Combinational Logic for a Two-Level Lift: 04-BS-4, December 2013, Q7

Question 7: Combinational Logic for a Two-Level Lift (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. A = person in car, B = car on 1st floor, C = car on 2nd floor, D/E = 1st/2nd floor corridor buttons, F/G = in-car buttons for 1st/2nd floor, H = doors closed. All gate types are permitted; assume no two sensors change simultaneously.

Find. combinational logic that (a) starts the upward move on the relevant corridor/wall call, and (b) starts the appropriate move on an in-car/in-platform button.

The eight sensors are Boolean. The two enabling rules — corridor buttons work only for an empty car ($A=0$) and in-car buttons only for an occupied car ($A=1$) — together with “doors closed” ($H=1$) and the floor sensor gate every command, so each output is a single product (AND) term.

(a) Corridor call to go up (1→2). Movement up is initiated by the 2nd-floor corridor button $E$, but only for an empty car on the first floor with the doors shut:$$\text{UP}_a=\overline{A}\cdot B\cdot H\cdot E.$$
Logic for part (a): UP = A′·B·H·E (empty car, on 1st floor, doors closed, 2nd-floor corridor call).
(b) In-car buttons. The in-car buttons act only with a person aboard ($A=1$) and doors closed. Pressing $G$ (go to 2nd) from the first floor drives up; pressing $F$ (go to 1st) from the second floor drives down:$$\text{UP}_b=A\cdot B\cdot H\cdot G,\qquad \text{DOWN}_b=A\cdot C\cdot H\cdot F.$$
Logic for part (b): occupied-car buttons — UP = A·B·H·G, DOWN = A·C·H·F.

Logic expressions
Action	Boolean expression
(a) Up on wall/corridor call	$\overline{A}\,B\,H\,E$
(b) Up on in-car/in-platform button	$A\,B\,H\,G$
(b) Down on in-car/in-platform button	$A\,C\,H\,F$

Topic: Combinational logic design from a specification.

Key relations: Boolean AND ($\cdot$), complement ($\overline{X}$); a “disable when $X$” rule → literal $\overline{X}$; an “enable when $X$” rule → literal $X$.

Why this works: The specification is a set of independent enabling conditions, so it maps directly onto AND gates; the “disable” clauses become the complemented literal $\overline{A}$ (or the asserted $A$). No minimisation is needed because each command is already a single minterm-style product.

Common pitfall: Omitting the doors-closed / gate-closed term $H$ (a safety interlock), or using $A$ instead of $\overline{A}$ for a corridor call (which would let a summoned car move while occupied).

Source: Mano & Ciletti, Digital Design, Ch. 2–4 (Boolean algebra, combinational design).

Question 7 of 7: Combinational Logic for a Two-Level Lift

Question 7: Combinational Logic for a Two-Level Lift (20 marks)