logic gates- sum of products

Thread Starter

fan_boy17

Joined Apr 17, 2012
12
cant seem to do this question. Don't understand what the sigma means in this


2 Find the sum of products expression for the following functions and
implement using NAND gates.

F(A,B,C) = Ʃ0,1,3,4,6,7
F(A,B,C,D) = Ʃ 0,1,2,3,7,8,9,11,12,15
F(P,Q,R,S) = Ʃ 0,4,5,6,7,8,9,10
 

MrChips

Joined Oct 2, 2009
30,714
Sigma means sum, i.e. OR
The numbers refer to the minterms
For F(A,B,C), this has three bits. Hence 8 minterms, 0 to 7, or 000 to 111 in binary
minterm 0 = A' B' C'
minterm 1 = A' B' C
etc
 

Thread Starter

fan_boy17

Joined Apr 17, 2012
12
i used karnaugh maps to do it and i get an answer of (not B. not c) + (A not. C) + BC

is that correct as the sum of products expression?
 

WBahn

Joined Mar 31, 2012
29,979
i used karnaugh maps to do it and i get an answer of (not B. not c) + (A not. C) + BC

is that correct as the sum of products expression?
You need to learn how to check your work. Here's how you can do it in this case:

not B . not C => ABC = x00 = 000 (0) or 100 (4)

A . not C => ABC = 1x0 = 100 (4) or 110 (6)

B . C => ABC = x11 = 011 (3) or 111 (7)

So you have implemented:

f(A,B,C) = SUM{0, 3, 4, 6, 7}

It would appear that case 1 is not covered. So let's check that specific case:

ABC = 1 = 001

not B . not C => F
A . not C => F
B . C => F

Sure enough, it isn't covered.

If you start checking your own work every opportunity you can, you will lose a LOT fewer points now and quite possibly avoid killing someone down the road.
 

WBahn

Joined Mar 31, 2012
29,979
i used karnaugh maps to do it and i get an answer of (not B. not c) + (A not. C) + BC
I just realized an inconsistency in the above notation, If the first term, it is written

not B. something

and in the second term it is written

A not. something

I had interpretted the expression as:

(B')(C') + (A)(C') + (B)(C)

But I don't know that this is what fan_boy17 meant. He might have meant


(B')(C') + (A')(C) + (B)(C)

Now, this doesn't cover all the terms, either.

There are two sets of two-input minterms that solve the static logic problem, but you must use all six together if you want to avoid static timing hazards. Depending on the course level you are taking, you may or may not have been introduced the concept of static and dynamic timing hazards and how to deal with them.
 

MrChips

Joined Oct 2, 2009
30,714
Good point.
I naturally assumed that the OP made a typo and meant to write A . not C = A . C'
In either case, there will still be a missing term.

A C' would include minterm6, minterm1 is missing.

A' C includes minterm1, minterm6 is missing.
 
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