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Help Me simplify: C*(A+B) + ~A*B


Stopping the clock without gating the clockCan I simplify this to a 2-to-1 multiplexer?Noobie at IC, need help with basic circuit that isn't returning correct valuesSimplify a logic expression - Where have I gone wrong?Design 5 bit counter with two control inputs (direction and stop)Need help with NOT Gate oscillator circuitHow to design a counter with an arbitrary sequenceFinding the K-map of an = comparatorWhy does a turbine with PMSG and full converter system draws reactive power when turbine is stopped or there is no wind?Need Help With Transistor Logic Circuit Design













2












$begingroup$


I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.









share









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 4




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    14 hours ago
















2












$begingroup$


I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.









share









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 4




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    14 hours ago














2












2








2





$begingroup$


I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.









share









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.







digital-logic electrical





share









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share



share








edited 13 hours ago









pipe

10.1k42656




10.1k42656






New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 14 hours ago









Alan KazemianAlan Kazemian

183




183




New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    14 hours ago














  • 4




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    14 hours ago








4




4




$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
14 hours ago




$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
14 hours ago










1 Answer
1






active

oldest

votes


















8












$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



C(A + B) + A'B = AC + BC + A'B =


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B 


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$













  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    13 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    11 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



C(A + B) + A'B = AC + BC + A'B =


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B 


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$













  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    13 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    11 hours ago
















8












$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



C(A + B) + A'B = AC + BC + A'B =


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B 


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$













  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    13 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    11 hours ago














8












8








8





$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



C(A + B) + A'B = AC + BC + A'B =


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B 


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$



I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



C(A + B) + A'B = AC + BC + A'B =


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B 


From here you will need to expand it and use the "OR absorption law" twice.







share|improve this answer














share|improve this answer



share|improve this answer








edited 11 hours ago

























answered 14 hours ago









Eugene Sh.Eugene Sh.

7,5681830




7,5681830












  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    13 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    11 hours ago


















  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    13 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    11 hours ago
















$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
13 hours ago




$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
13 hours ago












$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
11 hours ago




$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
11 hours ago










Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.










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