Question 7 of 7: Combinational Logic Design — Vehicle Alarm and Control

Question 7: Combinational Logic Design — Vehicle Alarm and 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 Boolean inputs A–G with the stated active-high meanings. "Any door open" is the complement of C (all-closed), and "belt not engaged" is the complement of D. "Key removed" is A'.

Find. Minimal combinational expressions and gate circuits for the alarm, engine-start, door-lock and door-unlock outputs.

Approach. Translate each English clause into a product (AND) term, OR the clauses for a given output, and factor common sub-expressions. All four outputs are purely combinational — no memory is required.

a) Alarm

The three triggering clauses are: ignition active with a fault, $B\,(C'+D')$; in motion with a fault, $G\,(C'+D')$; and key removed with lights on, $A'E$. Since the first two share the fault term $(C'+D')$, factor it:

$$\boxed{\text{Alarm}=(B+G)(C'+D') + A'E}$$

b) Engine start

All four enabling conditions must hold simultaneously — a single AND of ignition-active, all-doors-closed, belt-engaged and gear-in-Park:

$$\boxed{\text{Start}=B\,C\,D\,F}$$

c) Automatic door lock

Doors lock when they are closed and the car is moving:

$$\boxed{\text{Lock}=C\,G}$$

d) Automatic door unlock

Doors unlock when the car is stationary and the key has been removed:

$$\boxed{\text{Unlock}=G'\,A'}$$

These outputs are consistent: the engine-start term $BCDF$ requires exactly the "no-fault" conditions ($C=D=1$) that keep the alarm's $(C'+D')$ factor at 0, so a correctly-started engine never trips the ignition-active alarm branch. Lock and unlock are logical opposites on the motion variable, and unlock additionally requires the key removed, matching the specification.