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.
For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.
Name: ____________________ Date: ____________
1.Easy
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
1AB = 00, CD = 00
2AB = 00, CD = 01
3AB = 00, CD = 11
4AB = 00, CD = 10
5AB = 01, CD = 00
6AB = 01, CD = 01
7AB = 01, CD = 11
8AB = 01, CD = 10
9AB = 11, CD = 00
10AB = 11, CD = 01
11AB = 11, CD = 11
12AB = 11, CD = 10
13AB = 10, CD = 00
14AB = 10, CD = 01
15AB = 10, CD = 11
16AB = 10, CD = 10
2.Easy
The inputs of the logic circuit are A = 1, B = 0 and C = 1. Exactly one of its gates has the output 1. Click that gate.
A = 1B = 0C = 1
1N1
2N2
3G1
4G2
5G3
3.Medium
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 3, 4). How many literals L does it contain in total (a primed variable counts as one literal)?
F(A, B, C) = Σm(0, 3, 4)
L =
4.Medium
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(9, 10, 11, 12)
1AB = 00, CD = 00
2AB = 00, CD = 01
3AB = 00, CD = 11
4AB = 00, CD = 10
5AB = 01, CD = 00
6AB = 01, CD = 01
7AB = 01, CD = 11
8AB = 01, CD = 10
9AB = 11, CD = 00
10AB = 11, CD = 01
11AB = 11, CD = 11
12AB = 11, CD = 10
13AB = 10, CD = 00
14AB = 10, CD = 01
15AB = 10, CD = 11
16AB = 10, CD = 10
5.Hard
A synchronous decade (mod-10) counter counts 0, 1, …, 9 and then returns to 0. It starts at 8. Which count Q (in decimal) does it show after 19 clock pulses?
Q0 = 819N = 10
Q =
6.Hard
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 3 clock edges.
Q₃Q₂Q₁Q₀ = 11003
Q =
Answer key
1. AB = 10, CD = 10
2. G2
3. L = 5
4. AB = 11, CD = 00
5. Q = 7
6. Q = 9
Full step-by-step solutions: practise this topic on examproblems.com.