demorgans, (A+B).(A+C) into nand

Discussion in 'Homework Help' started by ninjaman, May 18, 2013.

May 18, 2013
306
1
hello
im new here and struggling with a question that i have for my btec. i dont get this stuff at all
i have to change this (A+B).(A+C)
to nand gates
i understand demorgans so far as, (A+B).(A+C) = (AB).(AC)
i then take a nand gate for A B and C, tye the inputs to get each one negated then the output from A goes into another gate with B and another with C. this double negates them. and i dont know if this is right or wrong as my tutor isnt helping much
i would really appreciate any help on this
cheers
simon

2. tubeguy Well-Known Member

Nov 3, 2012
1,157
197

May 18, 2013
306
1
i have three nand gates, with both inputs tied together for an or function. the output goes to two nand gates, B and C to a separate gate and A to both. this gives double negated AB.AC
is this right, i have recieved an email from my tutor and all he said is the only hint he can give is that there is a double negation somewhere.
any help would be appreciated. i have read through that site attached above and cant get anything from it.

4. WBahn Moderator

Mar 31, 2012
17,756
4,799

Why do you have AB.AC? How is this different from ABAC or A.B.A.C?

It would also help if you provided some kind of a diagram.

You stated that you knew that (A+B).(A+C) = (AB).(AC)

And tubeguy asked you how this could be true and you just ignored it.

Let's say A is False but B and C are both True. What is the left hand side? What is the right hand side?

May 18, 2013
306
1
i dont know how to get a diagram on here.
all i have is "make (A+B).(A+C) out of nand gates"
i dont know if B or C are true(they may be lies!?)
i have read through my course notes and dont understand it, i have read through the stuff on here and got bored because i didnt understand it.
i have watched videos on youtube and still dont understand it.
thankyou tubeguy for you help, i appreciate it.
i dont know if what i have put is right, it says this in the book so i wrote it as i understand it

6. WBahn Moderator

Mar 31, 2012
17,756
4,799
What is YOUR understanding of these equations?

If you have A+B, what Boolean operation do YOU understand that to be?

If you have AB, what Boolean operation do YOU understand that to be?

If you have A.B, what Boolean operation do YOU understand that to be?

If you have A and you want the Boolean inverse (or complement, or NOT) of that, what notation would YOU use?

We can't help you very much unless both you and we know what YOU mean by the notation YOU are using.

And if you are just taking (A+B).(A+C) from some sentence in a problem and you don't have any idea what is meant by it, then you have far more fundamental problems that we need to work on first. We are willing to do so, but let's first determine if it is necessary.

May 18, 2013
306
1
+ is or
. or AB is and
a line above is complement
A or B and A or C, make it into a nand gate. i dont know if any of this is true.
i put them through a nand making them all negated, i connected both inputs to make the nand gate an OR function, i believe this to be correct when only using nand gates. A B and C are all OR functions. the output fromthe A nand goes to two nand gates, one with B and the other with C. this will double negate them..... i think

Last edited: May 18, 2013

May 18, 2013
306
1
tried something and it didnt work

Last edited: May 18, 2013
9. WBahn Moderator

Mar 31, 2012
17,756
4,799
Keep up that attitude, and I won't.

Thank you. Since you can't represent a line above a word in ASCII text, many people use the . to mean negation. Since DeMorgan's involves negation, you were either using the . to mean negation or you weren't using any negation at all. Since you didn't have a valid result in either case, there was no way to tell except to drag out of you what you meant by the notation you were using.

This is very hard to follow. For instance, when you say "A B and C are all OR functions," I have no idea what you are talking about. A B and C are not functions at all, they are signals.

Let's go back to something you said in your first post:

i understand demorgans so far as, (A+B).(A+C) = (AB).(AC)

Make a truth table for the two sides:

X = (A+B).(A+C)
Y = (AB).(AC)

 A B C X Y 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

Are X and Y equal for all possible values of A, B, and C?

If yes, then your statement is correct.

If no, then your statement is incorrect and we need to not go any further until you get a handle on this step.

10. piratepaul Member

May 20, 2013
35
2
You need to change (A+B).(A+C) to NANDs ... (A+B).(A+B) is not NANDs, you can make any gate from NANDs, a NAND with its inputs tied together is a NOT.

( (AB).(AC) does not make sense, unless it is some strange code but....)

11. piratepaul Member

May 20, 2013
35
2
In fact (A+B).(A+C) = (AB) + (AC) Prob a bit late for the homework now.

12. WBahn Moderator

Mar 31, 2012
17,756
4,799
How do you figure this?

For the LHS, if A is HI, then both the left and right factors are HI and the result is HI, regardless of the state of B or C. But on the LHS, if B and C are both LO, then neither term is HI and the result is LO, regardless of the state of A.

Conversely, if A is LO, the RHS is LO regardless of B or C, while if B and C are HI, the LHS is HI regardless of the state of A.

Multiply it out:

(A+B).(A+C) = AA+AC+AB+BC = A+BC

13. piratepaul Member

May 20, 2013
35
2
A+BC ... blush ... you can make the circuit with NANDs.