Truth table with logical gates for a traffic light
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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
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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
yesterday
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
yesterday
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 yesterday
Niteesh Shanbog
519416
519416
New contributor
asked yesterday
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
yesterday
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
yesterday
add a comment |
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
yesterday
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
yesterday
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
yesterday
$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
yesterday
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
yesterday
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
yesterday
add a comment |
4 Answers
4
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
yesterday
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
yesterday
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
yesterday
add a comment |
$begingroup$
I will be honest and say, i am not all that good at designing logic diagrams, way too long since i had it in school. But there is one thing i remember: Logic Friday. Its a freeware program that is an immense help when designing such diagrams and solving the equations. Using Logic Friday, i got these results:
A - input 1
B - input 2
F3 - red
F4 - yellow
F5 - green
Logic input gave me this truthtable after minimizing:
F3 = A';
F4 = B;
F5 = A B';
And this diagram followed:
I know a program like this is no substitute for real knowledge, but it works for a home fiddler like me.
Good luck :)
$endgroup$
$begingroup$
thanks for the help :)
$endgroup$
– John wiliam
4 hours ago
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
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
yesterday
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
yesterday
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 yesterday
RenanRenan
4,35022244
4,35022244
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
yesterday
add a comment |
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
yesterday
1
1
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
yesterday
$begingroup$
thanks for the picture :)
$endgroup$
– John wiliam
yesterday
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
yesterday
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
yesterday
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 yesterday
answered yesterday
Andy akaAndy aka
245k11186425
245k11186425
$begingroup$
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
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
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
yesterday
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 yesterday
Tanner SwettTanner Swett
42437
42437
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
yesterday
add a comment |
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
yesterday
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
yesterday
$begingroup$
thanks for the advice :)
$endgroup$
– John wiliam
yesterday
add a comment |
$begingroup$
I will be honest and say, i am not all that good at designing logic diagrams, way too long since i had it in school. But there is one thing i remember: Logic Friday. Its a freeware program that is an immense help when designing such diagrams and solving the equations. Using Logic Friday, i got these results:
A - input 1
B - input 2
F3 - red
F4 - yellow
F5 - green
Logic input gave me this truthtable after minimizing:
F3 = A';
F4 = B;
F5 = A B';
And this diagram followed:
I know a program like this is no substitute for real knowledge, but it works for a home fiddler like me.
Good luck :)
$endgroup$
$begingroup$
thanks for the help :)
$endgroup$
– John wiliam
4 hours ago
add a comment |
$begingroup$
I will be honest and say, i am not all that good at designing logic diagrams, way too long since i had it in school. But there is one thing i remember: Logic Friday. Its a freeware program that is an immense help when designing such diagrams and solving the equations. Using Logic Friday, i got these results:
A - input 1
B - input 2
F3 - red
F4 - yellow
F5 - green
Logic input gave me this truthtable after minimizing:
F3 = A';
F4 = B;
F5 = A B';
And this diagram followed:
I know a program like this is no substitute for real knowledge, but it works for a home fiddler like me.
Good luck :)
$endgroup$
$begingroup$
thanks for the help :)
$endgroup$
– John wiliam
4 hours ago
add a comment |
$begingroup$
I will be honest and say, i am not all that good at designing logic diagrams, way too long since i had it in school. But there is one thing i remember: Logic Friday. Its a freeware program that is an immense help when designing such diagrams and solving the equations. Using Logic Friday, i got these results:
A - input 1
B - input 2
F3 - red
F4 - yellow
F5 - green
Logic input gave me this truthtable after minimizing:
F3 = A';
F4 = B;
F5 = A B';
And this diagram followed:
I know a program like this is no substitute for real knowledge, but it works for a home fiddler like me.
Good luck :)
$endgroup$
I will be honest and say, i am not all that good at designing logic diagrams, way too long since i had it in school. But there is one thing i remember: Logic Friday. Its a freeware program that is an immense help when designing such diagrams and solving the equations. Using Logic Friday, i got these results:
A - input 1
B - input 2
F3 - red
F4 - yellow
F5 - green
Logic input gave me this truthtable after minimizing:
F3 = A';
F4 = B;
F5 = A B';
And this diagram followed:
I know a program like this is no substitute for real knowledge, but it works for a home fiddler like me.
Good luck :)
answered 18 hours ago
FiskelordFiskelord
83
83
$begingroup$
thanks for the help :)
$endgroup$
– John wiliam
4 hours ago
add a comment |
$begingroup$
thanks for the help :)
$endgroup$
– John wiliam
4 hours ago
$begingroup$
thanks for the help :)
$endgroup$
– John wiliam
4 hours ago
$begingroup$
thanks for the help :)
$endgroup$
– John wiliam
4 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.
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$begingroup$
I remember doing a traffic light problem like this in one of my school labs.
$endgroup$
– user4574
yesterday
$begingroup$
@user4574 yep it is one of my assignments
$endgroup$
– John wiliam
yesterday