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Processor Architectures and Boolean Algebra: Question 8

Syllabus 15.2

Structured A2 6 marks

A logic circuit implements the Boolean expression Y = A.B + NOT A.C, where A, B and C are single-bit inputs.

(a) Complete the truth table for Y, for all eight combinations of A, B and C. [3]

A B C Y
0 0 0 ?
0 0 1 ?
0 1 0 ?
0 1 1 ?
1 0 0 ?
1 0 1 ?
1 1 0 ?
1 1 1 ?

(b) State the number of rows in your completed table for which Y = 1. [1]

(c) List the individual logic gates needed to build a circuit that implements Y directly from the expression above, without first simplifying it, stating how many of each gate are required. [2]

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Worked solution

Part (a): Completing the truth table

Y = A.B + NOT A.C is worked out for each row by first finding A.B and NOT A.C separately, then combining them with OR.

ABCNOT AA.BNOT A.CY = A.B + NOT A.C
0001000
0011011
0101000
0111011
1000000
1010000
1100101
1110101

Two rows checked explicitly:

  • Row A=0, B=0, C=1: NOT A = 1, so NOT A.C = 1.1 = 1; A.B = 0.0 = 0. Y = 0 + 1 = 1.
  • Row A=1, B=0, C=1: NOT A = 0, so NOT A.C = 0.1 = 0; A.B = 1.0 = 0. Y = 0 + 0 = 0.

So the completed Y column is: 0, 1, 0, 1, 0, 0, 1, 1 (for ABC = 000, 001, 010, 011, 100, 101, 110, 111). [3 marks: 1 for the correct Y values when A = 0 (rows 000–011: 0, 1, 0, 1), 1 for the correct Y values when A = 1 (rows 100–111: 0, 0, 1, 1), 1 for the fully correct table overall]

Part (b): Counting rows where Y = 1

Reading down the completed Y column (0, 1, 0, 1, 0, 0, 1, 1), the value 1 appears in rows ABC = 001, 011, 110 and 111.

That is 4 rows out of the 8 where Y = 1. [1 mark]

Part (c): Gates needed to build the circuit directly

Building Y = A.B + NOT A.C exactly as written, without simplifying it first, requires:

  • 1 NOT gate, to invert A and produce NOT A.
  • 2 two-input AND gates: one to compute A.B, and a separate one to compute NOT A.C (using the output of the NOT gate as one of its inputs).
  • 1 two-input OR gate, to combine the outputs of the two AND gates and produce Y.

That is 4 gates in total: 1 NOT, 2 AND, 1 OR. [2 marks: 1 for correctly including the NOT gate together with the two AND gates, 1 for the correct final OR gate combining them, with the correct total gate count]

Final answers

  • (a) Y: 0, 1, 0, 1, 0, 0, 1, 1 (for ABC = 000, 001, 010, 011, 100, 101, 110, 111)
  • (b) Y = 1 for 4 rows
  • (c) 1 NOT gate, 2 (two-input) AND gates, 1 (two-input) OR gate, 4 gates in total