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"" "" " " 289. First, the elemental equation is Z A B A B AB A B A B B , and thus Z A B A; answer; by items 14 ; 7 ; 17 : The above answer is accomplished by the network below.
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"" 290. (a) Z A B AB, answer. (b) First, by item (17), Z A B AB, which is produced by the network
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Note that Z A B AB does satisfy the given truth table. 291. (a) """ " " " Z A B C A B C A B C ABC, answer. Check: inspection of the answer shows that Z 1 (which denotes the presence of a signal on the output line) if any of the input signal combinations 000, 010, 101, and 111 appear on the input lines, and Z 0 for any other combination of input signals. "" " " (b) First note that Z A C B B AC B B , which, after applying items (14), (7), and (17), becomes Z A C AC, which can be generated by the hardware arrangement
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Note that in this case the signal B need not be connected to the network, because the state of signal B, 1 or 0, does not a ect the state of output signal Z. "" " " " " " " " " 292. (a) Z A B C D A B C D A B C D A B C D ABC D ABCD, answer. " " (b) One way to proceed is as follows. First, noting that the quantity A D factors from the rst three terms, and AC factors from the last three terms, we can take the following steps: " " " " " " Z A D B C B C BC AC B D B D BD then " "" " " " Z A D B C B C C AC B D B D D
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which, after applying items (14) and (7), becomes " " " " Z A D B B C AC B B D which, after applying items (17) and (16), becomes Z A D B C AC B D ; answer: (c)
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293. (a)
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"" " "" " " " Z A B C D A B C D A B C D AB C D ABCD, answer. "" " " "" (b) Z C A B D D A A BD A B D , then "" "" Z C A B BD A B D by items (14) and (7). Next, " " " Z C BD B A A D " " " " and since A A D A D (example 11 on p. 349, we have " " " " Z C BD B A D C BD B AD " " by item (18). Now, making use of the basic relationship X Y X Y, the last result becomes Z C BD B AD ; answer: (c)
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294. (a)
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Here, each block of information fed into the encoder will consist of 12 binary digits, each such block representing one of the 12 possible values of Vq , including Vq 0.
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Hence, letting A; B; C; . . . ; I; J; K, denote each input block to the encoder, these blocks will be in the form of 12 binary numbers from 0000 to 1011 (binary eleven). Thus the TRUTH TABLE for the encoder is as follows: Vq 0 1 2 3 4 5 6 7 8 9 10 11 (b) A 0 1 1 1 1 1 1 1 1 1 1 1 B 0 0 1 1 1 1 1 1 1 1 1 1 C 0 0 0 1 1 1 1 1 1 1 1 1 D 0 0 0 0 1 1 1 1 1 1 1 1 E 0 0 0 0 0 1 1 1 1 1 1 1 F 0 0 0 0 0 0 1 1 1 1 1 1 G 0 0 0 0 0 0 0 1 1 1 1 1 H 0 0 0 0 0 0 0 0 1 1 1 1 I 0 0 0 0 0 0 0 0 0 1 1 1 J 0 0 0 0 0 0 0 0 0 0 1 1 K 0 0 0 0 0 0 0 0 0 0 0 1 X4 0 0 0 0 0 0 0 0 1 1 1 1 X3 0 0 0 0 1 1 1 1 0 0 0 0 X2 X1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1
Close inspection of the above table will show that the output X digits will have the value 1 (signal is present ) only if the following Boolean equations are satis ed. (Otherwise it s understood that X 0, meaning the signal is not present for any other arrangements of the A; B; C; . . . ; I; J; K input signals.) " " " " " X1 1 A B C D E F G H I J K " " X2 1 B D F H J " " X3 1 D F F H X4 1 H
Thus, in terms of hardware, it would be necessary to provide 5 not, 7 and, and 3 or devices. 295. First note that as we go from RIGHT TO LEFT in the series of eq. (576) the values of the exponents increase by 1 from term to term. Let us show this more clearly by showing a few more terms at the right-hand end of the series, thus F z 1 z 1 z 2 z 3 z n z3 z n z2 z n z1 z n
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