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+11 votes

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

+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 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**

+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

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

Ref http://www.sukanta.info/wp-content/uploads/2012/05/Excess-3-code-to-bcd-converter3.pdf

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