EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: blinky87 on November 13, 2019, 09:32:21 pm
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Hi everyone
Practicing 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 B
or
A AND B AND ~C
I 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
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Hi everyone
Practicing 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 B
or
A AND B AND ~C
I 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
The 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.
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Thank you for your help!
I have amended the truth table;
https://imgur.com/a/W3cU43e
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Thank you for your help!
I have amended the truth table;
https://imgur.com/a/W3cU43e
You added the missing state. Now do the boolean analysis of the last state using the two expressions ORed with each other.
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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 CRUDE
A/B C' C
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0 0 | 0 0
0 1 | 1 1
1 1 | 1 0
1 0 | 0 0
Draw 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 rows
Draw 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 columns
Combine 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 (http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/karnaugh.html)
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My outcome is to use K-maps, I just didn;t want to overly complicate things just yet. One step at a time :D. 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.
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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)'
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Thank you for your help!
I have amended the truth table;
https://imgur.com/a/W3cU43e
You added the missing state. Now do the boolean analysis of the last state using the two expressions ORed with each other.
Looking back I realized that I had ~A inverted in my notes.