Question 3.4: Derive a truth table for the POS expression as below: F = (A......
Derive a truth table for the POS expression as below:
F= (A+B+C)(A+\bar C)(A+\bar B)
Step 1 There are three variables in the domain, so eight possible combinations of binary values of 3-input variables exist as listed in the left three columns of Table 3.4.3.
Step 2 Substitute each binary value into the POS expression. When ABC = 000, F = 0; when ABC = 001, F = 0; when ABC = 010, F = 0; when ABC = 011, F = 0, and so on. The resulting output values are listed in the output column of Table 3.4.3
Alternatively, you can convert the POS expression into the standard form.
The first term is already a standard sum term.
The second term has a missing variable B, so add B{\bar{B}} and apply the distributive law as below:
The third term has a missing variable C, so add {{C}}{\bar{C}} and apply the distributive law as below:
A+\bar{B}=A+\bar{B}+C\bar{C}=(A+\bar{B}+C)(A+\bar{B}+\bar{C})So the standard POS form of the original expression is as follows:
F=(A+B+C)(A+B+\overline{{{C}}})(A+\overline{{{B}}}+\overline{{{C}}})(A+\overline{{{B}}}+C)=\prod M(0,1,2,3)The binary values that make the standard sum terms (maxterms) in expression equal to 0 are A+B+C\,(M_{0})\colon\!000,A+B+\bar{C}\,(M_{1})\colon001,A+\bar{B}+C\,(M_{2})\colon010,\,A+\bar{B}+\bar{C}\,(M_{3})\colon011. For each of these binary values, a 0 is placed in the output column as shown in Table 3.4.3. For each of the remaining binary combinations, a 1 is placed in the output column.
| Table 3.4.3: Truth table. | |||
| Input | Output | ||
| A | B | C | F |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |