Boolean Logic and Logic Gates: Question 9
Syllabus 10.1, 10.2, 10.3
A warehouse extractor-fan controller has three inputs: H = 1 if a humidity sensor reads above the trigger threshold, otherwise 0; C = 1 if the air-conditioning unit is currently running, otherwise 0; and L = 1 if a staff member has pulled the manual override lever, otherwise 0. Inside the controller, the signals pass through three gates in sequence:
- Gate 1 is a NOT gate, with input C.
- Gate 2 is an AND gate, with inputs H and the output of Gate 1.
- Gate 3 is an OR gate, with inputs L and the output of Gate 2.
The output of Gate 3, labelled Vent, switches the extractor fan on whenever Vent = 1.
(a) State the logic expression for Vent in terms of H, C and L, and name the three gates in the order the signals pass through them. [2]
(b) Complete the truth table for Vent, showing all eight combinations of H, C and L. [4]
| H | C | L | Vent |
|---|---|---|---|
| 0 | 0 | 0 | |
| 0 | 0 | 1 | |
| 0 | 1 | 0 | |
| 0 | 1 | 1 | |
| 1 | 0 | 0 | |
| 1 | 0 | 1 | |
| 1 | 1 | 0 | |
| 1 | 1 | 1 |
(c) State whether the fan runs (Vent = 1) in the specific case where the humidity sensor reads high, the air-conditioning unit is already running, and the lever has not been pulled, and explain your answer using the truth table. [2]
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Worked solution
Part (a): Writing the logic expression
Gate 1 is a NOT gate taking C, so its output is NOT C.
Gate 2 is an AND gate combining H with the output of Gate 1, giving H AND (NOT C).
Gate 3 is an OR gate combining L with the output of Gate 2, giving the final output:
Vent = (H AND (NOT C)) OR L
The three gates, in the order the signals pass through them, are: NOT (applied to C), then AND (combining H with the inverted C), then OR (combining that result with L).
Part (b): Building the truth table row by row
For each row, first invert C, then AND the result with H, then OR that with L:
| H | C | L | NOT C | H AND (NOT C) | Vent = (H AND (NOT C)) OR L |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 |
Whenever H = 0, the AND gate’s output is always 0 regardless of C, so Vent simply follows L. Whenever H = 1 and C = 0, NOT C = 1, so the AND gate outputs 1, which is enough on its own to make the OR gate output 1 no matter what L is. Whenever H = 1 and C = 1, NOT C = 0, so the AND gate outputs 0, and Vent then follows L exactly as in the H = 0 rows.
So the completed table for Vent is:
| H | C | L | Vent |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Part (c): Applying the truth table to a specific case
The case described is H = 1 (humid), C = 1 (AC already running), L = 0 (lever not pulled). Looking up row (H = 1, C = 1, L = 0) in the completed table gives Vent = 0.
So the fan does not run in this case: even though it is humid, the air-conditioning unit is already running, so Gate 1 outputs NOT C = 0, which forces Gate 2’s output to 0 regardless of H, and since the lever has not been pulled (L = 0), Gate 3’s OR has nothing else to make it 1.
Final answers
- (a) Vent = (H AND (NOT C)) OR L; gates in order: NOT, AND, OR
- (b) Vent values in row order: 0, 1, 0, 1, 1, 1, 0, 1
- (c) The fan does not run (Vent = 0), because the AC already running cancels the humidity condition, and the lever has not been pulled to override it.