Canonical-Normal-Form
Question 1 |
Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit
Σm(4,6) | |
Σm(4,8) | |
Σm(6,8) | |
Σm(4,6,8)
|
Question 1 Explanation:
f= f1* f2 + f3
f1*f2 is intersection of minterms of f1 and f2
f= (f1*f2) + f3 is union of minterms of (f1*f2) and f3
Σm(1,6,8,15)= Σm(4,5,6,7,8) * f2 + Σm(1,6,15)
Options A, B and D have minterm m4 which result in Σm(1,4,6,15), Σm(1,4,6,8, 15) and Σm(1,4,6,8, 15)respectively and they are not equal to f.
Option C : If f2= Σm(6,8)
RHS: Σm(4,5,6,7,8) * Σm(6,8) + Σm(1,6,15)
=Σm(6,8) + Σm(1,6,15)
= Σm(1,6,8,15)
= f= LHS
f1*f2 is intersection of minterms of f1 and f2
f= (f1*f2) + f3 is union of minterms of (f1*f2) and f3
Σm(1,6,8,15)= Σm(4,5,6,7,8) * f2 + Σm(1,6,15)
Options A, B and D have minterm m4 which result in Σm(1,4,6,15), Σm(1,4,6,8, 15) and Σm(1,4,6,8, 15)respectively and they are not equal to f.
Option C : If f2= Σm(6,8)
RHS: Σm(4,5,6,7,8) * Σm(6,8) + Σm(1,6,15)
=Σm(6,8) + Σm(1,6,15)
= Σm(1,6,8,15)
= f= LHS
There is 1 question to complete.