In: Computer Science
1. write a truth table using this symbol: -->
2. write the inputs for the truth table to the left of the --> and write the outputs for the truth table to the right of the -->
3. write the compliment, or NOT using '
As an example:
The truth table for AND is written this way:
A B --> A AND B
0 0 --> 0
0 1 --> 0
1 0 --> 0
1 1 --> 1
or this way:
A B --> A AND B
a' b' --> 0
a' b --> 0
a b' --> 0
a b --> 1
As another example:
For a 3 input, 2 output truth table, the truth table is written this way:
A B C --> X Y
0 0 0 --> 0 0
0 0 1 --> 1 0
0 1 0 --> 1 0
0 1 1 --> 0 1
1 0 0 --> 1 0
1 0 1 --> 0 1
1 1 0 --> 0 1
1 1 1 --> 1 1
Part a) Write the truth table for the boolean algebra equation
F1 = A+BD+CD
Part b) Write the truth table for the boolean algebra equation
F2 = (A+B+C)•(A+B+D)•(A+C+D)
Part c) Explain in English how F1 and F2 are related. Write your explanation in 4 sentences or less.
(Note: a) is the truth table for logic represented using sum of
products b) is the truth table for logic represented using product
of sums. In general, any logic that can be written as a sum of
products can also be written as a product of sums)
Truth Table of F1=A+BD+CD
A | B | C | D | BD | CD | A+BD | ---> A+BD+CD |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Truth Table of F2=(A+B+C)(A+B+D)(A+C+D)
A | B | C | D | A+B+C | A+B+D | A+C+D | --> (A+B+C).(A+B+D)(A+C+D) |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
F1=A+BD+CD
F2=(A+B+C)(A+B+D)(A+C+D)
=(A+B+CD).(A+C+D) (BY DISTRIBUTIVE THEOREM (A+B+C)(A+B+D)=(A+B+CD))
=A+(B+CD).(C+D)
=A+BC+BD+CD+CD
=A+BC+BD+CD
= F1+BC
THERE IS ONE TERM DIFFERENCE BETWEEN F1 AND F2 THAT IS BC TERM