Logic Gates and Truth Tables

AND gives 1 only if both inputs are 1. OR gives 1 if either or both are 1. EOR (exclusive OR) gives 1 when exactly one of two inputs is 1. NOT inverts one input. NAND is NOT-AND; NOR is NOT-OR.

For input pairs 00, 01, 10, 11, the output sequences are AND: 0001; OR: 0111; EOR: 0110; NAND: 1110; NOR: 1000. These rows are a compact way to check that “either” has not been misread as “exactly one”.

AQA notation uses a dot for AND, plus for OR and an overbar for NOT. A plus in Boolean algebra is a logical operation rather than ordinary arithmetic: 1+1=1 in the OR sense.

Suppose a warning is required when a door is open and the system is armed, or whenever a panic switch is pressed. Define binary inputs D, A and P with explicit meanings. The expression is $Y=(D\cdot A)+P$. Build the AND stage first, then feed its output and P into an OR gate.

For each row that requires output 1, form an AND term that selects that exact row: use an input directly when its row value is 1 and its NOT when it is 0. OR those terms together. Simplify if useful, then test all rows against the original requirement.

For two inputs with output 1 only at 01 and 10, the terms are $\overline A B$ and $A\overline B$, giving EOR. Treat gates as functional blocks; their internal transistor circuits are not required. With three inputs there are eight rows, so checking only two example cases is insufficient.

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.