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

... [Show More] 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 [Show Less]