Question

2.18 For the Boolean function F = xyz + x'y'z + w xy + wx'y + wxy (a) Obtain the truth table of E (b) Draw the logic diagram, using the original Boolean expression. (c) Use Boolean algebra to simplify the function to a minimum number of literals (d) Obtain the truth table of the function from the simplified expression and show that it is the same as the one in part (a). (e) Draw the logic diagram from the simplified expression, and compare the total number of gates with the diagram of part (b).

Answer 1

Solution:

b.

will wxlly w x' DI xylz X wl'xy F Das yly xlly\'s N wxy,

F = xy’z + x’ y’ z + w’ xy + wy + wx’ y + wxy

a)

wxyz F 0000 0 0001 1 0010 0 0011 0 0100 0 0101 1 0110 1 01111 1000 0 1001 1 1010 1 1011 1 1100 0 1101 1 1110 1 1111 0

c)         F = xy’ z + x’ y’ z + w’ xy + wx’ y + wxy

= (x +x’) y’ z + y (w’ x + w(x + x’)

a + a’ = 1 => F = y’ z + y(w’ x + w)

a + a’ b = a + b => F = y’ z + y(w + x).

d)         Wxyz               F

00000 0001 1 0010 0 0011 0 0100 0 0101 1 0110 1 01111 1000 0 1001 1 1010 1 1011 1 1100 0 1101 1 1110 1 1111 0

Both are same

e)         For simplified expression clearly itrequires only a not gate, two 2-input AND gates and two 2-input ORgates