Question 7 of 7: Combinational Logic Design — Subway Door Control

Question 7: Combinational Logic Design — Subway Door Control (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Both the open-trigger and close-trigger conditions reduce, from the prose, to a single AND-of-four-literals condition — i.e. each truth table has exactly one row equal to 1 and the minimal realization is a 4-input AND gate with inverters on the literals that must be FALSE.

Open condition. The doors open only when: the train is not moving (A′), the doors are not already open — "no action if the doors are already open" (B′), the train is at a station (C), and the open button is pressed (D). Sensors E, F, G do not affect opening. $$ \text{Open} = A'\!\cdot\!B'\!\cdot\!C\!\cdot\!D. $$

Close condition. The doors close only when: the doors are currently open — "no action if the doors are closed" (B), the close button is pressed (E), the signal is green, i.e. NOT "not green" (G), and no obstruction has been detected (F′). The problem statement conditions closing on signal and obstruction only, not on train motion, so A does not appear. $$ \text{Close} = B\!\cdot\!E\!\cdot\!F'\!\cdot\!G. $$

Logic circuits (parts c, d). Because each truth table has exactly one 1-row, the minimal sum-of-products IS that single product term — no Karnaugh-map reduction is needed. Each is realized directly as inverters on the negated literals feeding a single 4-input AND gate.