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Boolean Expression help

NanyBany

May 15, 2014
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Can anyone help me simplify this Boolean expression, what I got is several different results, the most recent one is:
!C(!A + A!B)
20140625_104542.jpg

20140625_104044.jpg

My goal is to make a 7 segment display output the decimal number based on the combination of the 4 bit input. Ex 0000 will make the 7 segment display 0. But when this simplified Boolean expression is implemented, it doesn't work. I have no idea where I went wrong, please no K-maps. Thank you for any help much appreciated.

For some reason the truth table won't post upright :/ Sorry.
[Moderator's comment: I've fixed your image so it's the right way round. --KrisBlueNZ]
 
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Gryd3

Jun 25, 2014
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You would need a boolean expression for each segment, would you not?
Which segment is this for?
Please tell me which segment you are working on... as the first line appears to be searching for numbers 02346789.
 

NanyBany

May 15, 2014
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You would need a boolean expression for each segment, would you not?
Which segment is this for?
Please tell me which segment you are working on... as the first line appears to be searching for numbers 02346789.
The first segment 'T' it should be searching for numbers 02356789 not 4.
 

Gryd3

Jun 25, 2014
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sevensegment.gif

Would a binary to 7-segment decoder schematic help you out? I didn't dig too deep mind you, so I am unsure if your table matched this diagram.
It involves a little tracing.. but you can see the inverters, and gates and or gates. Just follow the traces backward from the display and you should be able to make out the logic from each segment.

*Sorry for the lazy answer, I just figured it would be easier to look up existing work than to map it out all over again ;)
If this is not what your after, I'll sit down and map something out.
 

NanyBany

May 15, 2014
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sevensegment.gif

Would a binary to 7-segment decoder schematic help you out? I didn't dig too deep mind you, so I am unsure if your table matched this diagram.
It involves a little tracing.. but you can see the inverters, and gates and or gates. Just follow the traces backward from the display and you should be able to make out the logic from each segment.

*Sorry for the lazy answer, I just figured it would be easier to look up existing work than to map it out all over again ;)
If this is not what your after, I'll sit down and map something out.

I appreciate the answer I really do, but I want to simplify and implement it all myself for an educational experience. I would like to know what I did wrong when I simplified the Boolean expression. And if anyone could, I would like to see their simplified solution of the Boolean expression.
 

Laplace

Apr 4, 2010
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It appears that you used the identity (!B!D+BD)=1 which is not true.
 

NanyBany

May 15, 2014
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It appears that you used the identity (!B!D+BD)=1 which is not true.

And you are awesome, thank you. That is exactly what I was looking for. What I think is that

(!B!D + BD) = B Xor D.

If not, then what does (!B!D + BD) = ?????? Thank you again for the help.
 

Harald Kapp

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B xor D is 1 if either B or D is 1, but not both are 1 or 0.
!B!D + BD is 1 if either both B and D are 0 or B and D are 1.

So (!B!D + BD) = B Xor D is not correct. You find the error yourself?
 

NanyBany

May 15, 2014
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B xor D is 1 if either B or D is 1, but not both are 1 or 0.
!B!D + BD is 1 if either both B and D are 0 or B and D are 1.

So (!B!D + BD) = B Xor D is not correct. You find the error yourself?

So it is XNOR then? I don't even know how to reference that in a Boolean expression. Would it be (B Xor D)'
 

Harald Kapp

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B Xnor D = NOT(B xor D), I assume that's what you mean by (B Xor D)'.
 
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