逻辑与计算机设计基础答案 chapter02

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2-2.*
a) X Y + XY + XY
= ( X Y + X Y ) + ( XY + XY ) = X (Y + Y ) + Y ( X + X ) = X +Y
a) F = (A + B)(A + B) b) F = ((V + W )X + Y )Z c) F = [W + X + (Y + Z )(Y + Z )][W + X + YZ + YZ ] d) F = ABC + (A + B)C + A(B + C )
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2-10.*
2-12.*
b)
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111 X
W 1 11 1
Z Σm(3Σ, m 4,( 3 5,, 4 7,,5 9, ,7 1, 39,, 1134,, 1 145,)15)
c)
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11 B
A
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1
D Σm(0Σ, m 2,( 0 6,, 2 7,,6 8, ,7 ,180,, 1 103,, 1 135,)15 )
= (Y + X )(Y + Y ) + XZ
= Y + X + XZ
= Y + ( X + X )( X + Z )
= X +Y +Z
d) XY + Y Z + XZ + XY + YZ
=
= XY + YZ ( X + X ) + XZ + XY + YZ = XY + XYZ + XYZ + XZ + XY + YZ = XY (1 + Z ) + XYZ + XZ + XY + YZ = XY + XZ (1 + Y ) + XY + YZ = XY + XZ + XY (Z + Z ) + YZ = XY + XZ + XYZ + YZ (1 + X ) = XY + XZ (1 + Y ) + YZ = XY + XZ + YZ
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c) XY + YZ + XZ = XY + YZ + XZ
X Y Z XY Y Z XZ XY + YZ + XZ XY YZ
0 0 0 1 1 1 1 1
XZ XY + YZ + XZ
000 0 0 0
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001 0 1 0
c) (A + BC + C D )(B + EF ) = (A + B + C )(A + B + D )(A + C + D )(B + EF ) = (A + B + C )(A + B + D )(A + C + D )(B + E )(B + F ) p.o.s. (A + BC + C D )(B + EF ) = A(B + EF ) + BC (B + EF ) + C D (B + EF ) = AB + AEF + BC EF + BC D + C DEF s.o.p.
Problem Solutions – Chapter 2
CHAPTER 2
© 2016 Pearson Education, Inc.
2-1.*
a) XYZ = X + Y + Z Verification of DeMorgan’s Theorem
XY
Z XYZ
XYZ
X +Y +Z
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F = XZ + WXY + WXY F = XZ + WXY + WXY
c)
C
XX
1 X1 1
1 A
1X B
1
XX
D PriPmreism=esAB=, CA,B A, CD,, A BDD, BD EssEenstsieanl t=iaCl , = ADC, AD
F = C + AD + (BD or AB) F = C + AD (BD or AB)
p.o.s. (C + D )(A + D )(A + B + C )
p.o.s. (C + D )(A + D )(A + B + C )
p.o.s. (A + B)(B + D )(B + C + D )
2-25.*
a)
Bபைடு நூலகம்
X1
A
1X1
C
PrimesP=riAmBe,s A=C ,A BBC, A, CA,BBCC, ABC EssentEiasls=enAtBia,l A=C ,A BBC, AC, BC
Problem Solutions – Chapter 2
Truth Tables a, b, c
XYZa
ABCb
0000
0001
0010
0011
0100
0100
0111
0111
1000
1000
1011
1010
1101
1100
1111
1111
WX Y Z c 00000 00010 00101 00110 01000 01010 01101 01110 10000 10010 10101 10110 11001 11011 11101 11111
a) Sum of Minterms: XYZ + XYZ + XYZ + XYZ Product of Maxterms: (X + Y + Z )(X + Y + Z )(X + Y + Z )(X + Y + Z )
b) Sum of Minterms: ABC + ABC + ABC + ABC Product of Maxterms: (A + B + C )(A + B + C )(A + B + C )(A + B + C )
a) (AB + C )(B + C D ) = AB + ABC D + BC = AB + BC s.o.p. = B(A + C ) p.o.s.
b) X + X (X + Y )(Y + Z ) = (X + X )(X + (X + Y )(Y + Z )) = (X + X + Y )(X + Y + Z ) p.o.s. = (1+ Y )(X + Y + Z ) = X + Y + Z s.o.p.
c) Prime = AB, AC , AD , BC , BD , C D Essential = AC , BC , BD
2-22.*
a) s.o.p. C D + AC + BD
b) s.o.p. AC + BD + AD
c) s.o.p. BD + ABD + (ABC or AC D )
c) WX (Z + YZ ) + X (W + WYZ ) = WXZ + WXYZ + WX + WXYZ = WXZ + WXZ + WX = WX + WX = X
d) (AB + AB)(CD + CD) + AC = ABCD + ABCD + ABCD + ABCD + A + C = ABC D + A + C = A + C + A(BC D ) = A + C + C (BD ) = A + C + BD
=
X +Y
1
2-7.* 2-9.*
Problem Solutions – Chapter 2
b) AB + BC + AB + BC
=
= ( AB + AB) + (BC + BC)
= B( A + A) + B(C + C)
B+ B =1
c) Y + XZ + XY
=
= Y + XY + XZ
endmodule
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c) Sum of Minterms: WXYZ + WXYZ + WXYZ +WXY Z +WXYZ +WXYZ + WXYZ Product of Maxterms: (W + X + Y + Z )(W + X + Y + Z )(W + X + Y + Z ) (W + X + Y + Z )(W + X + Y + Z )(W + X + Y + Z ) (W + X + Y + Z )(W + X + Y + Z )(W + X + Y + Z )
2-34.*
X1
N1
N2
X2
N3
N6
f
N4
X3
N5 X4
2-38.*
begin F <= (X and Z) or ((not Y) and Z);
end;
2-39.*
X1
N1
N2
X2
N3
N6
f
N4
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N5 X4
2-43.*
module circuit_4_53(X, Y, Z, F); input X, Y, Z; output F; assign F = (X & Z) | (Z & ~Y);
2-19.*
a) Prime = XZ , WX , XZ , WZ Essential = XZ , XZ
b) Prime = C D , AC , BD , ABD , BC Essential = AC , BD , ABD
F = AB + AC + BC F = AB + AC + BC
b)
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1 1X
X
X11
W
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X
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PrimesP=riXmZe,s X=Z ,X WZ,XXYZ,, WWXXYY,, WYXZY, W WYZ, WYZ EssentEiasls=enXtZial = XZ
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Problem Solutions – Chapter 2
2-27.*
X ⊕Y = XY + XY Dual(X ⊕ Y ) = Dual(XY + XY ) = (X + Y )(X + Y ) = XY + XY = XY + XY = X ⊕Y
2-29.*
The longest path is from input C or D. 0.073 ns + 0.073 ns + 0.048 ns + 0.073 ns = 0.267 ns
1 X +Y +Z X Y + XZ + YZ
a) XY + XYZ + XY = X + XYZ = (X + XY )(X + Z ) = (X + X )(X + Y )(X + Z ) = (X + Y )(X + Z ) = X + YZ
b) X +Y (Z + X + Z ) = X +Y (Z + XZ ) = X +Y (Z + X )(Z + Z ) = X +YZ + XY = (X + X )(X + Y ) + YZ = X + Y + YZ = X + Y
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2-15.* 2-18.*
Problem Solutions – Chapter 2
a)
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Z XXZZ + XXYY
b)
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C AA++CCB
c)
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C BB++CC
a) Y
1 X 111
Z
Σm(3Σ, m 5,( 3 6,, 5 7,)6, 7)
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b) X + YZ = ( X + Y ) ⋅ ( X + Z )
The Second Distributive Law
XY
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YZ
X + YZ X + Y
X+Z
(X + Y)(X + Z)
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