chapter  11
Boolean algebra and logic circuits
Pages 20

Problem 1. Derive the Boolean expression and construct a truth table for the switching circuit shown in Fig. 11.5.

The switches between 1 and 2 in Fig. 11.5 are in series and have a Boolean expression of B · A. The parallel circuit 1 to 2 and 3 to 4 have a Boolean expression of (B · A + B). The parallel circuit can be treated as a single switching unit, giving the equivalent of switches 5 to 6, 6 to 7 and 7 to 8 in series. Thus the output is given by:

Z = A · (B ·A+B) ·B The truth table is as shown in Table 11.2. Columns 1 and 2 give all the possible combinations of switches A and B. Column 3 is the and-function applied to columns 1 and 2, giving B · A. Column 4 is B, i.e. the opposite to column 2. Column 5 is the or-function applied to columns 3 and 4. Column 6 is A, i.e. the opposite to column 1. The output is column 7 and is obtained by applying the and-function to columns 4, 5 and 6.