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 March 27th, 2016, 09:00 PM #1 Member   Joined: Dec 2014 From: Queensland Posts: 59 Thanks: 2 Boolean algebra I'm trying to simplify a Boolean algebra equation so I can plot a simplified logic circuit. I'm down as far as B'(D+A'D) + A'BC'D' I'm wondering about the B'(D+A'D). Can this be B'(D+A'D) B'(1+A'D) B'A'D or B'(D+A'D) B'D Stuck on this problem also for another problem. C'D'(B'+A') can the A'+B' be treated as 1+1 so I can simplify by doing C'D'(1+A') or C'D'(0+A') to get just C'D'A'? Thanks Last edited by skipjack; March 28th, 2016 at 01:58 AM.
March 27th, 2016, 09:21 PM   #2
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Quote:
 Originally Posted by unistu or B'(D+A'D) B'D
This is correct. Try substituting in 0, then 1 for the D + A'D expression. You'll see that for both values, the answer is always D. Make sense?

March 27th, 2016, 09:29 PM   #3
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Quote:
 Originally Posted by Joppy This is correct. Try substituting in 0, then 1 for the D + A'D expression. You'll see that for both values, the answer is always D. Make sense?
Thanks, do you mean substituting like

B'(D+A'D) can be like
B'(1+(0*1)) then
B'(1+0)
B'D

Cheers

March 27th, 2016, 09:38 PM   #4
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Quote:
 Originally Posted by unistu can the A'+B' be treated as 1+1?
What makes you think you can treat it in this way? A and B both have two possible states, and together they have four possible states. The truth table for this expression (A' + B') will be:

A B = X
0 0 | T
0 1 | T
1 0 | T
1 1 | F

Now having seen the truth table, we can see that using a NAND gate with inputs A and B would give the same result.

Consider DeMorgans Theorem:

A' + B' = (AB)' and

A'B' = (A + B)'

Now you can see there is multiple ways to express original terms (probably more ways than i've written here):

C'D' (B' + A') = (AB)'C'D' = (B' + A') (C + D)'

Which one to use? Depends, maybe you have limited IC's and can only use NAND gates or NOR gates. This is a common exercise for teachers to give, and forces you to use your resources effectively and efficiently.

Last edited by Joppy; March 27th, 2016 at 10:00 PM.

March 27th, 2016, 09:54 PM   #5
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Quote:
 Originally Posted by unistu B'(D+A'D) can be like B'(1+(0*1)) then B'(1+0) B'D
Yes this is correct, but i'd refrain from doing it like this (my wording may have mislead you on post #2).

Notice how you've only considered one combination of D and A (you considered A to be 1 and D to be 1)? You need to consider all combinations!

Similar to what i did in a previous post, extract the expression of interest, and do up a quick truth table to see how it behaves.

A D = Z
0 0 | 0
0 1 | 1
1 0 | 0
1 1 | 1

See how our output Z mimics D? In other words, Z = D where Z is just an arbitrary output for the time being.

Last edited by Joppy; March 27th, 2016 at 10:02 PM.

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