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

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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 perlfaq1.pod, 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)

11010112 = 6B16 1740038 = 11111000000000112

101101112 = B716 67.248 = 110111.01012

10100.11012 = 14.D16 F3A516 = 11110011101001012

110110012 = 3318 AB3D16 = 10101011001111012

101111.01112 = 57.348 15C.3816 = 101011100.001112

102316 = 10000001000112 = 100438

7E6A16 = 1111110011010102 = 771528

ABCD16 = 10101011110011012 = 1257158

C35016 = 11000011010100002 = 1415208

9E36.7A16 = 1001111000110110.01111012 = 117066.3648

DEAD.BEEF16 = 1101111010101101.10111110111011112 = 157255.5756748

11010112 = 10710 1740038 = 6349110

101101112 = 18310 67.248 = 55.312510

10100.11012 = 20.812510 F3A516 = 6237310

120103 = 13810 AB3D16 = 4383710

71568 = 369410 15C.3816 = 348.2187510

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 = 11111012 348910 = 66418

20910 = 110100012 971410 = 227628

13210 = 10001002 2385110 = 5D2B16

72710 = 104025 5719010 = DF6616

143510 = 26338 6511310 = FE5916

1100010

110101

+ 11001

1001110

-------------------------

1011000

101110

+ 100101

1010011

--------------------------

111111110

11011101

+ 1100011

101000000

----------------------------------

11000000

1110010

+ 1101101

11011111

-----------------------------

1372

+ 4631

59A3

--------------------

4F1A5

+ B8D5

5AA7A

----------------------

F35B

+ 27E6

11B41

--------------------

1B90F

+ C44E

27D5D

---------------------

decimal + 18 + 115 +79 –49 –3 –100

signed-magnitude 00010010 01110011 01001111 10110001 10000011 11100100

two’s-magnitude 00010010 01110011 01001111 11001111 11111101 10011100

one’s-complement 00010010 01110011 01001111 11001110 11111100 10011011

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 (0001000000002) to 12,750

feet (0000111111112) in the middle of a transmission, possibly yielding a result of 25,500 feet

(0001111111112).

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

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

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