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" " " " ABC AB C AB C A B C : 278. Z A 1 B A, answer, by items (9) and (7). " 279. Z AAB 1 C AB, answer, by items (5), (9), and (7). 280. One way is to rst note that A A B A AB A 1 B A, so that the given expression simpli es to Z A B C ; answer: Or instead, if we wish, we can begin by rst taking the product of the two binomials; thus Z A AB AC BB BC A B C AB 1 C AB AC AB which, since AB AB AB, by item (6), gives Z AB AC A B C ; answer; as before: " 281. One procedure is to rst note that, by items (14) and (7), we have BC B C " " " B B C C and AC A C A C C A so that the given expression becomes "" "" Z A C B C A C C B "" " Now, by item (16), note that C C B C B, so that the last expression above becomes. " Z A B C; answer: 282. One way is to rst apply item (17) then item (5), thus getting "" " " "" " Z B BCDD B BCD " Now apply the generic form of item (16), that is, A AY A Y, then apply item (17). Doing this, the last answer above becomes " " Z B C D B C D; answer: Comment. We have thus found, by applying the laws of Boolean algebra, that " " B C D B D B C D It should be noted that, although the two sides of the equation represent physically di erent switching arrangements, both sides perform the same electrical switching operation. Thus the right-hand side does the same electrical switching job as the left-hand side, but has the advantage of being physically simpler. Also, from a practical viewpoint, it s important to note that the switching network represented by the right-hand side requires just one inverter (one transistor), while the electrically equivalent network represented by the left-hand side requires three inverters. " " 283. Note rst that ABC A BC A A BC BC, by items (14) and (7), and upon making use of this relationship the problem becomes """ Z BC A B C BC A B C """ because A B C A B C (from problem 276). Now apply, to the last expression
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" " above, the generic form of item (17), that is, X Y X Y , to get Z BC A B C BC A B C ; answer; by item 15 : Note that we ve greatly simpli ed the original switching network because now only two and circuits, one or circuit, and one not circuit are needed. 284. Using the result of problem 276, the given problem can be written as """ " "" " "" " " " " " Z A B C D A C D A C D B 1 A C D, by items (9) and (7). Hence, again from problem 276, Z A C D; answer: 285. By items (17) and (15), " A A B A B; answer: " " "" " "" 286. Z A A B C A B C B C A B C , by items (14) and (7); then
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" " " Z B C C A B C A , by item (16); thus Z B A C; answer; by item 18 : 287.
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288. We are asked to simplify the expression for the output signal Z in problem 287. To " do this, we can begin with the simpli cation A A BC A BC, by item (16), which puts Z into the form Z A B BC C D A B C D where we used the simpli cation B BC B 1 C B. Now apply items (17) and (15) to get Z A B CD; answer (not the quantity A or B and the quantity C and D). A sketch of the simpli ed network is shown below.
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