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# TEST BANK FOR Digital Design Principles and Practices 3rd Edition By John F. Wakerly

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**E X E R C I S E S O L U T I O N SINTRODUCTION11.2 Three definitions of “bit”:(1) A binary digit (p. 1).(2) Past tense of “bite” (p. 1).(3) A small amount (pp. 6, 10).1.3ASIC Application-Specific Integrated CircuitCAD Computer-Aided DesignCD Compact DiscCO Central OfficeCPLD Complex Programmable Logic DeviceDAT Digital Audio TapeDIP Dual In-line PinDVD Digital Versatile DiscFPGA Field-Programmable Gate ArrayHDL Hardware Description LanguageIC Integrated CircuitIP Internet ProtocolLSI Large-Scale IntegrationMCM Multichip Module2 DIGITAL CIRCUITSMSI Medium-Scale IntegrationNRE Nonrecurring EngineeringOK Although we use this word hundreds of times a week whether things are OK or not, we have probablyrarely wondered about its history. That history is in fact a brief one, the word being first recorded in1839, though it was no doubt in circulation before then. Much scholarship has been expended on theorigins of OK, but Allen Walker Read has conclusively proved that OK is based on a sort of joke.Someone pronounced the phrase “all correct” as “oll (or orl) correct,” and the same person or someoneelse spelled it “oll korrect,” which abbreviated gives us OK. This term gained wide currency by beingused as a political slogan by the 1840 Democratic candidate Martin Van Buren, who was nicknamedOld Kinderhook because he was born in Kinderhook, New York. An editorial of the same year, referringto the receipt of a pin with the slogan O.K., had this comment: “frightful letters . . . significant ofthe birth-place of Martin Van Buren, old Kinderhook, as also the rallying word of the Democracy of thelate election, ‘all correct’ .... Those who wear them should bear in mind that it will require their moststrenuous exertions ... to make all things O.K.” [From the American Heritage Electronic Dictionary(AHED), copyright 1992 by Houghton Mifflin Company]PBX Private Branch ExchangePCB Printed-Circuit BoardPLD Programmable Logic DevicePWB Printed-Wiring BoardSMT Surface-Mount TechnologySSI Small-Scale IntegrationVHDL VHSIC Hardware Description LanguageVLSI Very Large-Scale Integration1.4ABEL Advanced Boolean Equation LanguageCMOS Complementary Metal-Oxide SemiconductorJPEG Joint Photographic Experts GroupMPEG Moving Picture Experts GroupOK (see above)PERL According to some, it’s “Practical Extraction and Report Language.” But the relevant Perl FAQ entry,in perlfaq1.pod, says “never write ‘PERL’, because perl isn't really an acronym, apocryphal folkloreand post-facto expansions notwithstanding.” (Thanks to Anno Siegel for enlightening me on this.)VHDL VHSIC Hardware Description Language1.8 In my book, “dice” is the plural of “die.”2–1E X E R C I S E S O L U T I O N SNUMBER SYSTEMSAND CODES22.1 (a) (b)(c) (d)(e) (f)(g) (h)(i) (j)2.3 (a)(b)(c)(d)(e)(f)2.5 (a) (b)(c) (d)(e) (f)(g) (h)(i) (j)11010112 = 6B16 1740038 = 11111000000000112101101112 = B716 67.248 = 110111.0101210100.11012 = 14.D16 F3A516 = 11110011101001012110110012 = 3318 AB3D16 = 10101011001111012101111.01112 = 57.348 15C.3816 = 101011100.001112102316 = 10000001000112 = 1004387E6A16 = 1111110011010102 = 771528ABCD16 = 10101011110011012 = 1257158C35016 = 11000011010100002 = 14152089E36.7A16 = 1001111000110110.01111012 = 117066.3648DEAD.BEEF16 = 1101111010101101.10111110111011112 = 157255.575674811010112 = 10710 1740038 = 6349110101101112 = 18310 67.248 = 55.31251010100.11012 = 20.812510 F3A516 = 6237310120103 = 13810 AB3D16 = 438371071568 = 369410 15C.3816 = 348.21875102–2 DIGITAL CIRCUITS2.6 (a) (b)(c) (d)(e) (f)(g) (h)(i) (j)2.7 (a) (b) (c) (d)2.10 (a) (b) (c) (d)2.112.18Suppose a 3n-bit number B is represented by an n-digit octal number Q. Then the two’s-complement of B isrepresented by the 8’s-complement of Q.2.22 Starting with the arrow pointing at any number, adding a positive number causes overflow if the arrow isadvanced through the +7 to –8 transition. Adding a negative number to any number causes overflow if thearrow is not advanced through the +7 to –8 transition.12510 = 11111012 348910 = 6641820910 = 110100012 971410 = 22762813210 = 10001002 2385110 = 5D2B1672710 = 104025 5719010 = DF6616143510 = 26338 6511310 = FE59161100010110101+ 110011001110-------------------------1011000101110+ 1001011010011--------------------------11111111011011101+ 1100011101000000----------------------------------110000001110010+ 110110111011111-----------------------------1372+ 463159A3--------------------4F1A5+ B8D55AA7A----------------------F35B+ 27E611B41--------------------1B90F+ C44E27D5D---------------------decimal + 18 + 115 +79 –49 –3 –100signed-magnitude 00010010 01110011 01001111 10110001 10000011 11100100two’s-magnitude 00010010 01110011 01001111 11001111 11111101 10011100one’s-complement 00010010 01110011 01001111 11001110 11111100 10011011hj b4j + i 2j ×i = 03å=Therefore,B bi × 2ii – 04n 1 –å hi × 16ii = 0n 1 –å = =–B 24n bii = 04n 1 –å – × 2i 16n hi × 16ii = 0n 1 –å = = –EXERCISE SOLUTIONS 2–32.24 Let the binary representation of be . Then we can write the binary representation of as, where . Note that is the sign bit of . The value of isCase 1 In this case, if and only if , which is true if andonly if all of the discarded bits are 0, the same as .Case 2 In this case, if and only if , whichis true if and only if all of the discarded bits are 1, the same as .2.25 If the radix point is considered to be just to the right of the leftmost bit, then the largest number is andthe 2’s complement of is obtained by subtracting it from 2 (singular possessive). Regardless of the positionof the radix point, the 1s’ complement is obtained by subtracting from the largest number, which has all 1s(plural).2.28Case 1 First term is 0, summation terms have shifted coefficients as specified. Overflow if.Case 2 Split first term into two halves; one half is cancelled by summation term if. Remaining half and remaining summation terms have shifted coefficients as specified. Overflow if.2.32 001–010, 011–100, 101–110, 111–000.2.34 Perhaps the designers were worried about what would happen if the aircraft changed altitude in the middle of atransmission. With the Gray code, the codings of “adjacent” alitudes (at 50-foot increments) differ in only onebit. An altitude change during transmission affects only one bit, and whether the changed bit or the original istransmitted, the resulting code represents an altitude within one step (50 feet) of the original. With a binarycode, larger altitude errors could result, say if a plane changed from 12,800 feet (0001000000002) to 12,750feet (0000111111112) in the middle of a transmission, possibly yielding a result of 25,500 feet(0001111111112).**

## [Solved] TEST BANK FOR Digital Design Principles and Practices 3rd Edition By John F. Wakerly

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