# Digital Electronics help

Discussion in 'Homework Help' started by aniskazi, May 7, 2014.

1. ### aniskazi Thread Starter Member

May 3, 2014
44
0
Hello one more quesstion came up
here we go :

A candy bar manufacturing company requires a digital system that can take ingredients from 3 different containers in a proper proportion as shown in the figure below, and mix any 2 of them to make the final product. The mixing system has two outputs for indicating an invalid/ not allowable condition by a warning buzzer and normal condition by normal operation. The three ingredients are supplied by 3 different logic circuits. The circuits have following conditions:

The logic circuit Ia: should output HIGH when the ingredient A's quantity=8

The logic circuit Ib: should output HIGH when the ingredient B's quantity=4
The logic circuit Ic: should output HIGH when the ingredient C's quantity=2
The outputs of these three logic circuits are connected with mixing system (logic circuit M). The mixing system should combine any two of the ingredients. Therefore, if exactly 3 inputs of the mixing circuit are high the warning buzzer should turn on & normal operation of. otherwise normal operation indicator should remain on & warning oFf.
http://postimg.org/image/elu9451st/

This is the cirucit that came with the question

Truth table that I have made:

A B C Output
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 warning buzzer

now tell me what should I do ?

Last edited: May 9, 2014
2. ### WBahn Moderator

Mar 31, 2012
20,048
5,637
Your problem description makes no sense. Please reread it carefully and make sure that it is accurate and clear.

3. ### aniskazi Thread Starter Member

May 3, 2014
44
0
Well I want to know that my eq: that I have made is correct or not and than the circuit that I have made has only one output but the qs says that the circuit should have two outputs one about the warning buzzer and the second about the normal condition so I'm confused how can i get two outpust in my circuit.

4. ### WBahn Moderator

Mar 31, 2012
20,048
5,637
Getting more than one output is very straight forward -- just treat it as two problems. One problem is to generate the signal for the warning buzzer and the second is to generate the signal for the normal condition. But keep in mind the relationship between the two signals because you can probably generate one from the other in a pretty easy way.

I can't tell if your equation is right or not because your problem description is self-contradictory. It makes no sense.

5. ### aniskazi Thread Starter Member

May 3, 2014
44
0
Okay can you just tell me that my equation that I have made is correct or not?

6. ### tshuck Well-Known Member

Oct 18, 2012
3,527
675
It describes the relationship seen in the truth table regarding the 'output' variable, but remember, you can verify this yourself by expanding the equation into a truth table and comparing the reconstructed one to the other one you came up with.

With that said, I'm not sure what the goal is in including the mixing description - it seems rather unnecessary and convoluted, but provided your interpretation of it is correct, your table makes sense.

aniskazi likes this.
7. ### WBahn Moderator

Mar 31, 2012
20,048
5,637
What did I just say? I cannot tell if your equation is correct because the problem description upon which it is based is self-contradictory.

The problem statement says weakly states that the "normal" operation is if exactly two of the ingredients get mixed but then later it explicitly says that the warning buzzer should sound only if exactly three of the ingredients have a HI output and that it is normal otherwise. That means that "normal" includes the case when no ingredient is HI and also when only one ingredient is HI. Your truth table is consistent with the weakly stated description of normal operation but not with the explicitly stated description.

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8. ### aniskazi Thread Starter Member

May 3, 2014
44
0
Hello can you please check the question again I have changed it I have uploaded a circuit diagram that was with the question I thought it was no use to upload it but I was wrong please check it out

9. ### aniskazi Thread Starter Member

May 3, 2014
44
0

Hello can you please check the question again I have changed it I have uploaded a circuit diagram that was with the question I thought it was no use to upload it but I was wrong please check it out

10. ### tshuck Well-Known Member

Oct 18, 2012
3,527
675
Once again, editing the original post makes it hard to follow this thread. Please post any new information as a new post in the thread (don't modify the original post).

The diagram helps to clarify things (a little, anyway). Assuming the mixing container can not fit more than 14 units of ingredients, having all three ingredients runs the risk of spilling ingredients and requires some sort of warning to the operator. If there is anything from 0 to 12 units of ingredients, the mixer will mix (e.g. spin the whisk), expecting ingredients to be added/be there already. The ingredients seem to be dispensed by a machine that injects exactly the amount of ingredients that will cause its respective sensor to output a 1, making less than 14 units of ingredients for the mixing machine corresponding to, at most, two ingredients being mixed.

11. ### aniskazi Thread Starter Member

May 3, 2014
44
0
Sorry about the editing...
and well the total no: of inputs when I'll be making my truth table will be 8 or will it be 14 than?

12. ### tshuck Well-Known Member

Oct 18, 2012
3,527
675
It seems as though you are to assume the level sensors are already built and you are interfacing after this fact. If this is true (since wasn't given as part of the exercise), you are only responsible for the mixer controller as you've done thus far.

13. ### aniskazi Thread Starter Member

May 3, 2014
44
0
Okay so according to that my eq will be :

Normal buzzer:A' BC + AB' C + A B C'

Warning Buzzer : ABC

correct ?

14. ### tshuck Well-Known Member

Oct 18, 2012
3,527
675
So is the mixer in normal operation when there aren't any ingredients?

15. ### aniskazi Thread Starter Member

May 3, 2014
44
0
Well the question says "The mixing system should combine any two of the ingredients"

so if no ingredients are mixed nothing will happen no warning buzzer would go off nor the normal operation would be on that's just my point of view so that's why in my truth table when there are two HIGHS I assume it as normal and whenever there is only 1 HIGH I assume it as 0 and when all are HIGH than warning buzzer correct?

16. ### WBahn Moderator

Mar 31, 2012
20,048
5,637
Instead of applying your point of view, how about reading what the specs explicitly state: "Therefore, if exactly 3 inputs of the mixing circuit are high the warning buzzer should turn on & normal operation of. otherwise normal operation indicator should remain on & warning oFf."

17. ### WBahn Moderator

Mar 31, 2012
20,048
5,637
According to the specs given, yes.

The problem is clearly contrived to make the logic easy, even though it makes little physical sense. I guess we can contrive the reasoning and assume that if 0 or 1 ingredients are ready that the mixing operation won't happen and that this is normal -- the mixer is waiting for ingredients to become ready. But if all three are ready the mixer is too stupid to only take two of them and that is when the problems arise.

18. ### tshuck Well-Known Member

Oct 18, 2012
3,527
675
It seems that the wording implies it is in the normal state whenever the warning is not active...

19. ### aniskazi Thread Starter Member

May 3, 2014
44
0
So according to what you are saying the truth table will be like this

A B C NOR W
0 0 0 1 0
0 0 1 1 0
0 1 0 1 0
0 1 1 1 0
1 0 0 1 0
1 0 1 1 0
1 1 0 1 0
1 1 1 0 1

don't you think it will be too complicated like this.

20. ### WBahn Moderator

Mar 31, 2012
20,048
5,637
Too complicated????

It's become trivial!