There are two standard or canonical ways of expressing boolean functions:
We will focus on SOP.
Consider
Let's look at the truth table which corresponds to this function:
A | B | C | f(A,B,C) | |
m_{0} | 0 | 0 | 0 | 0 |
m_{1} | 0 | 0 | 1 | 1 |
m_{2} | 0 | 1 | 0 | 0 |
m_{3} | 0 | 1 | 1 | 1 |
m_{4} | 1 | 0 | 0 | 1 |
m_{5} | 1 | 0 | 1 | 1 |
m_{6} | 1 | 1 | 0 | 1 |
m_{7} | 1 | 1 | 1 | 1 |
Each row of the truth table corresponds to one of the 2^{n} = 8 possible minterms in n=3
variables.
The minterm code for n=3 is as follows:
m_{0} | 0 | 0 | 0 | |||
m_{1} | 0 | 0 | 1 | C | ||
m_{2} | 0 | 1 | 0 | B | ||
m_{3} | 0 | 1 | 1 | B | C | |
m_{4} | 1 | 0 | 0 | A | ||
m_{5} | 1 | 0 | 1 | A | C | |
m_{6} | 1 | 1 | 0 | A | B | |
m_{7} | 1 | 1 | 1 | A | B | C |
The function
can be put into canonical SOP form
algebraically as follows:
Any Boolean function can be expressed in canonical SOP form.
ANU Engineering - ENGN3213