1. P.50
1.1)
a) To Hexa
4* + 3X + 7* + 5* =
287.625/16 F
17/16 1
1/16 1
0
=
b) (B8E.F) To Octal
=
2958/8 6
369/8 1
46/8 6
5/8 5
0
=
C)
=
6/8 6
0
=
2. P.109
2.2)
a) X’Y’ + X’Y + XY = X’ + Y
X’(Y’ + Y) + XY
(X’ + X) + (X’ + Y) = X’ + Y
b) A’B + B’C’ + AB + B’C = 1
B(A’ + A) + B’(C’ + C)
= B + B’ = 1
c) Y + X’Z + XY’ = X + Y + Z
(Y + Y’) + (Y + X) + X’Z
Y + X + X’Z
Y + (X + X’) + (X + Z)
= Y + X + Z
D)
X’Y’ + Y’Z + XZ + XY + YZ = X’Y’ + XZ + YZ’
Z(Y’ + Y) + X’Y’ + XZ + XY
Z + X’Y’ + XZ + XY
= X’Y’ + XZ + YZ’
3. P.111
2.14)
a) F(A,B,C) = ∑
B’ B
A’ 1 0 1 1
A 1 0 0 0
C’ C C’
F = B’C’ + A’B
b) F(X,Y,Z) = ∑
F = X’Z + XY’
c) F(P,Q,R) = ∑
F = P’R + PR’
d) F(A,B,C) = ∑
B’ B
A’ 1 1 1 0
A 1 0 0 0
C’ C C’
F = B’C’ + A’C
Y’ Y
X’ 0 1 1 0
X 1 1 0 0
Z’ Z Z’
Q’ Q
P’ 0 1 1 0
P 1 0 0 1
R’ R R’
4. 2.15)
a) ACD’ + BC + ABCD’ + AB’C
C’ C
A’ 0 0 0 0 B’
0 0 1 1 B
A 0 0 1 1
0 0 1 1 B’
D’ D D’
∑
F = AC + BC
b) ABC’D + ACD’ + AB’C + C’D
C’ C
A’ 1 0 0 0 B’
1 0 0 0 B
A 1 1 0 1
1 0 1 1 B’
D’ D D’
∑
F = C’D’ + AC’B + D’CA + AD’C
c) ABC’D + ACD’ + AB’C + C’D
C’ C
A’ 0 1 1 0 B’
0 0 1 1 B
A 0 0 0 0
0 1 0 0 B’
D’ D D’
∑
F = A’BC + A’B’D + B’C’D
5. P.112
2.18)
a) XY’ + XZ’ + XYZ
Y’ Y
X' 0 0 0 0
X 1 1 1 1
Z’ Z Z’
∑
F = X
b) XYZ’ + WX’Y + WXY’ + W’X’Y + WXYZ
Y’ Y
W’ 0 0 1 1 X’
0 0 0 1
X
W 1 1 1 1
0 0 1 1 X’
Z’ Z Z’
∑
F = YZ’ + WX + W’X’Y + WX’Y
c) BD + AB’D + AB’C
Y’ Y
A’ 0 0 1 1 B’
0 0 0 1
B
A 1 1 1 1
0 0 1 1 B’
D’ D D’
∑
F = ACB’ + BD + AD
6. p.113
2.24)
a) F(A,B,C,D) = ∑ , J(A,B,C,D) = ∑
Y’ Y
A’ 1 1 0 X B’
0 0 1 X
B
A 0 1 1 0
X X 0 X B’
D’ D D’
ANSWER IS: A’B’C + ABD + BCD
b) F(A,B,C,D) = ∑ , J(A,B,C,D) = ∑
C’ C
A’ 0 0 X 1 B’
1 X 0 X
B
A 1 X 1 0
0 1 0 0 B’
D’ D D’
ANSWER IS: BC’ + ADC’ + ABD + A’B’C
c) F(A,B,C) = ∑ , J(A,B,C) = ∑
B’ B
A' X 1 X 1
A 1 0 X X
C’ C C’
ANSWER IS: A’ + C’
7. p.113
2.25)
a) F(A,B,C) = ∑ , J(A,B,C) = ∑
B’ B
A' X 0 1 0
A 0 1 X 1
C’ C C’
ANSWER IS: AB + AC + BC’
b) F(W,X,Y,Z) = ∑ , J(W,X,Y,Z) = ∑
Y’ Y
W’ 1 0 0 1 X’
1 1 X 0
X
W 0 X 1 1
1 0 X 0 X’
Z’ Z Z’
ANSWER IS: W’Y’Z’ + X’Y’Z’ + XZ + WXY + W’X’YZ’
c) F(A,B,C,D) = ∑ , J(A,B,C,D) = ∑
C’ C
A’ 0 0 X X B’
1 X 1 1
B
A 1 0 1 X
1 0 X X B’
D’ D D’
ANSWER IS: C + A’B + AC’D’