in Digital Logic edited by
23 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.

in Digital Logic edited by


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.

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5 Answers

27 votes
Best answer

There are two conditions for a function being self dual.

  1. it should be a neutral function. (no. of minterms = no. of max terms)
  2. no two mutually exclusive terms 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 $4$ minterms, 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

$$\overset{\textbf{Truth Table}}{\begin{array}{|cccc|cccc|} \hline \rlap{\textbf{Inputs}}&&&&\rlap{\textbf{Outputs}}\\ \hline \textbf{W} & \textbf {X} &\textbf {Y} & \textbf {Z} &  \textbf{A}&\textbf {B} & \textbf {C} &  \textbf{D} \\\hline0&0&1&1&0&0&0&0\\\hline0&1&0&0&0&0&0&1\\\hline0&1&0&1&0&0&1&0 \\\hline 0&1&1&0&0&0&1&1 \\\hline0&1&1&1&0&1&0&0 \\\hline 1&0&0&0&0&1&0&1\\\hline1&0&0&1&0&1&1&0\\\hline1&0&1&0&0&1&1&1 \\\hline 1&0&1&1&1&0&0&0 \\\hline1&1&0&0&1&0&0&1 \\\hline \end{array}}$$


edited by


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.

@Tendua awesome explanation but just for the sake of correction POS is asked and not SOP.


@Gate Fever No you are wrong. Xs-3 code is for BCD. So we need to just add 3 to 0 - 9 digits i.e. 0+3 to 9+3 = 3 to 12.

15 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

7 votes

part b

thanks :)

6 votes

(b) Excess-3 to BCD 

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

2 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


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