# exclusive or gate diagram help

Discussion in 'Homework Help' started by cloudhalo, Sep 28, 2013.

1. ### cloudhalo Thread Starter New Member

Sep 28, 2013
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ok I have this hw problem here it is:
They vote on a bylaw amendment for their company. If the bylaw has to be passed 66.67% must be cast for the motion. each shareholder has proportional votes corresponding to their share they own. they are given a box with a YES or NO switch. 3/4 of the votes must be yes for for the bylaw to pass, so create a circuit diagram and a truth table to interpret the voting.
these are the people who have a percentage of he share of the company

now ive done the truth table and basically when the inputs are 7,11,13, 14, 15 the out put is 1 or the bylaw passes. now i know this is correct because at least 3 of the 4 above must hit yes which is above 66.67%.
now i need HELP developing a circuit diagram, ive tried most of the combos with inverter, xor, nor, Nand, but none of it seems to work. also this hw is teaching us how to use exclusive or gates so i know the circuit diagram has to include one, but if u guys find a solution without using the exclusive or gate thats fine.

thanks

2. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
You were fine up to this point, but then you kinda went off the rails.

This is known as "design by happening", namely, try a bunch of random things and hope one of them happens to work.

"if u guys find a solution without using the exclusive or gate thats fine"?

Excuse me, this is YOUR homework.

We will help guide YOU so that YOU can find a solution to YOUR homework.

Start with the terms you (correctly) identified as being in the expression, namely {7,11,13,14,15}. Express each of those in terms of the input variables (A,B,C,D) where A=Adam, B=Bob, C=Charlie, and D=Don.

3. ### cloudhalo Thread Starter New Member

Sep 28, 2013
8
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well I already did the input variables so
7=/abcd
11=a/bcd
13=ab/cd
14=abc/d
15=abcd

the / before the variable means its inverted, but i dont know what to do from here because usually for the solution to the tables output for 1's, u would put like /a+b+c+d or /a*b*c*d
but i dont know whether its gonna be a AND gate, or an OR gate thats going to be part of the sum of the out puts, so its just plain /abcd with no AND or OR gate, i dont know what to do from here??
also ive tried almost every combo i can think of, like i stated before, i am truly stuck
when one logic diagram works for all the outputs, it doesn't work for every combo for the table-thats my issue

4. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
I would recommend not using the / to indicated inversion as it tends to be used by some people before the variable and by others after the variable. In text, the apostrophe is at least a bit less ambiguous.

7 = A'BCD
11 = AB'CD
13 = ABC'D
14 = ABCD'
15 = ABCD

Now, in normal algebra, XY and X*Y are the same thing, right? The same holds true in Boolean algebra. Two variables against each other with no operator indicated means that there is an implied AND operator between them.

I have no idea what you mean by "when one logic diagram works for all the outputs, it doesn't work for every combo for the table". Please post an exampe of a diagram that "works for all the outputs" but that "doesn't work for every combo for the table" so that we can figure out exactly what you are trying to say.

5. ### cloudhalo Thread Starter New Member

Sep 28, 2013
8
0
http://s1293.photobucket.com/user/Cloud_Halo/media/Photoon9-29-13at1042AM_zps18fa84f0.jpg.html?filters[user]=136728827&filters[recent]=1&sort=1&o=0
its basically 1 or gate 1 nor gate then the output for those goes into the exclusive or gate, they work for all the outputs that i stated before, but when putting in the output for 0, or some other number you get 1 as the output instead of 0. this is where im stuck, ive tried almost every combo i can think of, and im not getting anywhere, but i got a question.
for the outputs of
7 = A'BCD
11 = AB'CD
13 = ABC'D
14 = ABCD'
15 = ABCD
i just want to know if we can use boolean algebra to simplify the values for 7, 11, 13 14, 15, cause if u combine them you get AB'CD+ABC'D+ABCD'+A'BCD+ABCD, im using the + because im assuming the solution to the problem includes an xor gate as the final gate so the outputs of the previous gates have to go through it and become a +. if we can simplify which i do not think we can, then its gonna be slightly harder to make a circuit diagram.
back to the original problem, yep im truly stuck, dont know what to do for drawing a logic diagram that actually works for all combos(0-15) of the table

Last edited: Sep 29, 2013
6. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
And what was your reasoning for using this approach? If you were just trying random circuits, then that is "a happening".

This is correct. But what you say next makes no sense.

You use the + (which is nothing more than the logical OR operator) because you want the output to be True if ANY of those five terms is True. It has absolutely nothing to do with any assumptions about XOR gates.

What is the Boolean logic expression for A XOR B?

Do you see that pattern anywhere within the combined expression above?

7. ### cloudhalo Thread Starter New Member

Sep 28, 2013
8
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as for pattern i notice u can simplify 14 and 15, or abcd' and abcd, . im thinking 14 and 15 simplified is abc right, cause d+d'= 1?
then 11 and 13 can be ad cause c'+c'=1 and b'+b=1

that i think can become ad'??? cause c+c'=1 and b+b'=1, then a+a=a, so all u have left is ad'. i really dont know if this is right or if im following the bool rules correctly. please help meeee

8. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
It would be helpful if you developed your work in a systematic way. For instance

F = AB'CD+ABC'D+ABCD'+A'BCD+ABCD
F = AB'CD+ABC'D+ABCD'+(A'+A)BCD
F = AB'CD+ABC'D+ABCD'+BCD

Now, you are claiming that 11 and 13 can be simplified. Those two terms are

ABC'D+ABCD'

What, specifically, are you saying these two terms can be simplified to?

Going back to the original expression:

F = AB'CD+ABC'D+ABCD'+A'BCD+ABCD

Let's see if we can follow your first simplification a bit further. You know that X+X=X, but that works the other way and X=X+X, meaning that we can repeat terms without changing the result. So let's repeat ABCD three times and group one with each of the other four terms.

F = (AB'CD+ABCD) + (ABC'D+ABCD) + (ABCD'+ABCD) + (A'BCD+ABCD)

What does this simplify to?

In words, how could you describe the logic that this version represents (in terms of the logic that the problem is trying to describe)?

9. ### Alec_t AAC Fanatic!

Sep 17, 2013
7,023
1,453
Er, which is it?

10. ### WBahn Moderator

Mar 31, 2012
20,064
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If you look at the numbers, you'll see that in this particular case they are equivalent.

11. ### cloudhalo Thread Starter New Member

Sep 28, 2013
8
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at alec- its 3/4 cause if 2 just said yes it wouldnt go throught, but honestly i am stuck idk what to do

12. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
Let's try this again.

F = (AB'CD+ABCD) + (ABC'D+ABCD) + (ABCD'+ABCD) + (A'BCD+ABCD)

What does this simplify to?

13. ### cloudhalo Thread Starter New Member

Sep 28, 2013
8
0
ahh
now for the one you have up there it would simplify to:
F= (ACD)+(ABC)+(BCD)
F=ABCD+BCD
F=ABCD-final result
i understand why you simplified each term by ABCD because thats what they all have in common, but can you really do that?(Ive never learned that} a
here is my logic, if in the original expression F=A'BCD+AB'CD+ABC'D+ABCD'+ABCD, you simplify with ABCD with only one of the terms before it(it could be anyone) the two terms including ABCD will basically combine into one simplified term, so how can you use ABCD again with another term or all of them(which you did) at the same time to simplify it when it was already used with another term and simplified? or rather how could you simplifiy every term with ABCD? what theorm or law allowed you to do this?
is it like factoring, where you take out the all like variables each term has, or something?

Last edited: Sep 30, 2013
14. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
Let's examine 11 and 13 again

AB'CD+ABC'D

You are saying that this simplifies to just AD.

Does it? Let's say that A=True and B=True. Does that guarntee that AB'CD+ABC'D is True? What if B=False and C=False? For that matter, what if A, B, C, and D are all True?

Let's factor out the common factors from both:

What does the expression in parentheses simplify to?

I don't know how you are getting any of this.

We have

F = (AB'CD+ABCD) + (ABC'D+ABCD) + (ABCD'+ABCD) + (A'BCD+ABCD)

The first expression in parens becomes ACD(B'+B) which reduces to ACD. The third parens reduces to ABC and the fourth to BCD. What happened to the expression in the second set of parens?

How did you go from "(ACD)+(ABC)+(BCD)" to "ABCD+BCD"?

How did you go from "ABCD+BCD" to "ABCD"?

Ask if this last one makes sense. The final result requires that all four inputs be True. But the expression just before it is True as long as B, C, and D are True without regard to whether A is True.

This last one should go

ABCD+BCD = (A+1)BCD = BCD

But this is not the correct final answer because the work prior has problems.

I explained this in Post #8, "Let's see if we can follow your first simplification a bit further. You know that X+X=X, but that works the other way and X=X+X, meaning that we can repeat terms without changing the result. So let's repeat ABCD three times and group one with each of the other four terms."

If we have

F = AB + CD + CD

We can reduce this using the property that CD + CD = CD.

But we can reverse that and use the property that AB = AB + AB to write

F = AB + AB + CD + CD

Hence, we can take

F = AB'CD + ABC'D + ABCD' + A'BCD + ABCD

And use the fact that

ABCD = ABCD + ABCD + ABCD + ABCD to do the following:

F = AB'CD + ABC'D + ABCD' + A'BCD + (ABCD)

F = AB'CD + ABC'D + ABCD' + A'BCD + (ABCD + ABCD + ABCD + ABCD)

F = AB'CD + ABC'D + ABCD' + A'BCD + ABCD + ABCD + ABCD + ABCD

And then just rearrange terms to get

F = (AB'CD+ABCD) + (ABC'D+ABCD) + (ABCD'+ABCD) + (A'BCD+ABCD)

QED.

So use this last one as your starting point again and be a bit more careful.

I think you would be well served by explicitly showing each line of your work. Don't do anything in your head -- you haven't got this stuff down well enough to do that reliably.

15. ### cloudhalo Thread Starter New Member

Sep 28, 2013
8
0
AD(B'C+BC') simplifies to AD(1+1) cause B'+B=1 an same for C, idk if i should leave it as 1+1 or add them???
(ACD)+(ABC)+(BCD) i basically factor out AC
so AC(D+B)+BCD-final result
then hmmmmm i guess thats it right? i mean u could have factor out only c instead of A to get C(AD)+(AB)+BD, but how do i know which version is right?
and going by this my circuit diagram will have 2 or 3 AND gates with an OR gate.
now how did u come to the conclusion of "Ask if this last one makes sense. The final result requires that all four inputs be True. But the expression just before it is True as long as B, C, and D are True without regard to whether A is True."?
i thought i stated that at least 3 inputs must be true for the bylaw to pass, which means the final result doesn't require all 4 to be true. the final result will output 1 if all are true or 3 are true. please clear up my confusion on that part, an how u came to that conclusion?

Last edited: Oct 1, 2013
16. ### WBahn Moderator

Mar 31, 2012
20,064
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I'll ask again: Does it? Let's say that A=True and B=True. Does that guarantee that AB'CD+ABC'D is True? What if B=False and C=False? For that matter, what if A, B, C, and D are all True?

If all else fails, write a truth table for (B'C+BC').

Before you do that, ask if this is possibly correct. What if everyone except C except C votes in favor? The measure shouild pass, right? But does your expression above reflect that?

But your final result in the post where I made that comment was

Just because you WANT the output to be True if any 3 people vote for it doesn't mean that your solution IS True if any 3 people vote for it. In YOUR solution, if ANY person votes False, the result is False. You need to be able to look at what your solution DOES without being influenced by what it is SUPPOSED to do.

17. ### cloudhalo Thread Starter New Member

Sep 28, 2013
8
0
(ACD)+(ABC)+(BCD)
that cannot be simplified anymore, cause i used a k-map to get that. tanks for the help, now i can make the circuit diagram using a 4 input nand gate

18. ### WBahn Moderator

Mar 31, 2012
20,064
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Don't thank me, because you won't listen to me and so have a wrong result.

I ask you again: Ask if this is possibly correct. What if everyone except C except C votes in favor? The measure shouild pass, right? But does your expression above reflect that?