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Hello everybody!
Could you, please, help me with this:
1) [NOT(A)*NOT(B)] + [A*B] = ?
2) [A+B] * [NOT(A) + NOT(B)] = ?
I appreciate your help!
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Hi Alex!
Let NOT(A)*NOT(B) + A*B = Q
Now consider different possible valves of "A" and "B" as follows..
A=0, B=0
A=1, B=0
A=0, B=1
A=1, B=1 and calculate respective values of Q.
Make a table of ur values like this..
A B Q
0 0 1
etc..
This is called ur Truth Table!
Note: ~A=NOT(A)=0 when A=1 and ~A=NOT(A)=1 when A=0 and so on.
You will find that..
1) Q=A⊗B (X-NOR gate)
2) Q=A⊕B (X-OR gate)
Last edited by ZHero (2010-05-16 05:25:23)
If two or more thoughts intersect, there has to be a point!
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ZHero,
Thank YOU very much!
1) Q=A⊗B (X-NOR gate) - expected, but wasn't sure
2) Q=A⊕B (X-OR gate) - for me XOR gets that's how: A * NOT(B) + NOT(A) * B = A XOR B - am I right?
Really appreciate your help, ZHero
Last edited by AlexandrSN (2010-05-16 06:47:54)
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Hi AlexandrSN;
From my notes:
Welcome to the forum!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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A * NOT(B) + NOT(A) * B = A XOR B - am I right?
Yes Alex! You are RIGHT!
That's just the original expression [A+B]*[~A+~B].
When you do the AND operation "*", using the Associative Property, you'll get "A*~A + B*~B" both of which equal 0
and what's left is "A*~B + B*~A" which, you're right, is A⊕B i.e. A(XOR)B.
Glad to Help.......
Last edited by ZHero (2010-05-16 07:18:27)
If two or more thoughts intersect, there has to be a point!
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