’ {By Elimination Theorem with X = (A + B)’} = CD + A’B’ 2.25 (b) F(W, X, Y, Z)’ = [X + YZ(W + X’)]’ = [X + X’YZ + WYZ]’ = [X + YZ + WYZ]’ = [X + YZ]’ = X’Y’ + X’Z’ 2.26 (b) G = [(AB)'(B + C)]’C = (AB + B’C’)C = ABC 2.27 F = (V + X + W) (V + X + Y) (V + Z) = (V + X + WY)(V + Z) = V + Z (X + WY) By Distributive Law with X = V W Y 2.28 (a) X + Z + V F 2.28 (b) Beginning with the answer to (a): F = ABC + A’BC + AB’C + ABC’ = BC + AB’C + ABC’ (By Uniting Theorem) = C (B + AB’) + ABC’ = C (A+ B) + ABC’ (By Elimination Theorem) = AC + BC + ABC’ = AC + B (C + AC’) = AC + B (A + C) = AC + AB + BC F = A (B + C) + BC B C B C A + + B C F Alternate solutions: F = AB + C(A + B) A C + F = AC + B(A + C) F A B 2.29 (a) XYZ X+Y X’+Z 000 001 010 011 100 101 110 111 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 (X+Y) (X’+Z) 0 0 1 1 0 1 0 1 XZ X’Y XZ+X’Y 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 1 1 0 1 0 1 20 2.29 (b) XYZ X+Y Y+Z X’+Z 000 001 010 011 100 101 110 111 0 0 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 0 1 0 1 (X+Y) (Y+Z) (X’+Z) 0 0 1 1 0 1 0 1 © 2014 Cengage Learning. 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(X+Y) (X’+Z) 0 0 1 1 0 1 0 1 Unit 2 Solutions 2-29 (c) XYZ 000 001 010 011 100 101 110 111 XY 0 0 0 0 0 0 1 1 YZ 0 0 0 1 0 0 0 1 X’Z 0 1 0 1 0 0 0 0 XY+YZ+X’Z 0 1 0 1 0 0 1 1 XY+X’Z 0 1 0 1 0 0 1 1 2.29 (d) A B C W’XY+WZ 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 1 W’+Z 1 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 W+XY 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 000 001 010 011 100 101 110 111 A+C AB+C’ 0 1 0 1 1 1 1 1 1 0 1 0 1 0 1 1 (A+C) (AB+C’) 0 0 0 0 1 0 1 1 2.29 (e) WXYZ 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 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) = (1+Y’)(X+1)(X+Y’+Z’)(Z+X’)(Z+Y)(1) (Complementation Laws) = (1)(1)(X+Y’+Z’)(Z+X’)(Z+Y)(1) (Operations with 0 and 1) = (X+Y’+Z’)(Z+X’)(Z+Y) (Operations with 0 and 1) W’XY 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 WZ 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 G = (X +Y’ + Z’ )(X’ + Z)(Y + Z) 21 AB AC’ 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 (W’+Z)(W+XY) 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 1 (from the circuit) © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. AB +AC’ 0 0 0 0 1 0 1 1 Fundamentals of Logic Design 7th Edition Roth Solutions Manual Full Download: http://testbanklive.com/download/fundamentals-of-logic-design-7th-edition-roth-solutions-manual/ Unit 2 Solutions 22 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com