Quiz 1 February 11
X and Y are two Boolean functions of three binary variables a, b, c, as given
below.
X = a•b’ + a•c + b’•c, and Y = (a + b’)•(a + c)•(b’ + c).
(a) Obtain the expression for X in the First Canonical Form. [3]
(b) Obtain the expression for Y in the Second Canonical Form. [4]
(c) Show that X = Y. [3]
ANSWER
(a) X = a•b’•(c’ + c) + a•c•(b’ + b) + b’•c•(a’ + a) [1]
= a•b’•c’ + a•b’•c + a•b’•c + a•b•c + a’•b’•c + a•b’•c
= a’•b’•c + a•b’•c’ + a•b’•c + a•b•c. [2]
(b) Y = (a + b’ + c’•c)•(a + c + b’•b)•(b’ + c + a’•a) [1]
= (a + b’ + c’)•(a + b’ + c)•(a + c + b’)•(a + c + b)•(b’ + c + a’)•(b’ + c + a) [1]
= (a’ + b’ + c)•(a + b’ + c)• (a + b’ + c)•(a + b + c). [2]
(c) PoS is easier to expand because of our algebra has distributivity of
multiplication over addition only.
(a + b’)•(a + c)•(b’ + c) = (a + a•b’ + a•c + b’•c)•(b’ + c) [3]
= (a + b’•c )•(b’ + c) = a•b’ + a•c + b’•c.
Y = X.
