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 3 of 3.
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.
BONDs said:
1 decade ago
Shouldn't it be 6.
Because,
1. X+Z (let's say Q).
2. XY.
3. XQ.
4. YQ.
5 and 6. 2+3+4.
Because,
1. X+Z (let's say Q).
2. XY.
3. XQ.
4. YQ.
5 and 6. 2+3+4.
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.
Ramso said:
1 decade ago
The answer is 5, first it become xy+(x+y)(y+z)
It need 2 and gate and 3 or gate.
It need 2 and gate and 3 or gate.
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.
Shubhangi_n said:
1 decade ago
Answer is D as they want before simplification. Nidhya solved it after simplification.
Nandan said:
1 decade ago
Using K-map ,you get X+Y+Z, giving 2 gates. So answer is B
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.
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