Derive a Canonical SOP expression for a Boolean function F(X,Y,Z) repr

1.	Derive a Canonical SOP expression for a Boolean function F(X,Y,Z) represented by the following truth table:XYZF(X,Y,Z)00010011010001101001101011001111
Answer» F(X,Y,Z)= X’Y’Z’+X’Y’Z+XY’Z’+XYZ OR F(X,Y,Z)=∑(0,1,4,7)

Derive a Canonical SOP expression for a Boolean function F(X,Y,Z) represented by the following truth table:XYZF(X,Y,Z)00010011010001101001101011001111

Answer»

F(X,Y,Z)= X’Y’Z’+X’Y’Z+XY’Z’+XYZ

OR F(X,Y,Z)=∑(0,1,4,7)

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Derive a Canonical SOP expression for a Boolean function F(X,Y,Z) represented by the following truth table:XYZF(X,Y,Z)00010011010001101001101011001111

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