382 BOOLEAN ALGEBRA [CHAP. 15
EXAMPLE 15.11
(a) Suppose a logic circuitLhasn=4 input devicesA,B,C,D. The 2n= 24 =16-bit special sequences for
A,B,C,Dfollow:
A= 0000000011111111 ,C= 00110011001100110011
B= 0000111100001111 ,D= 01010101010101010101
That is:
(1) Abegins with eight 0’s followed by eight 1’s. (Here 2n−^1 = 23 =8.)
(2) Bbegins with four 0’s followed by four 1’s, and so on. (Here 2n−^2 = 22 =4.)
(3) Cbegins with two 0’s followed by two 1’s, and so on. (Here 2n−^3 = 21 =2.)
(4) Dbegins with one 0 followed by one 1, and so on. (Here 2n−^4 = 20 =1.)
(b) Suppose a logic circuitLhasn=3 input devicesA,B,C. The 2n= 23 =8-bit special sequences forA,B,
Cand their complementsA′,B′,C′are as follows:
A= 00001111 ,B= 00110011 ,C= 01010101
A′= 11110000 ,B′= 11001100 ,C′= 10101010
Figure 15-15 contains a three-step algorithm for finding the truth table for a logic circuitLwhere the output
Yis given by a Boolean sum-of-products expression in the inputs.
Fig. 15-15
EXAMPLE 15.12 Algorithm 15.5 is used to find the truth tableT=T (L)of the logic circuitLin Fig. 15-12
or, equivalently, of the above Boolean sum-of-products expression
Y=A·B·C+A·B′·C+A′·B
(1) The special sequences and their complements appear in Example 15.14(b).
(2) The products are as follows:
A·B·C= 00000001 ,A·B′·C= 00000100 ,A′·B= 00110000
(3) The sum isY=00110101.
Accordingly,
T( 00001111 , 00110011 , 01010101 )= 00110101
or simplyT (L)=00110101 where we assume the input consists of the special sequences.