Solution for Assessment-2
Part-1
1. Trurth table
| R | U | $C_a$ | $C_b$ | Y | S |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 | 0 |
2. Boolean Expression for Y and S
$$ Y = (R' U C_a' C_b') + (R' U C_a' C_b) + (R' U C_a C_b') + (R' U C_a C_b) \ + (R U' C_a' C_b') + (R U' C_a' C_b) + (R U' C_a C_b') + (R U' C_a C_b) \ + (R U C_a' C_b') + (R U C_a' C_b) + (R U C_a C_b') + (R U C_a C_b) $$
$$ S= (R' U' C_a' C_B)+(R' U' C_a C_b')+(R' U' C_a' C_B') $$
3. Minimised Boolean expression:
$$ Y=R+U $$
$$ S=R'U'(C_a+C_b) $$
OR $$ S=(R+U)'(C_a+C_b) $$
Or
$$ S=R'U'C_a+R'U'C_b $$
4.
$$ Y=R+U $$
$$ S=R'U'(C_a+C_b) $$
OR
$$ Y=R+U $$
$$ S=(R+U)'(C_a+C_b) $$
both are correct
5.
| Test Condition | Expected Outcomes | Value of (S) and (Y) | Justification | Screenshot of the simulation |
|---|---|---|---|---|
| $RUC_aC_b$ = 0001 | Brake does not apply and steering applies | Y=0 S=1 | Logic 4 | |
| $RUC_aC_b$ = 1110 | Brake applies and steering doesn't. | Y=1 S=0 | Logic 1/2 | |
| $RUC_aC_b$ = 1010 | Brake applies and steering doesn't. | Y=1 S=0 | Logic 1/2 |
6. Circut Conversion
To convert to a circut with only NOR or NAND, student should have done some reseach and relies on the following:
So student can use either or both, but the outcome should be the same. Therefore we need to check by using radom input for $R U C_a C_b$
Here is an exmaple:
Part-2
Task-1
Task-2
Task-3
Task-4
Task-5
