HW2ans
高处作业基本知识ppt课件.pptx
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结语 高处不胜寒，安全不能忘。为了高高兴兴上班来，平平安安回家
去，从事高处作业的人员一定要了解基本安全知识，熟悉高处作业安 全要求，不麻痹大意，不逾越红线。
更多EHS独家精只品有资工料作，安请全咨着询，“生安活应才管能家幸”福微着信！号：ansyingsj1
（7）摆动、立足处不是平面或只有很小的平面，即任一边小于500mm 的矩形平面、直径小于500mm的圆形平面或具有类似尺寸的其他形状的平面， 致使作业者无法维持正常姿势；
更多EHS（独8）家存精在有品毒资气料体或，空请气咨中含询氧“量安低于应0管.19家5的”作微业环信境号；：ansyingsj1
（9）可能会引起各种灾害事故的作业环境以及抢救突然发生的各种灾害 事故。
及时拆卸或采取固定措施。 拆卸下的物料及余料和废料
应及时清理运走，不得任意放置 或向下丢弃。
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Part 4：高处作业人员应知应会 （8）用于高处作业的防护设施，不得擅自拆除。
更多EHS独家精品资料，请咨询“安应管家”微信号：ansyingsj1
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Part 4：高处作业人员应知应会
（9）在雨、霜、雾、雪等 天气进行高处作业时，应采取防
坠落范围半径为半径，划成的
更进多行E的H作S业独。家精品资料，请咨询“安应管家与”水微平面信垂号直：的柱a形ns空y间in。gsj1
坠落高度基准面
通过可能坠落范围内最 低处的水平面。
可能坠落范围半径
为确定可能坠落范围而 规定的相对于作业位置的一 段水平距离。
2
目录
Contents
Part 1 何为高处作业 Part 2 高处作业分为哪几类 Part 3 如何对高处作业进行分级 Part 4 高处作业人员应知应会 Part 5 高处作业安全管理要点
CORONA 2.4GHz Spread Spectrum (FASST COMPATIBLE) R
CORONA 2.4GHz Spread Spectrum (FASST COMPATIBLE)Receiver Instruction Manual for R4FA‐SB and R6FA‐SB Compatibility:The CORONA 2.4GHz Spread Spectrum FASST Compatible Receiver is designed for use with FUTABA’s FASST 2.4GHz transmitters; including the 3PM,3PKS,3VCS,3GR,4PK(S),TM7, TM8, TM10, TM14 and the T6EX‐2.4G, 7C‐2.4G, 8FG, 10CG, 12FG. The R4FA‐SB and R6FA‐SB receivers supply a more useful mode for users. Both the R4FA‐SB and R6FA‐SB support FUTABA’s FASST air system and surface system. The R4FA‐SB supports 7‐channel with continuous PPM (positive and negative) output ,RSSI output and S.BUS output, R6FA‐SB supports 6 channel high speed PPM(HS) mode to optimize helicopter response control and S.BUS output.Under S.BUS output mode, both R4FA‐SB and R6FA‐SB supply 12 proportional channels and 2 DG channels. Therefore, the R4FA‐SB or R6FA‐SB becomes 14‐channel receivers when using S.BUS output.Specifications:Operating Current: 50mA maxOperating Voltage: 3.6 ~10VLatency: R4FA‐SB’s description14mS for independent 4 channel output and S.BUS output @ FASST multi‐channel mode21mS for Continuous PPM output and RSSI output@ FASST multi‐channel mode16mS for independent 4 channel output and S.BUS output @ FASST 7ch mode24mS for Continuous PPM output and RSSI output@ FASST 7ch mode14mS for independent 3 channel output@ FASST surface system C1 CODE modeR6FA‐SB’s description7mS for independent 6 channel (HS) output@ FASST multi‐channel mode14mS for independent 7 channel (LS) output and S.BUS output @ FASST multi‐channel mode 8ms for independent 6 channel (HS) output@ FASST 7ch mode16mS for independent 7 channel (LS) output and S.BUS output @ FASST 7ch mode14ms for independent 3 channel output@ FASST surface system C1 CODE modeSensitivity: about ‐100dBmOperation temperature:‐10~80 deg CSetup:Bind procedure:∙Turn on the FASST transmitter∙Connect the battery to the receiver while pressing the receiver’s F/S button.∙The Dual‐color LED’s will continuously cycle through the following:o Red LED light (searching radio signal)o Green LED light (acquired the radio signal)o Red LED off (bind ok)o Green LED flashes 10 times (ID stored in memory)o Green LED lights solid (normal operation)Note: FASST surface systems take a bit more time to complete the bind procedure.Fail‐safe setting:There are two ways to set the Failsafe setting on the CORONA 2.4GHz Spread Spectrum FASST Compatible Receiver;1.TX‐failsafe feature: This method sets the failsafe on the FASST transmitter and has priority (works onchannel 3 only under FASST 7ch mode or on multiple‐channels under FASST multi‐channel mode) while the receiver is on, just like FUTABA receivers (only available on FASST air system).2.RX‐failsafe feature: Turn on the FASST transmitter and then turn on the CORONA 2.4GHz Spread SpectrumReceiver, put all the sticks and switches to control inputs you want if the receiver looses signal and Press the F/S button down for about 5 ‐ 6 seconds while the Green LED lights solid (Rx in normal operation), then release the button. You will see the Red LED will flash for about 4 ‐ 5 seconds. (Note: The Red LED will FLASH high speed to indicate the RX‐failsafe is turned on OR FLASH low speed to indicate the RX‐failsafe is turned off). If you press the F/S button a second time while the Red LED is flashing, the receiver will change its RX‐failsafe status (on / off), then the LED will return to Green solid again. If you not press the F/S button, nothing will be changed and the LED will return to solid Green. If you want to cancel the RX‐failsafe feature (not just turn it off), you can do so by binding the receiver again. After binding operation the receiver will be back to factory settings without any failsafe feature.Note: If you do not set a failsafe setting, the receiver will hold all controls at the position of the last command received before signal was lost. When RX‐failsafe is turned on, the receiver will initiate the RX‐failsafe settings after loosing signal for over 1 second and the receiver will hold the last received positions until the failsafe takes over. When the RX‐failsafe and TX‐failsafe feature are both turned on, the receiver will use the TX‐failsafe command.We highly recommend you set failsafe feature before flying your models. An example of a minimal useful, failsafe setting would be to shut down the model’s throttle, so that it does not fly or drive away uncontrolled.Output mode setting (only available on FASST air system):Turn off the transmitter, connect the battery to the receiver, you will see the Red LED light flashing. The RED LED flashes at high speed to indicate the receiver is in the special output mode OR a Low speed indicates the receiver is under (LS) low speed PPM normal output mode, press the F/S button for 5‐6 seconds while the Green LED is off (Rx in signal searching status), then release the button. You will see the Green LED flash for about 4 ‐ 5 seconds. (Note: The Green LED will FLASH high speed under special mode OR FLASH low speed under normal output mode). If you press the F/S button a second time while the Green LED is flashing, the receiver will change its output mode status (special/normal), if you do not press the F/S button the output mode will not be changed and the Red LED will flash at its original speed.Note: Output mode function is described in the form below,R4FA‐SB R6FA‐SBnormal Ch1~CH4 independent PPM output normal Ch1~CH7 independent PPM outputCH1 Neg CPPM out(FUTABA trainer FUNC)^1 CH2 Pos CPPM out for special user^2 CH1~CH6 independent high speed(HS)PPM out for helicopter fast response controlCH3 RSSI PWM out for FPV()^3 specialCH4 S.BUS output for compact system specialCH7 S.BUS output for compact systemNote: ^1 refer the signal description picture below^2 refer the signal description picture below^3 refer the signal description picture belowRSSI PWM out define: Pulse width from about 900uS~ 2100uS to indicate RSSI strength from ‐100dBm~‐40dBm.Important Note: If you are using analog servos in your model you must keep your receiver under the factory settings (normal output mode) or your analog servo will get hot and possibly burn out. As well you cannot use a non S.BUS servo on a channel while S.BUS signal output present.LED status indicated under normal working status:RED LED GREEN LED Statusflash off No signal searchedoff solid Signal is very goodSometime flash solid Signal is not very goodflash flash Signal is weak。
cs61a2021fallhw02
cs61a2021fallhw02 from operator import add, mulsquare = lambda x: x * xidentity = lambda x: xtriple = lambda x: 3 * xincrement = lambda x: x + 1HW_SOURCE_FILE = __file__def product(n, term):"""Return the product of the first n terms in a sequence.n: a positive integerterm: a function that takes one argument to produce the term>>> product(3, identity) # 1 * 2 * 36>>> product(5, identity) # 1 * 2 * 3 * 4 * 5120>>> product(3, square) # 1^2 * 2^2 * 3^236>>> product(5, square) # 1^2 * 2^2 * 3^2 * 4^2 * 5^214400>>> product(3, increment) # (1+1) * (2+1) * (3+1)24>>> product(3, triple) # 1*3 * 2*3 * 3*3162""""*** YOUR CODE HERE ***"ans = 1for i in range(1, n + 1):ans = ans * term(i)return ansdef square(x):return x * xdef accumulate(merger, base, n, term):"""Return the result of merging the first n terms in a sequence and base.The terms to be merged are term(1), term(2), ..., term(n). merger is atwo-argument commutative function.>>> accumulate(add, 0, 5, identity) # 0 + 1 + 2 + 3 + 4 + 515>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 526>>> accumulate(add, 11, 0, identity) # 1111>>> accumulate(add, 11, 3, square) # 11 + 1^2 + 2^2 + 3^225>>> accumulate(mul, 2, 3, square) # 2 * 1^2 * 2^2 * 3^272>>> # 2 + (1^2 + 1) + (2^2 + 1) + (3^2 + 1)>>> accumulate(lambda x, y: x + y + 1, 2, 3, square)19>>> # ((2 * 1^2 * 2) * 2^2 * 2) * 3^2 * 2>>> accumulate(lambda x, y: 2 * x * y, 2, 3, square)576>>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)16"""ans = basefor i in range(1,n+1):ans = merger(ans,term(i))return ansdef summation_using_accumulate(n, term):"""Returns the sum: term(1) + ... + term(n), using accumulate.>>> summation_using_accumulate(5, square)55>>> summation_using_accumulate(5, triple)45"""return accumulate(add,0,n,term)def product_using_accumulate(n, term):"""Returns the product: term(1) * ... * term(n), using accumulate.>>> product_using_accumulate(4, square)576>>> product_using_accumulate(6, triple)524880"""return accumulate(mul,0,n,term)def accumulate_syntax_check():"""Checks that definitions of summation_using_accumulate andproduce_using_accumulate are each a single return statement.>>> # You aren't expected to understand the code of this test.>>> # Check that the bodies of the functions are just return statements.>>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".>>> import inspect, ast>>> [type(x).__name__ for x in ast.parse(inspect.getsource(summation_using_accumulate)).body[0].body] ['Expr', 'Return']>>> [type(x).__name__ for x in ast.parse(inspect.getsource(product_using_accumulate)).body[0].body]['Expr', 'Return']"""def zero(f):return lambda x: xdef successor(n):return lambda f: lambda x: f(n(f)(x))def one(f):"""Church numeral 1: same as successor(zero)""""*** YOUR CODE HERE ***"return lambda x: f(zero(f)(x))def two(f):"""Church numeral 2: same as successor(successor(zero))""""*** YOUR CODE HERE ***"return lambda x :f(zero(f)(f(zero(f)(x))))three = successor(two)def church_to_int(n):"""Convert the Church numeral n to a Python integer.>>> church_to_int(zero)>>> church_to_int(one)1>>> church_to_int(two)2>>> church_to_int(three)3""""*** YOUR CODE HERE ***"def add_church(m, n):"""Return the Church numeral for m + n, for Church numerals m and n.>>> church_to_int(add_church(two, three))5""""*** YOUR CODE HERE ***"def mul_church(m, n):"""Return the Church numeral for m * n, for Church numerals m and n.>>> four = successor(three)>>> church_to_int(mul_church(two, three))6>>> church_to_int(mul_church(three, four))12""""*** YOUR CODE HERE ***"def pow_church(m, n):"""Return the Church numeral m ** n, for Church numerals m and n.>>> church_to_int(pow_church(two, three)) 8>>> church_to_int(pow_church(three, two)) 9""""*** YOUR CODE HERE ***"。
机器学习第二章作业答案
Machine Learning HW214S051053-汪洋2.1Ans:Since all occurrence of “φ” for an attribute of the hypothesis results in a hypothesis which does not accept any instance, all these hypotheses are equal to that one where attribute is “φ”. So the number of hypothesis is 4*3*3*3*3*3 +1 = 973.With the addition attribute Watercurrent, the number of instances = 3*2*2*2*2*2*3 = 288, the number of hypothesis = 4*3*3*3*3*3*4 +1 = 3889.Generally, the number of hypothesis = 4*3*3*3*3*3*(k+1)+1.2.2Ans:start: S0={<Ø,Ø,Ø,Ø,Ø,Ø>}G0={<?,?,?,?,?,?>}Sunny, Warm, High, Strong, Warm, Same, Yes, No, Rainy, Cold, Normal, Cool, ChangeAdd Example 4 :<Sunny, Warm, High, Strong, Cool, Change, Yes>S1={<Sunny,Warm, High, Strong, Cool, Change>}G1={<?,?,?,?,?,?>}Add Example 3 :<Rainy, Cold, High, Strong, Warm, Change, No>S2={<Sunny,Warm, High, Strong, Cool, Change>}G2={<Sunny,?,?,?,?,?> <?,Warm,?,?,?,?> <?,?,?,?,Cool,?>}Add Example 2 :<Sunny, Warm, High, Strong, Warm, Same, Yes>S3={<Sunny,Warm, High, Strong, ?, ?>}G3={<Sunny,?,?,?,?,?> <?,Warm,?,?,?,?> <?,?,?,?,Cool,?>}Add Example 1 :<Sunny, Warm, Normal, Strong, Warm, Same, Yes>S4={<Sunny,Warm, ?, Strong, ?, ?>}G4={<Sunny,?,?,?,?,?> <?,Warm,?,?,?,?> }The final version space is the same because it contains all the hypotheses consistent with the set of training examples used. Since there is only one set of hypotheses which are consistent with any given set of training examples, the algorithm must arrive at that one set regardless of the order. The sum for the current order of training examples above is 14, counting the initial state. The minimum this sum can be for any set of training examples is: 2 * ( # examples + 1 )#examples stand for number of examplesFor our example set size, the minimum would be 10. Since the hypothesis representation in use can only generalize an attribute with ?, there can be only one hypothesis in S at any one time. Hypotheses never get added to S since the specific hypothesis space relies on conjunctive expressions. In other words, generalizing S cannot be done by using an or condition. Consider an attribute A with possible values b, c, and d. If b and c were consistent with all positive training examples but d was not, the hypothesis in S must represent A as ? instead of having two hypotheses, one specifying b and the other c. This is the natural bias of the representation which allows classification of instances not previously encountered. This leads to considering how tominimize G. The only time hypotheses are added to G is when a negative example is presented, as stated in the algorithm 1. Therefore, a general strategy would be to present all positive training examples before negative training examples.Following this strategy, we arrive at a sum of 11. 11 is the minimum for this set of training examples since in the final state G contains 2 hypotheses, one more than the absolute minimum calculated above.2.4Ans:(a) The S boundary is: S = (4,6,3,5) as 46,35≤≤≤≤. As is shown in the bellow graph1.x y(b) The G boundary is: G = (3,8,2,7) as 38,27x y≤≤≤≤. As is shown in the bellow graph1.(c) Suggest a query guaranteed to reduce the size of the version space, regardless of how thetrainer classifies it. Eg: (7,6)Suggest one that will not. Eg: (5,4)(d) The smallest number of training examples. 4pointsThe four points as following: (3,2,+) (5,9,+) (2,1,-) (6,10,-)Graph12.5Ans:(a)Step1: S0 {<(Q Q Q Q ), (Q Q Q Q)>} G0 {<(? ? ? ?), (? ? ? ?)>}Step2: S1 {<(male brown tall US), (female black short US)> G1 {<(? ? ? ?), (? ? ? ?)>}Step3: S2 {<(male brown ? ?), (female black short US)> G2 {<(? ? ? ?), (? ? ? ?)>}Step4: S3 {<(male brown ? ?), (female black short US)> G3 {<(male ? ? ?), (? ? ? ?)>，<? ? ? ?>,<? ? ? US>}Step5: S4 {<(male brown ? ?), (female ? short ?)> G4 {<(male ? ? ?), (? ? ? ?)>}(b) Suppose that each attribute in the hypothesis can be taken for two values. So the number ofhypotheses consistent with the subject for example is （2*2*2*2）*（2*2*2*2）= 25628=(c) The shortest sequence should be 8. So25628=We set the If there is only one training sample, then the hypothesis space is 256training sample for each of the attributes, so that each of the hypothesis space is halved. After the 8 training, it can converge to a single correct hypothesis.<female,blanck,short,Portuguese>,<female,blonde,tall,Indian><male,brown,short,Portuguese>,<female,blonde,tall,Indian><male,blanck,tall,Portuguese>,<female,blonde,tall,Indian><male,blanck,short,US>,<female,blonde,tall,Indian><male,blanck,short,Portuguese>,<male,blonde,tall,Indian><male,blanck,short,Portuguese>,<female,black,tall,Indian><male,blanck,short,Portuguese>,<female,blonde,short,Indian><male,blanck,short,Portuguese>,<female,blonde,tall,US>(d) To express all of the concepts in the language, we need to expand the hypothesis space, so that every possible hypothesis is included, so that the space is far greater than 256, and so can not get the final convergence, because every one of the training sample, the vote without any effect, so there is no way to see the sample classification. So there is no optimal query sequence.2.6Proof:Every member ofVS satisfies the right-hand side of expression.,H DLet h be an arbitrary member ofVS, then h is consistent with all training examples in D.H D,Assuming h does not satisfy the right-hand side of the expression, it means (s S)(g G)(g h s)(s S)(g G)(g h)(h s)⌝∃∈∃∈≥≥=⌝∃∈∃∈≥∧≥.Hence, there does not exist g from G so that g is more general or equal to h or there does not exist s from S so that h is more general or equal to s. If the former holds, it leads to an inconsistence according to the definition of G. If the later holds, it leads to an inconsistence according to the definition of S. Therefore, h satisfies the right-hand side of the expression.Notes: since we assume the expression is not fulfilled, this can be only be if S or G is empty, which can only be in the case of any inconsistent training examples, such as noise or the concept target is not member of H.。
Wien2K编译方法
Wien2K编译方法：以root用户登陆1.准备intel安装intel编译器和mkl库1.拷贝所有安装文件到opt目录2.解压缩相应的文件,有icc和ifort以及mkl的安装文件。
3.开始安装icc:进入解压后的目录,运行install。
4.除去询问激活方式步骤时选择”以后激活”外,其余全部采用默认安装5.同样的步骤安装ifort和mkl.如果询问是否覆盖时,选择是6.安装完成后,将准备好的licenses文件拷贝到/opt/intel/compiler2 安装pgi编译器(可选)cd /softtar xzvf pgi-workstation-complete-x64-901.tar.gzcd Setuptar xzvf pgilinux-901.tar.gz./install...选择安装路径安装单机版...cd ../Cracktar xjvf pgi_9.0-1_linux64.tar.bz2cp pgi_9.0-1_linux64_patcher 安装目录cd 安装目录./pgi_9.0-1_linux64_patchercd /soft/Crack/license.dat 安装目录vi ~/.bashrc 加入下面内容#--- for pgi9.01PATH=/opt/pgi/9.01/linux86-64/9.0/bin:$PATHexport PA THMANPATH=$MANPATH:/opt/pgi/9.01/linux86-64/9.0/manexport MANPATHLM_LICENSE_FILE=/opt/pgi/9.01/license.datexport LM_LICENSE_FILELD_LIBRARY_PATH=/opt/pgi/9.01/linux86-64/9.0-1/libso:$LD_LIBRARY_PATHexport LD_LIBRARY_PA TH保存退出后重新登录或者source ~/.bashrc3 安装mpich因为wien2k采用intel编译器，这里采用intel编译器来编译mpichtar xzvf mpich-1.2.7p.tar.gzcd mpich-1.2.7pexport CC=iccexport CXX=icpcexport FC=ifortexport F90=ifort./configure --with-device=ch_p4 --prefix=/opt/mpich/intel –rsh=ssh –cc=icc –c++=icpc –fc=ifort –f90=ifortmakemake install设置环境变量:vi ~/.bashrc#--- for mpichMPI=/opt/mpich/intelexport MPIPATH=$PATH:$MPI/binexport PA THMPI_LIB=$MPI/libMPI_INCLUDE=$MPI/includeexport MPI_LIBexport MPI_INCLUDE重新登陆用户或者source ~/.bashrc后再编译下面的.注意：以后如果用普通用户最好将此加入普通用户的.bashrc中。
ZEISS Otus 1.4 85 说明书
ZEISS Otus 1.4/85Technische Daten/Technical SpecificationsBrennweite/Focal length85 mm Blendenbereich/Aperture rangef/1.4 – f/16 Linsen / Gruppen/Lens elements / Groups 11 / 9Fokussierbereich/Focusing range0,8 m (31.50’’) - ∞ Arbeitsabstand/Free working distance0,65 m (25.59’’) - ∞ Bildfeld*/Angular field* (diag. / horiz. / vert.) 28.24° / 23.71° / 15.97° Bildkreisdurchmesser/Diameter of image field 43 mm (1.69″)Anlagemaß/Flange focal distanceZF.2: 46,50 mm (1.83″) ZE: 44,00 mm (1.73″)Objektfeld bei Naheinstellung* Coverage at close range (MOD)*278,85 mm x 185,61 mm (10.97‘‘ x 7.31‘‘) Abbildungsmaßstab bei Naheinstellung Image ratio at MOD1 : 7.7 Filterdurchmesser/Filter threadM86 x 1.00 Lage der Eintrittspupille (vor der Bildebene)Entrance pupil position ( in front of image plane) 90 mm (3.54’’) Drehwinkel des Fokussierrings (inf – MOD) Rotation angle of focusing ring (inf – MOD) 261 °Durchmesser max./Diameter max.ZF.2: 101 mm (3.98‘‘) ZE: 101 mm (3.98‘‘) Durchmesser des Fokussierrings Diameter of focusing ringZF.2: 92 mm (3.62‘‘) ZE: 92 mm (3.62‘‘) Länge (ohne Objektivdeckel)/Length (without lens caps) ZF.2: 122 mm (4.80‘‘) ZE: 124 mm (4.88‘‘) Länge (mit Objektivdeckeln)/Length (with lens caps) ZF.2: 138 mm (5.43‘‘) ZE: 141 mm (5.55‘‘) Gewicht/WeightZF.2: 1140g (2.51 lbs) ZE: 1200g (2.64 lbs)* bezugnehmend auf das 24x36mm Format/referring to 36 mm formatSonderglas / Special glas Asphären / Aspheric surfaceZEISS Otus 1.4/85Relative Beleuchtungsstärke/Relative Illuminance E [%]Die relative Beleuchtungsstärke zeigt die Abnahmeder Bildhelligkeit von der Mitte des Bildes zu denEcken. Angabe in Prozent.The relative illumination shows in percent thedecrease in image brightness from the imagecenter to edge.__ Blendenzahl: k = 1,4 / f-number = 1.4--- Blendenzahl: k = 4,0 / f-number = 4.0… Blendenzahl: k = 3,9 / f-number = 3.9Relative Verzeichnung/Relative DistortionV [%]Die Relative Verzeichnung zeigt die Abweichungder aktuellen von der idealen Bildhöhe.The relative distortion shows in percent thedeviation of the actual from the ideal imageheight.Angaben für unendlich.Data for infinity.ZEISS Otus 1.4/85MTF ChartsUnendlich / InfinityBlendenzahl: k = 1,4 / f-number = 1.4__ Sagittal … TangentialBlendenzahl: k = 4 / f-number = 4.0__ Sagittal … TangentialModulationsübertragung MTF als Funktion der Bildhöhe (u’) und Spaltorientierung. Weißes Licht. Ortsfrequenzen R=10, 20 und 40 Perioden/mm. // Modulation transfer MTF as a function of the image height (u´) and slit orientation. White light. Spatial frequencies R=10, 20 and 40 cycles/mm.ZEISS Otus 1.4/85MTF ChartsNaheinstellung / Short focusBlendenzahl: k = 1,4 / f-number = 1.4__ Sagittal … TangentialBlendenzahl: k = 4 / f-number = 4.0__ Sagittal … TangentialModulationsübertragung MTF als Funktion der Bildhöhe (u’) und Spaltorientierung. Weißes Licht. Ortsfrequenzen R=10, 20 und 40 Perioden/mm. // Modulation transfer MTF as a function of the image height (u´) and slit orientation. White light. Spatial frequencies R=10, 20 and 40 cycles/mm.ZEISS Otus 1.4/8504/17 · Änderungen vorbehalten/Subject to change.Carl Zeiss AG · /photoSchärfentiefe/Depth of Field (DOF)*Engravedf/1.4 f/2 f/2.8 f/4 f/5.6 f/8 f/11 f/16 INF 157 inf. 139 inf. 92 inf. 61 inf. 42 inf. 29 inf. 21 inf. 14.2 inf. 15 m 13.7 16.6 13.6 17.9 12.9 19.1 12.1 21 m 11.225 9.97 35 8.81 66 7.39 inf. 7 m 6.71 7.31 6.68 7.56 6.53 7.76 6.32 8.09 6.05 8.58 5.70 9.43 5.31 10.8 4.77 14.2 4 m 3.90 4.09 3.90 4.17 3.85 4.23 3.78 4.32 3.68 4.45 3.55 4.66 3.40 4.96 3.18 5.54 3 m 2.95 3.05 2.94 3.09 2.92 3.12 2.88 3.17 2.82 3.24 2.75 3.35 2.66 3.49 2.53 3.76 2.5 m 2.46 2.53 2.46 2.56 2.44 2.58 2.42 2.62 2.38 2.66 2.33 2.73 2.27 2.82 2.17 2.99 2 m 1.97 2.02 1.98 2.04 1.97 2.05 1.95 2.07 1.93 2.10 1.89 2.14 1.85 2.19 1.79 2.29 1.7 m 1.68 1.71 1.68 1.73 1.68 1.74 1.66 1.75 1.65 1.77 1.62 1.80 1.60 1.83 1.55 1.90 1.5 m 1.48 1.51 1.49 1.52 1.48 1.53 1.47 1.54 1.46 1.55 1.44 1.57 1.42 1.60 1.39 1.65 1.2 m 1.19 1.21 1.19 1.21 1.19 1.22 1.18 1.22 1.18 1.23 1.17 1.24 1.15 1.26 1.13 1.28 1 m 0.99 1.00 1.00 1.00 0.99 1.01 0.99 1.01 0.99 1.02 0.98 1.03 0.97 1.04 0.96 1.05 0.9 m 0.890.900.900.910.890.910.890.910.890.910.880.920.880.930.870.94* Schärfentiefetabelle für das 24x36mm Format, Zerstreuungskreis 0.033mm (D/1500), gerundet auf 0.01m //Depth-of-field table for sensor format 24x36mm, circle of confusion 0.033mm (D/1500), rounded to 0.01m。
02 ASSAB Equivalents (Chinese)
一胜百钢种及美,日及德国的钢材标准
Standards
SS W/Nr. ASSAB UDDEHOLM Sweden Germany 760 UHB 11 1650/1672 1.173 DF2/3 Arne 2140 1.251 Calmax Calmax 1.2358 … XW 5 Sverker 3 2312 (1.2436) XW 10 Rigor 2260 1.2363 XW 41/42 Sverker 21 2310 1.2379 ASSAB 88 Sleipner … … … … 1.2738 618 718HH Impax HH … 1.2738 Corrax Corrax … … Holdax Holdax … 1.2312 Ramax S Ramax S … … Stavax ESR* Stavax ESR* 2314 (1.2083) Alvar 14 Alvar 14 … 1.2714 Dievar* Dievar* … … Hotvar* Hotvar* … … 8407 2M Orvar 2M 2242 1.2344 8407 Supreme* Orvar Supreme* 2242 1.2344 QRO 90 Supreme* QRO 90 Supreme* … … Vidar Supreme* Vidar Supreme* 1.2343 ASP 23** Vanadis 23** 1.3344 2725 ASP 30/60** Vanadis 30/60** 2726/2727 (1.3207)/ 1.3241 Elmax** Elmax** … … Vanadis 4/6/10** Vanadis 4/6/10** … … * ESR = 电渣重熔 ** 粉未冶金(PM). 电渣加热(ESH) 牌号
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一。
已知X和Y都是取值于{0,1}的二进制随机变量，P(X=0)=p。
还已知P(X=Y|X)= 。
求概率P(Y=1)，熵H(X),H(Y),H(Y|X)以及互信息I(X;Y)。
假设 给定，p可变，求能使I(X;Y)最大的p。
答：
P(Y=1)=P(Y=1,X=1)+P(Y=1,X=0)
=P(Y=1|X=1)P(X=1)+P(Y=1|X=0)P(X=0)
=(1− )(1−p)+ p=1− −p+2 p
H(X)=−p log p−(1−p)log(1−p)
记µ=P(Y=1)=1− −p+2 p，则
H(Y)=−µlogµ−(1−µ)log(1−µ)=h2(µ)
H(Y|X)=−
1
a=0
1
b=0
P(Y=b,X=a)log P(Y=b|X=a)
=−
1
a=0
1
b=0
P(X=a)P(Y=b|X=a)log P(Y=b|X=a)
=
1
a=0
P(X=a)
b∈{a,¯a}
−P(Y=b|X=a)log P(Y=b|X=a)
=
1
a=0
P(X=a)h2( )=h2( )
I(Y;X)=H(Y)−H(Y|X)=h2(µ)−h2( )=h2(1−p− +2 p)−h2( )
今 是固定的，欲I(Y;X)最大化，必须h2(1−p− +2 p)最大。
二元的熵在等
概时最大，故此必须µ=1−p− +2 p=1
2，1
2
− =p(1−2 )，p=12−
1−2
=1
2。
二。
若{X i}是平稳二进制马氏序列，P(X i=1)=1
2，P(X i=X i−1|X i−1)=
0.9。
将{X i}两两组为一组，映射到{Y k},即(X1,X2)映射为Y1，(X3,X4)映射为Y2，……。
Y取值于{0,1,2,3}。
从(X2i−1,X2i)映射到Y i的规则是(0,0)→0, (0,1)→1，(1,1)→2，(1,0)→3。
求Y的概率分布并对Y进行Huffman编码。
答：
P(Y=0)=P(X2=0,X1=0)=P(X2=0|X1=0)P(X1=0)=0.9/2=0.45 P(Y=1)=P(X2=1,X1=0)=P(X2=1|X1=0)P(X1=0)=0.1/2=0.05 P(Y=2)=P(X2=1,X1=1)=P(X2=1|X1=1)P(X1=1)=0.9/2=0.45 P(Y=3)=P(X2=0,X1=1)=P(X2=0|X1=1)P(X1=1)=0.9/2=0.05如图所示，Huffman编码结果是（可以有其他答案）
0→(0)1→(110)2→(10)3→(111)
1
Figure1:Huffman编码
Y的熵是H(Y)=−2×0.45log20.45−2×0.05log20.05≈−1.47bit，Huffman编码结果的平均长度是¯N=1×0.45+2×0.45+3×(0.05+0.05)=1.65bit，效率是H(X)
¯N
≈89%。
三。
若八进制信源{X1,···,X L}(L非常大)的H∞(X)=0.3Det。
采用最好的压缩技术将{X1,···,X L}映射为独立等概的十进制序列{Y1,···,Y M}，问M最少是多少才能保证无失真复原出原序列？如果不采用任何压缩技术，M至少需要多少？如果只是把每个X i单独映射为一个十进制数字，M是多少？
答：信源总的熵是0.3L Det，1Det需要用1位十进制数，因此压缩后需要M= 0.3L位十进制数。
如果不压缩，则信源的全部可能状态数是8L=10lg8L≈100.9L，即需要0.9L位十进制数。
如果是逐符号映射，每个8进制符号需要1位十进制数，总共需要L位。
四。
称球问题。
一个YES/NO问题可以使我们获得1bit的信息，就是说，把全部可能结果等分为2后，可以排除一半结果。
假如我们构造一个三选一的问题，
则这个答案可以获得的信息量是log
3(3)T rt=log23bit(这里Trt是杜撰的单
位)。
假设有12个小球，其中11个重量相同，另有一个重量偏大或者偏小。
我们不确定第几号球异常，也不确定它是偏重还是偏轻。
一把天平能给出三个答案：左大右小，左小右大，左右相同。
问：(1)原问题的熵是多少bit？(2)用天平测量的话，至少需要多少次才能测出结果？(2)如果天平只有两种结果（谁大谁小，没有平衡），最小需要称几次？
答：问题中的不确定性包括，异常球的编号，熵是log
212bit；是偏重还是偏
轻，熵是1bit。
这两个不确定性相互独立，所以问题的总熵是H=log
212+1≈
4.585bits。
折合到以3为低的对数，是2.89Trt。
因此，如果天平能给出三种结果，则至少需要称3次；如果天平只能给出2种结果，至少需要称5次。
2。