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

## Digital Logic Design

Description: 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, Gate, Design
Showing pages  1  -  2  of  5
1
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
X YZ XYZ XYZ X+Y+Z
0 0 0 0 1 1
001 0 1 1
0 1 0 0 1 1
0 1 1 0 1 1
100 0 1 1
101 0 1 1
110 0 1 1
111 1 0 0
The Second Distributive Law
X Y Z YZ X+YZ X+Y X+Z (X+Y)(X+Z)
0 000000 0
001 0 0 0 1 0
0 100010 0
0 111111 1
100 0 1 1 1 1
101 0 1 1 1 1
110 0 1 1 1 1
111 1 1 1 1 1
X YZX
YYZXZXY+YZ+XZ XY YZ XZXY+YZ+XZ
0 00000 0 000 0
001010 1 001 1
0 10100 1 010 1
0 11100 1 001 1
100001 1 100 1
101010 1 100 1
110001 1 010 1
111000 0 000 0
XYZ X Y Z++=
XYZ+XY+()XZ+()=
XY YZ XZ++ XY YZ XZ++=
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2
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+++++
ACABCD++
FAB+()AB+()=
FVW+()XY+()Z=
FWX+YZ+()YZ+()+[]WX+YZ YZ++[]=
FABCAB+()CABC+()++=
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