16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Lecture 8 ast time: Multi-dimensional normal distribution f(x)= exp If a set of random variables X, having the multidimensional normal distribution is uncorrelated(the covariance matrix is diagonal, they are independent. The argument of the exponential becomes the sum over i of Thus, the distribution becomes a product of exponential terms in i If XY=0 The general moment of a multidimensional normal distribution, E X1X,2.X is known Laning and Battin. Random Processes in Automatic Control The Exponential Distribution Many components, especially electronic components, display constant percentage failure rates over long intervals Relati N(△t) rate of Time Ignore the " burn-in Start the time Approximation period axis here breaks down here Relative rate of failure vs time The familiar bathtub curve Failure rate= E(relative rate of failure) USing the random variable Tfor time to failure, and assuming a constant failure rate a (independent of time n) implies P(t1)=dt Page 1 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Lecture 8 Last time: Multi-dimensional normal distribution 1 x exp ⎢ ⎡ − 2 ( x − x ) M ( x − x ) ⎤ f () = ⎣ 1 T −1 ⎦ n ⎥ (2π ) 2 M If a set of random variables Xi having the multidimensional normal distribution is uncorrelated (the covariance matrix is diagonal), they are independent. The 2 argument of the exponential becomes the sum over i of xi . Thus, the 2 distribution becomes a product of exponential terms in i. If XY = 0 23 2 X Y = X Y 3 = 0 r 2 r n r The general moment of a multidimensional normal distribution, E 1 ⎡XX 2 ...Xn ⎦ ⎤, ⎣ 1 is known. • Laning and Battin. Random Processes in Automatic Control. The Exponential Distribution Many components, especially electronic components, display constant percentage failure rates over long intervals. Relative rate of failure vs time The familiar “bathtub” curve. Failure rate = E(relative rate of failure) Using the random variable T for time to failure, and assuming a constant failure rate λ (independent of time t) implies P t ( = t) λdt Page 1 of 6
16. 322 Stochastic Estimation and Control. Fall 2004 Prof. VanderⅤelde This relation alone defines a distribution for time to failure P(0=P() P(T>1) adt= f(odt 1-f(r)dr f(=x-if()dr Integral equation for/() df(o) =-f() Differential equation forf(o) f(=ce A(Odt=-te-L=c f(=he To find the moments of the distribution start with the characteristic function d叭(s)j ds(λ-js) d以(s)22 (-js)3 T I do(s) =-=t.= mean time to failure(mTtf dp(s)2 211 0r= 22m Thus in this case, o=T. The standard deviation is equal to the mean f(1)=-e Page 2 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde This relation alone defines a distribution for time to failure. ( T | t) P t ( Tt + dt AND T t) Pt = PT ( > t) f t dt λdt = () t 1 − ∫ f (τ τ )d 0 t f t( ) = − λλ τ f ( )dτ Integral equation for f t( ) ∫ 0 df () t = −λ f (t) Differential equation for f (t) dt f t() = ce −λt ∞ () −λt ∞ c ∫ f t dt = − c e = λ 0 λ 0 = 1 c = λ ft −λt () = λe To find the moments of the distribution, start with the characteristic function. ∞ () = −λt ∞ λ φ s ∫ e jstλe dt = λ e( js−λ )t = js − λ 0 λ − js 0 d s φ() λ j = 2 ds (λ − js) 2 2 d φ() 2 s λ j = 2 3 ds (λ − js) 1 d s φ( ) 1 T = = = t ≡ mean time to failure (MTTF) m j ds λ s=0 2 T 2 = 1 d φ(s) 2 = j 2 2 ds 2 λ s=0 1 = 1 = t 2 σ T 2 = 2 2 − 2 2 m λλλ Thus in this case, σ = T . The standard deviation is equal to the mean. T ⎛ ⎞ −⎜ ⎟ t e ⎝ ⎠ () = 1 tm ft tm Page 2 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Reliability over lifetime tr R(1)=P(T>1) dt Suppose a system contains n components all of which must operate if the system is to operate(no redundancy), and component failures are considered independent. This is called a simplex system For the system: R,(D)=P(T>1) =P(T1>1,2>1,…,Tn>D =P(T1>1)P(72>1).P(Tn>1) This is the same form for the system as for the individual components. Thus the system also has an exponential distribution of time to failure, with the indicated mean time to failure λ=∑ If all the components have the same mean time to failure In=In,I= Note the importance of part count n to system reliability Example: Lifetime systemm reliability Suppose we wish to achieve a 99% probability of successful system operation over a mission lifetime tl. Page 3 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Reliability over lifetime tl: ∞ − τ R tl () = PT ( > tl) = ∫ 1 e tm dτ t t m ∞ ⎛ ⎞ ⎛ ⎞ −⎜ ⎟ τ tl −⎜ ⎟ tm t ⎝ ⎠ m = − e = e ⎝ ⎠ τ =tl Suppose a system contains n components all of which must operate if the system is to operate (no redundancy), and component failures are considered independent. This is called a simplex system. For the system: Rt () = PT ( > t) s s ( t T2 = P T , 1 > > t,..., T > t) n = P T( ) ( ( 1 > t P T > t)...PT > t) 2 n ⎛ ⎞⎛ ⎞ ⎛ ⎞ −⎜ ⎟⎜ ⎟ −⎜ ⎟ t t t − ⎜ ⎟⎜ ⎟ ⎜ ⎟ mn = e ⎝ ⎠⎝ ⎠ t t m1 e tm2 ...e ⎝ ⎠ ⎛ ⎞ −⎜ ⎟ t ⎜ ⎟ tms = e ⎝ ⎠ This is the same form for the system as for the individual components. Thus the system also has an exponential distribution of time to failure, with the indicated mean time to failure. n 1 = ∑ 1 t i=1 t ms mi n λs = ∑λi i=1 If all the components have the same mean time to failure, t = t , 1, 2,..., i = n mi mc 1 n = t t ms mc 1 t = t ms mc n Note the importance of part count n to system reliability. Example: Lifetime system reliability Suppose we wish to achieve a 99% probability of successful system operation over a mission lifetime tl. Page 3 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. VanderⅤelde R,()=0.99 e(m)=0.99 0.99 =00 To achieve 99% reliability, the mTTf of a component must be at 100 times the mission lifetime Now suppose the system contains 1000 components of equal t 1011000 10010 105t 200,000 years. This is why we do not conduct high reliability, high6e For a planetary mission where t)=2 years, the component MTTf mu component count missions without redundancy Linearized Error Propagation Example: Orbiting spacecraft Orbiting spacecraft carrying lander Given an initial state: 5=56 Page 4 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde R t( l) = 0.99 s ⎛ ⎞ tl −⎜ ⎟ ⎜ ⎟ tms e ⎝ ⎠ = 0.99 tl 1 − = 0.99 tms tl = 0.01 tms t = 100t m l s To achieve 99% reliability, the MTTF of a component must be at 100 times the mission lifetime. Now suppose the system contains 1000 components of equal t . mc 1000 1 1 = ∑ = 1000 tms i=1 tmi tmi 1 t = tmc = 100t m l s 1000 t = 105 t m l c For a planetary mission where tl = 2 years, the component MTTF must be 200,000 years. This is why we do not conduct high reliability, high component count missions without redundancy. Linearized Error Propagation Example: Orbiting spacecraft Orbiting spacecraft carrying lander ⎡r Given an initial state: xˆ0 = 0 ⎤ ⎢ ⎥ ⎣v0 ⎦6 1× Page 4 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. VanderⅤelde Errors are zero mean ≌=xm-x Covariance matrix EoL=ee Lander deployment is impulsive(assume short burn, negligible change in DOS =+△ +6y-(M+△) e,+d1 日+J5y, where J E,=ee [e+Joye+8yJ'] Eo+JD/, if eSy=8ve,=0 Eo D=Vov: covariance matrix for velocity correction errors Now these errors must be propagated to the surface. A linearized description of error propagation is given by a transition matrix which relates perturbations in position and velocity components at the initial point to perturbations in position and velocity at any later point. For the present purpose, we may be interested only in the position perturbation at the end point. Direct sensitivity analysis or i=1.2 ax Page 5 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Errors are zero mean: e0 = x0actual − x0est. Covariance matrix: T × = e0 e [E0 ]66 0 Lander deployment is impulsive (assume short burn, negligible change in position): e = er0 r 1 = − v1nom ev1 v1true = v + ∆v + − δ v (v + ∆v) 0true 0est. = e + δ v v0 ⎡ er0 ⎤ e1 = ⎢ ⎥ ⎣ev0 + δ v ⎢ ⎦⎥ ⎡ ⎤ 0 = e0 + J v, where J = ⎢ ⎥ δ I ⎣ ⎦63 × E1 = e e T 1 1 T T T = [e0 + J v e ⎦ δ ]⎣ ⎡ 0 + δ v J ⎤ T T δ δ ve = E 0 = 0 0 + JDJ T , if e v T 0 = ⎡0 0 ⎤ = E0 + ⎢ ⎥ ⎣0 D⎦6 6× T D = δ v δ v : covariance matrix for velocity correction errors. Now these errors must be propagated to the surface. A linearized description of error propagation is given by a transition matrix which relates perturbations in position and velocity components at the initial point to perturbations in position and velocity at any later point. For the present purpose, we may be interested only in the position perturbation at the end point. Direct sensitivity analysis 6 ∂rsi δ r = ∑ δ xij , i = 1, 2 si j=1 ∂xij Page 5 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Often this sensitivity matrix can only be evaluated by simulation. If a small is introduced and the Or noted(i=1, 2), the ratios are finite-difference OI measures of the -s, which are the values comprising the jth column of s. thus 6 perturbed trajectories must be calculated and the end state differenced with the nominal end state to define S. Each perturbed trajectory defines one column of s e.= Or△r Linearized analysis Given the dynamic 文=f(x) F=1 v=a+g Consider the errors as a small perturbation around the nominal x(t) trajectory x=i,+ox=f( ≈f(xn)+x 文n=f(x) where 3= ax(1)=Φ(o2,1)6x() d(Lo, 1) 6()=dox(0)=F()x() =F()Φ(0,0)6x(t0) Therefore, for arbitrary Sx(to) (2,1) =F()Φ(to,0 Φ(,)=
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Often this sensitivity matrix can only be evaluated by simulation. If a small δ xij δ rsi is introduced and the δ r noted (i =1,2), the ratios are finite-difference si δ xij ∂rs measures of the i , which are the values comprising the jth column of S. Thus 6 ∂xij perturbed trajectories must be calculated and the end state differenced with the nominal end state to define S. Each perturbed trajectory defines one column of S. es = [S]2 6× e1 ∂r ∆r = si ≈ si Sij ∂ ∆xij xij Linearized analysis Given the dynamics x& = f x( ) r v & = v a & = + g Consider the errors as a small perturbation around the nominal x() t trajectory. x x & = & + δ x& = f x ( + δ x) n n df ≈ f ( ) x + δ x n dx x& = f ( ) x n n δ δ x F x , where Fij = ∂fi & = ∂x j () 0 xt , ( 0 ∂ =Φ(t t )δ x t ) δ &( ) = dΦ( , 0 x t ( ()δ x t t t ) δ x t0 ) = F t ( ) dt () 0 Ft , ( = Φ 0 (t t )δ x t ) Therefore, for arbitrary δ x( ) t0 dΦ( , ( ) 0 , t t ) = Φ(t t ) 0 F t dt Φ( , t t ) = I 00 Page 6 of 6