Digital logic: gates, K-maps & flip-flops worksheet with answers

Free printable Digital logic: gates, K-maps & flip-flops worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
ABCG15 nsN1: 1 nsG26 nsG35 nsF

The propagation delay of every gate is written under it. Find the worst-case propagation delay tpd from the inputs to the output F.

N1: 1 nsG1: 5 nsG2: 6 nsG3: 5 ns
tpd = ns
2.Easy
ABCN1N2G1N3G2G3F

For the logic circuit shown, how many of the 8 combinations of the inputs A, B and C give the output F = 1? Find N.

A, B, C → F
N =
3.Medium
ABCD000111100001111011111111

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, 4, 7, 9, 10, 12, 13, 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
4.Medium
ABCD000111100001111010110312140517160120130150141809011110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 1, 2, 4, 6, 7, 8, 10). How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 4, 6, 7, 8, 10)
N =
5.Hard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1001. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 3 clock edges.

Q₃Q₂Q₁Q₀ = 10013
Q =
6.Hard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1100. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 4 clock edges.

Q₃Q₂Q₁Q₀ = 11004
Q =

Answer key

  1. 1. tpd = 12 ns
  2. 2. N = 6
  3. 3. AB = 10, CD = 10
  4. 4. N = 4
  5. 5. Q = 3
  6. 6. Q = 12

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