16. 322 Stochastic Estimation and Control. Fall 2004 Prof. VanderⅤelde This relation alone defines a distribution for time to failure P(<TSI+dtT>0=P(<TSI+d AND T>) 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 616.322 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 < ≤ t + dt T > = 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