SuperMap 系列产品介绍.ppt

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1、: Without breaks, smooth, no interruptions; a range of values forming a line or curve without gaps or discontinuities: Broken down into pieces; opposite of continuous; a single part or point that can unambiguously be defined: Refer to something that is analogous, has a likeness, or is similar to som

2、ething else. Analog electronic signals are continuous, whereas voltage(电压), current(电流), or some other electrical parameter is used to represent a physical parameter. For instance, voltage changes may represent changes in temperature.: Refer to digits or numbers in any radix; digital electronics usu

3、ally involves binary or base-2 number systems. Digital signals are discrete, not continuousVc(0)=0V, Initial voltage at t0 is 0VVc=Va(1-e-t/RC)(a) Analog representation(b) Discrete representationDiscrete voltage values for time intervalAnalog volt meterDigital volt meter: - Reproducibility of Result

4、s - Ease of design, Flexibility and Functionality - Programmability - Speed, Economy and Steadily Advancing TechnologySSI (small-scale integration) contains 0-9 gatesMSI (medium-scale integration) contains 10-99gatesLSI (large-scale integration) contains 100 or more gatesVLSI (very large-scale integ

5、ration ) contains more than 1000 gatesSamsung LN40A750 40-Inch 1080p DLNA LCD HDTV ARM Powered Amazon Kindle2 e-book reader ARM PoweredFreescaleParrot : Zikmu Wireless speakers by Philppe Starck ARM926 ProcessorPalm PreARM Cortex A8Texas Instruments OMAP3430 Nintendo DSiARM9 and ARM7 ProcessorsKonic

6、a Minolta Magicolor 1650EN Printer ARM11 ProcessorZoran Corporation Quatro 4230 ProcessoriRiver Wave Home (Multimedia Communicator)ARM PoweredHP mv5150 Media Vault ProARM9 ProcessorrMarvell Orion (88F5182-A2) ProcessorAPSI M-480 PMPARM11 ProcessorTI OMAP2530 Dual-Core CPUTeleNav ShotgunARM926 Proces

7、sorSiRF Atlas IIIKangaroo TVARM926EJ-sFreescale i.MX21 ProcessorZeebo ConsoleARM11 Processor and ARM9 ProcessorQualcommDell Latitude E4200 NotebookARM Cortex M3Broadcom BCM5880w Digital systems represented information in a discrete form, and Number systems provide the means of quantifying informatio

8、n for processing by digital system. Positional Number SystemsNumber system conversionBinary CodesArithmeticw “base” 10 (radix is 10)w 10 digits: 0.9 536.15910=(5102)+(3101)+(6100)+(110-1)+(510-2)+(910-3) w In general, a decimal number with n digits to the left of the decimal point, and m digits to t

9、he right of the decimal point is written as:An-1 An-2 A1 A0 . A-1 A-2 A-m+1 A-mwhere Ai is a coefficient between 0.9, and i denotes the weight (=10i) of Ai.w “base” r (radix r)w r digitsw N = An-1 r n-1 + An-2 r n-2 + + A1 r + A0 + A-1 r -1 + A-2 r -2 + + A-m r -mMostSignificant Bit (MSB)LeastSignif

10、icantBit (LSB) Let r=radix or base of a number system; Let c=character from the character set of the radix; Let N=number to be represented in radix; Let n=the number of digits in the integer portion of N Let m=the number of digits in the fractional portion of Nw e.g. let r = 6(312.4)6 = 3 62 + 1 61

11、+ 1 60 + 4 6-1 = (116.66)10w Conversion from n-radix (any system with radix n) to decimal follows similar process as abovew Most common number systems for computers:Binary (r = 2)Octal (r = 8)Hexadecimal (r = 16)w Computers represent all data as “strings of bits”, each bit being either 0 or 1w “base

12、” 2, with 2 digits: 0 and 1e.g. (101101.10)2 = 125 + 024 + 123 + 122 + 021 + 120 + 12-1 + 02-2 (in decimal) = 32 + 0 + 8 + 4 + 0 + 1 + + 0 = (45.5)10Memorize at least through 216w “base” 8w with 8 digits: 0.7Eg.231.258=282 + 381+180+28-1+58-2(in decimal) =153.32812510w r = 16w Digits (convention): 0

13、.9, A, B, C, D, E, Fw A=10, B=11, , F = 15e.g1. (3FB)16 = 3162 + 15161 + 11160 (in decimal) = 768 + 240 + 11 = (1019)10e.g2. A59C.3A16 = (A163)+(5162)+(9161)+(C160)+(316-1)+(A16-2)Decimal，binary,octal,and hexadecimal number(011 010 101 000 . 111 101 011 100)2( 3 2 5 0 . 7 5 3 4 )8 Convert the binary

14、 number 010011110111.1101010102to octalSolution 010,011,110,111.110,101,0102=2367.6528( 6 A 8 . F 5 C )16( 0110 1010 1000 . 1111 0101 1100 )2 Convert the binary number 010011110111.1101010102to hexadecimalSolution 0100,1111,0111.1101,0101,00002=4F7.D5016E.g. Convert the numbers F37A.B216,735.58 to b

15、inarySolution F37A.B2=1111,0011,0111,1010.1011,00102 735.58 =111,011,101.1012Practice problem: convert 367.2368to binary then hexadecimalSolution 367.2368=011,110,111.010,011,1102011,110,111.010,011,1102=0,1111,0111.0100,1111,02=F7.4F1611001.0112 =(124)+(123)+(022)+(021)+(120) +(0 2-1)+(12-2)+(12-3)

16、=16+8+0+0+1+0+0.25+0.125=25.37510（LSB）（MSB）E.g. Convert 11910 to binary11910=11101112SolutionE.g. Convert .7510 to binaryMultiply .75 by 2 (.75)2=1.5 1(MSB)Multiply .75 by 2 (.5)2=1.0 1Multiply .75 by 2 (0)2=0.0 0(LSB)Solution.7510=.110=(1)2-1+(1)2-2w E.g. Convert 95.062510 to binary（con.）Solution F

17、irst, convert the integer part of the decimal number by successive division1.Devide 95 by 24.Devide 11 by 23.Devide 23 by 22.Devide 47 by 25.Devide 5 by 26.Devide 2 by 27.Devide 1 by 2(LSB)(MSB)9510=126+025+124+123+122+121+120w E.g. Convert 95.062510 to binary（con.）Second, convert the fraction1. (.6

18、25)2=0.125 0 (MSB)2. (.125)2=0.25 03. (.25)2=0.5 04. (.5)2=1.0 1 (LSB).062510=.0001295.062510=1011111.00012The conversion of decimal numbers to any other radix applies the successive division and successive multiplication algorithms Convert 23.62510 to octal (base 8)1.Divide 23 by 82.Divide 2 by 823

19、10=278 Convert the integer portion by successive division Convert te fraction portion by successive multiplication(.625)8=5.00 5 .62510=.58 Convert 324.25 to decimal352+251+450+25-1 =3(25)+2(5)+4(1)+2(1/5) =75+10+4+2/5 =89.410Practice problem: 1. Convert 345.26 to decimal 2. Convert 34.810 to base 5

20、 1. 137.333102. 114.45 Convert 9275.610 into BCD9=10012=00107=01115=01016=01109275.610=1001，0010，0111，0101.0110 in BCDTiming diagram illustrating potential error when two bits change sat the same time (American standard code for information interchange)(extended BCD interchange code)Partial listing

21、of seven-bit ASCII and eight-bit EBCDIC codes in hexadecimalSign magnitude and complementSign magnitude2s complement1s complement N = (s,an-1 . a0.a-1 . a-m)rsm, where s = 0 if N is positive and s = r -1 otherwise.E.g. N = -(15)10 In binary: N = -(15)10= -(1111)2 = (1, 1111)2sm In decimal: N = -(15)

22、10 = (9, 15)10smRadix complements (rs complements) Nr = rn - (N)r where n is the number of digits in (N)r.Diminished radix complements (r-1s complements) Nr-1 = rn - (N) r - 1u2s complement of (N)2 = (101001)2N2 = 26 - (101001)2 = (1000000)2 - (101001)2 = (010111)2 (N)2 + N2 = (101001)2 + (010111)2

23、= (1000000)2If we discard the carry, (N)2 + N2 = 0. Hence, N2 can be used to represent -(N)2. N2 2 = (010111)22 = (1000000)2 - (010111)2 = (101001)2 = (N)2.u2s complement of (N)2 = (1010)2 for n = 6N2 = (1000000)2 - (1010)2 = (110110)2.u10s complement of (N)10 = (72092)10N10 = (100000)10 - (72092)10

24、 = (27908)10.1.Copy the digits of N, beginning with the LSB and proceeding toward the MSB until the first nonzero digit, aj, has been reached2.Replace aj with r - aj.3.Replace each remaining digit aj , of N by (r - 1) - aj until the MSB has been replaced.Example: 10s complement of (56700)10 is (4330

25、0)10Example: 2s complement of (10100)2 is (01100)2.Example: 2s complement of N = (10110)2 for n = 8. Put three zeros in the MSB position and apply algorithm N = 00010110 N2 = (11101010)2 First replace each digit, ak , of (N)r by (r - 1) - ak and then add 1 to the resultant.w Example: Find 2s complem

26、ent of N = (01100101)2 . N = 01100101 10011010 Complement the bits +1 Add 1 N2 = (10011011)2w Example: Find 10s complement of N = (40960)10 N = 40960 59039 Complement the bits +1 Add 1 N2 = (59040)10w 2s complement number system Positive number : N = +(an-2, ., a0)2 = (0, an-2, ., a0)2cns, where .Ne

27、gative number: N = (an-1, an-2, ., a0)2 -N = an-1, an-2, ., a02 (twos complement of N), where . Example: Twos complement number system representation of (N)2 when (N)2 = (1011001)2 for n = 8: +(N)2 = (0, 1011001)2cns -(N)2 = +(N)22 = 0, 10110012 = (1, 0100111)2cns112nN 1021nNw Diminished radix compl

28、ement Nr-1 of a number (N)r is:Nr-1 = rn - (N)r - 1w Ones complement (r = 2):N2-1 = 2n - (N)2 - 1w Example: Ones complement of (01100101)2N2-1 = 28 - (01100101)2 - 1 = (100000000)2 - (01100101)2 - (00000001)2 = (10011011)2 - (00000001)2 = (10011010)2uFind the 2s complement of the binary number 11001

29、1uFind the 1s complement code for -1210uFind the 2s complement code for -1810(+X)+(+Y)(+X)+(-Y)(-X)+(+Y)(-X)+(-Y)(+X)-(+Y)(+X)-(-Y) (-X)-(+Y)(-X)-(-Y)+(X+Y)-(X+Y)+(X+Y)-(X+Y)+(X-Y)-(X-Y)+(X-Y)-(X-Y)-(X-Y)+(X-Y)-(X-Y)+(X-Y)+(X-Y)+(X-Y)+(X-Y)+(X-Y)OperationAddXYX=YXYSubtract 9+3=12 （-9）+（-3）=-12 01001

30、 10111 （-9）+ 00011 + 11101 （-3） 01100 1 10100 （-12） 9-3=6 9-（-3）=12 01001 01001 + 11101 + 00011 1 0 0110 01100 3-9=-6 00011 + 10111 11010 001101.1+ 001011.0 011000.101101.1 + 01011.0 11000.1w e.g. Add X and Y . Let X=-810 and Y=-1010, both in 2s complement 11000+ 10110(1)01110 111000+ 110110(1)10111

31、0w e.g. Subtract Y from X. Let X=+810=01000 and Y=-610=11010, X-Y=X+(-Y) 01000 = +810+ 00110 =(complement of -6 ) 01110 =+141001000=+810The 2s complement of -610(11010) is 00110w exe. Subtract Y from X. Let X=-1010=10110 and Y=-510=11011 10110 = -1010+ 00101 =(complement of -510 ) 11011 =-510The radix complement of 11011(-510) is 00101w Exe：perform the following operations by finding the radix complement of the subtrahend(减数) and adding the result to the minuend(被减数): 23.410 135.78 321.24 - 19.810 - 67.78 - 33.34