Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 17)
17.
How many gates would be required to implement the following Boolean expression after simplification? XY + X(X + Z) + Y(X + Z)
Discussion:
13 comments Page 2 of 2.
Nag said:
1 decade ago
@Otman your answer is correct but procedure is wrong.
1+something is 1.
1+something is 1.
EBoyones said:
1 decade ago
XY + X (X+Z) + Y (X+Z) = XY + XX + XZ + XY + YZ.
By close inspection there are two XY we retain just one of this form and XX = X.
= X + XY + XZ + YZ.
= X (1 + Y + Z) + YZ.
The form: (1 + Y + Z) = 1 for what ever value of Y and Z.
= X + YZ.
This Boolean expression can be built using only 2 gates which is an and gate and or gate respectively.
By close inspection there are two XY we retain just one of this form and XX = X.
= X + XY + XZ + YZ.
= X (1 + Y + Z) + YZ.
The form: (1 + Y + Z) = 1 for what ever value of Y and Z.
= X + YZ.
This Boolean expression can be built using only 2 gates which is an and gate and or gate respectively.
Kornelis said:
10 years ago
xy+x(x+z)+y(x+z) <=>
xy+xx+xz+yx+yz < => use aa = a, ab = ba && a+a = a.
xy+x+yz use <=> ab+a = a.
x+yz.
xy+xx+xz+yx+yz < => use aa = a, ab = ba && a+a = a.
xy+x+yz use <=> ab+a = a.
x+yz.
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