Find the minterm expansion of f(A, B, C, D) = (A + B + D')(A' + C)(C + D) Please tell me what rules you're applying? I keep getting stuck.

- anonymous

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- anonymous

Working it out.

- anonymous

Do you know where to start?

- anonymous

I tried to take (A' + C)(C + D) and convert it into x+yz form, getting (A'D + C). So altogether it's (A + B + D') (A'D + C). But from there, I'm not sure what to do
My my problem is I don't quite know how to have a good strategy for approaching these and I can't see where to apply formulas.

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- anonymous

I'm assuming I need to put it into SOP form.

- anonymous

Well, they're throwing you a bone here. This is in "product of sum" (POS) form. POS form expresses MAXTERMS. The minterms are everything that is NOT a maxterm. This is the strategy I would use. But here you need to have all variables.

- anonymous

I wouldn't put it into SOP form at all, actually. I'd get the maxterms, then just ...the minterms are everything the maxterms are not.

- anonymous

So then, would I convert these into binary form, find the numbers for the minterm and then...?

- anonymous

Well, first the products are incomplete. All three products need to have A, B, C, and D in some form.
So you can use the fact that XX' = 0 to add them in. Because when it comes out, it's as though they were never there.

- anonymous

So for example, with the first product (A + B + D'), you're missing "C".
So: (A + B + CC' + D')

- anonymous

Now distribute (A + B + C + D')(A + B+ C' + D')
See?

- anonymous

Oh, yeah, I forgot that step. But I mean after that, would I just find the actual numbers for the minterm by using the numbers that aren't there for the maxterm?

- anonymous

Yea, cancel out redundant maxterms, and express them as a decimal number. Those are your maxterms.

- anonymous

Then everything not present in the maxterms is a minterm.

- anonymous

Wouldn't I still have to simplify them algebraically? The numbers I mean. When I convert them back into variables?

- anonymous

Well, if you look at the first product only: (A + B + C + D')(A + B+ C' + D')
You have 1110 and 1100, this is 14 and 12.
So if there were NO other products your max terms would be:
\[\Pi M(12, 14)\]
Minterms would then be everything else.

- anonymous

Of course, you have two other products that are going to distribute into more products and some of them will cancel, but I don't see any crazy boolean algebra that you have to do here at all.

- anonymous

Ah okay. I'll work on it then.

- anonymous

And actually, that's not correct in my example. With maxterms, 1 is actually 0, so you have:
0001 and 0011, giving 1 and 3

- anonymous

So minterms would be everything 0 - 15 other than 1 and 3

- anonymous

Yep.

- anonymous

I see. I'll go ahead and jump on it. Thanks for the help. :)

- anonymous

No prob, let me know what you get there.

- anonymous

So far, I multiplied (A + B + CC' + D')(A' + BB' + C + DD')(AA' + BB' + C + D)
And that resulted in
AC + ACD + A'BC + A'BD + BC + BCD + A'CD' + CD'
and now I'm looking for consensus terms or something to get rid of. I want to use BC + BCD to get rid of BC, but I'm not sure if that's legal?

- anonymous

Overcomplicating.

- anonymous

Throw that out the window.

- anonymous

Lol okay

- anonymous

First term: (A + B + D') = (A + B + CC' + D') = (A + B + C + D')(A + B + C' + D')
=0001 0011
Second term: (A' + C) = (A' + BB' + C + DD') = (A' + B + C + D)(A' + B + C + D')(A' + B' + C + D)(A' + B' + C + D')
=1000 1001 1100 1101
Third Term: (C + D) = (AA' + BB' + C + D) = (A + B + C + D)(A + B' + C + D)(A' + B + C +D)(A' + B' + C + D)
=0000 0100 1000 1100
Every single one of those binary numbers there is a maxterm. Some of them repeat, and you can get rid of the redundant ones.

- anonymous

Looking at the first term, for example, do you see how the CC' distributes? We have to account for both the C and the C'.
That gives what we ORIGINALLY had (A + B + D'), but now we have two variants. One with a C and one with a C'.
Thats why we have (A + B + C + D')(A + B + C' + D')

- anonymous

Yeah, I see. I didn't need to distribute the way I did.

- anonymous

Nope, not at all, because its still in POS form. Distributing just makes it needlessly complicated. Finding maxterms and minterms is a first step in SIMPLIFYING a switching circuit to its simplest possible form.

- anonymous

Using boolean logic is brute force and may not result in a completely simplified circuit. This method is a first step in two methods that will ALWAYS give you the simplest switching circuit.

- anonymous

Is it ever the case when SOP form is considered simplified? I have no idea why I interpret "simplify" to mean "put it either in the smallest POS or SOP form"

- anonymous

I mean. Switch those around.

- anonymous

Well, ultimately these turn into a bunch of and/or gates and inverters and so forth. The simplest expression is the one that uses the least amount of them really. So you have the smallest number of or's, the smallest number of and's....and so forth.

- anonymous

Oh, I see.

- anonymous

So yes, either POS form can be simplified, or SOP can be. Doesn't matter. In this case there are an equal number of minterms to maxterms. But this does not mean much......yet

- anonymous

One more thing. Just in general, when simplifying boolean algebra overall, do you have any tips for how to approach problems that ask for simplifying something into POS or SOP respectively? It's just really hard for me to see, for example, when to call something X and where to group something into maybe Z or Z' or something. So I'm practicing, but... Yeah. Lol.

- anonymous

I start out to simply and end up complicating. Or start out trying to do POS and end up doing SOP. Stuff like that. I try to stick to the simplification theorems for each respectively but I notice there's times when using another method for half the problem is advantageous.

- anonymous

Well, the whole thing is when dealing w/ minterms and maxterms, the final expressions must ALL have ALL elements of the function.
To introduce the missing elements, when using maxterms (POS) you have to introduce them like they were never there. They have to be transparent. Thats why with maxterms, we use XX'. Because XX' = 0. So or'ing an expression with 0 has no effect. See?
When using minterms, we use X + X'. Because X + X' = 1. So when you "and" an expression with 1, it has no effect. See?

- anonymous

Right, I see.

- anonymous

So an SOP example might be, find the minterms of: F(A, B, C) = A'B' + AC
Well, there's two expressions there. The first one is missing C, the second is missing B.
So we say:
A'B'(C + C') + AC(B + B')
We get
A'B'C + A'B'C' + ABC + AB'C = 001 + 000 + 111 + 101
So the minterms are 0, 1, 5, 7

- anonymous

And there's no need to attempt anything like eliminating consensus terms, etc. in these sorts of problems.

- anonymous

Just use the number values.

- anonymous

No, I mean...in this case you're already given something in POS form, which is why I say they were throwing you a bone. If you're given something that's a mix, you may have to use some boolean trickery to get it into either form.

- anonymous

But once its in POS or SOP form, stop trying to simplify and start using what was used here.

- anonymous

Okay, that makes sense.

- anonymous

Very good. Digital circuits, I presume?

- anonymous

I'm not sure what the formal name of the class is. It's CpE 100 here. I'm a CS major, so this is pretty much my last class in this, lol.

- anonymous

Ah, then Digital Logic or something like that.

- anonymous

Computer Logic Design I
There we go.

- anonymous

Sounds about right. Good luck and enjoy :]

- anonymous

Lol, thanks. You helped a lot! :)

- anonymous

The maxterms are 0, 1, 3, 4, 8, 9, 12, 13
The minterms are 2, 5, 6, 7, 10, 11, 14, 15
btw ;]

- anonymous

Haha, thank you. :P

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