Logic gates and truth tables

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2014, 2022)
• Priority tier: T3
• Marks (count): 10 (1), 15 (1)
• Average solve time: ~5 min
• Difficulty mix: easy 2
• Section: B | Dominant type: derivation

Why This Chapter Matters

Logic circuit questions test Boolean algebra and gate-level circuit construction. The 2014 question (10 marks) asks for the circuit for a sum-of-products expression using only AND and OR gates — a direct realisation with no simplification needed. The 2022 question (15 marks) involves a more complex expression with NOT and asks for both the circuit and the truth table; critically, the expression simplifies dramatically (to $x+y$), and spotting that simplification earns the full 15 marks efficiently. Both questions are among the fastest in the paper once you know the procedure.

Minimum Theory

Gate types. AND gate: output 1 iff all inputs are 1. OR gate: output 1 iff at least one input is 1. NOT gate (inverter): output is the complement $\bar x$.

Sum-of-products (SOP) realisation. Any Boolean expression in disjunctive normal form (DNF) — a sum of product terms — is directly realised by one AND gate per product term and one OR gate combining all products. No simplification is needed unless specifically asked.

Hierarchy of evaluation. Parentheses first, then NOT, then AND, then OR. For $f=[x(\bar y+z)]+y$: evaluate $\bar y$, then $\bar y+z$, then $x(\bar y+z)$, then add $y$.

Key Boolean identities. $x\bar y+y=x+y$ (absorption–consensus): $x\bar y+y=(x+y)(\bar y+y)=(x+y)\cdot 1=x+y$. Also $x+xy=x$ (absorption); $1+x=1$.

Question Archetypes

logic-circuit (2 question(s); 2014, 2022)

Recognition Cues

• “Use only AND and OR logic gates to construct a circuit for …”
• “Find a combinatorial circuit for … and write its input/output table.”
• A Boolean expression over 2–4 variables is given.

Solution Template

1. Identify the expression’s structure. Draw the operator hierarchy (precedence: NOT > AND > OR).
2. Design the circuit layer by layer. First layer: NOT gates for any complements. Second layer: AND gates for product terms. Third layer: OR gates for sums.
3. List components. Count gates and state inputs/outputs.
4. Write the truth table. For $n$ variables, enumerate all $2^n$ rows. Fill intermediate columns (one per gate output), then the final output column.
5. Check for simplification using Boolean identities. State the simplified expression if it exists.

Worked Example

2014 Paper 2, 2014-P2-Q5d (10 marks)

Use only AND and OR logic gates to construct a logic circuit for the Boolean expression $z=xy+uv$.

Structure. $z=xy+uv$ is already in sum-of-products form: two AND terms combined by OR. No complements needed.

Circuit.

• AND gate 1: inputs $x,y$ — output $xy$.
• AND gate 2: inputs $u,v$ — output $uv$.
• OR gate: inputs $xy$, $uv$ — output $z=xy+uv$.

Total: 2 AND + 1 OR = 3 gates.

Truth table (selected rows; full table has 16 rows):

$\boxed{z=xy+uv \text{ — 2 AND gates into 1 OR gate}}$

2022 Paper 2, 2022-P2-Q6b (15 marks)

Find a combinatorial circuit for $f(x,y,z)=[x\cdot(\bar y+z)]+y$ and write its input/output table.

Operator hierarchy.

1. NOT gate: $y\to\bar y$.
2. OR gate: $\bar y,z\to\bar y+z$.
3. AND gate: $x,(\bar y+z)\to x(\bar y+z)$.
4. OR gate: $x(\bar y+z),y\to f$.

Components: 1 NOT + 2 OR + 1 AND = 4 gates.

Truth table:

Simplification. $f=x(\bar y+z)+y=x\bar y+xz+y$. Since $x\bar y+y=x+y$ (key identity): $f=(x+y)+xz=x(1+z)+y=x+y$.

$\boxed{f(x,y,z)=x+y \text{ (simplified — }z\text{ is irrelevant)}}$

The simplified circuit needs only 1 OR gate. The truth table confirms: $f=0$ only when $x=0$ and $y=0$.

Common Traps

• Not simplifying. The 2022 expression simplifies to $x+y$, eliminating $z$ entirely. Always apply Boolean absorption after drawing the circuit; the question implicitly rewards simplification.
• Key identity. $x\bar y+y=x+y$. This is a specific case of the consensus/absorption theorem; memorise it.
• AND vs. OR gate order. In sum-of-products, AND gates come before the OR gate. Reversing the order (OR gates feeding an AND) gives a product-of-sums (POS) circuit and produces wrong outputs.
• “Only AND and OR” constraint. In 2014, no NOT gates are available. The expression $xy+uv$ has no complemented literals, so this is not a problem — but always check before designing.

Marks-Aware Writing

A 10-mark answer must show: the circuit described (gate types, inputs, outputs) and a truth table with intermediate columns. A 15-mark answer must additionally show: the operator hierarchy of the expression, the full truth table for all $2^n$ input combinations, and the Boolean simplification with algebraic steps.

Practice Set

• 2014-P2-Q5d (10 m) — — Hint: SOP form, 2 AND + 1 OR; no complements needed.
• 2022-P2-Q6b (15 m) — — Hint: simplify via $x\bar y+y=x+y$; full truth table has 8 rows.