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Suppose I have a function defined as follows, such that it is expressed in POS form, with the product of the following max terms of 4 variables (A,B,C,D):

0,3,4,7,9,10,12,15

and I only have a 4 to 1 multiplexer, one inverter, and one 2-input gate of my choice.

What would be the approach going about solving a question like this? In addition, what would be the "correct" 2-input gate to use in this situation?

My idea:

Crunch out all possible combinations of inputs then take all possible gates possible for those particular inputs:

(A,B)
(B,C)

etc.

However I feel that takes too long and a rather cumbersome approach to take.

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  • \$\begingroup\$ I am assuming this is a homework question. I would be very confused if the chapter/chapters this belongs to do(es) not go through the basics of logic tables of one sort or another to solve this kind of problem. In stead of risking using and handing in methods not advised by your material, it would be wise to check that first. \$\endgroup\$ – Asmyldof Oct 5 '14 at 20:24
  • \$\begingroup\$ If you wrote out the full truth table, then wrote the boolean products for each row, and hence derive the sum of products, I'd expect the text book to give you techniques to simplify that boolean equation. If it doesn't, you could enter it into wolframalpha.com/examples/BooleanAlgebra.html and let it suggest simplifications, and see if any of those help. \$\endgroup\$ – gbulmer Oct 5 '14 at 20:28
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the question can be solved in another way... As, POS of F(A,B,C,D)=0,3,4,7,9,10,12,15.

convert to SOP.

SOP of F(A,B,C,D)=1,2,5,6,8,11,13,14.

Now when u write down all the minterms you can group two terms together like keeping A'B' common A'B'CD'+A'B'C'D i.e.sop(2,1) to be equal to A'B'(C exor D) similarly others will also group up to form terms of xnor

enter image description here

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  • \$\begingroup\$ Although this also is a good attempt, it does not meet the requirements listed by the OP. \$\endgroup\$ – Guill Oct 12 '14 at 7:55
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The solution to this problem cannot be found by just trying all possible logic gate . It might be a bit complicated to understand.

POS of F(A,B,C,D)=0,3,4,7,9,10,12,15.

convert to SOP.

SOP of F(A,B,C,D)=1,2,5,6,8,11,13,14.

when A is 1 give us total 8 possibilities out of which the 4 which we are given have the property of even number of 1's.(m8,m11,m13,m14).

when A is 0 C and D having odd number of 1's is the required condition.don't care what B is.(m1,m2,m5,m6).

the solution would be something like this:

enter image description here

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  • \$\begingroup\$ Although it is a good attempt, this circuit does not meet the requirements listed by the OP. \$\endgroup\$ – Guill Oct 12 '14 at 7:54

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