A burglar alarm system has four inputs: W = Window sensor triggered (1 = yes) D = Door sensor triggered (1 = yes) A = Alarm is armed (1 = yes) O = Override ...

(a) Write the Boolean expression for Z. [2]

The alarm sounds when all three conditions are met simultaneously: the alarm is armed (A), a sensor is triggered (W or D), and the override is not active (NOT O). The expression is:

Z = A AND (W OR D) AND (NOT O) [1 for AND with sensor OR, 1 for NOT O]

(b) Complete the truth table for A = 1, O = 0. [4]

With A = 1 and O = 0, NOT O = 1. The output depends entirely on whether any sensor fires:

When no sensor is triggered (row 1), W OR D = 0, so Z = 1 AND 0 AND 1 = 0 (alarm stays silent). In all other rows at least one sensor is active, so Z = 1 (alarm sounds). [1 per row]

(c) State the output Z when W=1, D=0, A=1, O=1. [2]

Z = 1 AND (1 OR 0) AND (NOT 1) = 1 AND 1 AND 0 = 0. [1]

In context: the window sensor has been triggered and the alarm is armed, but the override switch is active. The override disables the alarm, so it does not sound even though a breach has been detected. [1]

(d) Draw the logic circuit. [3]

The circuit requires three gates connected as follows:

1. A NOT gate takes input O and produces NOT O. [1]
2. An OR gate takes inputs W and D and produces W OR D. [1]
3. An AND gate (with three inputs, or two cascaded two-input AND gates) takes A, the output of the OR gate, and the output of the NOT gate, producing the final output Z. [1]

(e) Write the modified Boolean expression with a panic button P. [2]

The panic button must always sound the alarm regardless of other inputs. This is achieved by ORing P with the original expression:

Z = (A AND (W OR D) AND (NOT O)) OR P

When P = 1, the OR guarantees Z = 1 no matter what the other inputs are. When P = 0, the expression reverts to the original behaviour. [1 for OR with P, 1 for preserving original expression]