duality

Question

given F=∑m(0,2,4,6)

then find its F' and Fdual...in SOP form

Answer 1

$F(x,y,z)=\Sigma_m(0,2,4,6) = x'y'z'+x'yz'+xy'z'+xyz'$

Complement of Function, 

$F'(x,y,z)= (x+y+z)(x+y'+z)(x'+y+z)(x'+y'+z)$

$F'(x,y,z)= \Pi_M(0,2,4,6)$

$F'(x,y,z)=\Sigma_m(1,3,5,7)$

Dual of Function, 

$F_d(x,y,z)=(x'+y'+z')(x'+y+z')(x+y'+z')(x+y+z')$

$F_d(x,y,z) = M_7.M_5.M_3.M_1$

$F_d(x,y,z) =\Pi_M(1,3,5,7)$

$F_d(x,y,z) =\Sigma_m(0,2,4,6)$

Note:

In Dual of $f$, that is $f_d$, we replace AND $(.)$ by OR $(+)$, OR $(+)$ by AND $(.)$, $0$ by $1$ and $1$ by $0$ ONLY.

In Complement of $f$, that is $f'$, we need to replace all variable, say $x$ , by it complement variables , say $x'$, also.

Refer : https://gateoverflow.in/36297/computing-the-dual-of-the-boolean-function

 

Comments

But sir after minimizing function the result is C only then we get F' and F_d are same as C na