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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 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 650 Math Focus: Yet to find out.  
March 27th, 2016, 09:29 PM  #3 
Member Joined: Dec 2014 From: Queensland Posts: 59 Thanks: 2  
March 27th, 2016, 09:38 PM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 650 Math Focus: Yet to find out.  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 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 650 Math Focus: Yet to find out.  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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