Exam (elaborations) TEST BANK FOR Digital Design Principles and Practices 3rd Edition By John F. Wakerly (Solution Manual)

Exam (elaborations) TEST BANK FOR Digital Design Principles and Practices 3rd Edition By John F. Wakerly (Solution Manual) INTRODUCTION 1 1.2 Three definitions of “bit”: (1) A binary digit (p. 1). (2) Past tense of “bite” (p. 1). (3) A small amount (pp. 6, 10). 1.3 ASIC Application-Specific Integrated Circuit CAD Computer-Aided Design CD Compact Disc CO Central Office CPLD Complex Programmable Logic Device DAT Digital Audio Tape DIP Dual In-line Pin DVD Digital Versatile Disc FPGA Field-Programmable Gate Array HDL Hardware Description Language IC Integrated Circuit IP Internet Protocol LSI Large-Scale Integration MCM Multichip Module 2 DIGITAL CIRCUITS MSI Medium-Scale Integration NRE Nonrecurring Engineering OK Although we use this word hundreds of times a week whether things are OK or not, we have probably rarely wondered about its history. That history is in fact a brief one, the word being first recorded in 1839, though it was no doubt in circulation before then. Much scholarship has been expended on the origins 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 someone else spelled it “oll korrect,” which abbreviated gives us OK. This term gained wide currency by being used as a political slogan by the 1840 Democratic candidate Martin Van Buren, who was nicknamed Old Kinderhook because he was born in Kinderhook, New York. An editorial of the same year, referring to the receipt of a pin with the slogan O.K., had this comment: “frightful letters . . . significant of the birth-place of Martin Van Buren, old Kinderhook, as also the rallying word of the Democracy of the late election, ‘all correct’ .... Those who wear them should bear in mind that it will require their most strenuous exertions ... to make all things O.K.” [From the American Heritage Electronic Dictionary (AHED), copyright 1992 by Houghton Mifflin Company] PBX Private Branch Exchange PCB Printed-Circuit Board PLD Programmable Logic Device PWB Printed-Wiring Board SMT Surface-Mount Technology SSI Small-Scale Integration VHDL VHSIC Hardware Description Language VLSI Very Large-Scale Integration 1.4 ABEL Advanced Boolean Equation Language CMOS Complementary Metal-Oxide Semiconductor JPEG Joint Photographic Experts Group MPEG Moving Picture Experts Group OK (see above) PERL According to some, it’s “Practical Extraction and Report Language.” But the relevant Perl FAQ entry, in , says “never write ‘PERL’, because perl isn't really an acronym, apocryphal folklore and post-facto expansions notwithstanding.” (Thanks to Anno Siegel for enlightening me on this.) VHDL VHSIC Hardware Description Language 1.8 In my book, “dice” is the plural of “die.” 2–1 E X E R C I S E S O L U T I O N S NUMBER SYSTEMS AND CODES 2 2.1 (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) = 6B16 = = B716 67.248 = .01012 10100.11012 = 14.D16 F3A516 = = 3318 AB3D16 = .01112 = 57.348 15C.3816 = . = 00112 = 7E6A16 = = ABCD16 = = C35016 = = 9E36.7A16 = . = .3648 DEAD.BEEF16 = . = . = 10710 = = 18310 67.248 = 55. 10100.11012 = 20. F3A516 = = 13810 AB3D16 = 71568 = 15C.3816 = 348. 2–2 DIGITAL CIRCUITS 2.6 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) 2.7 (a) (b) (c) (d) 2.10 (a) (b) (c) (d) 2.11 2.18 Suppose a 3n-bit number B is represented by an n-digit octal number Q. Then the two’s-complement of B is represented 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 is advanced through the +7 to –8 transition. Adding a negative number to any number causes overflow if the arrow is not advanced through the +7 to –8 transition. 12510 = = 66418 20910 = = 13210 = = 5D2B16 72710 = = DF6616 = 26338 = FE5916 + 11001 ------------------------- + -------------------------- + ---------------------------------- + ----------------------------- 1372 + 4631 59A3 -------------------- 4F1A5 + B8D5 5AA7A ---------------------- F35B + 27E6 11B41 -------------------- 1B90F + C44E 27D5D --------------------- decimal + 18 + 115 +79 –49 –3 –100 signed-magnitude two’s-magnitude one’s-complement hj b4j + i 2j × i = 0 3å = Therefore, B bi × 2i i – 0 4n 1 – å hi × 16i i = 0 n 1 – å = = –B 24n bi i = 0 4n 1 – å – × 2i 16n hi × 16i i = 0 n 1 – å = = – EXERCISE SOLUTIONS 2–3 2.24 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 is Case 1 In this case, if and only if , which is true if and only if all of the discarded bits are 0, the same as . Case 2 In this case, if and only if , which is 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 and the 2’s complement of is obtained by subtracting it from 2 (singular possessive). Regardless of the position of the radix point, the 1s’ complement is obtained by subtracting from the largest number, which has all 1s (plural). 2.28 Case 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 a transmission. With the Gray code, the codings of “adjacent” alitudes (at 50-foot increments) differ in only one bit. An altitude change during transmission affects only one bit, and whether the changed bit or the original is transmitted, the resulting code represents an altitude within one step (50 feet) of the original. With a binary code, larger altitude errors could result, say if a plane changed from 12,800 feet (0002) to 12,750 feet (1112) in the middle of a transmission, possibly yielding a result of 25,500 feet (1112). X xn – 1xn – 2¼x1x0 Y xmxm – 1¼x1x0 m = n – d xm – 1 Y Y Y –2m – 1 xm – 1 xi × 2i i = 0 n 2 – å = × + The value of X is X –2n – 1 xn – 1 xi × 2i i = 0 n 2 – å = × + –2n – 1 xn – 1 Y 2m – 1 × xm – 1 xi × 2i i = m – 1 n 2 – å = × + + + –2n – 1 xn – 1 Y 2 × 2m – 1 xi × 2i i = m n 2 – å = × + + + (xm – 1 = 0) X = Y –2n – 1 × xn – 1 xi × 2i i = m n – 2 +å = 0 (xm¼xn – 1) xm – 1 (xm – 1 = 1) X = Y –2n – 1 × xn – 1 2 × 2m – 1 xi × 2i i = m n – 2 + +å = 0 (xm¼xn – 1) xm – 1 1.11¼1 D D B –bn – 1 × 2n – 1 bi × 2i i = 0 n 2 – å = + 2B –bn – 1 2n bi × 2i + 1 i = 0 n 2 – å = × + (bn – 1 = 0) bn – 2 = 1 (bn – 1 = 1) bn – 2 × 2n – 1 bn – 2 = 1 bn – 2 = 0 2–4 DIGITAL CIRCUITS 2.37 010 011 000 001 110 111 100 101 3–1 E X E R C I S E S O L U T I O N S DIGITAL CIRCUITS 3 3.1 The “probably” cases may cause damage to the gate if sustained. 3.3 A logic buffer is a non-linear amplifier that maps the entire set of possible analog input voltages into just two output votages, HIGH and LOW. An audio amplifier has a linear response over its specified operating range, mapping each input voltage into an output voltage that is directly proprtional to the input voltage. 3.6 From the American Heritage Electronic Dictionary (AHED), copyright 1992 by Houghton Mifflin Company: (1) A structure that can be swung, drawn, or lowered to block an entrance or a passageway. (2) a. An opening in a wall or fence for entrance or exit. b. The structure surrounding such an opening, such as the monumental or fortified entrance to a palace or walled city. (3) a. A means of access: the gate to riches. b. A passageway, as in an airport terminal, through which passengers proceed for embarkation. (4) A mountain pass. (5) The total paid attendance or admission receipts at a public event: a good gate at the football game. (6) A device for controlling the passage of water or gas through a dam or conduit. (7) The channel through which molten metal flows into a shaped cavity of a mold. (8) Sports. A passage between two upright poles through which a skier must go in a slalom race. (9) Electronics. A circuit with multiple inputs and one output that is energized only when a designated set of input pulses is received. (a) 0 (b) 1 (c) 0 (d) undefined (e) 1 (f) probably 1 (g) probably 0 (h) probably 0 3–2 DIGITAL CIRCUITS Well, definition (9) is closest to one of the answers that I had in mind. The other answer I was looking for is the gate of a MOS transistor. 3.14 A CMOS inverting gate has fewer transistors than a noninverting one, since an inversion comes “for free.” 3.15 Simple, inverting CMOS gates generally have two transistors per input. Four examples that meet the requirements of the problem are 4-input NAND , 4-input NOR , 2-in, 2-wide AND–OR–INVERT , and 2-in, 2-wide OR– AND–INVERT. 3.18 One way is that a romance could be sparked, and the designers could end up with a lot less time to do their work. Another way is that the stray perfume in the IC production line could have contaminated the circuits used by the designers, leading to marginal operation, more debugging time by the deidicated designers, and less time for romance. By the way, the whole perfume story may be apocryphal. 3.20 Using the maximum output current ratings in both states, the HIGH-state margin is 0.69V and the LOW-state margin is 1.02V. With CMOS loads (output currents less than 20 mA), the margins improve to 1.349V and 1.25V, respectively. 3.21 The first answer for each parameter below assumes commercial operation and that the device is used with the maximum allowable (TTL) load. The number in parentheses, if any, indicates the value obtained under a lesser but specified (CMOS) load. VOHmin 3.84V (4.4V ) VIHmin 3.15V VILmax 1.35V VOLmax 0.33V (0.1V) IImax 1mA IOLmax 4 mA (20 mA) IOHmax -4 mA (-20 mA) 3.22 Current is positive if it flows into a node. Therefore, an output with negative current is sourcing current. 3.23 The 74HC00 output drive is so weak, it’s not good for driving much: (a) Assume that in the LOW state the output pulls down to 0.33V (the maximum spec). Then the output current is , which is way more than the 4-mA commercial spec. (b) For this problem, you first have to find the Thévenin equivalent of the