Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 14)
14.
How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z)
Discussion:
28 comments Page 1 of 3.
Solwyslt said:
5 years ago
ANS D
READ THE QUESTION PROPERLY:
"Expression BEFORE simplification"
I'll suggest you to draw it with the gates.it is the best way for you to understand.
XY+X(X+Z)+Y(X+Z).
Solution:
XY - AND gate(2 inputs),
X+Z - OR gate(2 inputs),
X(X+Z)-AND gate(2 inputs),
Y(X+Z)-AND gate(2 inputs),
XY + X(X+Z) + Y(X+Z)-OR gate(3 inputs).
A total of 5 gates in all.
READ THE QUESTION PROPERLY:
"Expression BEFORE simplification"
I'll suggest you to draw it with the gates.it is the best way for you to understand.
XY+X(X+Z)+Y(X+Z).
Solution:
XY - AND gate(2 inputs),
X+Z - OR gate(2 inputs),
X(X+Z)-AND gate(2 inputs),
Y(X+Z)-AND gate(2 inputs),
XY + X(X+Z) + Y(X+Z)-OR gate(3 inputs).
A total of 5 gates in all.
(5)
Sasha said:
6 years ago
@All.
The Answer is 2 if you have simplified with boolean algebra down to x+ yz.
The Answer is 2 if you have simplified with boolean algebra down to x+ yz.
(5)
McallenNJB said:
7 years ago
Answer is 2.
XY+ X (X+Z) + Y (X+Z),
XY + X + XZ + XY + YZ,
(XY + X + XZ + XY) + YZ,
X (Y + 1 + Z + Y) + YZ,
X + YZ.
XY+ X (X+Z) + Y (X+Z),
XY + X + XZ + XY + YZ,
(XY + X + XZ + XY) + YZ,
X (Y + 1 + Z + Y) + YZ,
X + YZ.
(2)
Vinay gupta said:
8 years ago
Answer B is right. Because of after solving an expression, we got X+YZ so 2 gates required.
@Sukriti is right.
@Sukriti is right.
(1)
ABHISHEK MANKAR said:
9 years ago
XY + X + XZ + YX + YZ.
X(1 + Y + Z + Y) + YZ.
X.1 + YZ.
So, one AND gate & one OR gate are required.
X(1 + Y + Z + Y) + YZ.
X.1 + YZ.
So, one AND gate & one OR gate are required.
Ahlem said:
4 years ago
XY AND.
X+Z OR.
X (X+Z) AND.
Y (X+Z) AND.
XY+X (X+Z)+Y (X+Z) ONE OR.
The sum is 5.
X+Z OR.
X (X+Z) AND.
Y (X+Z) AND.
XY+X (X+Z)+Y (X+Z) ONE OR.
The sum is 5.
Mona said:
6 years ago
Assume that three input OR gate is used for ORing each operation so total 5 gates are required.
Rajan said:
6 years ago
The sixth gate could be the OR gate for adding the individual terms of the expression.
Somavardhan said:
7 years ago
For XY 1 AND gate x+z 1OR gate x (x+z) 1 and gate, why (x+z) 1and gate xy+x (x+z) +y (x+z) 1 3 input or gate.
Total 5 gates.
Total 5 gates.
Thabiso said:
7 years ago
Answer is E(5).
XY=1 AND gate,
(X+Z)= OR gate,
X.(X+Z)= AND gate,
Y.(X+Z)= AND gate,
XY+X(X+Z)+Y(X+Z)= the last OR gate.
XY=1 AND gate,
(X+Z)= OR gate,
X.(X+Z)= AND gate,
Y.(X+Z)= AND gate,
XY+X(X+Z)+Y(X+Z)= the last OR gate.
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