Truth table with logical gates for a traffic light Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Writing down logic of a circuitCreate a CMOS circuit from a logic functionSR Latch/Racing?What operation does this circuit perform?Output of XOR gate with high-impedance inputLogic Gates - Did I connect them correctly?NAND Gate Logic OptimizationBit counter using basic logic gatesBuild Truth Table given CircuitCreate a circuit with max 6 NAND gates
Antler Helmet: Can it work?
What would be the ideal power source for a cybernetic eye?
What is the longest distance a 13th-level monk can jump while attacking on the same turn?
How to deal with a team lead who never gives me credit?
I am not a queen, who am I?
What causes the vertical darker bands in my photo?
Gastric acid as a weapon
Is there a concise way to say "all of the X, one of each"?
Stars Make Stars
What is the musical term for a note that continously plays through a melody?
How to assign captions for two tables in LaTeX?
Why is "Captain Marvel" translated as male in Portugal?
Doubts about chords
ListPlot join points by nearest neighbor rather than order
Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?
How to motivate offshore teams and trust them to deliver?
Should I call the interviewer directly, if HR aren't responding?
WAN encapsulation
Do you forfeit tax refunds/credits if you aren't required to and don't file by April 15?
Did Kevin spill real chili?
Why does Python start at index 1 when iterating an array backwards?
Is 1 ppb equal to 1 μg/kg?
Is it true that "carbohydrates are of no use for the basal metabolic need"?
Does accepting a pardon have any bearing on trying that person for the same crime in a sovereign jurisdiction?
Truth table with logical gates for a traffic light
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Writing down logic of a circuitCreate a CMOS circuit from a logic functionSR Latch/Racing?What operation does this circuit perform?Output of XOR gate with high-impedance inputLogic Gates - Did I connect them correctly?NAND Gate Logic OptimizationBit counter using basic logic gatesBuild Truth Table given CircuitCreate a circuit with max 6 NAND gates
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I want to create this truth table:
So I tried this but I was not able to get the last row, which is:
While a =1 and b =1 ,yellow = 1 red = 0 and green = 0.
This is my design for it, but I could not make the last row.
Does anyone have a sample or a tip about what I can do to fix this?
Thanks!
logic-gates simulink
New contributor
$endgroup$
add a comment |
$begingroup$
I want to create this truth table:
So I tried this but I was not able to get the last row, which is:
While a =1 and b =1 ,yellow = 1 red = 0 and green = 0.
This is my design for it, but I could not make the last row.
Does anyone have a sample or a tip about what I can do to fix this?
Thanks!
logic-gates simulink
New contributor
$endgroup$
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
11 hours ago
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
6 hours ago
add a comment |
$begingroup$
I want to create this truth table:
So I tried this but I was not able to get the last row, which is:
While a =1 and b =1 ,yellow = 1 red = 0 and green = 0.
This is my design for it, but I could not make the last row.
Does anyone have a sample or a tip about what I can do to fix this?
Thanks!
logic-gates simulink
New contributor
$endgroup$
I want to create this truth table:
So I tried this but I was not able to get the last row, which is:
While a =1 and b =1 ,yellow = 1 red = 0 and green = 0.
This is my design for it, but I could not make the last row.
Does anyone have a sample or a tip about what I can do to fix this?
Thanks!
logic-gates simulink
logic-gates simulink
New contributor
New contributor
edited 13 hours ago
Niteesh Shanbog
513416
513416
New contributor
asked 13 hours ago
John wiliamJohn wiliam
163
163
New contributor
New contributor
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
11 hours ago
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
6 hours ago
add a comment |
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
11 hours ago
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
6 hours ago
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
11 hours ago
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
11 hours ago
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
6 hours ago
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can write the equations easily from the truth table
$$Red = bar A $$
$$Yellow = B$$
$$Green = A land bar B$$
They can be built the following circuit:
$endgroup$
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
YELLOW = $B$.
RED = $bar A$.
GREEN = $overline RED + YELLOW$
Purple is the implied logic for the red light, orange is the implied logic for the yellow light and green is for the green light.
One OR gate and two inverters would do the trick.
$endgroup$
$begingroup$
thanks i do not know know much about the gates i i were trying too much things thanks for helping :) i solve it
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
I do have a tip for you!
Work column by column, not row by row. First, create a circuit for the red light, ignoring the other two lights. Next, do the same for the yellow light. Finally, do the same for the green light. After you've created these three circuits, you can simply combine them into one.
$endgroup$
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
6 hours ago
add a comment |
Your Answer
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
);
);
John wiliam 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%2f432631%2ftruth-table-with-logical-gates-for-a-traffic-light%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can write the equations easily from the truth table
$$Red = bar A $$
$$Yellow = B$$
$$Green = A land bar B$$
They can be built the following circuit:
$endgroup$
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
You can write the equations easily from the truth table
$$Red = bar A $$
$$Yellow = B$$
$$Green = A land bar B$$
They can be built the following circuit:
$endgroup$
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
You can write the equations easily from the truth table
$$Red = bar A $$
$$Yellow = B$$
$$Green = A land bar B$$
They can be built the following circuit:
$endgroup$
You can write the equations easily from the truth table
$$Red = bar A $$
$$Yellow = B$$
$$Green = A land bar B$$
They can be built the following circuit:
answered 13 hours ago
RenanRenan
4,35022244
4,35022244
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
13 hours ago
add a comment |
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
13 hours ago
1
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
13 hours ago
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
YELLOW = $B$.
RED = $bar A$.
GREEN = $overline RED + YELLOW$
Purple is the implied logic for the red light, orange is the implied logic for the yellow light and green is for the green light.
One OR gate and two inverters would do the trick.
$endgroup$
$begingroup$
thanks i do not know know much about the gates i i were trying too much things thanks for helping :) i solve it
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
YELLOW = $B$.
RED = $bar A$.
GREEN = $overline RED + YELLOW$
Purple is the implied logic for the red light, orange is the implied logic for the yellow light and green is for the green light.
One OR gate and two inverters would do the trick.
$endgroup$
$begingroup$
thanks i do not know know much about the gates i i were trying too much things thanks for helping :) i solve it
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
YELLOW = $B$.
RED = $bar A$.
GREEN = $overline RED + YELLOW$
Purple is the implied logic for the red light, orange is the implied logic for the yellow light and green is for the green light.
One OR gate and two inverters would do the trick.
$endgroup$
YELLOW = $B$.
RED = $bar A$.
GREEN = $overline RED + YELLOW$
Purple is the implied logic for the red light, orange is the implied logic for the yellow light and green is for the green light.
One OR gate and two inverters would do the trick.
edited 12 hours ago
answered 13 hours ago
Andy akaAndy aka
244k11186425
244k11186425
$begingroup$
thanks i do not know know much about the gates i i were trying too much things thanks for helping :) i solve it
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
thanks i do not know know much about the gates i i were trying too much things thanks for helping :) i solve it
$endgroup$
– John wiliam
13 hours ago
$begingroup$
thanks i do not know know much about the gates i i were trying too much things thanks for helping :) i solve it
$endgroup$
– John wiliam
13 hours ago
$begingroup$
thanks i do not know know much about the gates i i were trying too much things thanks for helping :) i solve it
$endgroup$
– John wiliam
13 hours ago
add a comment |
$begingroup$
I do have a tip for you!
Work column by column, not row by row. First, create a circuit for the red light, ignoring the other two lights. Next, do the same for the yellow light. Finally, do the same for the green light. After you've created these three circuits, you can simply combine them into one.
$endgroup$
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
6 hours ago
add a comment |
$begingroup$
I do have a tip for you!
Work column by column, not row by row. First, create a circuit for the red light, ignoring the other two lights. Next, do the same for the yellow light. Finally, do the same for the green light. After you've created these three circuits, you can simply combine them into one.
$endgroup$
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
6 hours ago
add a comment |
$begingroup$
I do have a tip for you!
Work column by column, not row by row. First, create a circuit for the red light, ignoring the other two lights. Next, do the same for the yellow light. Finally, do the same for the green light. After you've created these three circuits, you can simply combine them into one.
$endgroup$
I do have a tip for you!
Work column by column, not row by row. First, create a circuit for the red light, ignoring the other two lights. Next, do the same for the yellow light. Finally, do the same for the green light. After you've created these three circuits, you can simply combine them into one.
answered 10 hours ago
Tanner SwettTanner Swett
42437
42437
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
6 hours ago
add a comment |
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
6 hours ago
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
6 hours ago
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
6 hours ago
add a comment |
John wiliam is a new contributor. Be nice, and check out our Code of Conduct.
John wiliam is a new contributor. Be nice, and check out our Code of Conduct.
John wiliam is a new contributor. Be nice, and check out our Code of Conduct.
John wiliam 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%2f432631%2ftruth-table-with-logical-gates-for-a-traffic-light%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
-logic-gates, simulink
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
11 hours ago
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
6 hours ago