Introduction to CS
Project 5: Logic Gates - Answers
1. Create a Truth Table for the following circuit:

D = NOT (A OR B)
E = B * C
Q = D OR E = (NOT (A OR B)) OR (B * C) |
| Inputs |
Outputs |
| A |
B |
C |
D |
E |
Q |
| 0 |
0 |
0 |
1 |
0 |
1 |
| 0 |
0 |
1 |
1 |
0 |
1 |
| 0 |
1 |
0 |
0 |
0 |
0 |
| 0 |
1 |
1 |
0 |
1 |
1 |
| 1 |
0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
1 |
0 |
0 |
0 |
| 1 |
1 |
0 |
0 |
0 |
0 |
| 1 |
1 |
1 |
0 |
1 |
1 |
|
2. Replace gates with NAND gates only - NAND Gate Synthesis

3. Eliminate redundancy in circuit

4. Create new Truth Table and verify output Q matches on both tables
- Q1 from first circuit and Q2 from the new replacement NAND circuit -
to "Prove" our results.
Inputs |
Outputs |
|
A |
B |
C |
D |
E |
F |
G |
Q1 |
Q2 |
|
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
|
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
|
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
|
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
|
1 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
|
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
|
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
|
1 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
- D = NOT A
- E = NOT B
- F = NOT(D * E)
- G = NOT(B * C)
- Q1 = NOT(F * G)
|