Help Me simplify: C*(A+B) + ~A*BStopping the clock without gating the clockCan I simplify this to a 2-to-1...
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Help Me simplify: C*(A+B) + ~A*B
Help Me simplify: C*(A+B) + ~A*B
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$begingroup$
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.
digital-logic electrical
edited 4 hours ago
pipe
10.1k42656
asked 5 hours ago
Alan KazemianAlan Kazemian
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.
$endgroup$
add a comment |
$begingroup$
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.
digital-logic electrical
edited 4 hours ago
pipe
10.1k42656
asked 5 hours ago
Alan KazemianAlan Kazemian
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.
$endgroup$
3
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago
add a comment |
$begingroup$
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.
digital-logic electrical
edited 4 hours ago
pipe
10.1k42656
asked 5 hours ago
Alan KazemianAlan Kazemian
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.
$endgroup$
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
digital-logic electrical
edited 4 hours ago
pipe
10.1k42656
asked 5 hours ago
Alan KazemianAlan Kazemian
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.
edited 4 hours ago
pipe
10.1k42656
asked 5 hours ago
Alan KazemianAlan Kazemian
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.
edited 4 hours ago
pipe
10.1k42656
edited 4 hours ago
pipe
10.1k42656
edited 4 hours ago
pipe
10.1k42656
10.1k42656
asked 5 hours ago
Alan KazemianAlan Kazemian
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.
asked 5 hours ago
Alan KazemianAlan Kazemian
183
asked 5 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.
3
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago
add a comment |
3
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago
3
3
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
5 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
3 hours ago
add a comment |
Your Answer
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1 Answer
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$begingroup$
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.
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
5 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
3 hours ago
add a comment |
$begingroup$
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.
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
5 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
3 hours ago
add a comment |
$begingroup$
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.
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
$endgroup$
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.
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
edited 3 hours ago
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
answered 5 hours ago
Eugene Sh.Eugene Sh.
7,5581830
7,5581830
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
5 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
3 hours ago
add a comment |
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
5 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
3 hours ago
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
5 hours ago
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
5 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
3 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
3 hours ago
add a comment |
Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.
Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.
Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago