Identify and use the standard symbols for logic gates

Cambridge IGCSE Computer Science 0478 – Boolean Logic

Learning objective

Identify and use the standard schematic symbols for the basic logic gates and move fluently between a textual description, a Boolean expression, a truth‑table and a circuit diagram.

1. Symbol key – “gate profile” (Cambridge schematic)

The table below shows the exact Cambridge symbol for each gate, its common name, the Boolean notation used in the syllabus and a concise verbal definition of the gate’s function.

Gate	Cambridge symbol	Boolean expression	Function (in words)
AND		\(A\!\cdot\!B\) (or \(AB\))	Outputs 1 only when **both** inputs are 1.
OR		\(A + B\)	Outputs 1 when **at least one** input is 1.
NOT		\(\overline{A}\)	Outputs the opposite (inverse) of the single input.
NAND		\(\overline{A\!\cdot\!B}\)	Outputs 0 only when **both** inputs are 1 (the opposite of AND).
NOR		\(\overline{A + B}\)	Outputs 1 only when **both** inputs are 0 (the opposite of OR).
XOR		\(A \oplus B\)	Outputs 1 when **exactly one** of the inputs is 1.
XNOR		\(\overline{A \oplus B}\)	Outputs 1 when the two inputs are **equal** (both 0 or both 1).

2. Truth tables for the two‑input gates

Each gate’s complete truth table is reproduced below for quick reference.

3. Using the symbols in Boolean expressions

• AND – write \(A\!\cdot\!B\) or simply \(AB\).
• OR – write \(A + B\).
• NOT – write \(\overline{A}\) (a bar over the variable) or !A in programming style.
• NAND – \(\overline{A\!\cdot\!B}\).
• NOR – \(\overline{A + B}\).
• XOR – \(A \oplus B\) (true when exactly one input is true).
• XNOR – \(\overline{A \oplus B}\) (true when the inputs are equal).

When drawing a circuit the same symbols are used; the Boolean expression is the “textual” counterpart of the diagram.

4. Converting between description, expression, truth table and circuit

1. English description → Boolean expression – replace “and” with “·”, “or” with “+”, “not”/“false” with an over‑bar.
2. Expression → Truth table – list all possible input combinations (2ⁿ rows for n inputs) and evaluate the expression.
3. Expression → Circuit diagram – use the Cambridge symbols that match each operator; combine them in the order dictated by parentheses.
4. Circuit diagram → Expression – read the diagram from left to right (or top to bottom), writing the corresponding Boolean operators.

5. Worked example – from a problem statement to a circuit

Problem statement:

“The output F should be true when A is true **and** either B is false **or** C is true.”

1. Translate the English description into a Boolean expression.
   • “A is true” → \(A\)
   • “B is false” → \(\overline{B}\)
   • “C is true” → \(C\)
   • “either B is false **or** C is true” → \(\overline{B}+C\)
   • “A is true **and** (…)” → \(A\!\cdot\!(\overline{B}+C)\)

   Hence \(F = A\!\cdot\!(\overline{B}+C)\).

2. Construct the truth table. (Three inputs → eight rows)

   A	B	C	\(\overline{B}\)	\(\overline{B}+C\)	F = A·(\(\overline{B}+C\))
   0	0	0	1	1	0
   0	0	1	1	1	0
   0	1	0	0	0	0
   0	1	1	0	1	0
   1	0	0	1	1	1
   1	0	1	1	1	1
   1	1	0	0	0	0
   1	1	1	0	1	1

3. Draw the circuit using Cambridge symbols.
   • NOT gate on input B → \(\overline{B}\).
   • OR gate combining \(\overline{B}\) and C → \(\overline{B}+C\).
   • AND gate combining the OR output with A → final output F.

   

6. Additional example – building a three‑input NAND gate

The syllabus only requires two‑input symbols, but a three‑input NAND can be constructed by cascading two‑input NAND gates.

1. First NAND: inputs A and B → \(X = \overline{A\!\cdot\!B}\).
2. Second NAND: inputs X and C → \(F = \overline{X\!\cdot\!C}= \overline{\overline{A\!\cdot\!B}\!\cdot\!C}\).
3. Using De Morgan’s law, the overall function simplifies to \(\overline{A\!\cdot\!B\!\cdot\!C}\) – a three‑input NAND.

7. Practice questions (exam‑style)

1. From description to expression – Write the Boolean expression for a circuit that outputs true only when both A and B are false.
   Answer: \(F = \overline{A}\!\cdot\!\overline{B}\) (or \(F = \overline{A + B}\)).
2. Identify the symbol – Which Cambridge schematic symbol corresponds to the expression \(\overline{A + B}\)?
   Answer: The **NOR** gate (OR symbol with a small circle on the output).
3. From expression to circuit (word description) – List the gates required to implement \(F = (A\!\cdot\!B) + (\overline{C})\).
   Answer: ONE AND gate (inputs A, B), ONE NOT gate (input C), ONE OR gate (inputs the AND output and the NOT output).
4. From circuit to expression – The circuit below uses a NOT gate on input A, an OR gate on inputs B and C, and an AND gate that combines the NOT output with the OR output. Write the Boolean expression for the overall output F.
   

   

   Answer: \(F = \overline{A}\!\cdot\!(B + C)\).
5. Complete the truth table – Fill in the missing column for the gate marked “XOR”.
   

   A	B	F (XOR)
   0	0	0
   0	1	1
   1	0	1
   1	1	0

6. Design challenge (multi‑gate) – Design a circuit that outputs 1 only when an odd number of the three inputs A, B, C are 1. Provide (i) the Boolean expression, (ii) the truth table, and (iii) a sketch using Cambridge symbols.
   Solution outline:
   • Expression: \(F = A\!\oplus\!B\!\oplus\!C\) (equivalently \((A\!\oplus\!B)\!\oplus\!C\)).
   • Truth table: 1 for rows 001, 010, 100, 111; 0 otherwise.
   • Circuit: two XOR gates in series – first XOR A and B, second XOR the result with C (use the standard XOR symbol).

8. Quick reference sheet (print‑friendly)

Copy the table below onto a single A4 sheet for fast revision before the exam.

Gate	Symbol	Boolean	Function (words)
AND		\(A\!\cdot\!B\)	1 only when both inputs are 1
OR		\(A + B\)	1 when at least one input is 1
NOT		\(\overline{A}\)	Inverse of the input
NAND		\(\overline{A\!\cdot\!B}\)	0 only when both inputs are 1
NOR		\(\overline{A + B}\)	1 only when both inputs are 0
XOR		\(A \oplus B\)	1 when exactly one input is 1
XNOR		\(\overline{A \oplus B}\)	1 when inputs are equal