111
Stochastic Structural Dynamics
Lecture-40
Dr C S ManoharDepartment of Civil Engineering
Professor of Structural EngineeringIndian Institute of ScienceBangalore 560 012 India
Problem solving session-4
2
20
Problem 40A sdof system driven by a filteredGaussian excitation is governed by the equations
2 ; 0 0; 0
Here is zero mean Gaussian white noise process
such that 2
x x x f t x x x
f f t
t
t t D
Set up the equations for time evolution of firsttwo order moments using Markov process approach.Consider the response in the steady state and evaluate
the response moments
3
20
1 2 2 1
1
2
3
1 22
2 2 1 3
3 3
2 ; 0 0; 0
0; 2
2
x x x f t x x x
f f t
t t t D t t
x xx xx f
x x
x x x xx x t
4
1 2
22 2 1 3
3 3
0
2
1 1 1
2
0 & 2
, , ; 0; 0
, ~ 1; ~ 1; ~
,
1,2
n n nt
j ijj i jj i j
dx x dt
dx x x x dt
dx x dB t
dB t dB t dB t D
dX t f X t t dt G X t t dB t t X X
X t f n dB t m G n md h X t tdt
h h hf X t GDGt X X X
Recall
5
2
1 1 1
22 2 1 3
1 2
2
3 23 3
,
1,2
, 2
2
n n nt
j ijj i jj i j
d h X t tdt
h h hf X t GDGt X X X
d h hh X t t X X X Xdt X X
h D hXX X
6
1 2
22 2 1 3
3 3
2
d X Xdtd X X X Xdtd X Xdt
7
21 1 2
2 22 2 1 3 2
23 3 3
2 21 2 2 2 1 3 1
1 3 2 3 3 1
22 3 2 1 3 3 3 2
2
2 2
2
2
2
d X X Xdtd X X X X Xdtd X X X Ddtd X X X X X X Xdtd X X X X X Xdtd X X X X X X X Xdt
8
0 0 0
1 0
2 0
3 0
2 2 21 2 3 1 2 1 3 2 3
Initial conditionsAssume that , & 0 = are all deterministic.
X 0
X 0
X 0
, , , , , 0@ 0
x x f f
x
x
f
X X X X X X X X X t
9
2
22 2 1 3 3 2
1 2 3 3
1 2
22 2 1 3
3 3
1 2 3
Steady state response analysissteady state
, 0
2 02
0
2 0
0
0
d h X t tdt
h h h D hX X X X XX X X X
d X Xdtd X X X Xdtd X Xdt
X X X
10
21 1 2
2 22 2 1 3 2
23 3 3
2 21 2 2 2 1 3 1
1 3 2 3 3 1
22 3 2 1 3 3 3 2
2 0
2 2 0
2 0
2 0
0
2 0
d X X Xdtd X X X X Xdtd X X X Ddtd X X X X X X Xdtd X X X X X Xdtd X X X X X X X Xdt
11
21
222
23
2
1 2
1 32
2 3
21
22
23
1 2
1 3
2 3
0 0 0 1 0 0 00 4 0 2 0 2 00 0 4 0 0 0
1 0 2 1 0 00 0 0 0 1 00 0 1 0 2 0
0 0 0 1 0 00 4
X
XDX
X XX XX X
X
X
X
X XX XX X
1
2
2
2
00 2 0 2 0
0 0 4 0 0 01 0 2 1 0 0
0 0 0 0 1 00 0 1 0 2 0
D
12
Problem 41In a study on reliability analysis of a cracked plate ithas become necessary to simulate a vector of six non-Gaussian random variables. The specification of theserandom variables is limited to the description of the 1storder pdf-s and the matrix of correlation coefficients.Develop a simulation procedure based on the Nataftransformation to simulate 5000 samples of the randomvariables. Estimate the 1st order PDF-s from the simulatedsample and perform the Kolmogorov-Smirnov test to verifyif the simulations have been performed satisfactorily.
13
1 2 3
4 5 6 2
The distribution of the basic random variables are as follows:~ 60,10 ; ~ 1,0.2 ; ~ 2,0.1 ;
~ 1 ; , ~ 33.0,0.47,3.5,0.3, 0.9
1.0 0 0 0 0 00 1.0 0 0 0 00 0 1.0 0 0 00 0 0 1.0 0 00 0 0 0 1.0 0.8350 0 0 0 0.835 1.0
X N X LN X LN
X EX X X N
1414
Partially specified non-Gaussian RVsNataf’s transformation
Let X1 and X2 be two random variables such that
•X1 and X2 are not completely specified•Knowledge on X1 and X2 is limited to first order pdfs and the covariance matrix.
Question: How to transform X to standard normal space?
1515
1
2
1 1
2 2
*1 2 1 2 12
1 11 1 1
1 2
2 22 2 2
2 1
1
1 1 1 2
2 1 1 2 2
2 2
Let( ) ( )
( )
with ~ (0,1), ~ (0,1) &
( ) ( ); 0
( ) ( ); 0
( ) 0( ) ( ) ( )
( ) ( ) ( )0( )
X
X
P X U
P X U
U N U N U Udx dxp x udu du
dx dxp x udu duu
p x u uJu p x p x
p x
1616
1 2
1 2
1 2
1 2
1 2
1 2
1 21 2 1 1 2 2
1 2
1 11 1 2 2
1 12 1 2 2
1 1 2 21 11 2
12 1 1 2 2 1 2 1 2
( , )( , ) ( ) ( )
( ) ( )
@ ( ) , ( )
( ) , ( )( ) ( )
( ) ( )
( )( ) ( , )
U UX X
X X
X X
X X
X X
p u up x x p x p x
u u
u P x u P x
P x u P xp x p x
P x P x
x x p x x dx dx
1 2
1 2
1 12 1 2 2
1 1 2 2 1 11 2
1 1 2 2 1 2
( ) , ( )( )( )
( ) ( )
( ) ( )
X X
X X
P x u P xx x
P x P x
p x p x dx dx
1717
1 2
1 1
1 1
1 1 2 1
1 2 X 1 X 1 1 1 1 2
1 1 *12 1 1 1 2 2 1 2 12 1 2
&
( )( ) , ,
X X
X X
P x z P x z
dx dx p x p x z z dz dz
P z P z z z dz dz
Substitute
18
1 2
*12
1 1 *12 1 1 2 2 2 1 2 12 1 2
*12
*12 1
( )( ) , ,
(1) Divide the range [-1,1] of into L divisions.
(2) For each value of solve the
X X
Lii
P z P z z z dz dz
Strategy for the determination of the unknown
12 1
12
*12 121
12
above equation
(numerically) and obtain the
corresponding values of . Note that
-1 1 1,2, , .
(3) Interpolate to obtain the value of
for which the target value of is r
Lii
i
Lii
i L
ealized.
1919
12
*12
12specified
1 1
1
*12 determined
1
2020
1 2
*12
1 1 *12 1 1 2 2 2 1 2 12 1 2
*
1 2
solve for by solving
( )( ) , ,
Simulate ~ 0, .
Simulate and us
X XP z P z z z dz dz
Z N
X X
Steps for simulation of 2 - dimensional Nataf random variables
Step 1
Step 2
Step 3
1
ing
; 1,2i X iX P U i
2121
1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
PD
F(x
)
simulation
target lognormal cdf
2222
0 2 4 6 8 10 120
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
PD
F(x
)
simulation
target exponential cdf
2323
Rho_equivalent_gaussianrho_t =1.0000 0 0 0 0 0
0 1.0000 0 0 0 00 0 1.0000 0 0 00 0 0 1.0000 0 00 0 0 0 1.0000 -0.88330 0 0 0 -0.8833 1.0000
2424
Msim=59.8787 1.0019 1.9990 1.0044 5.1723e-015 3.4961
Stdsim= 9.9726 0.1986 0.0998 1.0104 2.5838e-015 0.2982
Rhosim=1.0000 -0.0154 0.0093 -0.0135 -0.0013 0.0126-0.0154 1.0000 0.0206 0.0025 0.0014 -0.00680.0093 0.0206 1.0000 -0.0016 0.0018 -0.0058-0.0135 0.0025 -0.0016 1.0000 -0.0180 0.0116-0.0013 0.0014 0.0018 -0.0180 1.0000 -0.83450.0126 -0.0068 -0.0058 0.0116 -0.8345 1.0000
25
Problem 42A two dof system with cubic and hysteretic nonlinearstiffness characteristics is shown in the figure.
k1 k2 k3
c1 c2 c3
m1 m2
p1(t) p2(t)
u1(t) u2(t)
321
Left support
Right support
R1(t) R2(t)
1 2
3
Springs and have cubic force displacementcharacteristics and spring k is an inelasticspring modeled using Bouc's approach.
k k
26
31 1 1 1 2 1 2 1 1 1 1 2 1 2
32 1 2 1 1
32 2 2 2 1 3 2 2 2 1 2 2 1
3 2 3 2 2
12 2 2
The system is taken to be governed by the equations
1
| | | | | |n n
m u c u c u u k u u k u u
u u p t w t
m u c u u c u k u u u u
k u k z p t w t
z u z z u z Au
3
1 1 1 4
2 2 2 5
w t
p p w t
p p w t
27
5
1
2
1 2 1 2
3 1 2
1 2
are given to be a set of independent
white noise processes with .
The various system paramters are as follows:1.0kg, 1.5kg, 0.1kN / m, 0.2kN/m,0.15kN/m, 0.05,
2,
i i
i i i
w t
w t w t
m m k kk
1 2 0
1 2 3 4
5
4, 0.05, 0.5, 0.5, 1, 3,10, 20, 15.6 , 2 , 0.00420.01N, 0.02N, 0.01N, 10.0N and 20.0N
Using 1.5 order strong Taylor's scheme, develop a procedure to simulate samples of the
A nT s t s s
sytem response.Hence estimate the response moments (up to second order)and the first order pdf-s.
28
, 0 21
1
,
1
,0 ,0
, ,1
02 2
1 1
,
, 1 1
Recall
12
, ,2
2 1 2 ,
1 1 1 ,12 2
12
mn n n n j j nk k k k k
j
mj n j n j j j
k kj
j jj j j j
p
j i p j pi
p di
p k iki i k
d mi j lk k
i l j
x x a b W L a
L a Z b W Zt
W Z a a
i
L ati x
b b
2, ,
1
,d
j j i jki l i
ik k k
L bx x x
29
1 1 2 2 1 2
1 1 1 0 1 2 1 1 1 2 1 2 3 1 31
4 1 4 5 1 5
2 2 2 1 0 2 2 1 2 1 2 2 211
13 2 3 4 2 4 5 2 5
3 3 3 0 3 2 11
12
12
12
t
k k k k k k k
k k
k k k k k k
k k k
k k k k
x t u t u t u t u t z t p t p t
x x a L a L a Z L a Z L a Z
L a Z L a Z
x x a W L a L a Z L a Zm
L a Z L a Z L a Z
x x a L a L a
3 1 2 3 2 3 3 3
4 3 4 5 3 5
k k k
k k
Z L a Z L a Z
L a Z L a Z
30
4 4 4 2 0 4 2 1 4 1 2 4 221
23 4 3 4 4 4 5 4 5
5 5 5 3 0 5 2 1 5 1 2 5 21 3
3 5 3 4 5 4 5 5 5
6 6 6 4 0 6 2 1 6 1 2 6 21 4
12
12
12
k k k k k k
k k k
k k k k k k
k k k
k k k k k k
x x a W L a L a Z L a Zm
L a Z L a Z L a Z
x x a W L a L a Z L a Z
L a Z L a Z L a Z
x x a W L a L a Z L a Z
3 6 3 4 6 4 5 6 5
7 7 7 5 0 7 2 1 7 1 2 7 21 5
3 7 3 4 7 4 5 7 5
12
k k k
k k k k k k
k k k
L a Z L a Z L a Z
x x a W L a L a Z L a Z
L a Z L a Z L a Z
31
0 1 2 1 1 2 1 3 1 4 1 5 11
1
1 2 20 2 1 1 31 2 1 2
1
2 3 4 521 31 2 2 2 2
1 1 1 1
1 2 2 2 3 21 21 2 22
1 2 1
, , 0, 0, 0, 0
3 3
3
, ,
k k k k k k k
kk k k k
k k k kk k
k k k
L a a L a L a L a L a L am
aL a k k x x xm
a a a ac c k x x cm m m m
L a c c L a c L am m m
4 2 5 24
1
0,
, 0k kL a L am
32
0 3 4 1 3 2 3 3 3 4 3 5 32
2
1 220 4 3 12 2 2
2 2
3 4 523 12 3 2 2 3 3
2 1 2
1 4 2 41 22 2 32
2 1 2
3
, 0, , 0, 0, 0
3
3 1
, ,
k k k k k k k
k kk k k
k k kk k
k k
L a a L a L a L a L a L am
a aL a k x x cm m
a a ak k x x c c km m m
L a c L a c cm m m
L a
4 4 4 5 43 53
2 2
1 , 0,k k kk L a L am m
33
3
0 5 4 5 5 1 4 5
5 4 5 1 4 5 5 2 5 4 5 1 5
24 5 5 12
22
4 5 2 5 4 5 5 3
2
| | sgn | |
| || | 1 | | | | sgn | | sgn
2 | |2
1 | || | sgn 1 2 | | | | sgn
2
n nk k k k k k
n n nk k k k k k k k k k
nk k k
n nk k k k k k k
L a a x x x x A
a x x n x x x x nx x x
x x xm
n x x x n n x x x x
25
4 5 2 5 4 5 5 2 5
24 5 2 5 4 5 1 5
1 | || | sgn 2 1 | | | |
1 | | sgn 2 | |
n nk k k k k k k
n nk k k k k k
n x x x n x x x x
n n x x x nx x x
34
1 5 2 5 5 5 1 4 5222
3 5 4 5 1 4 5 5 2 5 4 5 2 53
4 5 5 5
0 6 6 1 6 2 6 3 6 4 6 51 4 1
0, | | sgn | | ,
| || | 1 | | | | sgn | | sgn ,
0, 0
, 0, 0, 0, ,
n nk k k k k k
n n nk k k k k k k k k
k k
k k k k k k
L a L a x x x x Am
L a x x n x x x x nx x x
L a L a
L a a L a L a L a L a L a
6
0 7 7 1 7 2 7 3 7 4 7 5 72 4 2
0
, 0, 0, 0, 0,
k
k k k k k k kL a a L a L a L a L a L a
35
time s
1u t
36
time s
1u t
37
time s
2u t
38
time s
2u t
39
time s
z t
40
time s
1p t
41
Stochastic structural dynamics
Uncertainty modeling•Probability•Random variables•Random processes
Propagation of uncertainty•Analytical procedures forLTI systems (time/frequency)•Markov vector approach
Reliability analysis•First passage•Extremes•Fatigue
Monte Carlo simulations•Gaussian/non-GaussianRandom variables & random processes•Variance reduction
Applications•Earthquake•Fatigue•High frequency analysis
42
What next?
•Structural system identification•Reliability model updating•Structural health monitoring
Further applications•Wind, waves, guideway uneveneness,…•Hazard and risk analysis
•Earthquake•Wind
•Performance based design•Design code development…..
Sensing + Computing
43
Discussion on mean square estimation
44
Introductory comments
Let X and Y be two random variables with a known jpdf. Assuming that in a particular experiment, the random variable Y can be measured and takes the value y. What can we say about the corresponding value, say x, of the unobservable variable X?
Suppose we make an estimate, say, x*, of the value of Xwhen Y=y, according to the rule x*=h(y). h(y)=unspecified function of y. The error of our estimate e=x-h(y).We can never hope to make e=0. Can we select h such that we minimize the expected value of some function of e?
45
Let x(t) and y(t) be two Gaussian random processes with a known joint pdf.Let it be assumed that we can observe y(t) and not x(t).Given the observation of a sample of y(t) for t in 0 to T, how to estimate the value of x(t) for some value of t?
Typical problem in dealing with random processes
46
Problem 1
Let Y be a random variable and c be a constant. We wish to estimateY by a constant.Find c such that E[(Y-c)2] is minimized.
YEdyyypcce
dyypcycYEe
Y
Y
)(0
)(22
47
Problem 2Let X and Y be two random variables. We wish to estimate Y by a function c(X).To find c(X) such that e=E{[Y-c(X)]2} is minimized.
2 2
2
2
0 0
2
( ) ( ) ( , )
( ) ( | ) ( )
( ) ( ) ( | )
would be a minimum if ( ) minimizes
( ) ( | ) for every fixed value of .
XY
Y X
X Y
Y
e E Y c X y c x p x y dxdy
y c x p y X x p x dxdy
p x y c x p y X x dy dx
e c x
y c x p y X x dy x
48
From solution of Problem 1 we have
( ) ( | ) |Yc x yp y X x dy E Y X x
Remarks•If Y=g(X), c(x)=E[g(X)|X=x]=g(x) & e=0.•If X and Y are independent, c(x)=<Y>=constant
49
2
22
2
2 2
min
Let ( )
0 0
0 0
Y X
Y X Y X
Y X X
XY XY X Y XY Y
X X X
Y X XY YXY Y
X
c X AX B
e E Y AX B
e E Y AX B B AB
e E Y AX A E Y A X
e E Y X A XA
r rA
re E Y r X
Linear MS estimation
2
22 21
X
YY XY X Y XY
X
E Y r X r
50
22min 1 XYY
X
YXYXYXY
X
YXY
re
rABrA
Let X and Y be Gaussian
X
XYXY
X
YXY rxrxXYExc
|)(
For normal random variables, linearand nonlinear ms estimation lead to identical results.
51
The orthogonality principle
2
0 0
Errordata
e E Y AX B
e E Y AX B XA
Y AX BX
Data is orthogonal to error
52
1 1 2 2
2 21 1 2 2
1
Let be a random variable &ˆ
be an estimatior of .
ˆ
Select 0 1, 2, ,
tn n
n n
ni i
i
S
S a X a X a X a XS
P E S S E S a X a X a X
Pa i na
General case of linear ms estimation
53
21 1 1 2 1 2 1 1
22 1 1 2 2 2 1 2
21 1 2 2 1
01 11 12 1 1
02 21 22 2 2
03
0 1 2
0
n n
n n
n n n n n
n
n
n n n nn n
SX a X a X X a X X
SX a X X a X a X X
SX a X X a X X a X
R R R R aR R R R aR
R R R R a
R RA A R
10R
54
1 1 2 2
2
1 1 2 2
1 01 2 02 0
ˆWe have 0 1,
ˆ 0
ˆ ˆ ˆ ˆ0
ˆ ˆ ˆ ˆ ˆ
ˆ ˆ
ˆ
i
n n
n n
n nt
E S S X i n
E S S a X a X a X
E S S S S S S
P E S S S S E S S S E S S S
P E S S S E S E SS
E SS E S a X a X a X
a R a R a R
A R
0
20
tP E S A R
55
21 2
21 2
21 2
21 2
0 0
( , , , )
( , , , ) ( , )
( , , , ) ( | ) ( )
( ) ( , , , ) ( | )
n
n SX
n S X
n SX
P E S g X X X
s g x x x p s x dsdx
s g x x x p s X x p x dsdx
p x s g x x x p s X x ds dx
Nonlinear estimation
1 2
is minimum when the second integrand is minimum for any
( , , , ) |n
Px
g x x x E S X x
56
1 1 2 2
1 1 2 2
We have
( ) 0 1,
ˆi.e., 0 1,
ˆ 0 for any 1,
n n i
i
n n i
E S a X a X a X X i n
E S S X i n
E S S c X c X c X c i n
General orthogonality principle
57
If ( ) is the nonlinear ms estimator of , the estimation error ( ) is orthogonal to any function ,linear or nonlinear function of data.
:
( ) ( ) ( ) ( ) ( , )
(
SX
g X SS g X w X
E S g X w X s g x w x p s x dsdx
s g x
Proof
) ( ) ( | ) ( )
( ) ( ) ( | ) ( )
( ) ( ) | ( )
( ) | ( ) ( )
( ) ( ) ( ) ( ) 0
S X
S X
X
X
X
w x p s x p x dsdx
w x s g x p s x ds p x dx
w x E S g X X x p x dx
w x E S X x g x p x dx
w x g x g x p x dx QED
58
If S and X are jointly normal it can be shown that linear and nonlinearestimations of S are equal.
59
Estimation of a random process
Let S(t) & X(ξ) be two random processes with .ba
Consider the problem of estimating S(t) for a fixed t in terms of X(ξ)specified for every ξ in an interval of finite or infinite length. X(ξ) = data available.
ba
60
2
1
1
1
ˆ( ) ( ) ( )
( ) ( ) ( )
Select ( ) such that is minimized.
ˆ( ) ( )
error ( ) ( )
Orthogonality principle
( ) ( ) ( )
b
a
b
a
n
k kk
m
k kk
m
k k jk
S t X h d
P E S t X h d
h P
S t h X
S t h X
E S t h X X
1
0 1,
( , ) ( ) ( , ) 1,m
SX j k XX k jk
j m
R t h R j m
61
0lim ( , ) ( ) ( , )
Smoothing: Prediction: ; , Forward prediction: Backward prediction: Filtering:
b
SX XXa
R t h R d
a t bX t S t t a b
t bt a
X t S t
62
2
2
Problem: Let ( ) be a stationary random process. Estimate ( ) in terms of .ˆ( ) ( )
( ) ( )
0 ( ) ( ) ( ) 0
( , ) ( )( , )
SS SS
SS S
S tS t S t
S t aS t
P E S t aS t
P E S t aS t S ta
R t t RaR t t
63
2
( ) ( ) ( ) ( )
( ) ( ) ( )
(0) ( )
( )(0)(0)
Let ( ) exp | |
exp | |exp | |
SS SS
SSSS
SS
SS
P E S t aS t S t aS t
E S t aS t S t
R aR
aRRR
R A
Aa
A
64
1 2
1 2
2
1 2
11
1 2
2
ˆExample : Let ( ) ( ) ( )
Note: ( ) ( ) ( ) ( ), ( )
( ) ( ) 0
( ) ( ) ( )
0 ( , ) ( , ) 0
( , ) ( )( , )
0 (
SS SS
SS SS
SS S
SS
S t a S t a S t
S t a S t a S t S t S t
E S t S t
P E S t a S t a S t
P R t t a R t ta
R t t RaR t t
P Ra
2 2
( , ), ) ( , ) 0
( , )SS
SSSS
R t tt t a R t t a
R t t
65
1 2 1 2
1 2
1 2
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
(0) ( ) ( , )SS SS SS
P E S t a S t a S t S t a S t a S t
E S t a S t a S t S t
R a R a R t t
66
2
2
Filteringˆ( ) ( )
( ) ( )
(0)0(0)
( ) ( ) ( )
(0)(0)(0)
SX
XX
SXSS
XX
S t aX t
P E S t aX t
RP aa R
P E S t aX t S t
RRR
67
Interpolation
t-NT t+NTλt t+T
To estimate s(t+ λ) in the interval t to t+T in terms of samples of S(t), S(t+kT), k=-N,-(N-1),…,0,1,2,…,N
S(t)
68
2
ˆ( ) ( ); 0
( ) ( )
0 ( ) ( ) ( ) ,
Set of 2 1 equations for
N
kk N
N
kk N
N
kk Nj
Nk k N
S t a S t kT T
P E S t a S t kT
P E S t a S t kT S t jT j N Na
N a
69
0
0 1
2
0 10
00
1
Quadrature
( )
ˆ (0) ( ) ( );
( ) (0) ( ) ( )
0 ( ) ( ) (0) ( )
( ) ( ) ( ) ( ) 0 0,1 equations in 1
b
N
b
N
b
k
n
Z S t dt
bZ a S a S T a S NT TN
P E S t dt a S a S T a S NT
P E S t S kT dt a E S S kTa
a E S T S kT a E S nT S kT j NN N
unknowns
70
Estimate present value of ( ) in terms of values of( ) for
( ) ( ) ( )
ˆ( ) ( ) ( )
ˆ( ) ( ) ( ) ,
( ) ( ) ( ) ( ) 0
( ) ( ) (SX XX
S tX
X t S t t
S t h X t d
S t S t X
E S t h X t d X t
R h R
Smoothing
)d
71
1
Probelm of dynamic state estimationProcess equation
Measurement equation
: 1 state evector: 1 process noise; iid sequence N(0,1): state transistion matrix: 1 measurem
k k k k
k k k k
k
k
k
k
x x w
z H xx nw n
n nz m
ent vector: relates states to measurements
: 1 measurement noise; iid N(0,1)k
k
H m nm
72
0: 1:
1:
| 1:
| | | 1:
Probelm of dynamic state estimationDetermine
|
|
|
|
k k
k k
k k k k
k k k k k k k k k
p x z
p x z
a x z
x a x a z
Kalman filter provides the exact solution to this problem
