Solutions Chapter 2, Digital Design, 4 Edition


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Solutions Chapter 2, Digital Design, 4th Edition

2.2) a) x b) (x + y)(x + y’) = x + xy + xy’ + yy’ = x c) xyz + x’y + xyz’ = xy + x’y = y d) (A + B’)’ (A’ + B’)’ = A’B’AB = 0 e) xy(z + z') + x'y(z + z')=xy1 + x'y1= xy + x'y = y(x+x’) = y f) xx' + xy' + xz + x'y + yy' + yz + x'z' + y'z' + zz'= 0 + xy' + xz + x'y + 0 + yz + x'z' + y'z' +0 =
(xy'+ x'y)+ (xz+ x'z') + (yz+y'z')= (xy)+(xz)’+(yz)’

2.3) a) ABC + A’B + ABC’ = AB + A’B = B b) x’yz + xz = (x + x’y)z = (x + y)z c) (x + y)’ (x’ + y’) = x’y’(x’ + y’) = y’(x’ + x’y’) = x’y’ + x’y’ = x’y’ d) xy + x(wz + wz’) = xy + xw = x(y + w) e) (BC’+A’D) (AB’+CD’) = (BC’AB’+BC’CD’+A’DAB’+A’DCD’) = (0 + 0 + 0 + 0) = 0 f) xx' + xz' + x'y' + y'z' + x'z' + z'z' = y'(x' + z')+ z' = x'y' +y’z’+z’ = x’y’+z’

2.4) a) A’C’ + ABC + AC’ = C’ + ABC = AB + C’ b) (x’y’+z)’+z+xy+wz = (x+y)z’+z+xy+wz = x+y+z+xy+wz = x + y + z c) A’B(D’ + C’D) + B(A + A’CD) = B(A’D’ + A’C’D + A + A’CD) = B(A’D’ + A + A’D) = B(A’ + A) = B d) (A’+C)(A’+C’)(A+B+C’D) = (A’+A’C+A’C’+CC’)(A+B+C’D) = A’(A + B + C’D) = A’B + A’C’D = A’(B + C’D) e) BD(AC+A’+AC’)=BD(A’+A(C+C’))=BD(A’+A1)=BD(A’+A)=BD1 = BD

2.8) F' = (wx + yz)' = (wx)'(yz)' = (w' + x')(y' + z') = w’y’ + w’z’ + x’y’ + x’z’
 FF' = (wx + yz) (w’y’+w’z’+x’y’+x’z’) = wxw’y’ + yzw’y’ + wxw’z’ + yzw’z’ + wxx’y’+ yzx’y’+ wxx’z’+ yzx’z’ Since xx’= 0, FF’ = 0+0+0+0+0+0+0+0 = 0
 F + F' = (wx + yz) + (wx + yz)'. Take s = wx + yz,then F + F' = s+s’. Since s+s’=1, F + F' = 1.

2.9)
2.15) 2.19) 2.20)
2.21) 2.22) 2.24)

a) (xy’+x’y)’= (xy’)’(x’y)’=(x’ + y)(x + y’)=xx’+x’y’+xy+yy’=x’y’+xy b) ((A’B+CD)E’+E)’=((A’B+CD)E’)’ E’=((A’B+CD)’+E)E’ = ((A’B+CD)’ E’+EE’=
((A’B)’(CD)’)’E’ = (A+B’)(C’+ D’)E’ = AC’E’+AD’E’+B’C’E’+B’D’E’ c) ((x' + y + z')(x + y')(x + z))' = (x' + y + z')'+(x + y')'+(x + z)'=xy'z+x'y+x'z'

T1 = A’B’C’ + A’B’C + A’BC’ = A’(B’C’+B’C+BC’)=A’(B’(C’+C)+BC’)=A’(B’+BC’)=A’(B’ + C’) T2 = A’BC + AB’C’ + AB’C + ABC’ + ABC = A’BC + A(B’C’ + B’C +BC’ + BC) = A’BC + A(B’ + B) = A’BC + A = A + BC

F(A,B,C,D) = Σ(1,3,5,7,9,11,13,15) = Π(0,2,4,6,8,10,12,14)

a) F(A, B, C, D) = Σ(3, 5, 9, 11, 15) F'(A, B, C, D) = Π (0,1,2,4,6,7,8,10,12,13,14)
b) F(x, y, z) = Π (0,1,2,4,7,9,12) F' = Σ (0,1,2,4,7,9,12)

a) F(x, y, z)=Σ(2,5,6)=Π((0,1,3,4,7) b) F(A,B,C,D) = Π (0,1,2,4,7,9,12) = Σ(3,5,6,8,10,11,13,14,15)

a) SOP: AB + BC b) SOP: x’ + y + z’

POS: B(A + C) POS: (x’ + y + z’)

XOR: (ab)=(a’+b)(a+b’)

COMPLEMENT: ((a’+b)(a+b’))’=(a’+b)’+(a+b’)’=ab’+a’b

DUAL of XOR: (a’b)+(ab’)= (a+b’)’+(a’+b)’=((a+b’)(a’+b))’ (same as XOR complement)

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Solutions Chapter 2, Digital Design, 4 Edition