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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.

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$Remark:$ self dual function is that function, if we calculate one time dual of that function then we get the original function back.

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

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why 0,1,2 are taken as dont care??

i got the reason why 13,14,15 are taken as dont care ,as 13+3=16 cant be represented in 4 bits, thats the reason for 14,15 too.
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@Tendua awesome explanation but just for the sake of correction POS is asked and not SOP.

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

(b) Excess-3 to BCD

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

part b

thanks :)

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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

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