Question 7 of 7: Combinational Logic Design – Elevator Control

Question 7: Combinational Logic Design – Elevator 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. Eleven Boolean inputs: $A$ (occupied), $B,C,D$ (position sensors: 1F/2F/3F), $E,F,G$ (corridor call buttons: 1F/2F/3F), $H,I,J$ (in-car buttons: 1F/2F/3F), $K$ (3F card-reader swipe). Rules: corridor buttons are active only when the car is empty ($A'$); a move whose destination is the 3rd floor additionally requires $K$; at most one position/button sensor is asserted at a time.

Find. Boolean expressions (and a gate-level circuit) for the four outputs: $UP1$ (up one floor), $DOWN1$ (down one floor), $UP2$ (up two floors), $DOWN2$ (down two floors).

Approach. For each output, enumerate every (current-floor, requesting-signal) pair that produces that specific motion, AND-gate each pair's conditions (gating corridor requests with $A'$ and in-car requests with $A$, and gating any 3rd-floor destination with $K$), then OR the pairs together.

1. a) Up one floor ($UP1$). This motion happens from 1F→2F (corridor call $F$ while empty, or in-car $I$ while occupied) or from 2F→3F (corridor call $G$ or in-car $J$, both requiring the card $K$ since the destination is the 3rd floor): $$UP1 = B\cdot(F\cdot A' + I\cdot A) \;+\; C\cdot K\cdot(G\cdot A' + J\cdot A)$$
2. b) Down one floor ($DOWN1$). From 2F→1F (corridor $E$ or in-car $H$) or from 3F→2F (corridor $F$ or in-car $I$) – no card needed, since the destination is not the 3rd floor: $$DOWN1 = C\cdot(E\cdot A' + H\cdot A) \;+\; D\cdot(F\cdot A' + I\cdot A)$$
3. c) Up two floors ($UP2$). Only possible from 1F→3F directly (corridor $G$ or in-car $J$), and the destination is the 3rd floor so $K$ is required: $$UP2 = B\cdot K\cdot(G\cdot A' + J\cdot A)$$
4. d) Down two floors ($DOWN2$). Only possible from 3F→1F (corridor $E$ or in-car $H$); leaving the 3rd floor needs no card: $$DOWN2 = D\cdot(E\cdot A' + H\cdot A)$$ Each output above was spot-checked against representative input vectors (e.g. an empty car requesting the 3rd floor with $K=0$ correctly produces no motion at all; the same request with $K=1$ correctly asserts only $UP2$), confirming both the corridor-disable-when-occupied rule and the card-gate-on-3F-destination rule.