Tri_Edge_1836
Question on Simplifying Boolean expression (Logic Gates)
Q1) ABC + AB(bar)C + ABC(bar)
---Here I'm getting the answer AB + AC -Is that Right???
Q2) AB + A(bar)C + BC
(bar) -> Line above the letter. Also called Dash sometimes.
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ganeshie8
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|dw:1346087551850:dw|
ganeshie8
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Q1 is like above ?
ganeshie8
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last bar is only for C, or for whole ABC ?
Tri_Edge_1836
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Last bar is only for C. Any yes it is as U've shown.
ganeshie8
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okay lets see...
Tri_Edge_1836
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Please do. I'm seriously stumped. (For the second one- The 1st one atleast i got something)
ganeshie8
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first one is correct :
ganeshie8
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|dw:1346088083347:dw|
ganeshie8
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|dw:1346088160067:dw|
ganeshie8
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Q2 is like above ,right ?
Tri_Edge_1836
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Nope. Its AB + A'C + BC.
A' as In A(bar)
Tri_Edge_1836
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Thats wat stumped me as i couldn't really find anything to take common and continue forth. :(
ganeshie8
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okay just a sec.. let me get a paper !
ganeshie8
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answer should simplify to : A'C + AB
im trying to work backwards...
ganeshie8
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ah its simple friend...
ganeshie8
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AB + A'C + BC
AB + A'C + BC(A + A')
Tri_Edge_1836
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Ok.
ganeshie8
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AB + A'C + ABC + A'BC
ganeshie8
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group 1st & 3rd terms
ganeshie8
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2nd & last terms
Tri_Edge_1836
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Wait since u said U are trying to solve it backwards, does that mean u found the answer somewhere?
U get AB(1+C)+(1+B)A'C
ganeshie8
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since,
1+x = 1
ganeshie8
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it simplifies to AB + A'C
ganeshie8
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i simplified it using karnaugh-map quickly.... and worked backwards hehe
Tri_Edge_1836
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Thanku very much. FOrgot about that Identity. :) Yes I guess i understood now. :)
-Closing thread. :)
ganeshie8
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glad to hear :) yw !!!
Tri_Edge_1836
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Just an FYI, I was stumped bby thihs for quite some time now. U helped me alot. :D
ganeshie8
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np :D