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+11 votes
  1. Is the $3\text{-variable}$ 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 $\text{excess-3}$ to $\text{BCD}$ converter.

asked in Digital Logic by Veteran (59.6k points)
edited by | 809 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

+15 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 $($$0-7$ are mutually exclusive $1-6, 2-5, 3-4$$)$ 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 Boss (15.9k points)
edited by
+13 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 Boss (15.9k points)
+4 votes

(b) Excess-3 to BCD 

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

answered by (239 points)
+3 votes

part b

thanks :)

answered by (441 points)
+1 vote

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 (335 points)

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