Answer on Question #43374 – Math - Discrete Mathematics
Expand the following Boolean functions into their canonical form:
i. f(X,Y,Z)=XY+YZ+X′Z+X′Y′
ii. f(X,Y,Z)=XY+X′Y′+X′YZ
Solution.
We will express each function as sum of minterms.
i. f(X,Y,Z)=XY+YZ+X′Z+X′Y′=XY(Z+Z′)+(X+X′)YZ+X′(Y+Y′)Z+X′Y′(Z+Z′)=XYZ+XYZ′+XYZ+X′YZ+X′YZ+X′Y′Z′=XYZ+XYZ′+X′YZ+X′Y′Z′
ii. f(X,Y,Z)=XY+X′Y′+X′YZ=XY(Z+Z′)+X′Y′(Z+Z′)+X′YZ=XYZ+XYZ′+X′Y′Z+X′Y′Z′+X′YZ.
Answer: i. f(X,Y,Z)=XYZ+XYZ′+X′YZ+X′Y′Z+X′Y′Z′
ii. f(X,Y,Z)=XYZ+XYZ′+X′Y′Z+X′Y′Z′+X′YZ.
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