boolean logic question - Page 1
Forum section: electronics / beginners

author: blinky87
author_rank: Newbie
timestamp: November 13, 2019, 09:32:21 pm
content: Hi everyonePracticing some Boolean logic which I want to simplify and make a circuit from eventually.The scenario I found states that Q can be equal to 1 providing one of both statements are true.~A AND BorA AND B AND ~CI can construct the truth table for the second argument, but since the first argument does not contain C, how is this represented in the truth table? Like this?https://imgur.com/a/avYZwJ8

author: Wimberleytech
author_rank: Super Contributor
timestamp: November 13, 2019, 09:40:56 pm
content: Quote from: blinky87 on November 13, 2019, 09:32:21 pmThe first argument generates a results that is ORed with a result of the second argument. Both arguments do not need to contain C. The layout of your truth table is fine but it is missing a state and also has an error.

author: blinky87
author_rank: Newbie
timestamp: November 13, 2019, 09:59:03 pm
content: Thank you for your help!I have amended the truth table;https://imgur.com/a/W3cU43e

author: Wimberleytech
author_rank: Super Contributor
timestamp: November 13, 2019, 10:01:42 pm
content: Quote from: blinky87 on November 13, 2019, 09:59:03 pmYou added the missing state. Now do the boolean analysis of the last state using the two expressions ORed with each other.

author: rstofer
author_rank: Super Contributor
timestamp: November 13, 2019, 10:56:07 pm
content: The Truth Table is the least usable way of describing the problem. You really should use Karnaugh Maps because the simplification is immediately apparent. This is going to be CRUDEA/B C' C------------0 0 | 0 00 1 |1 11 1 |101 0 | 0 0Draw an ellipse around the two 1's in the left column under C' and you get B & C' -- A is eliminated because it changes between the two rowsDraw an ellipse around the two 1's in the second row and you get A' & B -- C is eliminated because it changes between the two columnsCombine terms B & (A' + C')The Karnaugh Map is a lot like a Truth Table except that only one variable can change between adjacent rows or adjacent columns. That's why the minimization shows up naturally.http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/karnaugh.html

author: blinky87
author_rank: Newbie
timestamp: November 13, 2019, 11:07:43 pm
content: My outcome is to use K-maps, I just didn;t want to overly complicate things just yet. One step at a time. But yes, thank you!!!!...and thank you both for helping.The k-map result I got is A'B + BC'which is the same as yours I believe, however, you have factored out.I was just going through creating the truth table again and noticed I have Q = 1 when B & C. I'm struggling to work this out but I don't think I'm carrying the result through the combination circuit i.e. I am only taking into account the one output, Q.

author: onlooker
author_rank: Frequent Contributor
timestamp: November 13, 2019, 11:27:48 pm
content: Q = A'*B + A*B*C' = B*(A' + A*C') = B*[A'*(1+C') + A*C']= B*(A'*1 + A'*C' + A*C') = B*[A' + (A'+A)*C']= B*(A'+C') = B*(A*C)'

author: Wimberleytech
author_rank: Super Contributor
timestamp: November 14, 2019, 03:21:08 am
content: Quote from: Wimberleytech on November 13, 2019, 10:01:42 pmLooking back I realized that I had ~A inverted in my notes.