추정론 기말시험

Estimation Theory Final Test_Solution

December 8, 2010

1. A simplified spacecraft tracking problem is formulated by

c c c

c c c c

x w , w ~N(0, )

z x v , v ~N(0, q r

=

= +

).

)

(T10.2.1-1)

(a) Suppose that the measurements are taken every 0.5 second. Show that the discrete model for Eq. (T10.2.1-1) is given by

( 1) ( ) (

( ) ( ) ( ).

x k x k w

z k x k v k

+ = +

= +

k

)

(T10.2.1-2)

Determine the mean and variance of w k( ) and v k( ). (Solution) (10 points)

Given

, 0, 1, (0,

, (0, )

c c c c

x w A x G w A G w N q

z x v v N r

= ≡ + = =

= +

∼

∼

the discrete model is given by

( 1) ^{A Tc} ( ) ( ) ( ) ( ) ( ) ( )

x k+ =e x k +w k =x k +w k ≡ Ax k +Gw k (T2.1-1) (T2.1-2)

( ) ( ) ( ) ( ) ( )

z k =x k +v k ≡Hx k +v k where

(

⁰

( ) )

( ) 0, 0.5

( ) 0, 2 .

c c

T T

A T A

c c

w k N e G QG e d Tq q

v k N r r

T

τ τ τ = =

⎛ = ⎞

⎜ ⎟

⎝ ⎠

∫

∼

∼

(b) Find the steady-state Kalman filter solution for this problem assuming that and are white Gaussian and uncorrelated each other.

( )

w k v k( ) (Solutiom) (20 points)

Find P_∞ by solving the ARE

( )

( )

1

1

2

2

2

2 2

T T T

P A P P H HP H R HP A GQG P P P r P q

P q

P P r

−

∞ ∞ ∞ ∞ ∞

−

∞ ∞ ∞ ∞

∞ ∞

∞

⎡ ⎤

= ⎢⎣ − + ⎥⎦ +

⎡ ⎤

=⎣ − + ⎦+

= − +

+

T

2

2 2

P q

P r

∞

∞

+ =

2 0

2

P_∞ −qP_∞−rq=

2

4 4 .

q q

P_∞ = ± ⎛ ⎞⎜ ⎟⎝ ⎠ +rq Since , P_∞ >0

( )

2

1 2

4 4 4 16

q q

P_∞ = + ⎛ ⎞⎜ ⎟⎝ ⎠ +rq = q+ q + rq . (T2.1-3) And

( )

( )

( )

( )

1

1

2

2

2

2

1 16

4 .

1 16 2

4

K P H HP HT R

P P r

P

P r

q q rq

q q rq r

−

∞ ∞ ∞

−

∞ ∞

∞

∞

= +

= +

= +

+ +

=

+ + +

(T2.1-4)

The transfer function for the steady-state Kalman filter, H_SSKF( )z , is

[ ]

( )

¹

( ) 1

SSKF

H z H zI A I K H AK K

z K

− ∞

∞ ∞

∞

= − − =

− + (T2.1-5) Let q= =r 1 for comparison,

( ) 0.39

SSKF 0.61

H z

≈ z

− . (T2.1-6)

(c) Find the causal Wiener filter solution for this problem and compare it with the result of (b).

(Solution) (20 points)

Suppose w k( )and v k( )are white, then

S_W( )z =0.5q

From x k( + =1) x k( )+w( )k , ( ) ^{( )} ₁¹

( ) 1

X w to x

W

S z

H z

S z Z⁻

= =

− (T2.1-7) And

1

1

1

( ) ( ) ( ) ( )

1 1

0.5 1 1

0.5

(1 )(1 )

X W wto x wto x

S z S z H z H z

q z z

q

z z

−

−

−

=

= ⋅ ⋅

−

−

= − −

(T2.1-8)

Let,

^{( ) ( )}

( ) ( ) ( ) s k x k z k s k v k

=

= +

Then,

₁

1

( ) ( )

( ) ( ) ( )

0.5 2

(1 )(1 )

( ) ( ) 0.5

(1 )(1 )

S X

Z S V

SZ S

S z S z

S z S z S z

q r

z z

S z S z q

z z

−

−

=

= +

= +

− −

= =

− −

(T2.1-9)

( ¹) ¹ ₁^{( )} ( ) ( )

SZ Wiener

Z Z

S z

H z

S z S z

−

+ −

+

= ⎡⎢

⎣ ⎦

⎤⎥ (T2.1-10) (Note: Refer to Eq.(T2.1-7))

To compare with (b), let q=r=1, then

1 1 1 1 1

1 1 1 1

( ) 0.5 2

(1 )(1 )

0.5 2(1 )(1 )

(1 )(1 )

4.5 2 2

(1 )(1 )

1.22(1.64 )(1.64 )

(1 )(1 )

1.10(1.64 ) 1.10(1.64 ) (1 )

(1 )

( ) ( )

Z

Z Z

S z

z z

z z

z z

z z

z z

z z

z z

z z

z z

S z S z

−

−

−

−

−

−

−

−

−

+ −

= +

− −

+ − −

= − −

− −

= − −

− −

= − −

− −

= ⋅

−

−

≡

1 1

1

1 1

1 1

1 1

1 1

1

( ) 0.5 (1 )

1.10(1.64 )

( ) (1 )(1 )

0.45 (1 )(1.64 )

0.70 0.70

(1 ) (1 1.64 )

( ) 0.70

( ) 1

(1 ) 0.70

( )

1.10(1.64 ) (1 ) 0.39

1 0.61

SZ Z

SZ Z

Wiener

S z z

z

S z z z

z z

z Z

S z

S z z

H z z

z z

z

− −

−

− −

− −

+

− −

− −

−

= ⋅ −

−

− −

= − −

= −

− −

⎡ ⎤

⎢ ⎥ = −

⎣ ⎦

= − ⋅

− −

= −

( ) ^0.39

Wiener 0.61

H z

= z

− (T2.1-11)

Comparing Eqs. (T2.1-6) and (T2.1-11), we see that H_SSKF( )z =H_Wiener( )z . (d) Find the steady-state optimal smooth solution for this problem and compare

it with the result of (b).

(For (b), (c), and (d), let q= =r 1.) (Solution)

The forward filter is the same as (b), viz,

1 2

( 16

4 2

P q q r

K P

P r

∞

∞ ∞

∞

= + +

= +

) q

(T2.1-12)

The backward filter is given by

1 1 1

1 1 1

1 0.5 2

T T T T

S A I S G G S G Q G S A H R H

S S S

q r

− − −

∞ ∞ ∞ ∞

−

∞ ∞ ∞

⎡ ⎡ ⎤ ⎤

= ⎢⎣ − ⎣ + ⎦ ⎥⎦ +

⎡ ⎛ ⎞ ⎤

= −⎢ ⎜ + ⎟ ⎥ +

⎢ ⎝ ⎠ ⎥

⎣ ⎦

(T2.1-13)

Solve Eq.(T2.1-13) for S_∞ to obtain

1 16

1 1 ( 0

S 4 S

r q

∞ ∞

⎡ ⎤

= ⎢ + + ⎥ >

⎣ ⎦ )

The steady-state optimal smoother is obtained by

[ ]

[ ]

1 S

S S

K P S I P S

P I K P

−

∞ ∞ ∞ ∞

∞

= +

= −

For a simple comparison, let q=r=1. Then,

( )

( )

[ ]¹

1 1 17 1.28 4

1 1 17 1.28 4

1.28 1.28 1 1.28 1.28 0.62

(1 0.62) 1.28 0.49

S

S

P S K P

∞

∞

−

= + =

= + =

= × × + ×

=

= − × =

We can see that the steady-state error variance is reduced to 0.49 from 1.28.

2. Consider the following state and measurement equations

k k

k x w

x ⎥ +

⎦

⎢ ⎤

⎣

=⎡

+ 0 1

0 4

1 , ⎟⎟⎠,

⎞

⎜⎜⎝

⎛ ⎥

⎦

⎢ ⎤

⎣

⎡ 5 . 0 0

0 , 0 0

k ~

w x0 ~

(

0,P0

)

k k

k Hx v

z = + , v_k ~

( )

0,1 .

(a) Determine if the steady-state Kalman gain K is asymptotically stable when

[

^{0 3}

]

H = .

(Solution) (10 points)

4 0 0 0

, ,

0 0.5 0 1

0 0 0 0

0 0.707

0 0.5

A G I Q Q T

Q

⎡ ⎤

⎡ ⎤ Q

⎢ ⎥ ⎢ ⎥

= ⎢ ⎥ = =⎢ ⎥ =

⎢ ⎥

⎢ ⎥ ⎣ ⎦

⎣ ⎦

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥

= ⎢⎢⎣ ⎥⎥⎦ =⎢⎢⎣ ⎥⎥⎦

Reachability test for

(

^{A G Q,}

)

^.

0 0 0 0 0 0.707 0 0.707 1.

ρ⎡ ⎤

⎢ ⎥ =

⎢ ⎥

⎢ ⎥

⎣ ⎦

"

"

Therefore,

(

^{A G Q,}

)

is non-reachable.

Now, test detectability for

(

^{A H,}

)

. When , .

[0 3]

H =

1 1

2 2

4 0 4 3

0 3

0 1 0 1 3

l l

A LH l l

⎡ ⎤ ⎡

⎡ ⎤ −

⎡ ⎤ ⎤

⎢ ⎥ ⎢

⎢ ⎥

− = − ⎢ ⎥ = ⎥

⎢ ⎥ ⎢

⎢ ⎥ ⎣ ⎦ − ⎥

⎢ ⎥ ⎢

⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎥⎦

(

^{A H,}

)

is non-detectable since we can not move the eigenvalue 4.

According to Theorem 5-1 and 5-2 in the note, the steady-state error with the Kalman gain K is asymptotically unstable. In addition, it is not guaranteed that for every choice of there is a bounded limiting solution P.

P0

(b) What if ^{H =}

[

^{3 3}

]

^?

(Solution) (10 points) When H =[3 3],

1 1

2 2

4 0 4 3 3

3 3

0 1 3 1 3

l l

A LH l l

1 2

l . l

⎡ ⎤ ⎡

⎡ ⎤ − −

⎡ ⎤ ⎤

⎢ ⎥ ⎢

⎢ ⎥

− = − ⎢ ⎥ = ⎥

⎢ ⎥ ⎢

⎢ ⎥ ⎣ ⎦ − − ⎥

⎢ ⎥ ⎢

⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎥⎦

)

Two eigenvalues of A may be arbitrarily located by properly choosing and . Therefore,

(

is detectable. The steady-state error with the Kalman gain K is asymptotically unstable. However, for every choice of there is a bounded limiting solution P.

l1

P0

l2 A H,

3. The equation of motion of a pendulum hanging from a ceiling is given by the differential equation

2 2

( ) sin ( ) ( ), d t

t w t dt

θ + θ =

where is the angular position of the pendulum at time t. Discrete measurements of are given by , where the sampling

interval . Suppose that , , and

. Give the equations for the EKF that provides an estimate of

, , based on the measurements for . ( )t

θ

T (0, N

=1,2

( )t

θ z n( )=θ( )n +v n( ) ( ) ~ (0, 0.02) w t N

( )

z i =

=0.5 0.02)

,"

( ) ~ (0,1) v n N

1,2, , i " n (0) ~

θ ( )i

θ i

(Solution) (Kamen’s Problem 8.8)

4. Suppose that RV x is uniformly distributed on [-1, 1], and y=e^2x. (a) What is the mean of y, y?

(Solution) (5 points)

{ } { }

¹₁^{1 2 1}

(

^{2 2}

)

^1.813

2 4

x x

y E y E e e dx e e⁻

= = =

∫

− = − =

(b) What is the first-order approximation to y?

(Solution) (5 points) 1^st order approximation

( ) { } 1.

x x

y h x h E x

x ₌

+∂ =

∂

(c) What is the second-order approximation to y?

(Solution) (5 points) 2nd order approximation

{ ( )} ( )

{ }

{ }

2

2

2 2

1 2!

1 1

=1 1 4 1.666.

2 2

x x

x x x

y E h x h x E D h D h

h E x P

x ₌

= + +

+ ∂ = + ⋅ =

∂

5

3 ≈ (d) What is the unscented approximation to y?

(Solution) (5 points) Since

2 1

0, ^x 3

x = σ = and

(1) 1 (2) 1

,

3 3

x = x = − , we obtain

2 2

3 3

1 1.744.

u 2

y e e

⎛ − ⎞

= ⎜⎜ + ⎟⎟=

⎝ ⎠ .

(e) What is the variance of y? What is the unscented approximation to the variance of y?

(Solution) (10 points)

{ }

^{2 2}

{ }

⁴

^{( )}

²

var( )y =E y −y =E e ^x − 1.813 =6.8225 3.2885− =3.534..

( )

^{2 2 2 2}

2 2

( ) 3 3

1

1 1

var( ) 1.744 1.744 2.042.

2 2

i

u u

i

y h x y e e

−

=

⎡⎛ ⎞ ⎛ ⎞ ⎤

⎢ ⎥

⎡ ⎤

=

∑

⎣ − ⎦ = ⎢⎣⎜⎜⎝ − ⎟⎟⎠ +⎜⎜⎝ − ⎟⎟⎠ ⎥⎦≈