Question

Three light-emitting diodes (LEDs) [one Red, one Green, one Blue] turn on when a number 0–7 is passed through. Red turns on with even numbers, green turns on with odd numbers, blue turns on with multiples of 3. Zero means they are all off, seven means they are all on.

Set–up the appropriate truth table, simplify (minimally) using K-maps. Implement, using LogiSim, the simplified logic circuit with optimal number of logic gates.

Answer 1

Step-by-step explanation:

To set up the truth table, we need to determine the output states of the three LEDs (Red, Green, and Blue) for each input number from 0 to 7.

Input (Number) | Red LED | Green LED | Blue LED

0 | 1 | 0 | 0

1 | 0 | 1 | 0

2 | 1 | 0 | 0

3 | 0 | 1 | 1

4 | 1 | 0 | 0

5 | 0 | 1 | 0

6 | 1 | 0 | 0

7 | 1 | 1 | 1

Now let's simplify the circuit using K-maps. We will create K-maps for each LED and find the minimal expressions for each output.

Red LED K-map:

\ | 00 | 01 | 11 | 10 |

Blue LED K-map:

\ | 00 | 01 | 11 | 10 |

Green LED K-map:

\ | 00 | 01 | 11 | 10 |

From the K-maps, we can see that the minimal expressions for each LED are:

Red LED: R = A' (A' + B' + C)

Green LED: G = A + B

Blue LED: B = A' + C

Now, we can implement the simplified logic circuit using Logisim. The circuit will have three inputs (A, B, C) and three outputs (Red LED, Green LED, Blue LED).

Please note that I am an AI text-based model and cannot directly provide a Logisim implementation. However, you can use the simplified expressions obtained from the K-maps to design the circuit in Logisim using logic gates such as AND, OR, and NOT gates.