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)
RANU said:
1 decade ago
XY requires 1 AND gate.. X(X+Z) requires 1 OR gate & 1 AND gate ..similarly Y(X+Z) require only 1 gate i.e. AND gate since X+Z operation already done and finally XY + X(X + Z) + Y(X + Z) can use one 3 input OR gate for ORing each operation ,hence only 5 gates are required without simplification.
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).
Hema said:
1 decade 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.
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.
Usha said:
1 decade 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.
Nidhya said:
1 decade ago
My answer is C. on reducing the given exp we will get XY+XX+XZ+YX+YZ. here XY=YX; XX is nothing but X itself. X+XY+YX=X+XY=X(Y+1). so XZ - 1 gate YZ -1gate then for adding them - 2 gates. totally 4.
Rohit said:
1 decade 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.
Retheesh said:
1 decade ago
Answer is D.
Because it is required 4 gates for its individual operation and a final gate is required to make its sum.
So it is require toital- 5 gates.
Because it is required 4 gates for its individual operation and a final gate is required to make its sum.
So it is require toital- 5 gates.
Xin said:
1 decade ago
The problem didn't require simplification by just simply counting the operation.
1. XY
2. X+Z
3. X(X + Z)
4. (X + Z)
5. Y(X + Z)
So the answer is 5.
1. XY
2. X+Z
3. X(X + Z)
4. (X + Z)
5. Y(X + Z)
So the answer is 5.
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