Designing a Circuit for a Given Input/Output Table
Design a circuit for the following input/output table:
|
Inputs |
Outputs |
||
|
P |
Q |
R |
S |
|
1 |
1 |
1 |
0 |
|
1 |
1 |
0 |
1 |
|
1 |
0 |
1 |
0 |
|
1 |
0 |
0 |
0 |
|
0 |
1 |
1 |
0 |
|
0 |
1 |
0 |
0 |
|
0 |
0 |
1 |
0 |
|
0 |
0 |
0 |
1 |
First we construct a Boolean expression with this table as its truth table using disjunctive normal form. Rows 2 and 8 are the only rows with output.
Row 2 gives us P
Equivalently, application of the distributive law gives us:
Ø R Ù ((P Ù Q ) Ú (Ø P Ù Ø Q))
An equivalent circuit requiring one less gate.
Showing two circuits are equivalent
Using basic Boolean algebra properties (Theorem 1) one can simplify the Boolean expression for a circuit. This new, logically equivalent Boolean expression can then be used to construct a simpler circuit. Another technique to finding a simpler circuit is to find an equivalent circuit that uses fewer gates. Trial and error, intuition and insight are valid tools when pursuing this latter method.