Computer Science - Digital Computer Electronics - Discussion
Discussion Forum : Digital Computer Electronics - Section 1 (Q.No. 37)
37.
The simplified form of the Boolean expression (X + Y + XY)(X + Z) is
Discussion:
6 comments Page 1 of 1.
Manjeet said:
8 years ago
= (X+Y+XY) (X+Z)
= (X+Y(1+X)) (X+Z) // X+1=1
= (X+Y).(X+Z) // by distributive law (x+y).(x+z)=x + yz
= X + YZ.
= (X+Y(1+X)) (X+Z) // X+1=1
= (X+Y).(X+Z) // by distributive law (x+y).(x+z)=x + yz
= X + YZ.
(8)
Chary said:
1 decade ago
(X + Y + XY)(X + Z)
= X.X + X.Z + Y.X + Y.Z + XY.X + XY.Z
= X + XZ + XY + YZ + XY + XYZ
= x(1+Z) + XY + YZ(1+X)
= X + XY + YZ
= X(1+Y) + YZ
= X + YZ
Rules : X.X=X, X+X=X, 1+X=1, 1.X=1
= X.X + X.Z + Y.X + Y.Z + XY.X + XY.Z
= X + XZ + XY + YZ + XY + XYZ
= x(1+Z) + XY + YZ(1+X)
= X + XY + YZ
= X(1+Y) + YZ
= X + YZ
Rules : X.X=X, X+X=X, 1+X=1, 1.X=1
(6)
Shruti said:
8 years ago
Thank you.
(2)
Ritu kumari said:
9 years ago
Thanks @Chary.
(1)
Sandy said:
8 years ago
Thanks @Manjeet.
(1)
Rani k Kumar said:
3 years ago
Thanks @Manjeeth.
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