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
ABCD0001111000011110

Click the cell of the 4-variable Karnaugh map that belongs to minterm m11 (variables A, B, C, D with A the most significant bit).

m11
  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
2.Easy
ABCD0001111000011110

Click the cell of the 4-variable Karnaugh map that belongs to minterm m10 (variables A, B, C, D with A the most significant bit).

m10
  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
3.Medium
ABC00011110011011031204050716

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(0, 1, 2, 6). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(0, 1, 2, 6)
N =
4.Medium
ABCD000111100001111000111302041507060121131150141809011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 3, 5, 8, 10, 13, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 3, 5, 8, 10, 13, 15)
N =
5.Hard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKSHIFT

A 4-bit shift register Q₃Q₂Q₁Q₀ holds 0101. 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, 1, 0, 0, 1, 1 at the 6 edges. Give the final state Q = Q₃Q₂Q₁Q₀ as a decimal number.

Q₃Q₂Q₁Q₀ = 01010, 1, 0, 0, 1, 1
Q =
6.Hard
ABCD0001111000011110000103X214X507060120131151140809111010

For F(A, B, C, D) = Σm(4, 11, 14, 15) + d(2, 5), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(4, 11, 14, 15) + d(2, 5)
N =

Answer key

  1. 1. AB = 10, CD = 11
  2. 2. AB = 10, CD = 10
  3. 3. N = 2
  4. 4. N = 4
  5. 5. Q = 12
  6. 6. N = 3

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