1. The problem statement, all variables and given/known data
We have been given a task to develop a circuit which displays the square of a binary number on a 3 x 7 seq displays.

I have already gone through and done up the Karnaugh Maps for the task and have identified the minterms. However, I believe that these can still be simplified even more.

2. Relevant equations
Karnaugh Maps Output = A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'

3. The attempt at a solution
A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'
Factor out A' and D from minterms 1 and 2
A'D(b' + b) (c + c')
= A'D + A'BC'D + ABCD + B'C'D' + AB'C'
Factor out B and D from minterms 2 and 3
bd(a' + a)(c' + c)
= A'D + BD + B'C'D' + AB'C'

This is where I am getting stuck. Is it possible to further simplify the equation or is this the final solution???

 

Well, i think you didnt get the Karnaugh map correct because you simplified it again using the boolean algebra. This is not correct because if you do solve correct a Karnaugh map you get the simplest solution. However, i can't see a simpler solution to your final answer.

 

= A'D + BD + B'C'D' + AB'C'

Common factor simplification will reduce the gate count:

A'D + BD = D(A' + B)

This reduces from 2x 2I/P AND-gates and 1x 2I/P OR-gate to 1x 2I/P AND-gate and 1x 2I/P OR-gate

B'C'D' + AB'C' = B'C'(A + D')

This reduces from 2x 3I/P AND-gates and 1x 2I/P OR-gate to 1x 3I/P AND-gate and 1x 2I/P OR-gate.

So the simplified expression would be: D(A' + B) + B'C'(A + D')

Dave

 

Thanks alot for the help.

Also, I understand that the Karnaugh Maps usually produce the simplest forms but I saw that the equation could be further simplified because of the common factors.

So Am I right in thinking that the equation that Dave gave is correct or should I not try to further minimize the equation???

I am confused

 

Take a look at the material on boolean algebra in the AAC ebook.

hgmjr

 

Thanks alot for the help.

Also, I understand that the Karnaugh Maps usually produce the simplest forms but I saw that the equation could be further simplified because of the common factors.

So Am I right in thinking that the equation that Dave gave is correct or should I not try to further minimize the equation???

I am confused

You may be able to DeMorganise the expression to less gates, however this may not translate to a more simplified expression in terms of transistors or terms used. You would need to play around with DeMorgans Theorems to see if you can actually simplify the expression further - it isn't obvious from the above expression if it would work.

Your current expression utilises only one of each term and its compliment (actually there is no C); I would commend that this expression is fully simplified (note there are probably several simplifications of the original expression which would be equally simplified).

Dave

 

Dave, so your saying that my original equation should not be further simplified???

Also, thanks alot for the help guys.

I think you are correct in saying that the original form is already simplified and I am only complicating everything by overthinking

Once again thanks alot

If I have some more questions I will be sure to ask

 

Dave, so your saying that my original equation should not be further simplified???

Also, thanks alot for the help guys.

I think you are correct in saying that the original form is already simplified and I am only complicating everything by overthinking

Once again thanks alot

I cannot see that it can be logically simplified further. Like I said previously, you may be able to DeMorganise the expression to remove a term and its compliment, however this is not obvious nor would it yield a more simplified expression depending on how you make the judgement of simplification.

I would commend that your expression is in its simplified form.

If I have some more questions I will be sure to ask

No problems, that is why this site is here. Be sure to check the e-book and worksheets to help you with your studies.

Dave

 

Pariah,

Sorry I did respond to this sooner, but I just got back from vacation.

2. Relevant equations
Karnaugh Maps Output = A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'

I did the simplification using the Quine-McCluskey tabulation method http://www.cs.ualberta.ca/~amaral/courses/329/webslides/Topic5-QuineMcCluskey/sld001.htm , and verified that you cannot simplify the above expression any more. You should be able to verify that by using a K-map.

3. The attempt at a solution
A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'
Factor out A' and D from minterms 1 and 2
A'D(b' + b) (c + c')
= A'D + A'BC'D + ABCD + B'C'D' + AB'C'

Your Boolean algebra is in error. There is no way that Minterms(3,5) are going to reduce to a two variable term (A'D).

This is where I am getting stuck. Is it possible to further simplify the equation or is this the final solution???

Neither, you already had the simplest reduction.

See the last entry of this thread. http://forum.allaboutcircuits.com/showthread.php?t=12279&highlight=quine-mccluskey

Ratch