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
ABCN1: 1 nsG15 nsG28 nsG33 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: 8 nsG3: 3 ns
tpd = ns
2.Easy
ABCF000?001?010?011?100?101?110?111?

The function F = (A′·B′)·C is 1 for exactly one combination of the inputs. Click that row of the truth table.

F = (A′·B′)·C
  1. 1ABC = 000
  2. 2ABC = 001
  3. 3ABC = 010
  4. 4ABC = 011
  5. 5ABC = 100
  6. 6ABC = 101
  7. 7ABC = 110
  8. 8ABC = 111
3.Medium
ABC00011110010001131214150716

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(2, 3, 4, 5, 6). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(2, 3, 4, 5, 6)
N =
4.Medium
ABCD0001111000011110111111

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, 1, 9, 10, 11, 12)
  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
5.Hard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKDOWN

A 3-bit binary down counter starts at 1 (decimal, Q = 001). Which count Q (in decimal) does it show after 20 clock pulses?

Q0 = 120n = 3
Q =
6.Hard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKSHIFT

A 4-bit shift register Q₃Q₂Q₁Q₀ holds 0011. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀, Q₀ is lost) and the serial input enters Q₃. The serial input is 0, 0, 1, 0, 1 at the 5 edges. Give the final state Q = Q₃Q₂Q₁Q₀ as a decimal number.

Q₃Q₂Q₁Q₀ = 00110, 0, 1, 0, 1
Q =

Answer key

  1. 1. tpd = 11 ns
  2. 2. ABC = 001
  3. 3. N = 2
  4. 4. AB = 11, CD = 00
  5. 5. Q = 5
  6. 6. Q = 10

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