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Help with Boolean Algebra Simplification

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Emile_Uk

New Member
Hi,

If anyone can help me simplify the following two boolean expressions whilst listing the laws used at each stage then that would be hugely appreciated!

The two expressions are:

([P]*Q*[X]*Y)+([P]*Q*X*[Y])+(P*[Q]*X*[Y])+(P*[Q]*X*Y)+(P*Q*[X]*Y)+(P*Q*X*Y)

and

(P*[Q]*[X]*Y)+(P*Q*[X]*Y)+(P*Q*X*[Y])

Where * is AND and [] is an inverted input.

Any help would be greatly appreciated!

Cheers,

Emile
 

kinjalgp

Active Member
Solution...
1)
|P*Q*|X*Y + |P*Q*X*|Y + P*|Q*X*|Y + P*|Q*X*Y + P*Q*|X*Y + P*Q*X*Y
|P*Q*|X*Y + |P*Q*X*|Y + P*|Q*X*(|Y + Y) + P*Q*Y *(|X + X) [TAKING COMMON FACTORS]
|P*Q*|X*Y + |P*Q*X*|Y + P*|Q*X*(1) + P*Q*Y*(1) [A + |A = 1]
|P*Q*|X*Y + |P*Q*X*|Y + P*|Q*X + P*Q*Y [A*1 = A]

2)
P*|Q*|X*Y + P*Q*|X*Y + P*Q*X*|Y
P*|X*Y*(|Q + Q) + P*Q*X*|Y [TAKING COMMON FACTORS]
P*|X*Y*(1) + P*Q*X*|Y [A + |A = 1]
P*|X*Y + P*Q*X*|Y [A*1 = A]
 

kinjalgp

Active Member
My 2nd solution matches with that of yours but can you explain how you got the first one? I studied boolean algebra long back :)
 

Emile_Uk

New Member
Me Again!

Bandersnatch, your first simplification is out by one on the truth table i'm afraid, don't know what went wrong. I have been reading up on boolean algebra though and have managed to conjure up the following which gives the same output:

(P'*Q*(X@Y))+(P*Q'*X)+(P*Q*Y)

@ is an exclusive OR gate.

By the way, thanks for all the help you guys have given me, its really helped with my understanding of the subject.

Cheers,

Emile
 

Emile_Uk

New Member
Correction

Hello again,

I worked out where you went wrong Bandersnatch. I take it you used a Karnaugh Map to further simplify the expression.

The output of the map should be as follows:

(P'*Q*X*Y')+(Q*X'*Y)+(P*Q'*X)+(P*Q*Y)

You just missed out the negation on the first Y.

Thanks again for all of your help everyone, its much appreciated.

Cheers,

Emile
 

Agent 009

New Member
You could, instead of counting on error-prone method, use a software for the purpose of getting the correct answer, and then go back to your initial problem. LogicAid is such a program... But probably all programs use the Quine-McCluskey method, so I don't think, if you ever want to have a software showing up all steps, you'll ever find a software that could do it...
 

crust

Member
Off Topic, but if it is the same logicAid I am thinking of, I took the class from the guy that wrote the book and the software (Roth) while an undergrad at UT Austin, I received the software on a 5.25 inch floppy at the time for an old 486.
 

Agent 009

New Member
Hehe :lol: good one. But this version is a lot more 'sofisticated'. It's first a windows version, it can do quite good amount of operations like kmaps, minterms/maxterms expansion, boolean algebra simplification, and many other I dunno what. Well I think it doesn't fit on a 3.5" Floppy (1.44MB) or other 'small' media support, cause it was shipped with the textbook on a CD...
 

kinjalgp

Active Member
Anyone interested can download LogicAid from this location **broken link removed**
 
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