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+9 votes
  1. Is the 3-variable function $f= \Sigma(0,1,2,4)$ its self-dual? Justify your answer.
  2. Give a minimal product-of-sum form of the b output of the following excess-3 to BCD converter.


asked in Digital Logic by Veteran (69k points)
edited by | 635 views
plz answer part B also..
$Remark:$ self dual function is that function, if we calculate one time dual of that function then we get the original function back.

5 Answers

+14 votes
Best answer

there are two condition for a function being self dual.

1- it should be neutral function. (no. of minter = no . of max term)

2-no mutually two exclusive term should be there like (0-7 are mutaully exclusive 1-6, 2-5, 3-4) from these pairs only one should be there.
clearly there are 4 minterm, so number of minterms = no of maxterms.
and second condition is also satisfied. so it is a self dual function .

b) excess-3 to BCD

answered by Veteran (15.8k points)
edited by
+12 votes

checking self dual of a function.

f=Σ(0,1,2,4) = a'b'c' + a'b'c + a'bc' + ab'c' 

now write it in max term form (a'+b'+c')(a'+b'+c)(a'+b+c')(a+b'+c')

                                                        7             6             5           3

         =$\prod$(3,5,6,7) =$\sum$(0,1,2,4) = given function so its a self dual

answered by Veteran (15.9k points)
+4 votes

(b) Excess-3 to BCD 

Minimal product of sum (b4, b3, b2, b1)

answered by (225 points)
+3 votes


part b


thanks :)

answered by (423 points)
0 votes

a) It is a self dual function, because it satisfies both conditions of self dual fuction

            ->It should be neutral (i.e no. of minterms is equal to the no. of maxterms)

            ->The function does not contain two mutually exclusive term ( eg ABC mutually exclusive to A'B'C' , AB'C' mutually  excluseive to A'BC)

b) Excess-3 to BCD converter


answered by Junior (935 points)

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