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Agenda for understand req activity 1. Numbers 2. Decibels 3. Matrices 4. Transforms 5. Statistics 6. Software 3. Requirements 1 1. Numbers Significant digits Precision Accuracy 3. Requirements 1. Numbers 2 Significant digits (1 of 5) The significant digits in a number include the leftmost, non-zero digits to the rightmost digit written. Final answers should be rounded off to the decimal place justified by the data 3. Requirements 1. Numbers 3 Significant digits (2 of 5) number Examples digits implied range 251 25.1 0.000251 251x105 2.51x10-3 3 3 3 3 3 250.5 to 251.5 25.05 to 25.15 0.0002505 to 0.0002515 250.5x105 to 251.5x105 2.505x10-3 to 2.515x10- 2510 251.0 4 4 2509.5 to 2510.5 250.95 to 251.05 3 3. Requirements 1. Numbers 4 Significant digits (3 of 5) Example • 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 3. Requirements 1. Numbers 5 Significant digits (4 of 5) The 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 3. Requirements 1. Numbers 6 Significant digits (5 of 5) 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 3. Requirements 1. Numbers 7 Precision Precision refers to the degree to which a number can be expressed. Examples • 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 3. Requirements 1. Numbers 8 Accuracy Accuracy refers to the quality of the number. Examples • 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. 3. Requirements 1. Numbers 9 2. Decibels Definitions Common values Examples Advantages Decibels as absolute units Powers of 2 3. Requirements 2. Decibels 10 Definitions (1 of 2) The decibel, named after Alexander Graham Bell, is a logarithmic unit originally used to give power ratios but used today to give other ratios Logarithm of N • The power to which 10 must be raised to equal N • n = log10(N); N = 10n 3. Requirements 2. Decibels 11 Definitions (2 of 2) Power ratio • dB = 10 log10(P2/P1) • P2/P1=10dB/10 Voltage power • dB = 10 log10(V2/V1) • P2/P1=10dB/20 3. Requirements 2. Decibels 12 Common values dB 0 1 2 3 4 5 6 7 8 9 10 100 1000 3. Requirements ratio 1 1.26 1.6 2 2.5 3.2 4 5 6.3 8 10 20 30 2. Decibels 13 Examples 5000 = 5 x 1000; 7 dB + 30 dB = 37 dB 49 dB = 40 dB + 9 dB; 8 x 10,000 = 80,000 3. Requirements 2. Decibels 14 Advantages (1 of 2) Reduces the size of numbers used to express large ratios • 2:1 = 3 dB; 100,000,000 = 80 dB Multiplication in numbers becomes addition in decibels • 10*100 =1000; 10 dB + 20 dB = 30 dB The reciprocal of a number is the negative of the number of decibels • 100 = 20 dB; 1/100 = -20 dB 3. Requirements 2. Decibels 15 Advantages (2 of 2) Raising to powers is done by multiplication • 1002 = 10,000; 2*20dB = 40 dB • 1000.5 = 10; 0.5*20dB = 10 dB Calculations can be done mentally 3. Requirements 2. Decibels 16 Decibels as absolute units dBW = dB relative to 1 watt dBm = dB relative to 1 milliwatt dBsm = dB relative to one square meter dBi = dB relative to an isotropic radiator 3. Requirements 2. Decibels 17 Powers of 2 20 24 210 223 234 exact value approximate value 1 16 1024 8,388,608 17,179,869,184 1 16 1 x 1,000 8 x 1,000,000 16 x 1,000,000,000 2xy = 2y x 103x 3. Requirements 2. Decibels 18 3. Matrices Addition Subtraction Multiplication Vector, dot product, & outer product Transpose Determinant of a 2x2 matrix Cofactor and adjoint matrices Determinant Inverse matrix Orthogonal matrix 3. Requirements 3. Matrices 19 Addition C=A+B 1 -1 0 A= -2 1 -3 2 0 2 1 B= 0 -1 -1 -1 4 2 0 1 2 C= -2 1 -2 -1 5 -1 0 3 cIJ = aIJ + bIJ 3. Requirements 3. Matrices 20 Subtraction C=A-B 1 -1 0 A= -2 1 -3 2 0 2 1 B= 0 -1 -1 -1 4 2 0 1 0 0 1 C= -2 -3 -5 3 0 1 cIJ = aIJ - bIJ 3. Requirements 3. Matrices 21 Multiplication C=A-B 1 -1 0 A= -2 1 -3 2 0 2 1 B= 0 -1 -1 -1 4 2 0 1 C= 1 1 0 -5 -3 6 1 -2 0 cIJ = aI1 * b1J + aI2 * b2J + aI3 * b3J 3. Requirements 3. Matrices 22 Vector, dot product, & outer product A vector v is an N x 1 matrix Dot product = inner product = vT x v = a scalar Outer product = v x vT = N x N matrix 3. Requirements 3. Matrices 23 Transpose B=AT 1 -1 0 A= -2 1 -3 2 0 2 1 B= -1 0 -2 1 -3 2 0 2 bIJ = aJI 3. Requirements 3. Matrices 24 Determinant of a 2x2 matrix B = 1 -1 -2 1 = -1 2x2 determinant = b11 * b22 - bI2 * b21 3. Requirements 3. Matrices 25 Cofactor and adjoint matrices 1 -1 0 A= -2 1 -3 2 0 2 1 -3 0 2 B = cofactor = -1 0 -2 2 1 0 0 2 1 2 0 2 1 -1 2 0 -1 0 0 -3 1 -2 0 -3 1 -1 -2 1 C=BT = adjoint= 3. Requirements -2 -3 2 2 2 -2 -2 = 2 2 -2 3 3 -1 2 2 3 -2 2 3 -2 -2 -1 3. Matrices 26 Determinant determinant of A = 1 -1 0 1 -1 0 -2 1 -3 2 0 2 2 -2 -2 =4 =4 The determinant of A = dot product of any row in A times the corresponding row the adjoint matrix = dot product of any row or column in A times the corresponding row or column in the cofactor matrix 3. Requirements 3. Matrices 27 Inverse matrix B = A-1 =adjoint(A)/determinant(A) = 1 -1 0 -2 1 -3 2 0 2 3. Requirements 0.5 0.5 0.75 -0.5 0.5 0.75 -0.5 -0.5 -0.25 3. Matrices 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 28 Orthogonal matrix An orthogonal matrix is a matrix whose inverse is equal to its transpose. 1 0 0 0 0 cos sin -sin cos 3. Requirements 1 0 0 0 0 1 0 0 cos -sin = 0 1 0 sin cos 0 0 1 3. Matrices 29 4. Transforms Definition Examples Time-domain solution Frequency-domain solution Terms used with frequency response Power spectrum Sinusoidal motion Example -- vibration 3. Requirements 4. Transforms 30 Definition Transforms -- a mathematical conversion from one way of thinking to another to make a problem easier to solve problem in original way of thinking transform 3. Requirements solution in transform way of thinking 4. Transforms solution in original way of thinking inverse transform 31 Examples (1 of 3) problem in English solution in English English to algebra 3. Requirements solution in algebra 4. Transforms algebra to English 32 Examples (2 of 3) problem in English solution in English English to matrices 3. Requirements solution in matrices 4. Transforms matrices to English 33 Examples (3 of 3) problem in time domain Fourier transform solution in frequency domain inverse Fourier transform solution in time domain • Other transforms • Laplace • z-transform • wavelets 3. Requirements 4. Transforms 34 Time-domain solution We typically think in the time domain -- a time input produces a time output input output amplitude amplitude time 3. Requirements system 4. Transforms time 35 Frequency-domain solution (1 of 2) However, the solution can be expressed in the frequency domain. A sinusoidal input produces a sinusoidal output A series of sinusoidal inputs across the frequency range produces a series of sinusoidal outputs called a frequency response 3. Requirements 4. Transforms 36 Frequency-domain solution (2 of 2) input (sinusoids) output amplitude (dB) log frequency magnitude (dB) system log frequency phase (angle) 0 -180 log frequency 3. Requirements 4. Transforms 37 Terms used with frequency response Octave is a range of 2x Decade is a range of 10x amplitude (dB) power (dB) 20,10 6, 3 Slope = • 20 dB/decade, amplitude • 6 dB/octave, amplitude •10 dB decade, power • 3 dB decade, power 2 10 frequency 3. Requirements 4. Transforms 38 Power spectrum A power spectrum is a special form of frequency response in which the ordinate represents power g2-Hz (dB) log frequency 3. Requirements 4. Transforms 39 Sinusoidal motion Motion 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 = -2sin(t) • x-jerk = -3cos(t) • x-yank = 4sin(t) 3. Requirements 4. Transforms 40 Example -- vibration input g2-Hz (dB) log frequency transmissivity-squared amplitude (dB) log frequency output g2-Hz (dB) log frequency Output vibration is product of input vibration times the transmissivity-squared at each frequency 3. Requirements 4. Transforms 41 5. Statistics (1 of 2) Frequency distribution Sample mean Sample variance CEP Density function Distribution function Uniform Binomial 3. Requirements 5. Statistics 42 5. Statistics (1 of 2) Normal Poisson Exponential Raleigh Sampling Combining error sources 3. Requirements 5. Statistics 43 Frequency distribution Frequency distribution -- A histogram or polygon summarizing how raw data can be grouped into classes number n = sample size = 39 8 6 4 2 3. Requirements 2 4 5 60 61 62 7 4 3 22 63 64 65 66 67 height (inches) 68 5. Statistics 6 6 44 Sample mean N = xi i=1 N An estimate of the population mean Example = [ 2 x 60 + 4 x61 + 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 3. Requirements 5. Statistics 45 Sample variance N 2= i=1 (xi - )2 N-1 An estimate of the population variance = standard deviation Example 2 = [ 2 x (60 - )2 + 3. Requirements 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 46 CEP Circular error probable is the radius of the circle containing half of the samples 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*(x2 + y2)] CEP 3. Requirements 5. Statistics 47 Density function Probability that a discrete event x will occur Non-negative function whose integral over the entire range of the independent variable is 1 f(x) x 3. Requirements 5. Statistics 48 Distribution function Probability that a numerical event x or less occurs The integral of the density function F(x) 1.0 x 3. Requirements 5. Statistics 49 Uniform (1 of 2) f(x) = 1/(x2 - x1 ), x1 x x2 = 0 elsewhere F(x) = 0, x x1 = (x - x1 ) / (x2 - x1 ), x1 x x2 = 1, x > x2 Mean = (x2 + x1 )/2 Standard deviation = (x2 - x1 )/sqrt(12) 3. Requirements 5. Statistics 50 Uniform (2 of 2) Example • 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 3. Requirements 5. Statistics 51 Binomial (1 of 2) f(x) = n!/[(n-x)!x!]px (1-p)n-x where p = probability of success on a single trial Used when all outcomes can be expressed as either successes or failures Mean = np Standard deviation = sqrt[np(1-p)] 3. Requirements 5. Statistics 52 Binomial (2 of 2) Example • 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!)(0.12)(0.93) = 0.07 3. Requirements 5. Statistics 53 Normal (1 of 2) f(x) = 1/[sqrt(2)exp[-(x-)2/(2 2) F(x) = erf[(x-)/] + 0.5 Mean = Standard deviation = Can be derived from binomial distribution 3. Requirements 5. Statistics 54 Normal (2 of 2) Example • 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 3. Requirements 5. Statistics 55 Poisson (1 of 2) f(x) = e-x/x! (>0) • = average number of times that event occurs per period • x = number of time event occurs Mean = Standard deviation = sqrt() Derived from binomial distribution Used to quantify events that occur relatively infrequently but at a regular rate 3. Requirements 5. Statistics 56 Poisson (2 of 2) Example • 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 3. Requirements 5. Statistics 57 Exponential (1 of 2) F(x) = exp(- x) F(x) = 1 - exp(- x) Mean = 1/ Standard deviation = 1/ Used in reliability computations where = 1/MTBF 3. Requirements 5. Statistics 58 Exponential (2 of 2) Example • 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 3. Requirements 5. Statistics 59 Raleigh (1 of 2) f(r) = [1/(22) * exp[-r2/(2 2)] F(r) = 1 - exp[-r2/(2 2)] Mean = sqrt(/2) Standard deviation = sqrt(2) Derived from binomial distribution Used to describe radial distribution when uncertainty in x and y are described by normal distributions 3. Requirements 5. Statistics 60 Raleigh (2 of 2) Example • 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 22)] = 0.25 3. Requirements 5. Statistics 61 Sampling A frequent problem is obtaining enough samples to be confident in the answer N N>M M 3. Requirements 5. Statistics 62 Combining error sources (1 of 4) Variances from multiple error sources can be combined by adding variances Example xorig = standard deviation in original position = 1 m vorig = standard deviation in original velocity = 0.5 m/s T = time between samples = 2 sec xcurrent = error in current position = square root of [(xorig)2 + (vorig * T)2] = sqrt(2) 3. Requirements 5. Statistics 63 Combining error sources (2 of 4) When multiple dimensions are included, covariance matrices can be added P1 = covariance of error source 1 P2 = covariance of error source 2 P = resulting covariance = P1 + P2 When an error source goes through a linear transformation, resulting covariance is expressed as follows T = linear transformation TT = transform of linear transformation Porig = covariance of original error source P = T * P * TT 3. Requirements 5. Statistics 64 Combining error sources (3 of 4) Example of propagation of position xorig = standard deviation in original position = 2 m vorig = standard deviation in original velocity = 0.5 m/s T = time between samples = 4 sec xcurrent = error in current position xcurrent = xorig + T * vorig vcurrent = vorig T = 1 0.5 0 1 Pcurrent = T * P orig * TT = 3. Requirements Porig = 1 0 4 1 22 0 0 0.52 4 0 5. Statistics 0 0.25 1 4 0 1 = 16 4 4 0.25 65 Combining error sources (4 of 4) Example of angular rotation Xoriginal = original coordinates Xcurrent = current coordinates T = transformation corresponding to angular rotation y y’ T = cos -sin where = atan(0.75) sin cos Porig = 3. Requirements x 1.64 -0.48 -0.48 1.36 Pcurrent = T * P orig * TT = x’ 0.8 -0.6 0.6 0.8 1.64 -0.48 -0.48 1.36 5. Statistics 0.8 0.6 -0.6 0.8 = 2 0 0 1 66 6. Software Memory Throughput Language Development method 3. Requirements 6. Software 67 Memory (1 of 3) All general purpose computers shall have 50 percent spare memory capacity All digital signal processors (DSPs) shall have 25 percent spare on-chip memory capacity All digital signal processors shall have 30 percent spare off-chip memory capacity All mass storage units shall have 40 percent spare memory capacity All firmware shall have 20 percent spare memory capacity 3. Requirements 6. Software 68 Memory (2 of 3) There shall be 50 percent spare memory capacity reference capacity memory-used usage common less-common capacity 100 Mbytes 100 Mbytes memory-used 60 Mbytes 60 Mbytes spare memory 40 Mbytes 40 Mbytes percent spare 40 percent 67 percent pass/fail fail pass 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 3. Requirements 6. Software 69 Memory (3 of 3) Memory capacity is most often verified by analysis of load files Memory capacity is frequently tracked as a technical performance parameter (TPP) Contractors don’t 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 3. Requirements 6. Software 70 Throughput (1 of 5) All general purpose computers shall have 50 percent spare throughput capacity All digital signal processors shall have 25 percent spare throughput capacity All firmware shall have 30 percent spare throughput capacity All communication channels shall have 40 percent spare throughput capacity All communication channels shall have 20 percent spare terminals 3. Requirements 6. Software 71 Throughput (2 of 5) There shall be 100 percent spare throughput capacity reference capacity throughput-used usage common common capacity 100 MOPS 100 MOPS throughput-used 50 MOPS 50 MOPS spare throughput 50 MOPS 50 MOPS percent spare 50 percent 100 percent pass/fail fail pass There are two ways of interpreting of spare throughput capacity based on reference used as denominator 3. Requirements 6. Software 72 Throughput (3 of 5) Availability of spare throughput • Available at the highest-priorityapplication 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 3. Requirements 6. Software 73 Throughput (4 of 5) Throughput capacity is most often verified by test • Analysis -- not common • Time event simulation -- not common • Execution monitor -- common but requires instrumentation code and hardware 3. Requirements 6. Software 74 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 3. Requirements 6. Software 75 Language (1 of 2) No more than 15 percent of the code shall be in assembly language. • Useful for device drivers and for speed • Not as easily maintained 3. Requirements 6. Software 76 Language (2 of 2) Remaining code shall be in Ada • Ada is largely a military language and is declining in popularity • C++ growing in popularity Language is verified by analysis of code C++ is becoming the most popular programming language but assembly language may still need to be used 3. Requirements 6. Software 77 Development method Several methods are available • Structured-analysis-structured-design vs Hatley-Pirba • Functional vs object-oriented • Classical vs clean-room Generally 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 3. Requirements 6. Software 78