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) | |
| m0 | 0 | 0 | 0 | 0 |
| m1 | 0 | 0 | 1 | 1 |
| m2 | 0 | 1 | 0 | 0 |
| m3 | 0 | 1 | 1 | 1 |
| m4 | 1 | 0 | 0 | 1 |
| m5 | 1 | 0 | 1 | 1 |
| m6 | 1 | 1 | 0 | 1 |
| m7 | 1 | 1 | 1 | 1 |
Each row of the truth table corresponds to one of the 2n = 8 possible minterms in n=3
variables.
The minterm code for n=3 is as follows:
| m0 | 0 | 0 | 0 |
|
|
|
| m1 | 0 | 0 | 1 |
|
|
C |
| m2 | 0 | 1 | 0 |
|
B |
|
| m3 | 0 | 1 | 1 |
|
B | C |
| m4 | 1 | 0 | 0 | A |
|
|
| m5 | 1 | 0 | 1 | A |
|
C |
| m6 | 1 | 1 | 0 | A | B |
|
| m7 | 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