Bolean Function required

[The Gladtiator]

the problem : a majority gate is a digital circuit whose output is equal to 1 if the majority inputs are 1's. the output is 0 otherwise. by means of a truth table, find the bolean function implemented by a 3-input majority gate. ...... My Question is : what does he mean by " majority inputs are 1's " ??? the answer is F= xy+xz+yz , but I dont know how ?? some one explain pls ?

 

[David James]

Wikipedia: "A majority gate returns true if and only if more than 50% of its inputs are true".

 

[OldTechie]

The function says that when a majority of the inputs are 1, then the output is 1. Since there are 3 inputs, a majority vote says that at least 2 of them (any 2 inputs) must be 1. (A majority of 3 is 2). If there were 4 inputs, then you would have to test for any 3 inputs being 1 at the same time. And so on...

Dave M

 

[The Gladtiator]

So the truth table is this x y z f
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 0

Then F=x'yz+xy'z+xyz' how the answer is f=xy+xz+yz

 

[OldTechie]

So the truth table is this x y z f
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 0

Then F=x'yz+xy'z+xyz' how the answer is f=xy+xz+yz

You almost hav it right... Remember that the problem is for a MAJORITY of inputs being 1 to output a 1.
Is all 3 of the 3 inputs a majority?

Dave M

 

[The Gladtiator]

Is all 3 of the 3 inputs a majority?

Dave M

AHA ... thQ man I though it must me 2 inputs 1 and 1 is 0 to output 1 ... THQ

 

[OldTechie]

Oops... I didn't answer all of your question. Since you're only interested in the True inputs, the False inputs are normally ignored, unless the False input is important in the solution. If the question had been something like,
"Build a truth table for a majority of 3 inputs being 1, but at least one input being 0", then your truth table and the Boolean expression would be correct.
As it stands now, your solution is incorrect.

Dave M

 

[The Gladtiator]

"Build a truth table for a majority of 3 inputs being 1, but at least one input being 0"

Dave M

No No,, The problem says "whose output is equal to 1 if the majority inputs are 1's. the output is 0 otherwise." your answer is right

 

[OldTechie]

No No,, The problem says "whose output is equal to 1 if the majority inputs are 1's. the output is 0 otherwise." your answer is right

Yes,I know what the original question was... but carefully reread my last reply. I said "Oops... ...... If the question HAD BEEN something like, "Build a truth table for a majority of 3 inputs being 1, but at least one input being 0", then your truth table and the Boolean expression would be correct."

I was trying to lead you to a correct solution for the original question, and lead you to understand the importance of recognizing required states and those that can be ignored. As your last Boolean expression and truth table was posted, your solution is wrong. Why? Because your truth table reads:

x y z f
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 0
The last entry is wrong for the original question. Three 1 inputs is a majority of 1s (a majority of 3 means at least 2, not exactly 2), and thus satisfies any of the terms in the Boolean expression. The last line should be:
1 1 1 1

Hope that helped clarify my statements

Cheers,
Dave M