We can obtain the canonical expression for a given function applying minterms or maxterms.
x100001111x200110011x301010101f(x1,x2,x3)01001110Mintermm0=xˉ1xˉ2xˉ3m1=xˉ1xˉ2x3m2=xˉ1x2xˉ3m3=xˉ1x2x3m4=x1xˉ2xˉ3m5=x1xˉ2x3m6=x1x2xˉ3m7=x1x2x3MaxtermM0=x1+x2+x3M1=x1+x2+xˉ3M2=x1+xˉ2+x3M3=x1+xˉ2+xˉ3M4=xˉ1+x2+x3M5=xˉ1+x2+xˉ3M6=xˉ1+xˉ2+x3M7=xˉ1+xˉ1+xˉ3
Given the function f(x1,x2,x3) above, we can obtain its canonical expression in two ways:
f(x1,x2,x3)=∑(m1,m4,m5,m6)f(x1,x2,x3)=Π(m0,m2,m3,m7)
Example:
\hline
x_1 & x_2 & f(x_1,x_2) \\ \hline
0&0&1\\
\hline
0&1&1\\
\hline
1&0&0\\
\hline
1&1&1\\
\hline
\end{array}\\
f(x1,x2)=xˉ1xˉ2+xˉ1x2+x1x2f(x1,x2)=xˉ1(xˉ2x2)+x1x2f(x1,x2)=xˉ1+x1x2f(x1,x2)=xˉ1+x2