# DeMorgan’s Theorem-Digital Logic Design-Solution Manual, Exercises for Digital Logic Design and Programming. Punjab Engineering College

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This is solution manual for Digital Logic Design course. It was helpful for assignment Dr. Archan Singh gave us at Punjab Engineering College. It includes: Verification, DeMorgan, Theorem, Second, Distributive, Law, NAND...
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2-2.

a) X Y + XY + XY = X + Y

= (XY+ X Y ) + (X Y + XY)

= X(Y + Y) + Y(X + X) +

= X + Y

Verification of DeMorgan’s Theorem

XY Z XYZ XYZ X+Y+Z

0 0 0 0 1 1

0 0 1 0 1 1

0 1 0 0 1 1

0 1 1 0 1 1

1 0 0 0 1 1

1 0 1 0 1 1

1 1 0 0 1 1

1 1 1 1 0 0

The Second Distributive Law

XY Z YZ X+YZ X+Y X+Z (X+Y)(X+Z)

0 0 0 0 0 0 0 0

0 0 1 0 0 0 1 0

0 1 0 0 0 1 0 0

0 1 1 1 1 1 1 1

1 0 0 0 1 1 1 1

1 0 1 0 1 1 1 1

1 1 0 0 1 1 1 1

1 1 1 1 1 1 1 1

XY Z XY YZ XZ XY+YZ+XZ XY YZ XZ XY+YZ+XZ

0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 0 1 0 0 1 1

0 1 0 1 0 0 1 0 1 0 1

0 1 1 1 0 0 1 0 0 1 1

1 0 0 0 0 1 1 1 0 0 1

1 0 1 0 1 0 1 1 0 0 1

1 1 0 0 0 1 1 0 1 0 1

1 1 1 0 0 0 0 0 0 0 0

XYZ X Y Z+ +=

X YZ+ X Y+( ) X Z+( )⋅=

XY YZ XZ+ + XY YZ XZ+ +=

1

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Problem Solutions – Chapter 2

b) A B+ B C + AB + B C = 1

= (A B+ AB) + (B C + B C)

= B(A + A) + B(C + C)

= B + B

= 1

c) Y + X Z + X Y = X + Y + Z

= Y + X Y + X Z

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

= Y + X + X Z

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

= X + Y + Z

d) X Y + Y Z + XZ + XY + Y Z = X Y + XZ + Y Z

= X Y + Y Z(X + X) + XZ + XY + Y Z

= X Y + X Y Z + X Y Z + XZ + XY + Y Z

= X Y (1 + Z) + X Y Z +XZ + XY + Y Z

= X Y + XZ(1 + Y) + XY + Y Z

= X Y + XZ + XY (Z + Z)+ Y Z

= X Y + XZ + XY Z +Y Z (1 + X)

= X Y + XZ(1 + Y) + Y Z

= X Y + XZ + Y Z

2-7.

a) X Y + 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 + YZ + X Y Z = X + (YZ + X)(YZ + YZ) = X + Y(X + YZ)

= X + XY + YZ = (X + X)(X + Y) + YZ = X + Y + YZ = X + Y

c) WX(Z + YZ) + X(W + W YZ) = WXZ + WXYZ + WX + WXYZ

= WX + WXZ + WXZ = WX + WX = X

d)

=

=

= A + C + A(BCD)

= A + C + BCD

= A + C + C(BD)

= A + C + BD

2-9.

a)

b)

c)

d)

AB AB+( ) CD CD+( ) AC+ ABCD ABCD ABCD ABCD A C+ + + + +

A C ABCD+ +

F A B+( ) A B+( )= F V W+( )X Y+( )Z= F W X+ Y Z+( ) Y Z+( )+[ ] W X+ YZ YZ+ +[ ]= F ABC A B+( )C A B C+( )+ +=

2

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Problem Solutions – Chapter 2

2-10.

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: A B C + A B C + A B C + A B C

Product of Maxterms: (A + B + C)(A + B + C)(A + B + C)(A + B + C)

c) Sum of Minterms: 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

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)

2-12.

a) (AB + C)(B + CD) =AB + BC + ABCD= 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) = X + Y + Z s.o.p. and p.o.s.

c) (A + BC + CD)(B + EF) = (A + B + C)(A + B + D)(A + C + D)(B + E)(B + F) p.o.s.

(A + BC + CD)(B + EF) = A(B + EF) + BC(B + EF) + CD(B + EF)

= AB + AEF + BCEF + BCD + CDEF s.o.p.

2-15.

Truth Tables a, b, c

X Y Z a A B C b W X Y Z c

0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 1 0 0 0 1 1 0 0 0 1 0

0 1 0 0 0 1 0 0 0 0 1 0 1

0 1 1 1 0 1 1 1 0 0 1 1 0

1 0 0 0 1 0 0 0 0 1 0 0 0

1 0 1 1 1 0 1 0 0 1 0 1 0

1 1 0 1 1 1 0 0 0 1 1 0 1

1 1 1 1 1 1 1 1 0 1 1 1 0

1 0 0 0 0

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 1

X

Y

Z

A

B

C

a) b) c)

X Z + XY A + CB B + C

A

B

C

1

11

1 1

1

1 1 1 1

1

1

1

1

1

3

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Problem Solutions – Chapter 2

2-18.

2-19.

Using K-maps:

a) Prime = XZ, WX, X Z,W Z b) Prime = CD, AC, B D, ABD, B C c) Prime = AB, AC, AD, BC, BD, CD

Essential = XZ, X Z Essential = AC, B D, ABD Essential = AC,BC, BD

2-22.

Using K-maps:

a) s.o.p. CD + AC + B D b) s.o.p.AC + B D + A D c) s.o.p.B D + ABD + (ABC or ACD)

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.

2-28.

W

X

Y

Z

A

B

C

D

a) b) c)

X

Y

Z

Σm 3 5 6 7, , ,( ) Σm 3 4 5 7 9 13 14 15, , , , , , ,( ) Σm 0 2 6 7 8 10 13 15, , , , , , ,( )

11

1

1

11

1

1

1 1

1

1 11

1 1

1

11

1

Primes = AB, AC, BC, A B C Essential = AB, AC, BC F = AB + AC + BC

Primes = X Z, XZ, WXY, WXY, W Y Z, WYZ Essential = X Z F =X Z + WXY + WXY

Primes = AB, C, AD, BD Essential = C, ADF = C + AD + (BD or AB)

W

X

Y

Z

A

B

C

D

a) b) c)

A

B

C

1

1

1

1

1 1

1

1

1

1 1

1 1 1

1

1

X

X X

X

X X

X X

X

X X

AB

AB

C

DC

D AB

C D AB DC

A B

A B

C D C

D

A B

C D

A B

D C

4-input NAND from 2-input NANDs and NOTs

a)

4

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Problem Solutions – Chapter 2

2-30.

2-34.

2-37.

2-39.

4 × 0.5 = 2 ns

2-44.

P-Logic N-Logic

X Y NAND NOR X Y NAND NOR X Y NAND NOR

L L H H 0 0 1 1 1 1 0 0

L H H L 0 1 1 0 1 0 0 1

H L H L 1 0 1 0 0 1 0 1

H H L L 1 1 0 0 0 0 1 1

A B

C

D

A B

A B

C D

C D

b)

A

B

C

D

a) b)

W

X

Y

Z F A B C+ +( ) A C+( ) A D+( )= F W X+( ) W X+( ) Y Z+( ) Y Z+( )=

1 1

1 1 1 1

1

11

A C A D

1

1

B A

C 1 Y Z Y Z A

W X W X

X YXY XY+=

Dual (X Y )⊕ Dual XY XY+( )=

X Y+( ) X Y+( )=

XY XY+ X Y+( ) X Y+( )=

X Y+( ) X Y+( )=

16 inputs 16 inputs

6 inputs

5

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