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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. Tags algebra, boolean Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post JustForFun2000 Computer Science 2 September 26th, 2016 09:11 AM sailorman444 Algebra 1 January 9th, 2014 07:21 AM Shamieh Calculus 2 September 12th, 2013 10:38 AM Anamaria Applied Math 1 February 18th, 2011 11:16 AM MoZ Applied Math 8 January 28th, 2009 05:29 PM

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