16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Lecture 8
Last time: Multi-dimensional normal distribution
1
T
⎡ 1
⎤
f ( x) =
exp ⎢ − ( x − x ) M −1 ( x − x )⎥
n
⎣ 2
⎦
( 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
x2
argument of the exponential becomes the sum over i of i . Thus, the
2
distribution becomes a product of exponential terms in i.
If XY = 0
X 2Y 3 = X 2 Y 3 = 0
The general moment of a multidimensional normal distribution, E ⎡⎣ X 1r1 X 2r2 ...X nrn ⎤⎦ ,
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 ≤ t + dt | T > t) = λ dt
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16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
This relation alone defines a distribution for time to failure.
P(t < T ≤ t + dt AND T > t)
P(t < T ≤ t + dt | T > t) =
P(T > t)
f (t )dt λ dt =
t
1 − ∫ f (τ )dτ
0
t
f (t) = λ − λ ∫ f (τ )dτ
Integral equation for f (t)
0
df (t )
= −λ f (t)
dt
f (t ) = ce − λ t
∞
c
∫ f (t )dt = − λ e
Differential equation for f (t)
− λt ∞
0
0
=
c
λ
= 1
c = λ
f (t ) = λ e − λ t
To find the moments of the distribution, start with the characteristic function.
∞
φ ( s ) = ∫ e jst λ e − λt
dt =
0
λ
js − λ
e( js−λ )t
∞
0
=
λ
λ − js
dφ ( s )
λj
=
(λ − js) 2
ds
2λ j 2
d 2φ ( s )
=
(λ − js)3
ds 2
T =
T2 =
σ T2 =
1 dφ ( s)
1
= = tm ≡ mean time to failure (MTTF)
j ds s=0 λ
1 d 2φ (s)
2
= 2
2
2
j ds s=0 λ
2
λ
2
−
1
λ
2
=
1
λ2
= tm2
Thus in this case, σ T = T . The standard deviation is equal to the mean.
⎛ t ⎞
1 −⎜ t ⎟
f (t ) = e ⎝ m ⎠
tm
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16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Reliability over lifetime tl:
τ
∞
1 −
R(tl ) = P(T > tl ) = ∫ e tm dτ
t
t m
= −e
⎛τ ⎞
−⎜ ⎟
⎝ tm ⎠
∞
=e
⎛t ⎞
−⎜ l ⎟
⎝ tm ⎠
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:
Rs (t ) = P(Ts > t)
= P(T1 > t , T2 > t,..., Tn > t)
= P(T1 > t ) P (T2 > t)...P (Tn > t)
=e
=e
⎛ t
−⎜
⎜ tm
⎝ 1
⎞ ⎛ t
⎟ −⎜
⎟ ⎜ tm
⎠ ⎝ 2
⎛ t
−
⎜
⎜ tm
⎝ s
⎞
⎟
⎟
⎠
e
⎞
⎟
⎟
⎠
...e
⎛ t
−⎜
⎜ tm
⎝ n
⎞
⎟
⎟
⎠
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
=∑
tms i =1 tmi
n
λs = ∑ λi
i =1
If all the components have the same mean time to failure,
tmi = tmc , i = 1, 2,..., n
1
n
=
tms tmc
tms =
1
tm
n c
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.
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16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Rs (tl ) = 0.99
e
⎛ t
−⎜ l
⎜ tm
⎝ s
1−
⎞
⎟
⎟
⎠
= 0.99
tl
= 0.99
tms
tl
= 0.01
tms
tms = 100tl
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 tmc .
1 1000 1 1000
=∑ =
tms i =1 tmi
tmi
1
tm = 100tl
1000 c
tmc = 105 tl
tms =
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: x̂0 = ⎢ 0 ⎥
⎣ v0 ⎦ 6×1
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16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Errors are zero mean:
e0 = x0actual − x0est .
Covariance matrix:
[ E0 ]6×6 = e0 e0T
Lander deployment is impulsive (assume short burn, negligible change in
position):
er1 = er0
ev1 = v1true − v1nom
= v0true + ∆v + δ v − ( v0est . + ∆v )
= ev0 + δ v
⎡ er0 ⎤
e1 = ⎢
⎥
⎢⎣ ev0 + δ v ⎦⎥
⎡ 0⎤
= e0 + J δ v , where J = ⎢ ⎥
⎣ I ⎦ 6×3
E1 = e1 e1T
= [ e0 + J δ v ] ⎡⎣ e0T + δ v T J T ⎤⎦
= E0 + JDJ T , if e0δ v T = δ v T e0T = 0
⎡0 0 ⎤
= E0 + ⎢
⎥
⎣ 0 D ⎦ 6×6
D = δ vδ v T : 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
δ rs = ∑
i
j=1
∂rsi
∂xij
δ xij , i = 1, 2
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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
is introduced and the δ rsi noted (i =1,2), the ratios
measures of the
∂rsi
∂xij
δ rs
are finite-difference
δ xij
i
, 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.
es = [ S ]2×6 e1
Sij =
∂rsi
∂xij
≈
∆rsi
∆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&n + δ x& = f ( xn + δ x )
≈ f ( xn ) +
df
dx
δx
x&n = f ( xn )
δ x& =
Fδ x, where Fij =
∂f i
∂x j
∂x (t ) = Φ (t0 , t )δ x (t0 )
dΦ (t0 , t )
δ x (t0 ) = F (t )δ x (t )
dt
= F (t )Φ (t0 , t )δ x (t0 )
δ x& (t ) =
Therefore, for arbitrary δ x (t0 )
dΦ(t0 , t )
= F ( t )Φ
(t0 , t )
dt
Φ(t0 , t0 ) = I
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