Digital Electronics: practice problems and exam questions

Binary numbers, logic gates and Boolean algebra, Karnaugh maps, combinational building blocks such as multiplexers, flip-flops, counters and shift registers, and the conversion between analogue and digital signals. Exams combine short hand calculations with circuit and timing diagrams.

Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Number systemsbinary and hex to decimal · two's complement · bits, ranges and memorycalculations
  2. Digital logic: gates, K-maps & flip-flopsgates, truth tables and delays · Boolean algebra, K-maps and multiplexers · don't-cares, flip-flops and counterscalculations
  3. Data acquisition & samplingNyquist, LSB and data rates · aliasing, bits and UART frames · harmonics, SQNR and system designcalculations

Exam tips for Digital Electronics

  • Label the Karnaugh map in Gray code (00, 01, 11, 10) before you place a single 1.
  • Make groups as large as possible and let them wrap round the edges; use don’t-cares only when they help.
  • In timing diagrams, read every input exactly at the active clock edge and write Q after each edge in a table.

Worked examples

Number systemsEasy
place values2256B16E1

Convert the hexadecimal number 0x2BE to decimal. Find N.

0x2BEA = 10 … F = 15

Find N

Show hints
  1. Hex places are worth 1, 16, 256, 4096 from the right.
  2. Replace letters by their values (A = 10 … F = 15), multiply by the place value and add.
  3. 2 = 2, B = 11, E = 14.
Show the answer and the step-by-step solution

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Multiply each digit by its place value

2·256 + 11·16 + 14·1 = 702

Answer

0x2BE = 702
Digital logic: gates, K-maps & flip-flopsMedium
ABCD00011110000111101111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(0, 3, 4, 10, 13, 14, 15)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
Show hints
  1. Neighbouring cells differ in exactly one variable.
  2. The leftmost and rightmost columns are neighbours, and so are the top and bottom rows.
  3. Check the four neighbours (up, down, left, right, wrapping round) of every 1.
Show the answer and explanation

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m3: m2, m1, m7, m11 → 0
ABCD00011110000111101111111

The isolated 1

m3 = A′B′CD → AB = 00, CD = 11

Answer

AB = 00, CD = 11

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

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