Help Me simplify: C*(A+B) + ~A*BStopping the clock without gating the clockCan I simplify this to a 2-to-1...
Can you combine War Caster, whip, and Warlock Features to Eldritch Blast enemies with reach?
What to do when being responsible for data protection in your lab, yet advice is ignored?
Slow moving projectiles from a hand-held weapon - how do they reach the target?
Why would the Pakistan airspace closure cancel flights not headed to Pakistan itself?
A starship is travelling at 0.9c and collides with a small rock. Will it leave a clean hole through, or will more happen?
Program that converts a number to a letter of the alphabet
Why Normality assumption in linear regression
How to acknowledge an embarrassing job interview, now that I work directly with the interviewer?
If I sold a PS4 game I owned the disc for, can I reinstall it digitally?
Jumping Numbers
Why do members of Congress in committee hearings ask witnesses the same question multiple times?
A minimum of two personnel "are" or "is"?
It took me a lot of time to make this, pls like. (YouTube Comments #1)
Typing Amharic inside a math equation?
Does Windows 10's telemetry include sending *.doc files if Word crashed?
How would one buy a used TIE Fighter or X-Wing?
Is there some relative to Dutch word "kijken" in German?
Cryptic with missing capitals
How do I say "Brexit" in Latin?
Checking for the existence of multiple directories
Why doesn't "auto ch = unsigned char{'p'}" compile under C++ 17?
Does Improved Divine Strike trigger when a paladin makes an unarmed strike?
Why does a metal block make a shrill sound but not a wooden block upon hammering?
How to deal with an incendiary email that was recalled
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
$begingroup$
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
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
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
4 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
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
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.
edited 3 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 4 hours ago
Alan KazemianAlan Kazemian
133
133
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
4 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
4 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
4 hours ago
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
4 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.
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 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
2 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f425103%2fhelp-me-simplify-cab-ab%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 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
2 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.
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 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
2 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.
$endgroup$
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.
edited 2 hours ago
answered 4 hours ago
Eugene Sh.Eugene Sh.
7,5481830
7,5481830
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 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
2 hours ago
add a comment |
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 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
2 hours ago
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 hours ago
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 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
2 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
2 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.
Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f425103%2fhelp-me-simplify-cab-ab%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
4 hours ago