26 votes 26 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. Digital Logic gatecse-2001 digital-logic normal descriptive min-sum-of-products-form + – Kathleen asked Sep 14, 2014 edited Jul 10, 2019 by ajaysoni1924 Kathleen 4.0k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Vidhya Pate commented Mar 20, 2017 reply Follow Share plz answer part B also.. 0 votes 0 votes Manu Thakur commented Dec 14, 2017 reply Follow Share $Remark:$ self dual function is that function, if we calculate one time dual of that function then we get the original function back. 8 votes 8 votes Please log in or register to add a comment.
Best answer 32 votes 32 votes There are two conditions for a function being self dual. it should be a neutral function. (no. of minterms = no. of max terms) 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 $$\overset{\textbf{Truth Table}}{\begin{array}{|cccc|cccc|} \hline \rlap{\textbf{Inputs}}&&&&\rlap{\textbf{Outputs}}\\ \hline \textbf{W} & \textbf {X} &\textbf {Y} & \textbf {Z} & \textbf{A}&\textbf {B} & \textbf {C} & \textbf{D} \\\hline0&0&1&1&0&0&0&0\\\hline0&1&0&0&0&0&0&1\\\hline0&1&0&1&0&0&1&0 \\\hline 0&1&1&0&0&0&1&1 \\\hline0&1&1&1&0&1&0&0 \\\hline 1&0&0&0&0&1&0&1\\\hline1&0&0&1&0&1&1&0\\\hline1&0&1&0&0&1&1&1 \\\hline 1&0&1&1&1&0&0&0 \\\hline1&1&0&0&1&0&0&1 \\\hline \end{array}}$$ MAPS: Tendua answered Oct 2, 2015 edited Jul 12, 2019 by ajaysoni1924 Tendua comment Share Follow See all 3 Comments See all 3 3 Comments reply Gate Fever commented Nov 22, 2018 reply Follow Share 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. 3 votes 3 votes tusharp commented Mar 29, 2019 reply Follow Share @Tendua awesome explanation but just for the sake of correction POS is asked and not SOP. 3 votes 3 votes commenter commenter commented Oct 17, 2019 reply Follow Share @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. 3 votes 3 votes Please log in or register to add a comment.
15 votes 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 Umang Raman answered Sep 30, 2015 Umang Raman comment Share Follow See all 0 reply Please log in or register to add a comment.
7 votes 7 votes part b thanks :) Harit answered Jan 2, 2017 Harit comment Share Follow See all 0 reply Please log in or register to add a comment.
7 votes 7 votes (b) Excess-3 to BCD Minimal product of sum (b4, b3, b2, b1) ravi rajak 3 answered Oct 13, 2017 ravi rajak 3 comment Share Follow See all 0 reply Please log in or register to add a comment.