Download presentation

Presentation is loading. Please wait.

Published byKarina Gordan Modified over 4 years ago

1
3. Requirements1 Agenda for understand req activity r1. Numbers r2. Decibels r3. Matrices r4. Transforms r5. Statistics r6. Software

2
3. Requirements2 1. Numbers rSignificant digits rPrecision rAccuracy 1. Numbers

3
3. Requirements3 Significant digits (1 of 5) rThe significant digits in a number include the leftmost, non-zero digits to the rightmost digit written. rFinal answers should be rounded off to the decimal place justified by the data 1. Numbers

4
3. Requirements4 Significant digits (2 of 5) Examples numberdigitsimplied range 2513250.5 to 251.5 25.1325.05 to 25.15 0.00025130.0002505 to 0.0002515 251x10 5 3250.5x10 5 to 251.5x10 5 2.51x10 -3 3 2.505x10 -3 to 2.515x10 -3 2512. 42511.5 to 2512.5 251.04250.95 to 251.05 1. Numbers

5
3. Requirements5 Significant digits (3 of 5) rExample There shall be 3 brown eggs for every 8 eggs sold. A set of 8000 eggs passes if the number of brown eggs is in the range 2500 to 3500 There shall be 0.375 brown eggs for every egg sold. A set of 8000 eggs passes if the number of brown eggs is in the range 2996 to 3004 1. Numbers

6
3. Requirements6 Significant digits (4 of 5) rThe implied range can be offset by stating an explicit range There shall be 0.375 brown eggs (±0.1 of the set size) for every egg sold. A set of 8000 eggs passes if the number of brown eggs is in the range 2200 to 3800 There shall be 0.375 brown eggs (±0.1) for every egg sold. A set of 8000 eggs passes only if the number of brown eggs is 3000 1. Numbers

7
3. Requirements7 Significant digits (5 of 5) r A common problem is to inflate significant digits in making units conversion. Observers estimated the meteorite had a mass of 10 kg. This statement implies the mass was in the range of 5 to 15 kg; i.e, a range of 10 kg. Observers estimated the meteorite had a mass of 22 lbs. This statement implies a range of 21.5 to 22.5 lb; i.e., a range of 1 pound 1. Numbers

8
3. Requirements8 Precision rPrecision refers to the degree to which a number can be expressed. rExamples Computer words The 16-bit signed integer has a normalized precision of 2 -15 Meter readings The ammeter has a range of 10 amps and a precision of 0.01 amp 1. Numbers

9
3. Requirements9 Accuracy rAccuracy refers to the quality of the number. rExamples Computer words The 16-bit signed integer has a normalized precision of 2 -15, but its normalized accuracy may be only ±2 -3 Meter readings The ammeter has a range of 10 amps and a precision of 0.01 amp, but its accuracy may be only ±0.1 amp. 1. Numbers

10
3. Requirements10 2. Decibels rDefinitions rCommon values rExamples rAdvantages rDecibels as absolute units rPowers of 2 2. Decibels

11
3. Requirements11 Definitions (1 of 2) rThe decibel, named after Alexander Graham Bell, is a logarithmic unit originally used to give power ratios but used today to give other ratios rLogarithm of N The power to which 10 must be raised to equal N n = log 10 (N); N = 10 n 2. Decibels

12
3. Requirements12 Definitions (2 of 2) rPower ratio dB = 10 log 10 (P 2 /P 1 ) P 2 /P 1 =10 dB/10 rVoltage power dB = 20 log 10 (V 2 /V 1 ) V 2 /V 1 =10 dB/20 2. Decibels

13
3. Requirements13 Common values dBratio 01 11.26 21.6 32 42.5 53.2 64 75 86.3 9810 20100 301000 2. Decibels

14
3. Requirements14 Examples r5000 = 5 x 1000; 7 dB + 30 dB = 37 dB r49 dB = 40 dB + 9 dB; 8 x 10,000 = 80,000 2. Decibels

15
3. Requirements15 Advantages (1 of 2) rReduces the size of numbers used to express large ratios 2:1 = 3 dB; 100,000,000 = 80 dB rMultiplication in numbers becomes addition in decibels 10*100 =1000; 10 dB + 20 dB = 30 dB rThe reciprocal of a number is the negative of the number of decibels 100 = 20 dB; 1/100 = -20 dB 2. Decibels

16
3. Requirements16 Advantages (2 of 2) rRaising to powers is done by multiplication 100 2 = 10,000; 2*20dB = 40 dB 100 0.5 = 10; 0.5*20dB = 10 dB rCalculations can be done mentally 2. Decibels

17
3. Requirements17 Decibels as absolute units rdBW = dB relative to 1 watt rdBm = dB relative to 1 milliwatt rdBsm = dB relative to one square meter rdBi = dB relative to an isotropic radiator 2. Decibels

18
3. Requirements18 Powers of 2 exact valueapproximate value 2 0 11 2 4 1616 2 10 10241 x 1,000 2 23 8,388,6088 x 1,000,000 2 34 17,179,869,18416 x 1,000,000,000 2 xy = 2 y x 10 3x 2. Decibels

19
3. Requirements19 3. Matrices rAddition rSubtraction rMultiplication rVector, dot product, & outer product rTranspose rDeterminant of a 2x2 matrix rCofactor and adjoint matrices rDeterminant rInverse matrix rOrthogonal matrix 3. Matrices

20
3. Requirements20 Addition c IJ = a IJ + b IJ 1 -1 0 -2 1 -3 2 0 2 1 -1 -1 0 4 2 -1 0 1 A=B= 2 -2 -1 -2 5 -1 1 0 3 C= C=A+B 3. Matrices

21
3. Requirements21 Subtraction c IJ = a IJ - b IJ 1 -1 0 -2 1 -3 2 0 2 1 -1 -1 0 4 2 -1 0 1 A=B= 0 0 1 -2 -3 -5 3 0 1 C= C=A-B 3. Matrices

22
3. Requirements22 Multiplication c IJ = a I1 * b 1J + a I2 * b 2J + a I3 * b 3J 1 -1 0 -2 1 -3 2 0 2 1 -1 -1 0 4 2 -1 0 1 A=B= 1 -5 -3 1 6 1 0 -2 0 C= C=A*B 3. Matrices

23
3. Requirements23 Transpose b IJ = a JI 1 -1 0 -2 1 -3 2 0 2 1 -2 2 -1 1 0 0 -3 2 A=B= B=A T 3. Matrices

24
3. Requirements24 Vector, dot product, & outer product rA vector v is an N x 1 matrix rDot product = inner product = v T x v = a scalar rOuter product = v x v T = N x N matrix 3. Matrices

25
3. Requirements25 Determinant of a 2x2 matrix 2x2 determinant = b 11 * b 22 - b I2 * b 21 B= 1 -1 -2 1 = -1 3. Matrices

26
3. Requirements26 Cofactor and adjoint matrices 1 -1 0 -2 1 -3 2 0 2 A= 1 -3 0 2 -1 0 0 2 -1 0 0 -3 -2 -3 2 2 1 0 2 2 1 0 -2 -3 -2 1 2 0 1 -1 2 0 1 -1 -2 1 2 -2 -2 2 2 -2 3 3 -1 =B = cofactor = 2 2 3 -2 2 3 -2 -2 -1 C=B T = adjoint= 3. Matrices - - - -

27
3. Requirements27 Determinant 1 -1 0 -2 1 -3 2 0 2 determinant of A = The determinant of A = dot product of any row in A times the corresponding column of the adjoint matrix = dot product of any row (or column) in A times the corresponding row (or column) in the cofactor matrix The determinant of A = dot product of any row in A times the corresponding column of the adjoint matrix = dot product of any row (or column) in A times the corresponding row (or column) in the cofactor matrix 1 -1 0 =4 2 -2 = 4 3. Matrices

28
3. Requirements28 Inverse matrix B = A -1 =adjoint(A)/determinant(A) = 0.5 0.5 0.75 -0.5 0.5 0.75 -0.5 -0.5 -0.25 1 -1 0 -2 1 -3 2 0 2 0.5 0.5 0.75 -0.5 0.5 0.75 -0.5 -0.5 -0.25 1 0 0 0 1 0 0 0 1 = 3. Matrices

29
3. Requirements29 Orthogonal matrix rAn orthogonal matrix is a matrix whose inverse is equal to its transpose. 1 0 0 0 cos sin 0 -sin cos 1 0 0 0 cos -sin 0 sin cos 1 0 0 0 1 0 0 0 1 = 3. Matrices

30
3. Requirements30 4. Transforms rDefinition rExamples rTime-domain solution rFrequency-domain solution rTerms used with frequency response rPower spectrum rSinusoidal motion rExample -- vibration 4. Transforms

31
3. Requirements31 Definition rTransforms -- a mathematical conversion from one way of thinking to another to make a problem easier to solve transform solution in transform way of thinking inverse transform solution in original way of thinking problem in original way of thinking 4. Transforms

32
3. Requirements32 Examples (1 of 3) English to algebra solution in algebra algebra to English solution in English problem in English 4. Transforms

33
3. Requirements33 Examples (2 of 3) English to matrices solution in matrices matrices to English solution in English problem in English 4. Transforms

34
3. Requirements34 Examples (3 of 3) Fourier transform solution in frequency domain inverse Fourier transform solution in time domain problem in time domain Other transforms Laplace z-transform wavelets 4. Transforms

35
3. Requirements35 Time-domain solution rWe typically think in the time domain -- a time input produces a time output 4. Transforms system time amplitude time amplitude inputoutput

36
3. Requirements36 Frequency-domain solution (1 of 2) rHowever, the solution can be expressed in the frequency domain. rA sinusoidal input produces a sinusoidal output rA series of sinusoidal inputs across the frequency range produces a series of sinusoidal outputs called a frequency response 4. Transforms

37
3. Requirements37 Frequency-domain solution (2 of 2) 4. Transforms system log frequency amplitude (dB) log frequency magnitude (dB) inputoutput log frequency phase (angle) 0 -180 (sinusoids) log frequency phase (angle) 0 0

38
3. Requirements38 Terms used with frequency response rOctave is a range of 2x rDecade is a range of 10x 4. Transforms amplitude (dB) power (dB) frequency 6, 3 210 20,10 Slope = 20 dB/decade, amplitude 6 dB/octave, amplitude 10 dB decade, power 3 dB decade, power

39
3. Requirements39 Power spectrum rA power spectrum is a special form of frequency response in which the ordinate represents power 4. Transforms g 2 -Hz (dB) log frequency

40
3. Requirements40 Sinusoidal motion rMotion of a point going around a circle in two-dimensional x-y plane produces sinusoidal motion in each dimension x-displacement = sin( t) x-velocity = cos( t) x-acceleration = - 2 sin( t) x-jerk = - 3 cos( t) x-yank = 4 sin( t) 4. Transforms

41
3. Requirements41 Example -- vibration Output vibration is product of input vibration times the transmissivity-squared at each frequency Output vibration is product of input vibration times the transmissivity-squared at each frequency 4. Transforms g 2 -Hz (dB) log frequency g 2 -Hz (dB)[amplitude (dB)] 2 input transmissivity-squared output

42
3. Requirements42 5. Statistics (1 of 2) rFrequency distribution rSample mean rSample variance rCEP rDensity function rDistribution function rUniform rBinomial 5. Statistics

43
3. Requirements43 5. Statistics (1 of 2) rNormal rPoisson rExponential rRaleigh rSampling rCombining error sources 5. Statistics

44
3. Requirements44 2 Frequency distribution rFrequency distribution -- A histogram or polygon summarizing how raw data can be grouped into classes height (inches) number 2 2 4 6 8 4566743 n = sample size = 39 2 60 61 62 63 64 65 66 67 68 5. Statistics

45
3. Requirements45 Sample mean r = x i rAn estimate of the population mean rExample = [ 2 x 60 + 4 x 61 + 5 x 62 + 7 x 63 + 4 x 64 + 6 x 65 + 6 x 66 + 3 x 67 + 2 x 68 ] / 39 = 2494/39 = 63.9 5. Statistics N i=1 N

46
3. Requirements46 Sample variance r 2 = (x i - ) 2 rAn estimate of the population variance r = standard deviation rExample 2 = [ 2 x (60 - ) 2 + 4 x (61 - ) 2 + 5 x (62 - ) 2 + 7 x (63 - ) 2 + 4 x (64 - ) 2 + 6 x (65 - ) 2 + 6 x (66 - ) 2 + 3 x (67 - ) 2 + 2 x (68 - ) 2 ]/(39 - 1] = 183.9/38 = 4.8 = 2.2 5. Statistics N-1 i=1 N

47
3. Requirements47 CEP rCircular error probable is the radius of the circle containing half of the samples r If samples are normally distributed in the x direction with standard deviation x and normally distribute in the y direction with standard deviation y, then CEP = 1.1774 * sqrt [0.5*( x 2 + y 2 )] CEP 5. Statistics

48
3. Requirements48 Density function rProbability that a discrete event x will occur rNon-negative function whose integral over the entire range of the independent variable is 1 f(x) x 5. Statistics

49
3. Requirements49 Distribution function rProbability that a numerical event x or less occurs rThe integral of the density function F(x) x 1.0 5. Statistics

50
3. Requirements50 Uniform (1 of 2) rf(x) = 1/(x 2 - x 1 ), x 1 x x 2 = 0 elsewhere r F(x) = 0, x x 1 = (x - x 1 ) / (x 2 - x 1 ), x 1 x x 2 = 1, x > x 2 rMean = (x 2 + x 1 )/2 rStandard deviation = (x 2 - x 1 )/sqrt(12) 5. Statistics

51
3. Requirements51 Uniform (2 of 2) rExample If a set of resistors has a mean of 10,000 and is uniformly distributed between 9,000 and 11,000, what is the probability the resistance is between 9,900 and 10,100 ? F(9900,10100) = 200/2000 = 0.1 5. Statistics

52
3. Requirements52 Binomial (1 of 2) rf(x) = n!/[(n-x)!x!]p x (1-p) n-x where p = probability of success on a single trial r Used when all outcomes can be expressed as either successes or failures rMean = np rStandard deviation = sqrt[np(1-p)] 5. Statistics

53
3. Requirements53 Binomial (2 of 2) rExample 10 percent of a production run of assemblies are defective. If 5 assemblies are chosen, what is the probability that exactly 2 are defective? f(2) = 5!/(3!2!)(0.1 2 )(0.9 3 ) = 0.07 5. Statistics

54
3. Requirements54 Normal (1 of 2) rf(x) = 1/[ sqrt(2 )exp[-(x- ) 2 /(2 2 ) rF(x) = erf[(x- )/ ] + 0.5 rMean = rStandard deviation = rCan be derived from binomial distribution 5. Statistics

55
3. Requirements55 Normal (2 of 2) rExample If the mean mass of a set of products is 50 kg and the standard deviation is 5 kg, what is the probability the mass is less than 60 kg? F(60) = erf[(60-50)/5] + 0.5 = 0.97 5. Statistics

56
3. Requirements56 Poisson (1 of 2) rf(x) = e - x /x!( >0) = average number of times that event occurs per period x = number of time event occurs rMean = rStandard deviation = sqrt( ) rDerived from binomial distribution rUsed to quantify events that occur relatively infrequently but at a regular rate 5. Statistics

57
3. Requirements57 Poisson (2 of 2) rExample The system generates 5 false alarms per hour. What is the probability there will be exactly 3 false alarms in one hour? = 5 x = 3 f(3) = e -5 (5) 3 /3! = 0.14 5. Statistics

58
3. Requirements58 Exponential (1 of 2) rF(x) = exp(- x) rF(x) = 1 - exp(- x) rMean = 1/ rStandard deviation = 1/ rUsed in reliability computations where = 1/MTBF 5. Statistics

59
3. Requirements59 Exponential (2 of 2) rExample If the MTBF of a part is 100 hours, what is the probability the part will have failed by 150 hours? F(150) = 1 - exp(- 150/100) = 0.78 5. Statistics

60
3. Requirements60 Raleigh (1 of 2) rf(r) = [1/(2 2 ) * exp[-r 2 /(2 2 )] rF(r) = 1 - exp[-r 2 /(2 2 )] rMean = sqrt( /2) rStandard deviation = sqrt(2) rDerived from normal distribution rUsed to describe radial distribution when uncertainty in x and y are described by normal distributions 5. Statistics

61
3. Requirements61 Raleigh (2 of 2) rExample If uncertainty in x and y positions are each described by a normal distribution with zero mean and = 2, what is the probability the position is within a radius of 1.5? F(1.5) = 1 - exp[-(1.5) 2 /(2 x 2 2 )] = 0.25 5. Statistics

62
3. Requirements62 Sampling rA frequent problem is obtaining enough samples to be confident in the answer 5. Statistics N M N>M

63
3. Requirements63 Combining error sources (1 of 4) rVariances from multiple error sources can be combined by adding variances r Example 5. Statistics x orig = standard deviation in original position = 1 m v orig = standard deviation in original velocity = 0.5 m/s T = time between samples = 2 sec x current = error in current position = square root of [(x orig ) 2 + (v orig * T) 2 ] = sqrt(2)

64
3. Requirements64 Combining error sources (2 of 4) rWhen multiple dimensions are included, covariance matrices can be added rWhen an error source goes through a linear transformation, resulting covariance is expressed as follows 5. Statistics P 1 = covariance of error source 1 P 2 = covariance of error source 2 P = resulting covariance = P 1 + P 2 T = linear transformation T T = transform of linear transformation P orig = covariance of original error source P = T * P * T T

65
3. Requirements65 Combining error sources (3 of 4) 5. Statistics rExample of propagation of position x orig = standard deviation in original position = 2 m v orig = standard deviation in original velocity = 0.5 m/s T = time between samples = 4 sec x current = error in current position x current = x orig + T * v orig v current = v orig 1 4 0 1 T =P orig = 2 0 0 0.5 2 P current = T * P orig * T T = 1 4 0 1 1 0 4 1 4 0 0 0.25 = 8 4 4

66
3. Requirements66 Combining error sources (4 of 4) 5. Statistics rExample of angular rotation X original = original coordinates X current = current coordinates T = transformation corresponding to angular rotation cos -sin sin cos T = where = atan(0.75) P orig = 1.64 -0.48 -0.48 1.36 P current = T * P orig * T T = 0.8 -0.6 0.6 0.8 = 2 0 0 1 1.64 -0.48 -0.48 1.36 0.8 0.6 -0.6 0.8 x y x y

67
3. Requirements67 6. Software rMemory rThroughput rLanguage rDevelopment method 6. Software

68
3. Requirements68 Memory (1 of 3) rAll general purpose computers shall have 50 percent spare memory capacity rAll digital signal processors (DSPs) shall have 25 percent spare on-chip memory capacity rAll digital signal processors shall have 30 percent spare off-chip memory capacity rAll mass storage units shall have 40 percent spare memory capacity rAll firmware shall have 20 percent spare memory capacity 6. Software

69
3. Requirements69 Memory (2 of 3) rThere shall be 50 % spare memory capacity referencecapacitymemory-used usagecommonless-common capacity100 Mbytes100 Mbytes memory-used60 Mbytes60 Mbytes spare memory40 Mbytes40 Mbytes percent spare40 percent67 percent pass/failfailpass There are at least two ways of interpreting the meaning of spare memory capacity based on the reference used as the denominator in computing the percentage There are at least two ways of interpreting the meaning of spare memory capacity based on the reference used as the denominator in computing the percentage 6. Software

70
3. Requirements70 Memory (3 of 3) rMemory capacity is most often verified by analysis of load files rMemory capacity is frequently tracked as a technical performance parameter (TPP) rContractors dont like to consider that firmware is software because firmware is often not developed using software development methodology and firmware is not as likely to grow in the future Memory is often verified by analysis, and firmware is often not considered to be software Memory is often verified by analysis, and firmware is often not considered to be software 6. Software

71
3. Requirements71 Throughput (1 of 5) rAll general purpose computers shall have 50 percent spare throughput capacity rAll digital signal processors shall have 25 percent spare throughput capacity rAll firmware shall have 30 percent spare throughput capacity rAll communication channels shall have 40 percent spare throughput capacity rAll communication channels shall have 20 percent spare terminals 6. Software

72
3. Requirements72 Throughput (2 of 5) rThere shall be 100 % spare throughput capacity referencecapacitythroughput-used usagecommoncommon capacity100 MOPS100 MOPS throughput-used50 MOPS50 MOPS sparethroughput 50 MOPS50 MOPS percent spare50 percent100 percent pass/failfailpass There are two ways of interpreting of spare throughput capacity based on reference used as denominator There are two ways of interpreting of spare throughput capacity based on reference used as denominator 6. Software

73
3. Requirements73 Throughput (3 of 5) rAvailability of spare throughput Available at the highest-priority- application level -- most common Available at the lowest-priority-application level -- common Available in proportion to the times spent by each segment of the application -- not common Assuming the spare throughput is available at the highest-priority-application level is the most common assumption Assuming the spare throughput is available at the highest-priority-application level is the most common assumption 6. Software

74
3. Requirements74 Throughput (4 of 5) rThroughput capacity is most often verified by test Analysis -- not common Time event simulation -- not common Execution monitor -- common but requires instrumentation code and hardware 6. Software

75
3. Requirements75 Throughput (5 of 5) Execution of a code segment that uses at least the number of spare throughput instructions required -- not common but avoids instrumentation Instrumenting the software to monitor runtime or inserting a code segment that uses at least the spare throughput are two methods of verifying throughput Instrumenting the software to monitor runtime or inserting a code segment that uses at least the spare throughput are two methods of verifying throughput 6. Software

76
3. Requirements76 Language (1 of 2) rNo more than 15 percent of the code shall be in assembly language. Useful for device drivers and for speed Not as easily maintained 6. Software

77
3. Requirements77 Language (2 of 2) rRemaining code shall be in Ada Ada is largely a military language and is declining in popularity C++ growing in popularity rLanguage is verified by analysis of code C++ is becoming the most popular programming language but assembly language may still need to be used C++ is becoming the most popular programming language but assembly language may still need to be used 6. Software

78
3. Requirements78 Development method rSeveral methods are available Structured-analysis-structured-design vs Hatley-Pirba Functional vs object-oriented Classical vs clean-room rGenerally a statement of work issue and not a requirement although customer prefers a proven, low-risk approach Customer does not usually specify the development method Customer does not usually specify the development method 6. Software

Similar presentations

OK

STATISTICS POINT ESTIMATION Professor Ke-Sheng Cheng Department of Bioenvironmental Systems Engineering National Taiwan University.

STATISTICS POINT ESTIMATION Professor Ke-Sheng Cheng Department of Bioenvironmental Systems Engineering National Taiwan University.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Ppt on different types of dance forms in africa Ppt on life and works of william shakespeare Ppt on writing letters Ppt on mobile phone based attendance tracking system Ppt on google company profile Ppt on network switches Ppt on directors under companies act 1956 Make appt online Ppt on animal cell and plant cell Ppt on omission of articles in spanish