Floating Point Representation - Transforming Numerical …浮点表示法转换的数值…
浮点数表示方法
浮点数表示方法浮点数表示法（Floating point representation）是用来表示实现有限位数范围内的实数值的表示法。
它利用一个移位浮点技术，使得移位整数和移位小数可以被合成成范围内的实数值。
1、特点：（1）数据格式灵活，能表示实数。
（2）能反映数据中的技术定格比较弱，有一定误差。
（3）表示范围广泛，但数据精度较差。
（4）为提高精度，采用双精度（double precision）或多精度（multi precision）浮点格式。
2、格式：（1）浮点数表示法通常由三个部分组成：浮点符号、整数部分和小数部分。
（2）示例：+0.100，其中“+”表示正数；“0”表示整数部分；“100”表示小数部分。
3、应用：（1）工程计算：飞行器的设计，图像处理软件工程等（2）科学计算：天文计算，气象学分析，量子力学等（3）计算机系统：操作系统，存储器管理，语言编译等。
（4）其他应用：时钟，位图，图形工具，材料工程计算等。
4、编码：（1）采用二进制编码：所有的数据都被表示为二进制.（2）指数编码：以进位计算法实现指数位编码，在数据表示中占据比较重要位置。
（3）尾数编码：尾数部分在许多浮点运算系统中扮演着十分重要的角色，它将实现精度提高。
5、运算：（1）加减法：两个数字相加减，异号相加减，同号相加减（2）乘法：浮点乘法可以看作由两部分组成，即放大和缩小（3）除法：浮点除法也可以看作是缩小和放大的过程。
6、算术系统：采用浮点数表示法的算术系统，有助于改善有限位数运算效率和精度，执行特殊型计算机指令（如浮点数运算）。
7、缺点：（1）数据精度较低，数据的浮点误差有时可能很大。
（2）单精度浮点数的表示范围有限，双精度浮点数也受制于计算机封装计算机指令的有限位性，所以对使用的运算系统的精度有一定的影响。
（3）浮点乘法和除法的执行效率较慢。
汇川plc浮点数表示标准
汇川plc浮点数表示标准The representation of floating-point numbers in Huichuan PLC follows the standard set by the IEEE 754 format. This format is a widely accepted standard for representing floating-point numbers in binary, providing a consistent way to store and manipulate real numbers in digital systems.汇川PLC中浮点数的表示遵循IEEE 754格式所设定的标准。
这种格式是用于以二进制方式表示浮点数的广泛接受的标准，提供了一种一致的方式来存储和操作数字系统中的实数。
In the IEEE 754 format, a floating-point number is represented using three basic components: the sign bit, the exponent, and the mantissa. The sign bit determines whether the number is positive or negative, the exponent specifies the power of 2 by which the mantissa should be multiplied, and the mantissa holds the significant digits of the number.在IEEE 754格式中，浮点数使用三个基本组件来表示：符号位、指数和尾数。
符号位决定了数字是正数还是负数，指数指定了尾数应该乘以的2的幂次，尾数保存了数字的有效数字。
fortran03
Exponent
Mantissa (Fraction)
20
2-1
2-23
0 10000000 10000000000000000000000
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Floating Point Representation
Example: B=2 and E=128
x sMBe
x = +1/4
E
x s1 M1B s1 M1B
e1 E
s2 M 2 B
e2 E
s2 M 2 B ? 2 )B e1 E (s1 M1 s2 M
e1 E
e 2 e1
B
e1 E
Example: B=2 and E=128
1/4 0 10000000 01000000000000000000000 3/4 0 10000000 11000000000000000000000 01/13/12 11
Consider to represent a real number x as
x sMBe
E
where s is the sign bit, M is the mantissa with t bits, B is the base (2, 8, 10, or 16, …), e is the exponent, and E is the exponential bias.
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Binary 1.0 100.0 101.0 0.1 0.01 0.001 0.11
= = = = = = =
0.1 • 21 0.1 • 23 0.101 • 23 0.1 • 20 0.01 • 20 0.001 • 20 0.11 • 20
PCM脉冲编码中英文翻译1份
英文原文Pulse-code modulationPulse-code modulation (PCM) is a digital representation of an analog signal where the magnitude of the signal is sampled regularly at uniform intervals，then quantized to a series of symbols in a numeric (usually binary) code．PCM has been used in digital telephone systems and 1980s-era electronic musical keyboards．It is also the standard form for digital audio in computers and the compact disc "red book" format．It is also standard in digital video，for example，using ITU-R BT．601．Uncompressed PCM is not typically used for video in standard definition consumer applications such as DVD or DVR because the bit rate required is far too high．ModulationIn the diagram，a sine wave (red curve) is sampled and quantized for pulse code modulation．The sine wave is sampled at regular intervals，shown as ticks on the x-axis．For each sample，one ofthe available values (ticks on the y-axis) is chosen by some algorithm (in this case，the floor function is used)．This produces a fully discrete representation of the input signal (shaded area) that can be easily encoded as digital data for storage or manipulation．For the sine wave example at right，we can verify that the quantized values at the sampling moments are 7，9，11，12，13，14，14，15，15，15，14，etc．Encoding these values as binary numbers would result in the following set of nibbles:0111，1001，1011，1100，1101，1110，1110，1111，1111，1111，1110，etc．These digital values could then be further processed or analyzed by a purpose-specific digital signal processor or general purpose CPU．Several Pulse Code Modulation streams could also be multiplexed into a larger aggregate data stream，generally for transmission of multiple streams over a single physical link．One technique is called time-division multiplexing，or TDM，and is widely used，notably in the modern public telephone system．Another technique is called Frequency-division multiplexing，where the signal is assigned a frequency in a spectrum，and transmitted along with other signals inside that spectrum．Currently，TDM is much more widely used than FDM because of its natural compatibility with digital communication，and generally lower bandwidth requirements．There are many ways to implement a real device that performs this task．In real systems，such a device is commonly implemented on a single integrated circuit that lacks only the clock necessary for sampling，and is generally referred to as an ADC (Analog-to-Digital converter)．These devices will produce on their output a binary representation of the input whenever they are triggered by a clock signal，which would then be read by a processor of some sort．DemodulationTo produce output from the sampled data，the procedure of modulation is applied in reverse．After each sampling period has passed，the next value is read and a signal is shifted to the new value．As a result of these transitions，the signal will have a significant amount of high-frequency energy．To smooth out the signal and remove these undesirable aliasing frequencies，the signal would be passed through analog filters that suppress energy outside the expected frequency range (that is，greater than the Nyquist frequency fs/2)．Some systems use digital filtering to remove some of the aliasing，converting the signal from digital to analog at a higher sample rate such that the analog filter required for anti-aliasing is much simpler．In some systems，no explicit filtering is done at all; as it's impossible for any system to reproduce a signal with infinite bandwidth，inherent losses in the system compensate for the artifacts-or the system simply does not require much precision．The sampling theorem suggests that practical PCM devices，provided a sampling frequency that is sufficiently greater than that of the input signal，can operate without introducing significant distortions within their designed frequency bands．The electronics involved in producing an accurate analog signal from the discrete data are similar to those used for generating the digital signal．These devices are DACs (digital-to-analog converters)，and operate similarly to ADCs．They produce on their output a voltage or current (depending on type) that represents the value presented on their inputs．This output would then generally be filtered and amplified for use．LimitationsThere are two sources of impairment implicit in any PCM system: Choosing a discrete value near the analog signal for each sample ( quantization error ) Between samples no measurement of the signal is made; due to the sampling theorem this results in any frequencyabove or equal to( Fs being the sampling frequency) being distorted or lost completely ( aliasing error). (One half the sampling frequencies are known as the Nyquist frequency.)Digitization as part of the PCM processIn conventional PCM，the analog signal may be processed (eg by amplitude compression)before being digitized．Once the signal is digitized，the PCM signal is usually subjected to further processing (eg digital data compression)．PCM with linear quantization is known as Linear PCM (LPCM)．Some forms of PCM combine signal processing with coding．Older versions of these systems applied the processing in the analog domain as part of the A/D process; newer implementations do so in the digital domain．These simple techniques have been largely rendered obsolete by modern transform-based audio compression techniques．DPCM encodes the PCM values as differences between the current and the predicted value．An algorithm predicts the next sample based on the previous samples，and the encoder stores only the difference between this prediction and the actual value．If the prediction is reasonable，fewer bits can be used to represent the same information．For audio，this type of encoding reduces the number of bits required per sample by about 25% compared to PCM．Adaptive DPCM (ADPCM) is a variant of DPCM that varies the size of the quantization step，to allow further reduction of the required bandwidth for a given signal-to-noise ratio．Delta modulation is a form of DPCM which uses one bit per sample．In telephony，a standard audio signal for a single phone call is encoded as 8000 analog samples per second，of 8 bits each，giving a 64 kbit/s digital signal known as DS0．The default signal compression encoding on a DS0 is either μ-law (mu-law) PCM (North America and Japan) or A-law PCM (Europe and most of the rest of theworld)．These are logarithmic compression systems where a 12 or 13-bit linear PCM sample number is mapped into an 8-bit value．This system is described by international standard G．711．An alternative proposal for a floating point representation，with 5-bit mantissa and 3-bit radix，was abandoned．Where circuit costs are high and loss of voice quality is acceptable，it sometimes makes sense to compress the voice signal even further．An ADPCM algorithm is used to map a series of 8-bit µ-law or A-law PCM samples into a series of 4-bit ADPCM samples．In this way，the capacity of the line is doubled．The technique is detailed in the G．726 standard．Later it was found that even further compression was possible and additional standards were published．Some of these international standards describe systems and ideas which are covered by privately owned patents and thus use of these standards requires payments to the patent holders．Some ADPCM techniques are used in Voice over IP communications．Encoding for transmissionPulse-code modulation can be either return-to-zero (RZ) or non-return-to-zero (NRZ)．For a NRZ system to be synchronized using in-band information，there must not be long sequences of identical symbols，such as ones or zeroes．For binary PCM systems，the density of 1-symbols is called ones-density．Ones-density is often controlled using precoding techniques such as Run Length Limited encoding，where the PCM code is expanded into a slightly longer code with a guaranteed bound on ones-density before modulation into the channel．In other cases，extra framing bits are added into the stream which guarantee at least occasional symbol transitions．Another technique used to control ones-density is the use of a scrambler polynomial on the raw data which will tend to turn theraw data stream into a stream that looks pseudo-random，but where the raw stream can be recovered exactly by reversing the effect of the polynomial．In this case，long runs of zeroes or ones are still possible on the output，but are considered unlikely enough to be within normal engineering tolerance．In other cases， the long term DC value of the modulated signal is important，as building up a DC offset will tend to bias detector circuits out of their operating range．In this case special measures are taken to keep a count of the cumulative DC offset，and to modify the codes if necessary to make the DC offset always tend back to zero．Many of these codes are bipolar codes，where the pulses can be positive，negative or absent．In the typical alternate mark inversion code，non-zero pulses alternate between being positive and negative．These rules may be violated to generate special symbols used for framing or other special purposes．中文译文脉冲编码调制Pulse-code modulation脉冲编码调制（Pulse-code modulation，PCM）是一种模拟讯号的数码化方法。
软考 浮点数表示方法
软考浮点数表示方法Floating-point numbers are a critical component in computer systems, allowing for the representation of real numbers with fractional parts. 浮点数是计算机系统中的一个关键组件，允许实数表示带有小数部分的数字。
The representation of floating-point numbers follows the IEEE 754 standard, which defines the format for single-precision (32-bit) and double-precision (64-bit) floating-point numbers. 浮点数的表示遵循IEEE 754标准，该标准定义了单精度（32位）和双精度（64位）浮点数的格式。
In a floating-point number, three components are used to encode the value: the sign bit, the exponent, and the mantissa (or significand). 在浮点数中，使用三个组件来编码值：符号位、指数和尾数（或有效数字）。
The sign bit determines whether the number is positive or negative, with 0 representing a positive number and 1 representing a negative number. 符号位确定数字是正数还是负数，其中0表示正数，1表示负数。
The exponent is used to scale the value represented by the mantissa, allowing for a wide range of values to be expressed in a compact format. 指数用于缩放由尾数表示的值，允许用紧凑的格式表示各种范围的值。
What Every Computer Scientist Should Know About Floating-Point Arithmetic
2550 Garcia Avenue Mountain View, CA 94043 U.S.A.
Part No: 800-7895-10 Revision A, June 1992
iii
Exception Handling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rounding Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Errors In Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theorem 14 and Theorem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
GDSII文件格式
GDSII format INDEX1.introduction2.bachus nauer forms3.GDSII BNF4.Record header5.Data types6.record types overview7.record types description8.example file example1.text presentation of GDSII file in KEYformat2.hex presentation of same file3.GDSII fileGDSII Stream format is the standard file format for transfering/archiving 2D graphical design data. It contains a hiearchy of structures, each structure containing elements (boundary/polygon,path/polyline, text,box, structure references, structure array references). The elements are situated on layers. It is a binary format that is platform independent, because it uses internally defined formats for its data types. While reading GDSII files, the GDSII internal data types (like reals, integers etc.) need to be converted to the platform/CAE package datatypes that are used.The GDSII format is a sequential list of records, each record contains a header to tell what information is in the record.The order of the record needs to be according to the GDSII BNF, because of this strict organization it is relativly easy to parse. The maximum number of vertixes is officially only 200 x,y pairs, but many packages can read up to the absolute maximum of 64k/2=32k, simple because this is the maximum record lenght that can be specified (two bytes).The format is hard to read, since it is binary, for that viewers are available to view (boolean) the contents as ASCII. Also an ASCII format has been developed (KEY format) which is more than just a text representation. It is possible to convert GDSIIformat to KEYformat and back. KEYformat has extended the basic primitives to contain cicrles, arcs, polygons/polylines with arc segments.The Bachus Nauer Form uses the following symbols:The following is the Bachus Naur Form of the GDSI format, the words in capital are the names of RECORDSintroductionBachus Nauer FormsSymbol Name Symbol Meaning Double Colon ::"Is composed of."Square brackets [ ]An element which can occor zero or one time.Braces{ }Choose one of the elements within the braces.Braces with an asteriks{ }*The elements within the braces can occur zero or more times.Braces with a plus { }+The elements within braces must occur one or more times.Angle brackets < >These elements are further defined as a seperate entitie in the syntax list.Vertical bar |OrGDSII BNF<stream format>::=HEADER BGNLIB LIBNAME [REFLIBS ] [FONTS ][ATTRTABLE ] [GENERATIONS ] [<FormatType>]UNITS {<structure>}* ENDLIB<FormatType>::=FORMAT | FORMAT {MASK }+ ENDMASKS <structure>::=BGNSTR STRNAME [STRCLASS ] {<element>}* ENDSTR<element>::={<boundary> | <path> | <sref> | <aref> | <text> | <node> | <box>}{<property>}* ENDEL<boundary>::=BOUNDARY [ELFLAGS ] [PLEX ] LAYER DATATYPE XY<path>::=PATH [ELFLAGS ] [PLEX ] LAYER DATATYPE [PATHTYPE ][WIDTH ] XYThe Stream format output file is composed of variable length records. Record length is measured in bytes. The minimum record length is four bytes. Within the record, two bytes (16 bits) is a word. The 16 bits in a word are numbered 0 to 15, left to right.The first four bytes of a record compose the recordheader. The first two bytes of the recordheader contain a count (in eight-bit bytes) of the total record length, so the maximum length is 65536 (64k). The next record starts immediately after the last byte of the previous record.The third byte of the header is the record type. The fourth byte of the header identifies the type of data contained within the record. The fifth until count bytes of a record contain the data.The fourth byte in the record header contains the data type for the rest of the record. The record length is used to find the number of items of the specified datatype.1.Bit Array:A bit array is a word which uses the value of a particular bit or group of bits to represent data. A bit array allows oneword to represent a number of simple pieces of information.2.Two-Byte Signed Integer:2-byte integer = 1 word 2s-complement representation. The range of two-byte signed integers is -32,768 to 32,767. The following is a representation of a two-byte integer, where S is the sign and M is the magnitude.<sref>::=SREF [ELFLAGS ] [PLEX ] SNAME [<strans>] XY<aref>::=AREF [ELFLAGS ] [PLEX ] SNAME [<strans>] COLROW XY <text>::=TEXT [ELFLAGS ] [PLEX ] LAYER <textbody><node>::=NODE [ELFLAGS ]. [PLEX ] LAYER NODETYPE XY <box>::=BOX [ELFLAGS ] [PLEX ] LAYER BOXTYPE XY<textbody>::=TEXTYPE [PRESENTATION ] [PATHTYPE ] [WIDTH ] [<strans>] XY STRING <strans>::=STRANS [MAG ] [ANGLE ]<property>::=PROPATTR PROPVALUERecord headerBitnr 0123456789101112131415Word1Total Record length in bytes Word2Record Type Data typeWord3Data until Word n (total record length/2)Data TypesData Type Value No Data 0Bit Array1Two Byte Signed Integer 2Four Byte Signed Integer 3Four Byte Real 4 (not used)Eight Byte Real 5ASCII string 6smmmmmmm mmmmmmmmThe following are examples of two-byte integers:00000000 00000001 = 100000000 00000010 = 200000000 10001001 = 13711111111 11111111 = -111111111 11111110 = -211111111 01110111 = -1373.Four-Byte Signed Integer:4-byte integer = 2 word 2s-complement representationThe range of four-byte signed integers is -2,147,483,648 to 2,147,483,647.The following is a representation of a four-byte integer, where S is the sign and M is themagnitude.smmmmmmm mmmmmmmm mmmmmmmm mmmmmmmmThe following are examples of four-byte integers:00000000 00000000 00000000 00000001 = 100000000 00000000 00000000 00000010 = 200000000 00000000 00000000 10001001 = 13711111111 11111111 11111111 11111111 = -111111111 11111111 11111111 11111110 = -211111111 11111111 11111111 01110111 = -1374.Four-Byte Real4-byte real = 2-word floating point representation(See 5.)5.Eight-Byte Real8-byte real = 4-word floating point representationFor all non-zero values:A floating point number has three parts: the sign, the exponent, and the mantissa.The value of a floating point number is defined as:(Mantissa) x (16 raised to the true value of the exponent field).The exponent field (bits 1-7) is in Excess-64 representation.The 7-bit field shows a number that is 64 greater than the actual exponent.The mantissa is always a positive fraction >=1/16 and <1. For a 4-byte real, themantissa is bits 8-31. For an 8-byte real, the mantissa is bits 8-63.The binary point is just to the left of bit 8.Bit 8 represents the value 1/2, bit 9 represents 1/4, etc.In order to keep the mantissa in the range of 1/16 to 1, the results of floating pointarithmetic are normalized. Normalization is a process where by the mantissa isshifted left one hex digit at a time until its left FOUR bits represent a non-zeroquantity. For every hex digit shifted, the exponent is decreased by one. Since themantissa is shifted four bits at a time, it is possible for the left three bits of thenormalized mantissa to be zero. A zero value, also called true zero, is representedby a number with all bits zero.The following are representations of 4-byte and 8-byte reals, where S is the sign, E is the exponent, and M is the magnitude. Examples of 4-byte reals are included in the following pages, but 4-byte reals are not used currently. The representation of the negative values of real numbers is exactly the same as the positive, except that the highest order bit is 1, not 0. In the eight-byte real representation, the first four bytes are exactly the same as in the four-byte real representation. The last four bytes contain additional binary places for more resolution.4-byte real:SEEEEEEE MMMMMMMM MMMMMMMM MMMMMMMM8-byte real:SEEEEEEE MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMMMMMMMMMM MMMMMMMMExamples of 4-byte real:Note: In the first six lines of the following example, the 7-bit exponent field = 65. The actual exponent is 65-64=1.01000001 00010000 00000000 00000000 = 101000001 00100000 00000000 00000000 = 201000001 00110000 00000000 00000000 = 311000001 00010000 00000000 00000000 = -111000001 00100000 00000000 00000000 = -211000001 00110000 00000000 00000000 = -301000000 10000000 00000000 0000000 = 0 .501000000 10011001 10011001 1001100 = 1 .601000000 10110011 00110011 0011001 = 1 .701000001 00011000 00000000 00000000 = 1.501000001 00011001 10011001 10011001 = 1.601000001 00011011 00110011 00110011 = 1.700000000 00000000 00000000 00000000 = 001000001 00010000 00000000 00000000 = 101000001 10100000 00000000 00000000 = 1001000010 01100100 00000000 00000000 = 10001000011 00111110 00000001 00000000 = 100001000100 00100111 00010000 00000000 = 1000001000101 00011000 01101010 00000000 = 1000006.ASCII StringA collection of ASCII characters, where each character is represented by one byte. All oddlength strings must be padded with a null character (the number zero), and the byte count for the record containing the ASCII string must include this null character. Stream read-inprograms must look for the null character and decrease the length of the string by one if the null character is present.The following table gives an overview of all the record that are used within a GDSII file. Record Types OverviewNr.Code Mnemonic Data Type description00002HEADER Two-Byte SignedIntegerversion number10102BGNLIB Two-Byte SignedIntegerbegin of library, last modification date andtime20206LIBNAME Two-Byte SignedIntegername of library30305UNITS Eight-Byte Real user and database units 40400ENDLIB No Data end of library50502BGNSTR Two-Byte SignedIntegerbegin of structure + creation and modificationtime60606STRNAME ASCII string name of structure70700ENDSTR No Data end of structure80800BOUNDARY No Data begin of boundary element90900PATH No Data begin of path element100A00SREF No Data begin of structure reference element 110B00AREF No Data begin of array reference element 120C00TEXT No Data begin of text element130D02LAYER Two-Byte SignedIntegerlayer number of element140E02DATATYPE Two-Byte SignedIntegerDatatype number of element150F03WIDTH Four-Byte SignedIntegerwidth of element in db units161003XY Four-Byte SignedIntegerlist of xy coordinates in db units171100ENDEL No Data end of element181206SNAME ASCII string name of structure reference191302COLROW Two-Byte SignedIntegernumber of colomns and rows in arrayreference211500NODE No Data begin of node element221602TEXTTYPE Two-Byte SignedIntegertexttype number231701PRESENTATION Bit Array text presentation, font 251906STRING ASCII string character string for text element261A01STRANS Bit Array array reference, structure reference and text transform flags271B05MAG Eight Byte Real magnification factor for text and references 281C05ANGLE Eight Byte Real rotation angle for text and references311F06REFLIBS ASCII string name of referenced libraries322006FONTS ASCII string name of text fonts definition filesRecords are always an even number of bytes long. The first four bytes of a record are the record header . If a record contains ASCII string data and the ASCII string is an odd number of bytes long, the data is padded with a null character. This paragraph lists the record types with a brief description of each. The descriptions include the record name and a four-digit number in brackets. The first two numbers within the brackets are the record type, and the last two numbers in brackets are the data type. All record numbers are expressed in hexadecimal. Contains two bytes of data representing the Stream version number. Contains the last modification time of a library (two bytes each for year, month, day, hour, minute, and second), the time of last access (same format), and marks the beginning of a library. 332102PATHTYPE Two-Byte Signed Integertype of PATH element end ( rounded, square)342202GENERATIONSTwo-Byte SignedInteger number of deleted structure ?????352306ATTRTABLE ASCII string attribute table, used in combination with element properties 382601ELFLAGS Two-Byte Signed Integertemplate data422A02NODETYPE Two-Byte Signed Integernode type number for NODE element 432B02PROPATTR Two-Byte Signed Integerattribute number 442C06PROPVALUE ASCII string attribute name452D00BOX No Databegin of box element 462E02BOXTYPE Two-Byte Signed Integerboxtype for box element 472F03PLEX Four-Byte Signed Integerplex number 503202TAPENUM Two-Byte Signed IntegerTape Number 513302TAPECODE Two-Byte Signed IntegerTape code 543602FORMAT Two-Byte Signed Integerformat type 553706MASKASCII string list of layers 563800ENDMASKS No Dataend of MASKRecord types descriptionHEADER0002Two-Byte Signed Integer1BGNLIB0102Two-Byte Signed IntegerBit 0123456789101112131415word1l C (hex) # of bytes in recordword201 (hex) 02 (hex)word3year (lastmodification time)Contains a string which is the library name. The library name must follow UNIX filenameconventions for length and valid characters. The library name may include the file extension (.sf or db in most cases). Contains two eight-byte real numbers. The first number is the size of a database unit in user units. The second number is the size of a database unit in meters. For example, if you create a library with the default units (user unit = 1 micron and 1000 database units per user unit), the first numberis .001, and the second number is 1E-9. Typically, the first number is less than 1, since you use more than 1 database unit per user unit. To calculate the size of a user unit in meters, divide the second number by the first. Marks the end of a library. Contains the creation time and last modification time of a structure (in the same format as the BGNLIB record), and marks the beginning of a structure. Contains a string which is the structure name. A structurename may be up to 32 characters long. Legal structurename characters are:z A through Z z a through z z 0 through 9 z Underscore (_) z Question mark (?) zDollar sign ($)Marks the end of a structure.word4month word5day word6hour word7minute word8secondword9year (last access time)word10month word11day word12hour word13minute word14second2LIBNAME0206ASCII String3UNITS0305Eight-Byte Real4ENDLIB0400No Data5BGNSTR0502Two-Byte Signed Integer6STRNAME0606ASCII String7ENDSTR0700No DataMarks the beginning of a boundary element. Marks the beginning of a path element. Marks the beginning of an Sref (structure reference) element.Marks the beginning of an Aref (array reference) element.Marks the beginning of a text element.Contains two bytes which specify the layer. The value of the layer must be in the range of 0 to 255.Contains two bytes which specify the datatype. The value of the datatype must be in the range of 0 to 255.Contains four bytes which specify the width of a path or text lines in database units. A negative value for width means that the width is absolute, that is, the width is not affected by the magnification factor of any parent reference. If omitted, zero is assumed.zContains an array of XY coordinates in database units. Each X or Y coordinate is four bytes long. Path elements may have a minimum of 2 and a maximum of 200 coordinates. Boundary and border elements may have a minimum of 4 and a maximum of 200 coordinates. The first and last coordinates of a boundary or border must coincide. z A text, or Sref element may have only one coordinate.zAn Aref has exactly three coordinates. In an Aref, the first coordinate is the array reference point (origin point). The other two coordinates are already rotated, reflected as specified in the STRANS record (if specified). So in order to calculate the intercolomn and interrow spacing, the coordinates must be mapped back to their original position, or the vector lenght (x1,y1-> x3,y3) must be divided by the number of row etc. . The second coordinate locates a position which is displaced from the reference point by the inter-column spacing times the number of columns. The third coordinate locates a position which is displaced from the reference point by the inter-row spacing times the number of rows. For an example of an array lattice see the next picture.8BOUNDARY 0800No Data 9PATH0900No Data10SREF0A00No Data11AREF0B00No Data12TEXT0C00No Data13LAYER0D02Two-Byte Signed Integer14DATATYPEOEO2Two-Byte Signed Integer15WIDTH0F03Two-Byte Signed Integer16XY 1003Two-Byte Signed IntegerAref rotated -30 degrees.z A node may have from one to 50 coordinates.z A box must have five coordinates, with the first and last coordinates being the same.17ENDEL1100No DataMarks the end of an element.18SNAME1206ASCII stringContains the name of a referenced structure.See also STRNAME.19COLROW1302Two-Byte Signed IntegerContains four bytes. The first two bytes contain the number of columns in the array. The third and fourth bytes contain the number of rows. Neither the number of columns nor the number of rows may exceed 32,767 (decimal), and both are positive. See also AREF.21NODE1500No DataPresent Marks the beginning of a node22TEXTTYPE1602Two-Byte Signed IntegerContains two bytes representing texttype. The value of the texttype must be in the range 0 to 255.23PRESENTATION1701Bit ArrayContains one word (two bytes) of bit flags for text presentation. Bits 10 and 11, taken together as a binary number, specify the font (00 means font 0, 01 rneans font 1, 10 means font 2, and 11 means font 3). Bits 12 and 13 specify the vertical justification (00 means top, 01 means middle, and 10 means bottom). Bits 14 and 15 specify the horizontal justification (00 means left, 01 means center, and 10 means right). Bits 0 through 9 are reserved for future use and must be cleared. If this record isomitted, then top-left justification and font 0 are assumed. The following shows a PRESENTATION record.Contains a character string, up to 512 characters long, for text presentation.Contains two bytes of bit flags for Sref, Aref, and text transforrnation. Bit 0 (the leftmost bit)specifies reflection. If bit 0 is set, the element is reflected about the X-axis before angular rotation. For an Aref, the entire array is reflected, with the individual array members rigidly attached. Bit 13 flags absolute magnification. Bit 14 flags absolute angle. Bit 15 (the rightmost bit) and all remaining bits are reserved for future use and must be cleared. If this record is omitted, the element is assumed to have no reflection, non-absolute magnification, and non- absolute angle.The following shows a STRANS record.Eight-Byte Real Contains a double-precision real number (8 bytes), which is the magnification factor. If this record is omitted, a magnification factor of one is assumed.Eight-Byte Real Contains a double-precision real number (8 bytes), which is the angular rotation factor. The angle of rotation is measured in degrees and in the counterclockwise direction. For an Aref , the ANGLE rotates the entire array (with the individual array members rigidly attached) about the array reference point. For COLROW record information, the angle of rotation is already inlcuded in the coordinates. If this record is omitted, an angle of zero degrees is assumed.Contains the names of the reference libraries. This record must be present if any reference libraries are bound to the working library. The name of the first reference library starts at byte 5 (immediately following the record header) and continues for 44 bytes. The next 44 bytes contain the name of the second library. The record is extended by 44 bytes for each additional library (up to 15) which is bound for reference. The reference library names may include directory specifiers (separated with "/") and an extension (separated with "."). If either the first or second library is not named, its place is filled with nulls. Bit 0123456789101112131415word16 (hex) # of bytes in record word217 (hex) 01 (hex)word3unused font number vertical presentaion horizontal presentation 25STRING 1906ASCII String26STRANS 1A01Bit ArrayBit 0123456789101112131415word16 (hex) # of bytes in record word21A (hex)01 (hex)word3reflectionunused absolute magnification absolute angle unused 27MAG 1B05Eight Byte Real28ANGLE 1C05Eight Byte Real31REFLIBS 1F06ASCII StringContains the names of the textfont definition files. This record must be present if any of the four fonts have acorresponding textfont definition file. This record must not be present if none of the fonts have a textfont definition file. The textfont filename of font 0 starts the record, followed by the textfont files of the remaining three fonts.Each filename is 44 bytes long. The filename is padded withnulls if the name is shorter than 44 bytes. The filename is null if no textfont definitioncorresponds to the font. The textfont filenames may include directory specifiers (separated with "/" and an extension (separated with ".").This record contains a value that describes the type of path endpoints. The value isz 0 for square-ended paths that endflush with their endpointsz 1 for round-ended pathsz 2 for square-ended paths that extend a half-width beyond their endpointsIf not specified, a Path-type of 0 is assumed.The following picture shows the pathtypes32FONTS 2006ASCII String33PATHTYPE 2102Two-Byte Signed IntegerPathtype 0 produces a square-ended path, ending flush withthedigitized endpoints. This isthe de-fault pathtype if none isspecifiedPathtype 1 produces a round-ended path. The two endsaresemicircular with center atthedigitized endpoints.Pathtype 2 produces a square-ended path. The ends of thepathextend beyond thedigitized end-points by one-half the path width.This record contains a value to indicate the number of copies of deleted or back-up structures to retain. This numbermust be at least 2 and not more than 99. If the GENERATION record is omitted, a value of 3 is assumed.Contains the name of the attribute definition file. This record is present only if an attribute definition file is bound to the library. The attribute defenition filename may include directory specifiers (separated with "/") and an extension (separated with "."). Maximum record size is 44 bytes.Unrelesed FeatureUnrelesed FeatureContains two bytes of bit flags. Bit 15 (the rightmost bit)specifies Template data. Bit 14 specifies External data(also referred to as Exterior data). All other bits are currently unused and must becleared to 0. If this record isomitted, all bits are assumed to be 0. The following shows an ELFLAGS record.For additional information on Template data, consult the GDSII Reference Manual. For additional information on External data, consult the CustomPlus User's Manual.(Unreleased feature)(Unreleased feature)(Unreleased feature)Contains two bytes which specify nodetype. The value ofthe nodetype must be in the range of 0 to 255. 34GENERATIONS 2202Two-Byte Signed Integer35ATTRTABLE 2306Two-Byte Signed Integer36STYPTABLE 2502Two-Byte Signed Integer37STRTYPE2502Two-Byte Signed Integer 38ELFLAGS 2601Bit ArrayBit 0123456789101112131415word16 (hex) # of bytes in record word226 (hex) 01 (hex)word3unused external data template data39ELKEY 2703Two-Byte Signed Integer40LINKTYPE 28Two-Byte Signed Integer41LINKKEYS 29Two-Byte Signed Integer42NODETYPE 2A02Two-Byte Signed Integer43PROPATTR2B02Two-Byte Signed IntegerContains two bytes which specify the attribute number. The attribute number is an integer from 1 to 127. Attribute numbers 126 and 127 are reserved for the user integer and userstring (CSD) properties which existed prior to Release 3.0.44PROPVALUE2C06ASCII stringContains the string value associated with the attribute named in the preceding PROPATTR record. Maximumlength is 126 characters. The attribute-value pairs associated with any one element must all have distinct attribute numbers. Also, the total amount of property data that may be associated with any one element is limited: thetotal length of all the strings, plus twice the number of attribute-value pairs, must not exceed 128 (or 512 if the element is an Sref, Aref, contact, nodeport, or node). For example, if a boundary element uses property attribute2 with property value "metal," and property attribute 10 with property value "property," the total amount of property data is 18 bytes. This is 6 bytes for "metal" (odd-length strings must be padded with a null) + 8 for "property" + 2 times the 2 attributes (4) = 18.45BOX2D00No DataMarks the beginning of a box element.46BOXTYPE2E02Two-Byte Signed IntegerContains two bytes which specify boxtype. The value of the boxtype must be in the range of 0 to 255.47PLEX2F03Two-Byte Signed IntegerA unique positive number which is common to all elementsof the plex to which this element belongs. The head of the plex is flagged by setting the seventh bit; therefore, plexnumbers should be small enough to occupy only the right-most 24 bits. If this record is not present, the element is not a plex member Applies to Pathtype 4.Contains four bytes which specify indatabase units the distance a path outline begins before orafter the last point of the path. Value can be negative.50TAPENUM3202Two-Byte Signed IntegerContains two bytes which specify the number of the current reel of tape for a multi-reel Stream file. For the first tape, the TAPENUM is 1: for the second tape, the TAPENUM is 2. For each additional tape, increment the TAPENum by one.51TAPECODE3302Two-Byte Signed IntegerContains 12 bytes. This is a unique 6-integer code which iscommon to all the reels of multi-reel Stream file. It verifies that the correct reels are being read.52STRCLASS3401Two-Byte Signed IntegerNot used53RESERVED3503Two-Byte Signed IntegerThis record type was used for NUMTYPES but was not required.54FORMAT3602Two-Byte Signed IntegerDefines the format of a Stream tape in two bytes. The possible values are:1.for GDSII Archive format2.for GDSII Filtered format3.for EDSM Archive format4.for EDSHI Filtered forrnatAn Archive Stream file contains elements for all the layers and data types. In an Archive Stream file, the FORMAT record is followed immediately by the UNITS record. A file which does not have the FORMAT record is assumed to be an Archive file.A Filtered Stream file contains only the elements on the layers and with the datatypes you specify during creation ofthe Stream file. The list of layers and datatypes specified appear in MASK records. At least one MASK record must immediately follow the FORMAT record. The MASK records are terminated with the ENDMASKS record.55MASK3706ASCII string(Required for and present only in FilteredStream file. )Contains the list of layers and datatypes specified by the user when creating the file. At least one MASK record must immediately follow the FORMAT record. More than one MASK record may occur. The last MASK record is followed by the ENDMASK record.In the MASK list, datatypes are separated from the layers with a semicolon. Individual layers or datatypes are sepa-rated with a space. A range of layers or datatypes is specified with a dash.An example MASK list looks like this: 1 5 -7 10 ; 0- 25556ENDMASKS3800NoData(Required for and present only in FilteredStream file.)Marks the end of the MASK records. The ENDMASKS record must follow the last MASK record. ENDMASKS is immediately followed by the UNITS record.。
数值计算基础 英文
数值计算基础英文The foundation of numerical computation involves the fundamental principles and techniques used in performing mathematical calculations using numerical values. It encompasses various mathematical operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction. Additionally, it includes concepts related to numerical methods, numerical analysis, and computational mathematics.Numerical computation serves as the basis for various fields such as engineering, physics, computer science, and finance. It is essential for solving complex mathematical problems and simulating real-world phenomena through computational models. Understanding the basics of numerical computation is crucial for accurately representing, analyzing, and interpreting quantitative data in diverse applications.In numerical computation, it is important to considerthe precision of numerical values, the impact of rounding errors, and the selection of appropriate algorithms for different types of calculations. Additionally, the representation of numbers in different formats such as floating-point representation and fixed-point representation is a key aspect of numerical computation.Moreover, the study of numerical computation involves the analysis of convergence, stability, and accuracy of numerical algorithms. It also encompasses the utilization of computational tools and software for efficient and reliable numerical calculations.Overall, a strong foundation in numerical computation is essential for professionals and students in various fields to effectively tackle mathematical problems, optimize algorithms, and make informed decisions based on numerical analysis. It provides the necessary skills to leverage computational resources for problem-solving and decision-making in diverse disciplines.。
C语言中浮点数的表示范围浅析
C语言中浮点数的表示范围浅析Abstract：The float is a data type in C language，but its in standard C and did not give a specific of description：that is the number of storage format and the scope of representation.Part of classically C language programming tutorial gives a range of floating-point represent，but there is not rigorous and need to bined IEEE754 standards，provided analyzation and conclusions in C language in internal storage floating-point format.Keywords：C language；floating-point；scope of representation1 引言（Introduction）浮点数运算是科学计算必须面对的问题，由于计算机内部本身不能精确地处理某些整数或小数，因此在运算时可能存在较大的误差，运算结果将直接影响到系统的可靠性和安全性等。
C语言因功能强大、程序设计灵活且支持底层应用，在科学计算、数据处理等领域中得到了广泛应用，但C语言在浮点数运算方面也存在数据表示的不精确性等问题。
经典C语言并没有对浮点数专门说明，国内很多教材虽述及浮点数，但也只是给出表示范围，对于浮点数的解释尚不够充分，描述尚不够严谨，因此学生在对浮点数的学习过程中经常存在这样或那样理解上的困惑。
这里就浮点数的表示范围，结合IEEE754做进一步的分析，为以后浮点数教学和学习给出参考[1]。
Floating Point Representation - Transforming Numerical …浮点表示法转换的数值…-PPT精选文档
5.7 41 50110.112 11 1 0.1012 12 05 11 1.1021 1 10 21
We have the representation as
001011101
Sign of the Sign of the
number
exponent
mantissa
. V ( a 1 ) s l 1 u m 2 e 2 e ' 127
13
Example#1
11010001010100000000000000000000
Sign
Biased
(s) Exponent (e’)
Mantissa (m)
Va l1 u s 1 e .m 2 2 e ' 127
e Actual range of
126e127
Special Exponents and Numbers
e 0
all zeros
e255 all ones
s
e
m Represents
0 all zeros all zeros
0
1 all zeros all zeros 0 all ones all zeros 1 all ones all zeros
Floating Point Representation
Major: All Engineering Majors
Authors: Autar Kaw, Matthew Emmons
Transforming Numerical Methods Education for STEM Undergraduates
1 .0 .. 0 0 .2 .2 1 ..2 2 6 .1 1 8 30 8
If the floating-point number representation on a certain
CS 3510 Fall 2009______________________________Exam #3print Last Name, First NameMatching (2 points each)1. Identify whether each of the following is based on combinational logic ( C ) or sequential logic ( S ).Fill-in-the-blanks Questions (2 points each)2. ______________________ contains the address of an instruction to be fetched.3. ______________________ contains the instruction most recently fetched.4. ______________________ contains the address of a location in memory.5. ______________________ contains a word of data to be written to memory or the word mostrecently read.Very Short Essay Questions (3 points each) – only use the space provided6. Explain the purpose of the decoder in a memory subsystem. (See Fig. 5.2, page 740)7. How are the lines grouped in a BUS? Which are the types of buses used in a computer system?8.What are the differences between bandwidth and transfer rate from the CPU to memory.9. Briefly explain the purpose of a replacement algorithm in cache memory. Is cache constructed with SRAM or DRAM? Explain.10. What comprise the “datapath” ?Short Problems1. If the floating-point number representation on a certain system has a sign bit, a 3-bit exponent and a 4-bit significand, what is the largest positive and the smallest positive number that can be stored on this system if the storage is normalized ? – 5 points∙There is no excess-n approach on the exponent∙Exponents use two’s complement notation∙Exponents of all zeros and all ones are allowed2. If the floating-point number representation on a certain system has a sign bit, a 5-bit exponent using excess-16 and an 8-bit normalized mantissa, (a) show how the computer would represent the numbers 100.0 and 0.25, (b) show how the computer would add the two floating-point numbers in part a by changing 0.25 so the two numbers are both expressed using the same power of 2. – 6 points3. Investigate the operation of the following circuit. Assume an initial state of 0000. Determine the outputs (the Qs) as the clock ticks eight times (1 to 8). – 5 pointsProblems1. A cache memory has a total of 64 blocks, each consisting of 128 words. Main memory contains 4096 blocks. For direct mapping, calculate the number of bits in main memory address, and the size of the Tag, Block, and Word fields. (8 points)2. Solve problem 5.18, page 808 (364). (7 points)3. Complete the truth table for the following sequential circuit: - 6 points4. Complete the truth table for the following sequential circuit: - 6 points5. Design a sequential circuit that counts up and down in even numbers from 0 to6. When the input equals 0, the circuit counts down and vice versa for an input of 1. The circuit produces a 1 for its output when ever the number is greater than 1 and less than 5.(a)What is the state diagram of the sequential circuit? Be sure and illustrate in the state diagramthe input and output as the circuit transitions between states – use the variable x for input and z for output. – 4 points(b)Create a state table from the state diagram – be sure and illustrate present state, next state andoutput – 3 points(c)Redo the state table using assignments – 3 points(d)Determine the next-state and output equations – 3 points(e)Draw the sequential circuit using JK-flip-flops, AND gates, OR gates and NOT gates – 5 points Questions on Processor (Questions 1 to 6 are 5 points each. Question 7 is 10 points.)1.Why is the MFC step needed when reading from or writing to main memory ?2.Which component is responsible for controlling and synchronizing the various functions andcomponents in the ISP?3.Briefly explain how the ISP allows registers to write and read to and from each other.4.Briefly explain how the ALU and associated MUX are used in incrementing the PC?5.Explain the differences between RISC and CISC processors. How many registers should beincluded in the processor architecture?6.Are the memory operands fetched during the interpretation of an instruction or during theexecution of the instruction? Explain.7.Write an assembly language program with the following specification. The program is to copyan array to a different (and specified location). The main section of code calls a routine(function) passing three arguments: the number of bytes, the source array, and the destination array. The routine will copy the list of values from the source location to the destination location. The first parameter is the total number of data items to copy. Each data item is four bytes.。
导论名词解释
导论概念Hardware The physical elements of a computing system (printer, circuit boards,wires, keyboard…)Software The programs that provide the instructions for a computer to executeAbstraction A mental model that removes complex details（Early History of computing）：Abacus An early device to record numeric valuesBlaise Pascal Mechanical device to add, subtract, divide & multiply Joseph Jacquard Jacquard’s Loom, the punched cardCharles Babbage Analytical EngineAda Lovelace First Programmer, the loopAlan Turing Turing Machine, Artificial Intelligence TestingHarvard Mark I, ENIAC, UNIV AC IEarly computers launch new era in mathematics, physics, engineering and economics1st Generation ：Vacuum Tubes Large, not very reliable, generated a lot of heat Magnetic Drum Memory device that rotated under a read/write headCard Readers -》Magnetic Tape DrivesSequential auxiliary storage devices2nd GenerationTransistor Replaced vacuum tube, fast, small, durable, cheap Magnetic Cores Replaced magnetic drums, information available instantlyMagnetic Disks Replaced magnetic tape, data can be accessed directly 3rd GenerationIntegrated Circuits Replaced circuit boards, smaller, cheaper, faster, more reliable.Transistors Now used for memory constructionTerminal An input/output device with a keyboard and screen4rd GenerationLarge-scale Integration Great advances in chip technologyPCs, the Commercial Market, WorkstationsPersonal Computers were developed as new companies like Apple and Atari came into being. Workstations emerged.Moore’s Law"The number of transistorsincorporated in a chip will approximately double every24 months."Parallel ComputingComputers rely on interconnected central processing units that increaseprocessing speed.NetworkingWith the Ethernet small computers could be connected and share resources. A file server connected PCs in the late 1980s.ARPANET and LANs ->InternetMachine LanguageComputer programs were written in binary (1s and 0s)Assembly Languages and translatorsPrograms were written in artificial programming languages and were then translated into machine languageProgrammer ChangesProgrammers divide into application programmers and systems programmersHigh Level LanguagesUse English-like statements and make programming easier. Fortran, COBOL, Lisp are examples.Systems Software–utility programs,–language translators,–and the operating system, which decides whichprograms to run and when.•Separation between Users and HardwareComputer programmers began to write programs to be used by people who did not know how to programStructured ProgrammingPascal, C, C++New Application Software for UsersSpreadsheets, word processors, database management systems5th Gen SoftwareMicrosoftThe Windows operating system, and other Microsoft application programs dominate the marketObject-Oriented DesignBased on a hierarchy of data objects (i.e. Java)World Wide WebAllows easy global communication through the InternetNew UsersToday’s user needs no computer knowledgeLec.2::Number SystemNatural NumbersZero and any number obtained by repeatedly adding one to it.Negative NumbersA value less than 0, with a –signIntegersA natural number, a negative number, zeroRational NumbersAn integer or the quotient of two integersThe base of a number determines the number of digits and the value of digit positionsPosition is key, carry values are used:Lec3::Data Representation IHere is a formula that you can use to compute the negative representation Neg(i) = 10^k - i (k is the number of the digits)•This representation of negative numbers is called the ten’s complement（补）.Bit：A bit is the basic unit of information in computing and digital communications. A bit canhave only one of two values, and may therefore be physically implemented with a two-statedevice. The most common representation of these values are 0 and 1. •byte–8 bits. (c types maybe int8_t, uint8_t, char )•integer–A natural number, a negative numberTwo’s ComplementTo make it easier to look at long binary numbers, we make the number line vertical.Negative (I) = 2k –I where k the number of digits•Overflowoccurs when the value that we compute cannot fit into the number of bits we have allocated for the result. For example, if each value is stored using eight bits, adding 127 to 3 overflows.Scientific notationA form of floating-point representation in which the decimal point is kept to the right of the leftmost digit.Computers are finite. Computer memory and other hardware devices have only so much room to store and manipulate a certain amount of data. The goal, is to represent enough of the world to satisfy our computational needs and our senses of sight and sound.Analog dataA continuous representation, analogous to the actual information itrepresents.Digital dataA discrete representation, breaking the information up into separate elements.A mercury thermometer is an analog device. The mercury rises in a continuous flow in the tube in direct proportion to the temperature. Computers, cannot work well with analog information. So we digitize information by breaking it into pieces and representing those pieces separately.Why do we use binary?Modern computers are designed to use and manage binary values because the devices that store and manage the data are far less expensive and far more reliable if they only have to represent on of two possible values.A character setis a list of characters and the codes used to represent each one.ASCII stands for American Standard Code for Information Interchange. Note that the first 32 characters in the ASCII character chart do not have a simple character representation that you could print to the screen.The amount of data that is used to represent a color is called the color depth.•HiColoris a term that indicates a 16-bit color depth. Five bits are used for each number in an RGB value and the extra bit is sometimes used to represent transparency.TrueColorindicates a 24-bit color depth. Therefore, each number in an RGB value gets eight bits.Digitizing a picture is the act of representing it as a collection of individual dots called pixels.•The number of pixels used to represent a picture is called the resolution.•The storage of image information on a pixel-by-pixel basis is called a rastergraphicsformat. Several popular raster file formats including bitmap (BMP). Keyword Encoding Frequently used words are replaced with a single character.run-length encodinga sequence of repeated characters is replaced by a flag character, followed by the repeated character, followed by a single digit that indicates how many times the character is repeated.Huffman EncodingAn important characteristic of any Huffman encoding is that no bit stringused to represent a character is the prefix of any other bit string used to represent a character.To digitize the signal we periodically measure the voltage of the signal and record the appropriate numeric value. The process is called sampling.In general, a sampling rate of around 40,000 times per second is enough to create a reasonable sound reproduction.–MP3 employs both lossy and lossless compression.Data compression Reduction in the amount of space needed to store a piece of data.•Compression ratio The size of the compressed data divided by the size of the original data.• A data compression techniques can be–lossless, which means the data can be retrieved without any loss of the original information,–lossy, which means some information may be lost in the process of compaction.Gates and CircuitsGate（门）A device that performs a basic operation on electrical signals•Circuits（电路）Gates combined to perform more complicated tasksTransistor A device that acts, depending on the voltage level of an input signal, either as a wirethat conducts electricity or as a resistor that blocks the flow of electricity A transistor has three terminals–A source –A base –An emitter, typically connected to a ground wireA NOT gate is sometimes referred to as an inverter（反相器）because it inverts the input valueBoolean expressions（布尔表达式）Expressions in Boolean algebra, a mathematical notation for expressing two valued logicLogic diagram（逻辑图）A graphical representation of a circuit Each type of gate is represented by a specific graphical symbol•Truth table（真值表）A table showing all possible input value and the associated output valuesTwo general categories In a combinational circuit, the input values explicitly determine the outputIn a sequential circuit, the output is a function of the input values as well as the existing state of the circuitcircuit equivalenceThat is, both circuits produce the exact same output for each input value combinationAt the digital logic level, addition is performed in binary•Addition operations are carried out by special circuits called, appropriately, addersA circuit that computes the sum of two bits and produces the correct carry bit is called a half adder•A circuit called a full adder takes the carry-in value into account Memory is a collection of cells,each with aunique physical address. RAM stands for Random Access Memory Inherent in the idea of being able to access each location is the ability to change the contents of each location•ROM stands for Read Only Memory The contents in locations in ROM cannot be changed•RAM is volatile, ROM is not This means that RAM does not retain its bit configuration when the power is turned off, but ROM doesMost modern ALUs have a small amount of special storage units called registers.Control unit The organizing force in the computerThe instruction register (IR)contains the instruction that is being executedThe program counter (PC) contains the address of the next instruction to be executedALU and the control unit called the Central Processing Unit, or CPUA CPU cache (缓存)is a cache used by the central processing unit of a computer toreduce the average time to access memory.Input UnitA device through which data and programs from the outside world are entered into the computerOutput unit A device through which results stored in the computer memory are made available to the outside worldMachine language （机器语言）The instructions built into the hardware of a particular computerThe relationship between the processor and the instructions it can carry out is completely integratedCompiler （编译）A program that translates a high-level language program into machine codeInterpreter （解释）A translating program that translates and executes the statements in sequenceThe divide-and-conquer （算法设计基本思想之一，“分治”）approachcan be applied over and over again until each sub task is manageable.Algorithm A set of instructions for solving a problem or sub problem in a finite amount of time using a finite amount of data The instructions must be unambiguousObject A thing or entity that makes sense within the context of the problem.A group of similar objects is described by an object class, or class• A class contains fields that represent the properties and behaviors of the class–A field can contain data value(s) and/or methods (subprograms) Information Hiding The practice of hiding the details of a module with the goal of controlling access to the details of the module.–Abstraction A model of a complex system that includes only the details essential to the viewer.Data abstraction Separation of the logical view of data from their implementation.Procedural abstraction Separation of the logical view of actions from their implementation.Control abstraction Separation of the logical view of a control structure from its implementation.Abstract data typeA data type whose properties (data and operations)are specified independently of any particular implementation. Application level:View of the data within a particular problem Logical level:An abstract view of the data values (the domain) and the set of operations to manipulate them.Implementation level:A specific representation of the structure to hold the data items and the coding of the operations in a programming language.Data structures :The implementation of a composite data fields in an abstract data type . Containers :Objects whole role is to hold and manipulate other objects.Array:An array is a named collection of homogeneous items in which individual items are accessed by their place within the collection. An item’s place within the collection is calledan index.Linked implementation:An implementation based on the concept of a node.Node:A node is made up of two pieces of information--the item that the user wants in the list, and a pointer to the next node in the list.Generic data type (or class):A data type or class in which the operations are specified but the type or class of the objects being manipulated is not.Binary search:A sequential search of a list begins at the beginning of the list and continues until the item is found or the entire list has been searched.A binary search looks for an item in a list using a divide-and-conquer strategy.Stack:A stack is an abstract data type in which accesses are made at only one end.Queue:A Queue is an abstract data type in which items are entered at one end and removed from the other end.Application software:Software written to address specific needs—to solve problems in the real world.System software :Software that manages a computer system at a fundamental levelIt provides the tools and an environment in which application software can be created and run.Operating system:An operating system manages computer resources,provides an interface through which a human can interact with the computer,allows an application program to interact with these other system resources. Multiprogramming :The technique of keeping multiple programs in mainmemory at the same time that compete for access to the CPU so that they can execute.Memory management :The process of keeping track of what programs are in memory and where in memory they reside.Process :A program in execution.Timesharing system:A system that allows multiple users to interact with a computer at the same time.Multiprogramming:A technique that allows multiple processes to be active at once, allowingprogrammers to interact with the computer system directly, while still sharing its resources.Logical address (sometimes called a virtual or relative address) :A value that specifies a generic location, relative to the program but not to the reality of main memory.Physical addres s :An actual address in the main memory device.Fixed partitions :Main memory is divided into a particular number of partitions. Dymanic partitions :Partitions are created to fit the needs of the programs.Base register:A register that holds the beginning address of the current partition. Bounds register:A register that holds the length of the current partition.Three ways to allocate to a new program:First fit :Allocate program to the first partition big enough to hold it Best fit:Allocated program to the smallest partition big enough to hold it Worst fit : Allocate program to the largest partition big enough to hold it. Paged memory technique:A memory management technique in which processes are divided into fixed-size pages and stored in memory frames when loaded into memory.Frame :A fixed-size portion of main memory that holds a process pagePage:A fixed-size portion of a process that is stored into a memory frame Page-map table (PMT) :A table used by the operating system to keep track of page/frame relationships.Demand paging :An important extension of paged memory managementPage swap :The act of bringing in a page from secondary memory, which often causes another page to be written back to secondary memoryCPU Scheduling :The act of determining which process in the ready state should be moved to the running state.Nonpreemptive scheduling(非抢占式调度):The currently executing process gives up the CPU voluntarilyPreemptive scheduling(抢占式调度):The operating system decides to favor another process, preempting the currently executing process Turnaround time(时间片):The amount of time between when a process arrives in the ready state the first time and when it exits the running state for the last time.File :A named collection of related data.File system:The logical view that an operating system provides so that users can manage information as a collection of filesDirectory :A named group of files.Text file:A file in which the bytes of data are organized as characters from the ASCII or Unicode character setsBinary file :A file that contains data in a specific format, requiring interpretation Sequential access :Information in the file is processed in order, and read and write operations move the current file pointer as far as needed to read or write the dataDirect access :Files are conceptually divided into numbered logical records and each logical record can be accessed directly by numberDirectory tree :A logical view of a file system; a structure showing the nested directory organization of a file systemRoot directory :The directory at the highest level.Path :A text designation of the location of a file or subdirectory in a file system. Absolute path :A path that begins at the root and specifies each step down the tree until it reaches the desired file or directoryRelative path:A path name that begins at the current working directory.软件工程：是指导计算机软件开发和维护的工程学科。
IEEE标准754中英文
如有你有帮助，请购买下载，谢谢！几乎所有计算机都支持二进制数据表示，即能直接识别二进制数据表示并具 有相应的指令系统。
通常采用的二进制定点数据表示主要有：符号数值、反码、 补码以及带偏移增值码四种形式，其中最常用的是补码形式，这些都已在计算机 组成原理课程中做了详细讨论，这里不再阐述。
二进制浮点数的表示，由于不同机器所选的基值、尾数位长度和阶码位长度不 同，因此对浮点数表示有较大差别，这就不利于软件在不同计算机间的移植。
美 国 IEEE（电子及电子工程师协会）为此提出了一个从系统结构角度支持浮点数 的表示方法， 称之为 IEEE 标准754(IEEE，1985)，当今流行的计算机几乎都采 用这一标准。
IEEE 754在标识符点数时， 每个浮点数均由3个部分组成：符号位 S，指数部 分 E 和尾数部分 M。
浮点数可采用以下四种基本格式： (1)单精度格式(32位)：E=8位，M=23位。
(2)扩展单精度格式：E≥11位，M≥31位。
(3)双精度格式(64)位：E=11位，M=52位。
(4)扩展双精度格式(64位)：E≥15位，M≥63位。
其中，单精度格式(32位)中的阶码为8位， 另有一位尾数的符号位 S，处在最高 位。
如图，浮点数的分数部分与有效位部分两者是不同的， 由于 IEEE754标准 约定在小数点左部有一位隐含位，从而使其有效位实际有24位，这样便使尾数的 有效值变为1M。
阶码部分采用移码表示， 移码值为127，从而使阶码值的范围 由原来的1到254，经移码后变为-126到+127。
IEEE 754标准的单精度和双精度浮点数表示格式。
其中，阶码值0和255分别用 来表示特殊数值：当阶码值为255时，若分数部分为0，则表示无穷大；若分数部 分不为0，则认为这是一个‘非数值’。
当阶码和尾数均为0时则表示该数值为0， 因为非零数的有效位总是≥1，因此特别约定，这表示为0。
当阶码为0， 尾数不 为0时，该数绝对值较小， 允许采用比最小规格化数还要小的数表示。
吉林大学计算机导论复习重点
Chapter 1 Computer and DataKnowledge point:1.1The computer as a black box.1.2von Neumann model1.3The components of a computer: hardware, software, and data.1.4The history of computers.REVIEW QUESTIONS1.How is computer science defined in this book?A：Issues related to the computer.2.What model is the basis for today’s computers? ( Knowledge point 1.2)A：The von Neumann model.3.Why shouldn’t you call a computer a data processor? ( Knowledge point 1.1)A：Computer is general-purpose machine. it can do many different types of tasks.4.What does a programmable data processor require to produce output data? ( Knowledgepoint 1.1)A：The input data and the program.5.What are the subsystems of the von Neumann computer model? ( Knowledge point 1.2) A：Memory, arithmetic logic unit, control unit, and input/output.6.What is the function of the memory subsystem in von Neumann’s model? ( Knowledgepoint 1.2)A：Memory is the storage area. It is where programs and data are stored during processing.7.What is the function of the ALU subsystem in von Neumann’s model? ( Knowledge point1.2)A：ALU is where calculation and logical operations take place.8.What is the function of the control unit subsystem in von Neumann’s model? ( Knowledgepoint 1.2)A：It controls the operations of the memory, ALU, and the input/output subsystem.9.What is the function of the input/output subsystem in von Neumann’s model?( Knowledge point 1.2)A：The input subsystem accepts input data and the program from outside the computer; the output subsystem sends the result of processing to the outside.pare and contrast the memory contents of early computers with the memory contentsof a computer based on the von Neumann model? ( Knowledge point 1.2)A：Computer based on the von Neumann model stores both the program and its corresponding data in the memory. Early computers only stored the data in the memory.11.How did the von Neumann model change the concept of programming? ( Knowledgepoint 1.2)A：A program in the von Neumann model is made of a finite number of instructions. The instructions are executed one after another.12.The first electronic special-purpose computer was called c( Knowledge point1.4)a. Pascalb. Pascalinec. ABCd. EDV AC13.One of the first computers based on the von Neumann model was called d( Knowledge point 1.4)a. Pascalb. Pascalinec. ABCd. EDV AC14.The first computing machine to use the idea of storage and programming was calledd( Knowledge point 1.4)a. the Madelineb. EDV ACc. the Babbage machined. the Jacquard loom15.d separated the programming task from the computer operation tasks.( Knowledge point 1.3)a. Algorithmsb. Data processorsc. High-level programming languagesd.Operating systems30. According to the von Neumann model, can the hard disk of today be used as input or output? Explain. ( Knowledge point 1.2)A：Yes. When the hard disk stores data that results from processing, it is considered an output device; when you read data from the hard disk, it is considered an input device. 32. Which is more expensive today, hardware or software? ( Knowledge point 1.3)A：Software.Chapter 2 Data RepresentationKnowledge point:2.1 Data Types.2.2 Data inside the Computer.2.3 Representing Data.2.4 Hexadecimal and Octal notation.REVIEW QUESTIONS five types of data that a computer can process. ( Knowledge point2.1)A：Numbers, text, images, audio, and video.2.How does a computer deal with all the data types it must process? ( Knowledge point 2.2) A：All data types are transformed into bit pattern.3.4.What is the difference between ASCII and extended ASCII? ( Knowledge point 2.3)A：ASCII is a bit pattern made of 7 bits and extended ASCII is a bit pattern made of 8 bits.5.What is EBCDIC? ( Knowledge point 2.3)A：Extended Binary Coded Decimal Interchange Code.6.How is bit pattern length related to the number of symbols the bit pattern can represent?( Knowledge point 2.3)A：The relationship is logarithmic.7.8.9.What steps are needed to convert audio data to bit patterns? ( Knowledge point 2.3)A：Sampling, Quantization, and Coding.10.What is the relationship between image data and video data? ( Knowledge point 2.3)A：Video is a representation of images in time.34. A company has decided to assign a unique bit pattern to each employee. If the company has 900 employees, what is the minimum number of bits needed to create this system of representation? How many patterns are unassigned? If the company hires another 300employees, should it increase the number of bits? Explain your answer. ( Knowledge point 2.3)A：log2900≈10,210-900=124,Yes, 900+300>210Chapter 3 Number RepresentationKnowledge point:3.1 Convert a number from decimal, hexadecimal, and octal to binary notation and vice versa.3.2 Integer representation: unsigned, sign-and-magnitude, one’s complement, and two’s complement.3.3 Excess system.3.4 Floating-point representation.REVIEW QUESTIONS5. What are three methods to represent signed integers? (Knowledge point 3.2)A：Sign-and-Magnitude, One’s Complement, and Two’s Complement.9. Name two uses of unsigned integers. ( Knowledge point 3.2)A：Counting and Addressing.10. What happens when you try to store decimal 130 using sign-and-magnitude representation with an 8-bit allocation? ( Knowledge point 3.2)A：Overflow.11. Compare and contrast the representation of positive integers in sing-and-magnitude, one’s complement, and two’s complement. ( Knowledge point 3.2)A：The representation of positive integers in sing-and-magnitude, one’s complement, and two’s complement is the same.14. Compare and contrast the range of numbers that can be represented in sign-and-magnitude, one’s complement, and two’s complement. ( Knowledge point 3.2)A：Sign-and-Magnitude range –(2N-1-1)～+(2N-1-1)One’s Complement range –(2N-1-1)～+(2N-1-1)Two’s Complement range –(2N-1)～+(2N-1-1)16. What is the primary use of the Excess_X system? ( Knowledge point 3.3)A：The primary use of the Excess_X system is in storing the exponential value of a fraction.17. Why is normalization necessary? ( Knowledge point 3.4)A：A fraction is normalized so that operations are simpler.Chapter 4 Operation On BitsKnowledge point:4.1 Arithmetic operations.4.2 Logical operations.4.3 Mask.4.4 Shift operations.REVIEW QUESTIONS3. What happens to a carry form the leftmost column in the final addition? ( Knowledge point4.1)A：The carry is discarded.5. Define the term overflow. ( Knowledge point 4.1)A：Overflow is an error that occurs when you try to store a number that is not within the range defined by the allocation.8. Name the logical binary operations. ( Knowledge point 4.2)A：NOT, AND, OR, and XOR.10. What does the NOT operator do? ( Knowledge point 4.2)A：It inverts bits.(it changes 0 to 1 and 1 to 0)11. When is the result of an AND operator true? ( Knowledge point 4.2)A：Both bits are 1.12. When is the result of an OR operator true? ( Knowledge point 4.2)A：Neither bit is 0.13. When is the result of an XOR operator true? ( Knowledge point 4.2)A：The two bits are not equal.17. What binary operation can be used to set bits? What bit pattern should the mask have? ( Knowledge point 4.3)A：OR. Use 1 for the corresponding bit in the mask.18. What binary operation can be used to unset bits? What bit pattern should the mask have? ( Knowledge point 4.3)A：AND. Use 0 for the corresponding bit in the mask.19. What binary operation can be used to flip bits? What bit pattern should the mask have? ( Knowledge point 4.3)A：XOR. Use 1 for the corresponding bit in the mask.Chapter 5 Computer OrganizationKnowledge point:5.1. three subsystems that make up a computer5.2. functionality of each subsystem5.3. memory addressing and calculating the number of bytes5.4. addressing system for input/output devices.5.5. the systems used to connect different components together.Review questions:1. What are the three subsystems that make up a computer?(Knowledge point 5.1) Answer: the CPU, main memory, and the input/output (I/O) subsystem.2. What are the parts of a CPU? (Knowledge point 5.1)Answer: The CPU performs operations on data and has a ALU, a control unit, and a set of registers.3. What‘s the function of the ALU? (Knowledge point 5.2)Answer: The ALU performs arithmetic and logical operations.Exercises:78. A computer has 64MB of memory. Each word is 4 bytes. How many bits are needed toaddress each single word in memory? (Knowledge point 5.3)Solution:The memory address space is 64 MB, that is 2 raised to the power 26. The size of each word in bytes is 2 raised to the power 2. So we need 24(subtract 2 from 26) bits to address each single word in memory.79. How many bytes of memory are needed to store a full screen of data if the screen is made of 24 lines with 80 characters in each line? The system uses ASCII code, with each ASCII character store as a byte. (Knowledge point 5.3)Solution:The quantity of bytes in a full screen is 1920 (24*80) while the system uses ASCII code with each ASCII character store as a byte. So we need 1920 bytes of memory to store the full screen of data.87. A computer uses isolated I/O addressing. Memory has 1024 words. If each controller has 16 registers, how many controllers can be accessed by this computer? (Knowledge point 5.4) Solution:Memory has 1024 words. So the address space is 1024. Each controller has 16 registers. Then we get 64 (divide 16 by 1024)controllers which can be accessed by this computer.88. A computer uses memory-mapped I/O addressing. The address bus uses 10 lines. If memory is made of 1000 words, how many four-register controllers can be accessed by this computer? (Knowledge point 5.4)Solution:The address bus uses 10 lines. So, the address space is 1024(2 raised to the power 10). The memory is made of 1000 words and each controller has four registers. Then we get (1024-1000)/4 = 6 four-register controllers which can be accessed by this computer.Chapter 6 Computer NetworksKnowledge point:6.1. OSI model6.2. TCP/IP protocol6.3. three types of networks6.4. connecting devices6.5. client-server modelReview questions:2. Name the layers of the OSI model? (6.1)Answer: Physical layer, Data link layer, Network layer, Transport layer, Session layer, Presentation layer and Application layer.3. Name the layers of the TCP/IP protocol suite. (6.2)Answer: The layers of the TCP/IP protocol suite are: physical and data-link layers network layer, transport layer, and application layer.8. What are the three common topologies in LANs? Which is the most popular today? (6.3) Answer: bus topology, star topology, ring topology, star topology9. Name four types of network connecting devices. (6.4)Answer: the four types of network connecting devices are repeater, bridge, router and gateway.Chapter 7 Operating SystemsKnowledge point:7.1. the definition of an operating system7.2. the components of an operating system7.3. Memory Manager7.4. Process manager7.5. deadlockReview questions:4. What are the components of an operating system? (7.2)Answer: An operating system includes: Memory Manager, Process Manager, Device Manager and File Manager13. What kinds of states can a process be in? (7.4)Answer: ready state, running state, waiting state.15. If a process is in the running state, what states can it go to next? (7.4)Answer: ready state, waiting state.What’s the definition of an operating system? (7.1)Answer: An operating system is an interface between the hardware of a computer and user(programs or humans) that facilitates the execution of other programs and the access to hardware and software resources.What are the four necessary conditions for deadlock? (7.5)Answer: mutual exclusion, resource holding, no preemption and circular waiting.51. A multiprogramming operating system uses paging. The available memory is 60 MB divided into 15 pages, each of 4MB. The first program needs 13 MB. The second program needs 12MB. The third program needs 27 MB. How many pages are used by the first program? How many pages are used by the second program? How many pages are used by the third program? How many pages are unused? What is the total memory wasted? What percentage of memory is wasted? (7.3)Answer:Each page is 4MB. The first program needs 13 MB. It is obviously that 4*3<13<4*4. So the first program uses 4 pages and wastes 3(16-3) MB. The second program needs 12 MB. It is obviously that 12=4*3. So the second program uses 3 pages and wastes 0 MB. The third program needs 27 MB. It is obviously that 4*6<27<4*7. So the first program uses 7 pages and wastes 1(28-27) MB. There are 1(15-4-3-7) page unused. There are totally 4(3+0+1) MB memory wasted. The percent of memory wasted is 4/(60-4*1)=7%.Chapter 8 AlgorithmsKnowledge point:8.1. the concepts of an algorithm and a subalgorithm8.2. three constructs for developing algorithms8.3. basic algorithms8.4. tools for algorithm representation8.5. recursionReview questions:1. What is the formal definition of an algorithm? (8.1)Answer: An ordered set of unambiguous steps that produces a result and terminates in a finite time.2. Define the three constructs used in structured programming. (8.2)Answer: The three constructs in structured programming are Sequence, Decision and Repetition.10. What are the three types of sorting algorithms? (8.3)Answer: bubble sort, selection sort and insertion sort.12 What is the purpose of a searching algorithm? (8.3)Answer: The purpose is to find the location of a target among a list of objects.13. What are the two major types of searches? How are they different? (8.3)Answer: sequential search and binary search. The difference is whether the list is ordered or not.55. A list contains the following elements. Using the binary search algorithm, trace the steps followed to find 20. At each step, show the values of first, last and mid.3, 7, 20, 29, 35, 50, 88, 200 (8.3)Solution:index 0---1---2---3---4---5---6---73, 7, 20, 29, 35, 50, 88, 200First=0, Last=7, Mid=(0+7)/2=3The data is D(3)=29, 20 smaller than the D(3), so remove the data from index 3 to 7. Change the new point First=0, Last= mid-1=2, and Mid=(0+2)/2=1The data is D(1)=7, 20 bigger than the D(1), so remove the data from index 0 to 1. Change the new point First=mid+1=2, Last=2, and Mid=(2+2)/2=2The data is D(2)=20, we find the data 20 in index=258. Write a recursive algorithm to find the combination of n objects taken k at a time using following definition.C(n,k)=1 , if k=0 or n=kC(n,k)=C(n-1,k)+(n-1,k-1) , if n>k>0 (8.5)Solution:A: CInput : n and kIf(k==0 or n==k)Then return 1End ifIf(n>k and k>0)Then Return C(n-1,k)+(n-1,k-1)End ifEnd。
固定点与浮点
固定点与浮点数字信号处理技术可以分为两类-定点和浮点运算。
这些名称是指用于存储和处理数据的数字表示的格式。
定点DSP设计代表和操纵整数-通过至少16位，收益率高达65,536个可能的位模式（2 -正面和负面的整数16）。
浮点DSP代表和操纵通过类似的方式，以科学记数法，其中一个数字代表一个尾数和一个指数（例如，A×2 32位的最低有理数B，其中'A'是尾数'B'是指数），收益率高达4,294,967,296个可能的位模式（2 32）。
“定点”是指在数字代表相应的方式，用一个固定位数的号码后，有时会之前，小数点。
与浮点表示，小数点的位置可以“浮动”的相对数量的有效位数。
例如，用一个统一的小数点位置约定的定点表示可以代表的数字123.45，1234.56，12345.67，等，而浮点表示，除了代表等1.234567，123456.7，0.00001234567，12345670亿等，可以支持浮点值比定点范围更广，能够代表非常小的数字和非常大的数字。
定点表示法，相邻数字之间的差距总是等于一个值，而在浮点表示法，相邻数字之间的差距不是均匀分布的- 任何两个数字之间的差距是价值约一万次，比小（ANSI / IEEE标准754标准格式）数字，大数字和小小的数字之间的差距之间的巨大差距。
动态范围和精度固有的浮点计算的幂保证更大的动态范围-最大和最小的可表示的数字-这是特别重要时，处理非常大的数据集或数据集的范围可能是不可预知的。
因此，浮点处理器，非常适合计算密集型应用。
同样重要的是在精确的范围内考虑固定和浮点格式- 数字之间的差距的大小。
每次DSP产生一个新的号码，通过数学计算，这个数字必须四舍五入到最接近的值，可以通过使用格式存储。
四舍五入和/或截断数字信号处理过程中自然产生的量化误差或'噪音' - 实际模拟值和量化的数字值之间的偏差。
由于相邻数字之间的差距可以是定点加工相比，浮点处理时，舍入误差可以更明显。
modbus tcp 04读浮点数例子
modbus tcp 04读浮点数例子Modbus TCP 04 Read Floating-Point Number: ExampleIn this article, we will discuss an example scenario of reading floating-point numbers using Modbus TCP protocol (Function Code 04). Modbus TCP is a widely used communication protocol in industrial automation systems.The Modbus TCP 04 function code is used for reading input registers, which are 16-bit data units storing integers or floating-point numbers. This function code allows us to retrieve the value of input registers from a Modbus TCP server.To read floating-point numbers using Modbus TCP 04, we need to follow these steps:1. Establish a TCP connection with the Modbus TCP server using the appropriate IP address and port number.2. Send a Modbus TCP request packet specifying the function code 04 and the starting address of the input register from which we want to read the floating-point number.3. Handle the Modbus TCP response packet received from the server. The response packet will contain the requested floating-point number, represented as a 32-bit value. This value can be interpreted as per the IEEE 754 standard for floating-point representation.4. Convert the received 32-bit value into a floating-point number using suitable programming techniques. Most programming languages provide libraries or functions to handle the conversion according to the IEEE 754 standard.5. Display or use the obtained floating-point number as per the requirements of your application.It's important to note that the starting address provided in the Modbus TCP request packet must be valid and within the range supported by the Modbus TCP server.Additionally, ensure that the TCP connection is established successfully and that there are no network or server-side issues preventing the communication.By implementing the above steps, you can read floating-point numbers using Modbus TCP 04 function code in your industrial automation systems. This capability allows for real-time monitoring and control of sensors or devices that provide floating-point data.Modbus TCP is widely supported by various devices and software platforms, making it a versatile and reliable choice for industrial communication. Proper implementation and adherence to the Modbus TCP specifications will ensure accurate and efficient communication between your application and the Modbus TCP server.In conclusion, the Modbus TCP 04 function code is an indispensable tool for reading floating-point numbers in Modbus TCP communication. By following the outlined steps and handling the data correctly, you can effectively retrieve and utilize floating-point values in your industrial automation applications.。
modbus rtu 浮点型数据 高低位顺序
modbus rtu 浮点型数据高低位顺序Modbus RTU is a popular communication protocol used in industrial automation systems. It is a widely used protocol for connecting industrial electronic devices and equipment. One common use of Modbus RTU is for transmitting floating point data, which presents a unique challenge due to the high and low byte order. In this article, we will explore how Modbus RTU handles floating point data and the considerations that need to be taken into account when working with floating point data in this protocol.Floating point data is a type of data representation that is used to store and transmit real numbers in a wide range of magnitudes. In Modbus RTU, floating point data is typically transmitted using two consecutive 16-bit registers, which contain the high and low order bytes of the floating point number. This presents a challenge, as the order in whichthese bytes are transmitted and received must be managed to ensure that the correct value is interpreted.The high and low byte order in Modbus RTU is defined by the endianness of the protocol. Endianness refers to the order in which bytes are arranged in memory or transmitted over a network. In Modbus RTU, the high byte is transmitted first, followed by the low byte. This means that when working with floating point data in Modbus RTU, the high byte of the floating point number will be transmitted first, followed by the low byte.To correctly interpret floating point data in Modbus RTU, it is important to be aware of the byte order and to ensure that the transmitted data is received and interpreted in the correct order. This typically involves reading the two consecutive 16-bit registers and then rearranging the bytes to correctly interpret the floating point number. This process can be managed by using appropriate conversionalgorithms in the software that is interfacing with the Modbus RTU device.Another consideration when working with floating point data in Modbus RTU is the representation of the floating point number itself. Modbus RTU does not have a native data type for floating point numbers, so they must be represented using the available data types, such as 16-bit or 32-bit integer values. This means that the floating point number must be converted to and from these integer representations when transmitting and receiving data over Modbus RTU.When converting a floating point number to an integer representation for transmission over Modbus RTU, it is important to consider the precision and range of the floating point number. The integer representation must be able to accurately represent the floating point number without losing precision or exceeding the range of the integer data type. Similarly, when receiving data over Modbus RTU, the integerrepresentation must be correctly converted back to a floating point number to ensure that the original value is accurately interpreted.In conclusion, working with floating point data in Modbus RTU requires careful consideration of the byte order and the representation of the floating point number. By understanding the byte order of Modbus RTU and using appropriate conversion algorithms, it is possible to accurately transmit and interpret floating point data in this protocol. Additionally, by carefully managing the conversion between floating point numbers and integer representations, it is possible to ensure that the precision and range of the floating point number are maintained when working with Modbus RTU.。
浮点表示法尾数
浮点表示法尾数
随着计算机科学的发展，浮点表示法尾数（floating-point representation of fractions）已成为现代计算机的重要元素。
这种表示形式的特点是，它可以用一定的方式使数值表示为有限的符号序列，并且在计算机表示数值时，不会出现舍入误差或舍入失真。
浮点表示法尾数以有限的字符序列存储大量数据。
它以一定的方式将数值表示为含有小数点的有理数、科学计数法、非规则表达式或比例缩放的数值。
浮点表示法尾数的概念源自科学计算，它可以提供精确的结果，并且容易实现。
例如，有计算器可以用科学计算器来计算任意大小的数字和运算结果。
尽管浮点表示法尾数可以实现精确的计算，但它也有一定限制。
例如，由于有限的字符序列，它不能准确表示任意大小的数值。
此外，它也不能对数值进行精确的运算。
因此，在使用浮点表示法尾数时，应格外注意，以避免出现误差或失真的情况。
浮点表示法尾数不仅应用于计算机领域，还在其他领域得到广泛应用。
例如，它可用于金融计算、科学计算和精密测量。
浮点表示法尾数还可以用来提供流畅的视频和图像表示，运用浮点表示法尾数的精确度和计算的容易性，可以大大增加视频和图像的真实感。
随着科技的进步，浮点表示法尾数以它的精确度和有效性受到广泛认可。
它已经成为计算机和信息处理领域的一种重要工具。
回顾以上，浮点表示法尾数是一种有效而可靠的计算工具，可以
用来表示有理数、科学计数法、非规则表达式或比例缩放的数值，并且可以大大减少舍入误差或失真的情况。
它已经成为计算机和信息处理领域的重要元素，并广泛应用于金融计算、科学计算和精密测量等领域。
浮点表示法尾数
浮点表示法尾数浮点表示法尾数（FloatingPointRepresentationSignificands）是一种使用数值计算机程序中最常用的表示法，它允许用户在十进制数和二进制数之间进行转换。
浮点表示法尾数主要用于数据转换和计算，以及图形显示，以完成计算机程序中的各种数值计算任务。
浮点表示法尾数的表示格式由两部分组成，即符号（Sign）与尾数（Mantissa），它们被称为浮点表示法的“符号尾数”形式。
符号尾数形式除了用来表示小数之外，还常用于标识负数，表示基数和指数等。
从符号尾数形式中可以推导出浮点表示法尾数的构成：-号：给出了数字的正负状态，用“+”表示正，用“-”表示负。
-数：表示一个数字的有效数字位，又称小数部分，可以是整数或小数。
尾数的表示方法又分为小数表示法、十进制表示法和二进制表示法，并且每种表示法都有自己的指数（Exponent），即尾数后面跟随的一个或几个数字。
小数表示法采用定点运算，即在尾数中加以指数修饰，其形式为： m x 10^e其中m表示尾数，e表示指数，数字10表示基数，e表示指数。
此外，小数表示法还允许使用正负修饰符。
在十进制表示法中，m表示尾数，e表示指数，e的值可以是正也可以是负，表示尾数的指数。
计算机程序经常使用的十进制表示法形式为：m x 10^e其中m表示小数位，e表示指数，e可以是正也可以是负。
这时，尾数的值就由m x 10^e来表示，e的值越大，表示的尾数也就越大。
二进制表示法也按照前面的形式表示：m x 2^e其中m表示尾数，e表示指数，e可以是正也可以是负，表示尾数指数。
不同的是，指数是根据二进制数进行计算的，而不是十进制数，所以e的值可以比十进制的更大。
因此，在浮点表示法尾数的表示中，通常将指数表示为x 2^e，也就是将小数表示为m x 2^e，而不是m x 10^e，这样可以大大提高计算能力。
在使用浮点表示法尾数时，还需要注意以下几点：- 点数的有效位不是无限的，必须限定其有效位数；- 定点运算时，指数e的值不可以太大，否则会引起运算错误； - 浮点表示法尾数的表示中，基数可以由十进制改为二进制，也可以由二进制改为十进制；- 点表示法尾数的表示方法也可以做为图形显示的工具，如在实验中可以使用它来模拟复杂的数字图形。
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8
Relative Error and Machine Epsilon
The absolute relative true error in representing a number will be less then the machine epsilon
/topics/floatingpoint_represe ntation.html
THE END
y m2e
sign of number 0 for ve,1for- ve
m mantissa12 m 102
1 is not stored as it is always given to be 1.
eintegexr ponent
5
validlab/goldberg/paper.pdf
IEEE-754 Format Single Precision
32 bits for single precision
00000000000000000000000000000000
Sign
Biased
(s) Exponent (e’)
Mantissa (m)
• Standardizes representation of floating point operations on different computers.
One Great Reference
What every computer scientist (and even if you are not) should know about floating point arithmetic!
5.7 41 50110.112 11 1 0.1012 12 05 11 1.1021 1 10 21
We have the representation as
001011101
Sign of the Sign of the
number
exponent
mantissa
exponent
6
Example 0.02813 021.110 20 25 1.110 20 2011 20
10 bit word (sign, sign of exponent, 4 for exponent, 4 for mantissa)
0101101100
Sign of the number
Sign of the exponent
? ???????????????????????????????
Sign
Biased
(s) Exponent (e’)
Mantissa (m)
5 .5 8 1 1 3 0 0 1 4 1 1 .? 2 ?
15
Exponent for 32 Bit IEEE-754
8 bits would represent 0e255
Bias is 127; so subtract 127 from representation
127e128
16
Exponent for Special Cases
e Actual range of
1e254 e 0and e255are reserved for special numbers
IEEE-754 Floating Point Standard
• Standardizes representation of floating point numbers on different computers in single and double precision.
Machine epsilon
ma c2 h23 1.1 91 0 7
19
Additional Resources
For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, s, related physical problems, please visit
Floating Point Representation
Major: All Engineering Majors
Authors: Autar Kaw, Matthew Emmons
Transforming Numerical Methods Education for STEM Undergraduates
-0
0 or 1 all ones non-zero NaN
IEEE-754 Format
The largest number by magnitude
1 .1 ...1 2 . 2 .1 .2 .3 7 .4 1 0 30 8
The smallest number by magnitude
1 .0 .. 0 0 .2 .2 1 ..2 2 6 .1 1 8 30 8
Example
9 bit-hypothetical word
▪the first bit is used for the sign of the number, ▪the second bit for the sign of the exponent, ▪the next four bits for the mantissa, and ▪the next three bits for the exponent
6/26/2020
1
Floating Point Representation
Floating Decimal Point : Scientific Form
25.678is writteans2.5678102 0.00367is8writteans3.678103 25.678is writteans2.5678102
e Actual range of
126e127
Special Exponents and Numbers
e 0
all zeros
e255 all ones
s
e
m Represents
0 all zeros all zeros
0
1 all zeros all zeros 0 all ones all zeros 1 all ones all zeros
exponent
mantissa
1.1100 2 201102 0.0274375
a
0.02832 0.0274375 0.02832
0.034472 24 0.0625
9
IEEE 754 Standards for Single Precision Representation
111 .1012 02 (1 000 1)2 0 1 0 0 20 0 1 1 .62 2 1 5 6 12 27 1 1 .6 2 2 3 5 5 5 .5 8 1 1 3 0 0
14
Example#2
Represent -5.5834x1010 as a single precision floating point number.
Machine Epsilon
Defined as the measure of accuracy and found by difference between 1 and the next number that can be represented
7
Example
Ten bit word
▪Sign of number ▪Sign of exponent ▪Next four bits for exponent ▪Next four bits for mantissa
0 0 0 0 0 0 0 0 0 0 110
Next number
0 0 0 0 0 0 0 0 0 1 1.00 2 01.1 0612 0
3
Example
The form is
signmanti1s0esxaponent
or
m10e
Example: For
2.5671802
1
m 2.5678
e2
4
Floating Point Format for l 1 u m 2 e 2 e ' 127
13
Example#1
11010001010100000000000000000000
Sign
Biased
(s) Exponent (e’)
Mantissa (m)
Va l1 u s 1 e .m 2 2 e ' 127