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 2 of 3.
SUDHANSHU said:
8 years ago
x.x = x USING AND GATE.
x+x =x USING OR GATE.
question is XY + X(X + Z) + Y(X + Z).
= XY + XX+ XZ +YX +YZ.
( XY + YX ) will be ( XY) & (X.X) will be (X).
= XY + X + XZ + ZY.
So there will be 4 GATEs and 1 extra GATE for Addition.
So total number of gate is 5.
OPTION (D).
x+x =x USING OR GATE.
question is XY + X(X + Z) + Y(X + Z).
= XY + XX+ XZ +YX +YZ.
( XY + YX ) will be ( XY) & (X.X) will be (X).
= XY + X + XZ + ZY.
So there will be 4 GATEs and 1 extra GATE for Addition.
So total number of gate is 5.
OPTION (D).
Sukriti said:
9 years ago
XY + X(X + Z) + Y(X + Z),
= XY + XX + XZ + YX + YZ,
= XY + X + XZ + YZ Where (XX = X, XY = YX),
= X(1 + Y) + XZ + YZ (1+Y=1),
= X(1 + Z) + YZ (1+Z=1),
= X + YZ.
Therefore two logic gates are required. ie 1 AND gate for YZ and 1 OR gate for X + YZ.
= XY + XX + XZ + YX + YZ,
= XY + X + XZ + YZ Where (XX = X, XY = YX),
= X(1 + Y) + XZ + YZ (1+Y=1),
= X(1 + Z) + YZ (1+Z=1),
= X + YZ.
Therefore two logic gates are required. ie 1 AND gate for YZ and 1 OR gate for X + YZ.
SUKHPREET SINGH said:
9 years ago
X.Y = AND GATE
X+Z = OR GATE
X(X+Z) = AND GATE
Y(X+Z) = AND GATE
X.Y+X(X+Z)+Y(X+Z) = OR GATE
5 GATES REQUIRED
X+Z = OR GATE
X(X+Z) = AND GATE
Y(X+Z) = AND GATE
X.Y+X(X+Z)+Y(X+Z) = OR GATE
5 GATES REQUIRED
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.
Abhishek said:
9 years ago
Without simplification also gates will do. x+z gate can be used twice.
Rohit said:
10 years ago
Here, Answer is D.
AND gate (1st) = xy.
OR gate (1st) = x+z.
AND gate (2nd) = x(x+z).
AND gate (3rd) = y(x+z).
OR gate (2nd) = Final expression.
So, no. of total gates = 5.
AND gate (1st) = xy.
OR gate (1st) = x+z.
AND gate (2nd) = x(x+z).
AND gate (3rd) = y(x+z).
OR gate (2nd) = Final expression.
So, no. of total gates = 5.
Usha said:
10 years ago
The question asked was no. of gates required before simplification. So,
For XY-1 AND GATE.
For X+Z-1 OR GATE.
For X(X+Z)-1 AND GATE.
For X+Z-1 OR GATE.
For Y(X+Z)-1 AND GATE.
And for overall addition 1 OR GATE required. So answer will be 6.
For XY-1 AND GATE.
For X+Z-1 OR GATE.
For X(X+Z)-1 AND GATE.
For X+Z-1 OR GATE.
For Y(X+Z)-1 AND GATE.
And for overall addition 1 OR GATE required. So answer will be 6.
Hema said:
10 years ago
5 gates only I didn't see properly its not x+z it was X(X+Z). So, 5 gates is answer.
Hema said:
10 years ago
Perfectly the answer is C means 4 gates.
Inputs are xyz.
xy-1 AND gate.
z and x to OR gate gives x+z.
Then, x+z and y to AND gate gives why (x+z).
Finally, xy, (x+z) and y(x+z) should connect to OR gate.
So totally 4 gates i.e., 2 OR gates and 2 AND gates.
Inputs are xyz.
xy-1 AND gate.
z and x to OR gate gives x+z.
Then, x+z and y to AND gate gives why (x+z).
Finally, xy, (x+z) and y(x+z) should connect to OR gate.
So totally 4 gates i.e., 2 OR gates and 2 AND gates.
Ram93 said:
1 decade ago
P:xy - 2 input AND.
Q:x+z - 2 input OR.
R:xQ - 2 input AND.
S:yQ - 2 input AND.
P+R+S - 3 input OR.
Total logic gates 5.
Q:x+z - 2 input OR.
R:xQ - 2 input AND.
S:yQ - 2 input AND.
P+R+S - 3 input OR.
Total logic gates 5.
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