Question 15
A boom gate at a railway level crossing has two train sensors, one on each side of the crossing. The boom must be lowered when either sensor detects a train. The crossing operator can also lower the boom using a manual switch, but the manual switch only works when the maintenance key is not inserted in the control box.
Key
Input A = Train sensor 1 (train detected = 1)
Input B = Train sensor 2 (train detected = 1)
Input C = Manual switch (on = 1)
Input D = Maintenance key (inserted = 1)
E and F are intermediate input/output signals to the logic gates.
Output X = Boom (boom lowered = 1)
Construct a logic gate circuit for the boom gate, clearly labelling all inputs and outputs. Complete columns F and X of the corresponding truth table on the next page.
Truth table for the boom gate
| Inputs | Intermediate signals | Output | ||||
|---|---|---|---|---|---|---|
| A | B | C | D | E | F | X |
| 0 | 0 | 0 | 0 | 0 | ||
| 0 | 0 | 0 | 1 | 0 | ||
| 0 | 0 | 1 | 0 | 0 | ||
| 0 | 0 | 1 | 1 | 0 | ||
| 0 | 1 | 0 | 0 | 1 | ||
| 0 | 1 | 0 | 1 | 1 | ||
| 0 | 1 | 1 | 0 | 1 | ||
| 0 | 1 | 1 | 1 | 1 | ||
| 1 | 0 | 0 | 0 | 1 | ||
| 1 | 0 | 0 | 1 | 1 | ||
| 1 | 0 | 1 | 0 | 1 | ||
| 1 | 0 | 1 | 1 | 1 | ||
| 1 | 1 | 0 | 0 | 1 | ||
| 1 | 1 | 0 | 1 | 1 | ||
| 1 | 1 | 1 | 0 | 1 | ||
| 1 | 1 | 1 | 1 | 1 | ||
Two ways to lower the boom, and one key that locks out the switch
A train passes over both sensors, then the operator uses the manual switch, first with the maintenance key out and then with it inserted. Watch the 1s travel through the gates to the boom.
An orange, heavier wire carries 1. Each box shows the value on its line. The sensors reach the boom through two ORs, and the switch has to get past the AND, which the key closes.
Test the circuit, then complete columns F and X
Switch the inputs on and off, or press a row of the table to load its inputs. The circuit shows E, F and X for that row. Then press the F and X cells to fill them in (each press goes blank, 0, 1) and check your columns.
An orange, heavier wire carries 1, and each circle shows the value on its line. Press a circle on the left to switch that input.
The shaded row, with the orange bar, is the row the circuit is showing.
Work through the model response one mark at a time
Try each mark yourself before you reveal it. The model response builds the sketch for Marks 1 to 6, then completes the table for Marks 7 and 8.
Boolean expression: X = (A OR B) OR (C AND NOT D). Gates: two 2-input OR gates, one AND gate, one NOT gate.
A and B feed OR1. Its output feeds OR2, whose output drives the boom.
The parts this mark adds are drawn in orange with a heavier line.
Both sensors go through OR1 then OR2 to the boom.
Standard gate symbols must be used. Gate names (OR1, OR2, AND, NOT) are not required.
Accept any logically equivalent circuit for the first and second bullets (e.g. a single 3-input OR gate for train sensor 1, train sensor 2 and F). Where E is not used, award the sixth bullet for correct labels on F and the boom output.
Marker: provides a correctly applied logic gate circuit including train sensor 1 and train sensor 2 $\Rightarrow$ OR1 $\Rightarrow$ OR2 $\Rightarrow$ boom
C feeds the AND gate, whose output feeds OR2.
The parts this mark adds are drawn in orange with a heavier line.
The manual switch goes through the AND gate then OR2 to the boom.
Accept any logically equivalent circuit for the first and second bullets (e.g. a single 3-input OR gate for train sensor 1, train sensor 2 and F). Where E is not used, award the sixth bullet for correct labels on F and the boom output.
Marker: manual switch $\Rightarrow$ AND $\Rightarrow$ OR2 $\Rightarrow$ boom
D passes through the NOT gate into the AND gate’s second input.
The parts this mark adds are drawn in orange with a heavier line.
The key is inverted by the NOT gate before the AND gate.
Accept an equivalent arrangement for F (e.g. NOR of NOT C and D) when it produces the same F column.
Marker: maintenance key $\Rightarrow$ NOT $\Rightarrow$ AND
Inputs labelled “A: train sensor 1 (train detected = 1)” and “B: train sensor 2 (train detected = 1)”.
The labels this mark adds are in orange.
A and B carry their names and on-states.
Labels must include the input or output name and its on-state, e.g. 'D: maintenance key (inserted = 1)'. Letters alone do not earn the label marks.
Allow FT marks for an incorrect logic gate circuit when awarding the label marks.
Marker: correctly labels train sensor 1 and train sensor 2 inputs
Inputs labelled “C: manual switch (on = 1)” and “D: maintenance key (inserted = 1)”.
The labels this mark adds are in orange.
C and D carry their names and on-states.
Labels must include the input or output name and its on-state, e.g. 'D: maintenance key (inserted = 1)'. Letters alone do not earn the label marks.
Allow FT marks for an incorrect logic gate circuit when awarding the label marks.
Marker: correctly labels manual switch input and maintenance key input
OR1 output labelled E, AND output labelled F, OR2 output labelled “X: boom (lowered = 1)”.
The labels this mark adds are in orange.
E, F and the boom output (with on-state) are all labelled.
Accept any logically equivalent circuit for the first and second bullets (e.g. a single 3-input OR gate for train sensor 1, train sensor 2 and F). Where E is not used, award the sixth bullet for correct labels on F and the boom output.
Labels must include the input or output name and its on-state, e.g. 'D: maintenance key (inserted = 1)'. Letters alone do not earn the label marks.
Allow FT marks for an incorrect logic gate circuit when awarding the label marks.
Marker: correctly labels intermediary inputs/outputs E and F, and the boom output
Column F: 1 only when C = 1 and D = 0.
| A | B | C | D | E | F | X |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | |
| 0 | 0 | 0 | 1 | 0 | 0 | |
| 0 | 0 | 1 | 0 | 0 | 1 | |
| 0 | 0 | 1 | 1 | 0 | 0 | |
| 0 | 1 | 0 | 0 | 1 | 0 | |
| 0 | 1 | 0 | 1 | 1 | 0 | |
| 0 | 1 | 1 | 0 | 1 | 1 | |
| 0 | 1 | 1 | 1 | 1 | 0 | |
| 1 | 0 | 0 | 0 | 1 | 0 | |
| 1 | 0 | 0 | 1 | 1 | 0 | |
| 1 | 0 | 1 | 0 | 1 | 1 | |
| 1 | 0 | 1 | 1 | 1 | 0 | |
| 1 | 1 | 0 | 0 | 1 | 0 | |
| 1 | 1 | 0 | 1 | 1 | 0 | |
| 1 | 1 | 1 | 0 | 1 | 1 | |
| 1 | 1 | 1 | 1 | 1 | 0 |
F is 1 in rows 0010, 0110, 1010 and 1110, the four rows where C = 1 and D = 0.
F = C AND NOT D in all 16 rows.
Column F is marked against the signal the student has labelled F in their circuit.
F $=$ C AND (NOT D). X $=$ E OR F.
Marker: correctly completes column F
Column X: 1 when E = 1 or F = 1.
| A | B | C | D | E | F | X |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 0 | 1 |
The outlined cell is row 0010, the only row where E = 0 but X = 1, because F = 1 there.
X = E OR F in all 16 rows.
Allow FT mark for column X based on values for column F.
F $=$ C AND (NOT D). X $=$ E OR F.
Marker: correctly completes column X
Boolean expression: X = (A OR B) OR (C AND NOT D). Gates: two 2-input OR gates, one AND gate, one NOT gate.
A and B feed OR1. Its output feeds OR2, whose output drives the boom. Mark 1 C feeds the AND gate, whose output feeds OR2. Mark 2 D passes through the NOT gate into the AND gate’s second input. Mark 3
Inputs labelled “A: train sensor 1 (train detected = 1)” and “B: train sensor 2 (train detected = 1)”. Mark 4 Inputs labelled “C: manual switch (on = 1)” and “D: maintenance key (inserted = 1)”. Mark 5 OR1 output labelled E, AND output labelled F, OR2 output labelled “X: boom (lowered = 1)”. Mark 6
Column F: 1 only when C = 1 and D = 0. Mark 7 Column X: 1 when E = 1 or F = 1. Mark 8
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- No NOT on the key. Feeding D straight into the AND gate means the manual switch works only when the key is inserted. Your F column would have 1s in rows 0011, 0111, 1011 and 1111 instead.
- Putting the NOT D across the whole output. ANDing NOT D with the whole output (sensors and switch) means an inserted key also blocks the train sensors, so a train could arrive with the boom up during maintenance.
- Labelling with letters only. Writing “A” or “D” on a line earns no label marks. Each label needs the name and the on-state, for example D: maintenance key (inserted = 1).
- Copying E into column X. X is 1 wherever E = 1, but also in the row where F = 1 and there is no train: C = 1, D = 0, which is row 0010. Copying E loses that row and the mark.
The past QCAA questions this one was modelled on
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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