The big idea: A truth table lists every possible set of inputs and what the circuit outputs for each one.
It is the complete answer. Once the table is filled in, there is nothing left to wonder about.
How many rows: Two inputs give 4 rows, three give 8, four give 16 — the count doubles with every extra input.
The rule is 2 to the power of the number of inputs.
Write the inputs the same way every time: Count upwards in binary: 000, 001, 010, 011, 100, 101, 110, 111.
The rightmost column alternates every row, the next every two rows, the next every four. Follow that pattern and you cannot miss a combination.
1 — Name the inputs
Each condition in the description that is either true or false becomes one input with a letter: armed (A), door open (D), motion (M).
2 — Turn the sentence into an expression
"The alarm sounds when it is armed and a door opens or motion is seen" becomes S = A AND (D OR M).
"Unless" and "not" become NOT; "either" usually means OR.
3 — List every combination
Three inputs give 2 × 2 × 2 = 8 rows, counted up in binary from 000 to 111.
4 — Fill the output row by row
Apply the expression to each row. Then check one row against the description in words: armed, no door, no motion must give S = 0.
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Do one gate at a time: Give every gate in the circuit its own column, then work left to right.
Never try to jump from the inputs straight to the final output — that is where marks are lost.
X = (A AND B) OR NOT C. Step through all eight input combinations and watch the table fill in.
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1 — Set up the columns
One column per input, then one per gate, then the final output.
For this circuit: A, B, C, then the AND gate, then the NOT gate, then X.
2 — Fill the input columns
Three inputs, so eight rows, counted upwards in binary from 000 to 111.
Do this before touching any gate.
3 — Work out each gate column
Go down the AND column first, using only A and B. Then the NOT column, using only C.
Each entry is a one-second decision, because you are looking at one gate.
4 — Combine for the output
The OR column takes the two gate columns you already filled in — not the original inputs.
The working columns are what make the last step easy.
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A table is also an expression: Write · for AND, + for OR, and NOT for inversion.
So the circuit above is X = (A·B) + NOT C. Table, circuit and expression are three views of one thing.
Algebraic simplification
- Apply Boolean rules to shorten the expression
- A + A·B = A — if A alone is enough, the rest is redundant
- A·A = A and A + A = A
- A·(NOT A) = 0 and A + (NOT A) = 1
- Fast when you spot the pattern, easy to miss one
Karnaugh map
- Draw the outputs in a grid, then read the answer off
- Neighbouring squares differ by one variable only
- Circle groups of 1s in sizes 1, 2, 4 or 8
- A group spanning both values of a variable drops it
- Slower to draw, but it finds the simplification for you
Using a Karnaugh map
Put A down the side and B, C across the top, labelled 00, 01, 11, 10 — that order is what makes neighbours differ by one variable.
Write each row's output into its square.
Circle and read
Circle the biggest groups of adjacent 1s you can, in sizes 1, 2, 4 or 8. Groups may overlap, and may wrap around the edges.
For each circle, keep the variables that stay the same across it and drop the ones that change. OR the circles together.
Why simplify at all
A shorter expression is a circuit with fewer gates. That is less hardware, less power, and less to go wrong.
That is the reason the exam asks for it.
How this is tested — you must produce the whole table with working columns, not just the final answer. It comes up two ways:
Paper 1 Section A
- Construct a truth table from a circuit or a description, 3-6 marks
- Predict the output for stated inputs
- Write or simplify the Boolean expression
Paper 1 Section B — case study
- Build the table for a rule in the case study
- Say what happens when one input changes
The classic trap: Missing rows. Three inputs means eight rows, always — and they must be counted upwards in binary. A table with six rows loses the marks for the ones that are not there.
A pump runs when the tank is not full and either the manual switch is on or the timer is active. Construct a truth table for this rule, using F (full), M (manual) and T (timer).
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