Simplifying Circuits #
Here’s a truth table:
| A | B | Out |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
If we take the 1 terms and write this in SoP form:
AB + A’B
We can apply boolean algebra:
- AB + A’B
- = A(B+B')
- = A(1)
- = A
There’s a trick to speed that up with visual pattern matching, called a K-map (Karnaugh Map).
| A' | A | |
|---|---|---|
| B' | 0 | 1 |
| B | 0 | 1 |
Look for even-size rectangles of 1’s and circle them.
And the rectangle is in the A column, so our simplified expression is just A.
| A | B | C | Out |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
Grey code - each adjacent cell differs by one bit.
| A’B' | A’B | AB | AB' | |
|---|---|---|---|---|
| C | 0 | 0 | 0 | 0 |
| C' | 0 | 1 | 1 | 1 |
-> AC’ + BC'
Let’s consider a k-map of ABC that will resolve to just A.
| A’B' | A’B | AB | AB' | |
|---|---|---|---|---|
| C | 0 | 0 | 1 | 1 |
| C' | 0 | 0 | 1 | 1 |
Let’s consider a k-map of ABC that will resolve to just B'.
| A’B' | A’B | AB | AB' | |
|---|---|---|---|---|
| C | 1 | 0 | 0 | 1 |
| C' | 1 | 0 | 0 | 1 |
This gives us a two-by-two square of 1’s that wraps around the edge.
We can go to 4 variables with a 4x4 table. More still works, but it gets annoying.
