Digital logic

Digital systems represent information with two discrete levels, 0 and 1. To design or analyse them you must be able to convert between number systems, know how logic levels and gates behave, simplify Boolean expressions, and turn a truth table into a minimal circuit using Sum-of-Products, Product-of-Sums and Karnaugh maps. These skills are tested through short conversion and simplification questions.

Key concepts

• Number systems:

System	Base	Digits	Example
Binary	2	0, 1	1011₂ = 11₁₀
Octal	8	0–7	13₈ = 11₁₀
Decimal	10	0–9	11
Hexadecimal	16	0–9, A–F	B₁₆ = 11₁₀

• Decimal to base-r: repeatedly divide by r and read remainders bottom-up; for fractions, multiply by r and read integer parts top-down.
• Binary ↔ octal: group 3 bits; binary ↔ hex: group 4 bits, starting from the binary point.
• Codes: BCD (each decimal digit in 4 bits; 1010–1111 invalid), Gray code (only one bit changes between successive values; used in encoders and K-map ordering), Excess-3, ASCII (7-bit).
• Logic levels: Positive logic means high voltage = 1. For TTL, a valid input low is ≤ 0.8 V and input high ≥ 2.0 V; output low ≤ 0.4 V and output high ≥ 2.4 V (supply 5 V). The difference between guaranteed output and required input levels is the noise margin (0.4 V for TTL). Other characteristics: fan-out, propagation delay, power dissipation.
• Logic gates:

Gate	Expression	Output is 1 when
AND	$AB$	all inputs 1
OR	$A+B$	any input 1
NOT	$\bar A$	input 0
NAND	$\overline{AB}$	any input 0
NOR	$\overline{A+B}$	all inputs 0
XOR	$A\oplus B$	inputs differ (odd number of 1s)
XNOR	$\overline{A\oplus B}$	inputs equal

NAND and NOR are universal gates: any logic function can be built from either one alone.

• Boolean algebra laws: identity ($A+0=A$, $A\cdot1=A$), null ($A+1=1$, $A\cdot0=0$), idempotent ($A+A=A$), complement ($A+\bar A=1$, $A\bar A=0$), involution ($\bar{\bar A}=A$), commutative, associative, distributive ($A+BC=(A+B)(A+C)$), absorption ($A+AB=A$, $A(A+B)=A$), and $A+\bar AB = A+B$.
• De Morgan's theorems: $\overline{A+B}=\bar A\bar B$ and $\overline{AB}=\bar A+\bar B$. "Break the bar, change the sign."
• Duality: Swap AND↔OR and 0↔1 to get the dual of any valid identity.
• Sum-of-Products (SOP): OR of AND terms; each row of the truth table where output = 1 gives a minterm $m_i$. Canonical form: $F=\sum m(\dots)$. Implemented naturally with AND-OR or NAND-NAND.
• Product-of-Sums (POS): AND of OR terms; each row with output = 0 gives a maxterm $M_i$. $F=\prod M(\dots)$. Implemented with OR-AND or NOR-NOR.
• Karnaugh map: A grid arranged in Gray-code order so that adjacent cells differ by one variable. An n-variable map has $2^n$ cells. Group 1s (for SOP) or 0s (for POS) in rectangles of 1, 2, 4, 8… cells, making groups as large as possible; groups may wrap around edges and overlap. A group of $2^k$ cells eliminates $k$ variables. "Don't care" (X) cells may be treated as 1 or 0, whichever gives a simpler result.

High-yield points

• NAND and NOR are universal gates.
• De Morgan: $\overline{A+B} = \bar A\bar B$, $\overline{AB} = \bar A + \bar B$.
• $A + AB = A$ and $A + \bar AB = A + B$.
• TTL: input low ≤ 0.8 V, input high ≥ 2.0 V; noise margin 0.4 V.
• A K-map group of $2^k$ cells eliminates $k$ variables.
• $n$ variables give $2^{2^n}$ possible Boolean functions.

Formulas

$N_{10} = \sum_i d_i\, r^{i}$ $d_i$ = digit at position $i$, $r$ = base (radix).

Number of possible Boolean functions of $n$ variables: $2^{2^n}$.

Worked example

Simplify $F(A,B,C)=\sum m(0,2,4,5,6)$.

1. Truth-table 1s at 000, 010, 100, 101, 110.
2. K-map: cells 0, 2, 4, 6 all have $C=0$ → group of 4 gives $\bar C$.
3. Cells 4 and 5 ($A=1,B=0$) → group of 2 gives $A\bar B$.
4. $F=\bar C + A\bar B$.
5. Check row 101: $\bar C=0$, $A\bar B=1$ → F=1 ✓; row 011 (m3): $\bar C=0$, $A\bar B=0$ → F=0 ✓.

Conversion check: $(2F)_{16} = 2\times16+15 = 47_{10} = 101111_2$.

Exam tips

• NAND and NOR are universal; XOR is not.
• 1's complement inverts bits; 2's complement = 1's complement + 1 (covered further in arithmetic circuits).
• A K-map group of 8 in a 4-variable map leaves one literal.
• Minterms come from output 1; maxterms from output 0. $\sum m(\dots)$ and $\prod M(\dots)$ of the same function use complementary index sets.
• XOR of an odd number of 1s is 1 — used for parity generators.

Exam-style questions

From Calmon's question bank, which has 29 questions on this topic.

1. Convert (312)₈ into decimal:

• A. (200)₁₀
• B. (202)₁₀
• C. (204)₁₀
• D. (206)₁₀

Answer: B. Converting (312)₈ to decimal gives 3x64 + 1x8 + 2x1 = 202₁₀.

2. Which of the following does Boolean algebra NOT have?

• A. Complement of each element
• B. Identity element for the AND operation
• C. Multiplicative inverse
• D. Distributive law of AND over OR

Answer: C. Boolean algebra has complements ($a + \bar{a} = 1$), identities (1 for AND, 0 for OR) and distributive laws, but no multiplicative inverse (no $a^{-1}$ with $a \cdot a^{-1} = 1$ for $a = 0$).

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