Find the sum-of-products expansions the Boolean function F(x, Y, 2) that equals 1 if and only if
a)x = 0.
b) xy = 0.
c) x +y = 0.
d)xyz = 0.
a)
x‾(yz+y‾z+yz‾+y‾z‾)=x‾yz+x‾y‾z+x‾yz‾+x‾y‾z‾\overline{x}(yz+\overline{y}z+y\overline{z}+\overline{y}\overline{z})=\overline{x}yz+\overline{x}\overline{y}z+\overline{x}y\overline{z}+\overline{x}\overline{y}\overline{z}x(yz+yz+yz+yz)=xyz+xyz+xyz+xyz
b)
xy‾=x‾+y‾\overline{xy }= \overline {x}+\overline{y}xy=x+y
then:
xy‾z+xy‾z‾=(x‾+y‾)z+(x‾+y‾)z‾=x‾z+y‾z+x‾z‾+y‾z‾\overline{xy}z+\overline{xy}\overline{z}=(\overline {x}+\overline{y})z+(\overline{x}+\overline{y})\overline{z}=\overline{x}z+\overline{y}z+\overline{x}\overline{z}+\overline{y}\overline{z}xyz+xyz=(x+y)z+(x+y)z=xz+yz+xz+yz
c)
x+y‾=x‾⋅y‾\overline{x +y}=\overline{x}\cdot \overline{y}x+y=x⋅y
x‾⋅y‾⋅z+x‾⋅y‾⋅z‾\overline{x}\cdot \overline{y}\cdot z+\overline{x}\cdot \overline{y}\cdot \overline{z}x⋅y⋅z+x⋅y⋅z
d)
xyz‾=xy‾+z‾=x‾+y‾+z‾\overline{xyz}=\overline{xy}+\overline{z}=\overline{x}+\overline{y}+\overline{z}xyz=xy+z=x+y+z
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