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Truth table with logical gates for a traffic light

Writing down logic of a circuitSR 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 OptimizationCreate a circuit with max 6 NAND gates

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

$begingroup$

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

share|improve this question

edited Apr 19 at 15:45

John wiliam

asked Apr 15 at 11:46

John wiliamJohn wiliam

1855 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I remember doing a traffic light problem like this in one of my school labs.
  $endgroup$
  – user4574
  Apr 15 at 13:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user4574 yep it is one of my assignments
  $endgroup$
  – John wiliam
  Apr 15 at 18:14
  

  
  
  
  

add a comment
 | 

3

$begingroup$

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

share|improve this question

edited Apr 19 at 15:45

John wiliam

asked Apr 15 at 11:46

John wiliamJohn wiliam

1855 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I remember doing a traffic light problem like this in one of my school labs.
  $endgroup$
  – user4574
  Apr 15 at 13:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user4574 yep it is one of my assignments
  $endgroup$
  – John wiliam
  Apr 15 at 18:14
  

  
  
  
  

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

share|improve this question

edited Apr 19 at 15:45

John wiliam

asked Apr 15 at 11:46

John wiliamJohn wiliam

1855 bronze badges

$endgroup$

share|improve this question

edited Apr 19 at 15:45

John wiliam

asked Apr 15 at 11:46

John wiliamJohn wiliam

1855 bronze badges

share|improve this question

edited Apr 19 at 15:45

John wiliam

asked Apr 15 at 11:46

John wiliamJohn wiliam

1855 bronze badges

asked Apr 15 at 11:46

John wiliamJohn wiliam

1855 bronze badges

asked Apr 15 at 11:46

John wiliamJohn wiliam

1855 bronze badges

4 Answers
4

active

oldest

votes

3

$begingroup$

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.

share|improve this answer

edited Apr 15 at 12:12

answered Apr 15 at 11:53

Andy akaAndy aka

254k1111 gold badges196196 silver badges452452 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 15 at 11:59
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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:

share|improve this answer

answered Apr 15 at 11:57

RenanRenan

4,37122 gold badges2222 silver badges4444 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the picture :)
  $endgroup$
  – John wiliam
  Apr 15 at 12:00
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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.

share|improve this answer

answered Apr 15 at 14:18

Tanner SwettTanner Swett

45933 silver badges88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the advice :)
  $endgroup$
  – John wiliam
  Apr 15 at 18:13
  

  
  
  
  

add a comment
 | 

1

$begingroup$

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 :)

share|improve this answer

answered Apr 16 at 6:00

FiskelordFiskelord

3866 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the help :)
  $endgroup$
  – John wiliam
  Apr 16 at 19:13
  

  
  
  
  

add a comment
 | 

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3

$begingroup$

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.

share|improve this answer

edited Apr 15 at 12:12

answered Apr 15 at 11:53

Andy akaAndy aka

254k1111 gold badges196196 silver badges452452 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 15 at 11:59
  
  

  
  
  
  

add a comment
 | 

3

$begingroup$

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.

share|improve this answer

edited Apr 15 at 12:12

answered Apr 15 at 11:53

Andy akaAndy aka

254k1111 gold badges196196 silver badges452452 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 15 at 11:59
  
  

  
  
  
  

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.

share|improve this answer

edited Apr 15 at 12:12

answered Apr 15 at 11:53

Andy akaAndy aka

254k1111 gold badges196196 silver badges452452 bronze badges

$endgroup$

share|improve this answer

edited Apr 15 at 12:12

answered Apr 15 at 11:53

Andy akaAndy aka

254k1111 gold badges196196 silver badges452452 bronze badges

answered Apr 15 at 11:53

Andy akaAndy aka

254k1111 gold badges196196 silver badges452452 bronze badges

answered Apr 15 at 11:53

Andy akaAndy aka

254k1111 gold badges196196 silver badges452452 bronze badges

2

$begingroup$

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:

share|improve this answer

answered Apr 15 at 11:57

RenanRenan

4,37122 gold badges2222 silver badges4444 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the picture :)
  $endgroup$
  – John wiliam
  Apr 15 at 12:00
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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:

share|improve this answer

answered Apr 15 at 11:57

RenanRenan

4,37122 gold badges2222 silver badges4444 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the picture :)
  $endgroup$
  – John wiliam
  Apr 15 at 12:00
  

  
  
  
  

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:

share|improve this answer

answered Apr 15 at 11:57

RenanRenan

4,37122 gold badges2222 silver badges4444 bronze badges

$endgroup$

share|improve this answer

answered Apr 15 at 11:57

RenanRenan

4,37122 gold badges2222 silver badges4444 bronze badges

answered Apr 15 at 11:57

RenanRenan

4,37122 gold badges2222 silver badges4444 bronze badges

answered Apr 15 at 11:57

RenanRenan

4,37122 gold badges2222 silver badges4444 bronze badges

2

$begingroup$

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.

share|improve this answer

answered Apr 15 at 14:18

Tanner SwettTanner Swett

45933 silver badges88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the advice :)
  $endgroup$
  – John wiliam
  Apr 15 at 18:13
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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.

share|improve this answer

answered Apr 15 at 14:18

Tanner SwettTanner Swett

45933 silver badges88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the advice :)
  $endgroup$
  – John wiliam
  Apr 15 at 18:13
  

  
  
  
  

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.

share|improve this answer

answered Apr 15 at 14:18

Tanner SwettTanner Swett

45933 silver badges88 bronze badges

$endgroup$

share|improve this answer

answered Apr 15 at 14:18

Tanner SwettTanner Swett

45933 silver badges88 bronze badges

answered Apr 15 at 14:18

Tanner SwettTanner Swett

45933 silver badges88 bronze badges

answered Apr 15 at 14:18

Tanner SwettTanner Swett

45933 silver badges88 bronze badges

1

$begingroup$

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 :)

share|improve this answer

answered Apr 16 at 6:00

FiskelordFiskelord

3866 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the help :)
  $endgroup$
  – John wiliam
  Apr 16 at 19:13
  

  
  
  
  

add a comment
 | 

1

$begingroup$

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 :)

share|improve this answer

answered Apr 16 at 6:00

FiskelordFiskelord

3866 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the help :)
  $endgroup$
  – John wiliam
  Apr 16 at 19:13
  

  
  
  
  

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 :)

share|improve this answer

answered Apr 16 at 6:00

FiskelordFiskelord

3866 bronze badges

$endgroup$

share|improve this answer

answered Apr 16 at 6:00

FiskelordFiskelord

3866 bronze badges

answered Apr 16 at 6:00

FiskelordFiskelord

3866 bronze badges

answered Apr 16 at 6:00

FiskelordFiskelord

3866 bronze badges