(a) Write the Boolean expression that corresponds to the given truth table as a sum-of-products. INPUT OUTPUT A B C D Z 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 Z =
- (a)Write the Boolean expression that corresponds to the given truth table as a sum-of-products. INPUT OUTPUT A B C D Z 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 Z = [3]
- (b)(i)Complete the Karnaugh map (K-map) for the given truth table. [Figure 7.1][2]
- (b)(ii)Draw loop(s) around appropriate group(s) of 1s in the K-map to produce an optimal sum-of-products.[2]
- (b)(iii)Write the Boolean expression from your answer to part b(ii) as a simplified sum-of-products. Z = [2]
- (b)(iv)Write the simplified Boolean expression for your answer to part b(iii). Z = [1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [3] One mark per two correct products (Max 3)
- (Z =) A̅B̅CD + A̅BCD̅ + A̅BC̅D̅ + A̅BCD + ABCD̅ + ABCD
- [2] One mark for every two correct rows or columns (Max 2)
- AB\CD | 00 | 01 | 11 | 10
- ----------|----|----|----|----
- 00 | 0 | 0 | 1 | 0
- 01 | 0 | 0 | 1 | 1
- 11 | 0 | 0 | 1 | 1
- 10 | 0 | 0 | 1 | 0
- [2] One mark for correct loop (Max 2)
- [Figure 7.2 with loops]
- • Loop around (11,00), (11,01), (11,11), (11,10) - simplifies to AB
- • Loop around (10,01), (10,11) - simplifies to A B̅ D
- • Loop around (00,11), (01,11) - simplifies to A̅ CD
- [2] One mark per correct marking point (Max 2)
- • A B // A D
- + A D // + AB
- (Z =) A B + A D // A D + A B
- [1] (Z =) A (B + D) // A (D + B)
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