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04-BS-4 · May 2017

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, 04-BS-4 Electric Circuits and Power — May 2017. Closed book; one double-sided aid sheet permitted; any five of the seven questions constitute a complete paper (all seven are answered here as a complete study resource).

Reference texts: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed. (circuit analysis, transients, AC steady state, magnetic circuits); Boylestad & Nashelsky, Electronic Devices and Circuit Theory (rectifiers); Mano, Digital Design (combinational logic).

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.

Given. Seven binary sensors A–G (train moving, doors open, at station, open-button, close-button, obstruction, signal-green) and the stated open/close business rules, including the two "no action if already in that state" guards.

Find. The OPEN truth table and gate design (parts a, c), and the CLOSE truth table and gate design (parts b, d).

Approach. Translate each English rule directly into a Boolean AND of literals (complementing wherever the rule says "not" or "already"), then read the truth table straight off that single product term — a 1-minterm function needs only an AND gate with the appropriate input inversions, no simplification required.

  1. OPEN logic (parts a, c). Doors open when: not moving ($A'$), at a station ($C$), open button pressed ($D$), AND doors not already open ($B'$, the "no action if already open" guard): $$ OPEN = \boxed{A'\cdot B'\cdot C\cdot D}. $$ Sensors E, F, G do not appear in OPEN — the close-side rules do not gate the open function.
  2. CLOSE logic (parts b, d). Doors close when: currently open ($B$, the "no action if already closed" guard), close button pressed ($E$), signal green ($G$), AND no obstruction ($F'$): $$ CLOSE = \boxed{B\cdot E\cdot F'\cdot G}. $$ A and C/D do not appear — closing is independent of train motion/station/open-button state once the door is open and the close button is pressed.
Truth table
ABCDOPEN
00000
00010
00100
00111
01000
01010
01100
01110
10000
10010
10100
10110
11000
11010
11100
11110
Truth table
BEFGCLOSE
00000
00010
00100
00110
01000
01010
01100
01110
10000
10010
10100
10110
11000
11011
11100
11110
Door-Open Logic: OPEN = A'·B'·C·DABCDOPEN
OPEN gate: 4-input AND with A, B inverted
Door-Close Logic: CLOSE = B·E·F'·GBEFGCLOSE
CLOSE gate: 4-input AND with F inverted
Final results
FunctionExpression
OPENA'·B'·C·D
CLOSEB·E·F'·G
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