系统辨识与自适应控制作业_1

)
So from the from the data set we may easily get six numberical values of (a12 b21 − a22 b11 ) , b11 , (a21 b11 − a11 b21 ) , b21 , (a11 a22 − a12 a21 ) , (−a11 − a22 )
in the best case. If we want to get all the values in θT = [a11
a12 a21 a22 b11 b21 ], we must have (a11 a22 − a12 a21 ) ̸= 0 a12 a11 b11 (a12 b21 − a22 b11 ) ̸= 0 i.e. ̸= ̸= a22 a21 b21 (a21 b11 − a11 b21 ) ̸= 0 b11 a12 b21 − a22 b11 ̸= a21 b11 − a11 b21 b21
you cannot complete the calculations , indicate how you would approach the problem(reference:Godfrey and DiStefano,1985). Solution: (a) Proof: let θT = [a11 a12 a21 a22 b11 b21 k1 k2 k3 ] and a11 a12 0 A(θ) = a21 a22 0 0 1 0 b11 k1 B (θ) = b21 K (θ) = k2 0 k3 [ ] C (θ) = 0 0 1 then the transfer function can be verify as follows: y (t) = G(q, θ)u(t) + H (q, θ)e(t) = C (θ)[qI − A(θ)]−1 B (θ)u(t) + {C (θ)[qI − A(θ)]−1 K (θ) + I }e(t) −1 b a a 0 11 11 12 [ ] b21 u(t)+ = 0 0 1 qI − a21 a22 0 0 1 0 0 −1 a11 a12 0 k1 [ ] k2 e(t) + e(t) 0 0 1 qI − a21 a22 0 0 1 0 k3 b21 q + (a21 b11 − a11 b21 ) = 3 u(t)+ q + (−a11 − a22 ) q 2 + (a11 a22 − a12 a21 ) q ( q 3 + (k3 − a22 − a11 ) q 2 + (k2 + a11 a22 − a12 a21 − a11 k3 − a22 k3 ) q + q 3 + (−a11 − a22 ) q 2 + (a11 a22 − a12 a21 ) q ) (a21 k1 − a11 k2 + a11 a22 k3 − a12 a21 k3 ) e(t) 3 q + (−a11 − a22 ) q 2 + (a11 a22 − a12 a21 ) q From the above equation we can see that if we just want to know the answer of this question, we needn’t consider the form of H (q, θ), so we just define p = q −1 , and collect the G(q, θ) in p. (a21 b11 − a11 b21 ) p3 + b21 p2 (a11 a22 − a12 a21 ) p2 + (−a11 − a22 ) p + 1 First,the den of G(θ) lacks one item; and second, if we tranfer it into equations like (4.14) in the book, we can get na = 2, nb = 2, nk = 3.So at most, we can only get 4 of 6 parameters(not including the parameters of e(t)). So we can never get all of the coefficience in θT = [a11 a12 a21 a22 b11 b21 k1 k2 k3 ] from the above equation, in the other words,the six parameters aij ,bij are not identifiable. (b) Proof: AS Shown in Question (a),we just change [ ] C (θ ) = 0 0 1 2
. So we can get ( ) (a11 b21 − a21 b11 ) · a12 b21 2 − a21 b11 2 + a11 b11 b21 − a22 b11 b21 ̸= 0 And then the model’s Gram matrix Wc is . . 2 Wc = [ B . .AB . .A B ] b11 a11 b11 + a12 b21 (a11 2 + a12 a21 ) b11 + b21 (a11 a12 + a12 a22 ) = b21 a21 b11 + a22 b21 (a11 a21 + a21 a22 ) b11 + b21 (a22 2 + a12 a21 ) 0 b21 a21 b11 + a22 b21 If the model is controllable, the det of the the model’s Gram matrix Wc mustn’t be zero. ) ( a21 2 b11 3 + (a21 a22 b21 − 2 a11 a21 b21 ) b11 2 + a11 2 b21 2 − 1 a22 a11 b21 2 − 1 a12 a21 b21 2 b11 + a11 a12 b21 3 ) ( (a11 b21 − a21 b11 ) · a12 b21 2 − a21 b11 2 + a11 b11 b21 − a22 b11 b21 Then we can see that only if the model is controllable, all six parameters may be globally identifiable at values. 3
ProbΒιβλιοθήκη Baiduem 2
4E.5 A state-space model of ship-steering dynamics can be given as follows: v (t) a11 a12 0 v (t) b11 d r(t) = a21 a22 0 r(t) + b21 u(t) dt h(t) 0 1 0 0(t) h(t) where u(t) is the rudder angle,v (t) the sway velocity ,r(t) the tuning rate, and h(t) the heading angle. (a) Suppose only u(t) and y (t) = h(t) are measured. Show that the six parameters aij ,bij are not identifiable. [ ] v (t) (b) Try also to show that if u(t) and y (t) = are measured then all six pah(t) rameters are globally identifiable at values such that the model is controllable. If 1
)
Then we just change G(q, θ) to G(p, θ) ( (a b −a
12 21
2 22 b11 ) p +b11 p 2 (a11 a22 −a12 a21 ) p +(−a11 −a22 ) p+1 (a21 b11 −a11 b21 ) p3 +b21 p2 (a11 a22 −a12 a21 ) p2 +(−a11 −a22 ) p+1
to C (θ ) =
[
1 0 0 0
0 1
]
Also we can get the Matrix of Transfer Functions G(θ) ( b q +(a b −a b
11 12 21
G(θ) = and ( H (θ ) =
q 2 +(−a11 −a22 ) q +(a11 a22 −a12 a21 ) b21 q +(a21 b11 −a11 b21 ) q 3 +(−a11 −a22 ) q 2 +(a11 a22 −a12 a21 ) q
Problem 3
Consider a weighted cost function V (θ, t) = 1 ET W E = 2 the least-square estimations is ˆ = (ΦT W Φ)−1 ΦT W Y θ Solution: From the description of the question, we can obviously know that W T = W , because w1 0 · · · 0 0 w2 · · · 0 W = . . . . . .. . . . . . 0 0 · · · wn so 2V (θ, t) = E T W E = (Y − Φθ)T W (Y − Φθ) = Y T W Y − Y T W Φθ − θT ΦT W Y + θT ΦT W Φθ let A = ΦT W Φ, AB = ΦT W Y, so B = (ΦT W Φ)−1 ΦT W Y 2V (θ, t) = Y W Y − Y W Φ(Φ W Φ) Φ W Y
System Identification & Adptive Control Assignment 1
Pengju Gu StudentID:13031168
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June 15, 2014
Problem 1
4E.1 Consider the ARX model structure y (t) + a1 y (t − 1) + · · · + ana y (t − na ) = b1 u(t − 1) + · · · + bnb u(t − nb ) + e(t) where b1 is known to be 0.5.Write the corresponding predictor in the linear regression form (4.13). Solution: Since b1 = 0.5, hence we get y (t) = −a1 y (t − 1) − · · · − ana y (t − na ) + 0.5u(t − 1) + b2 u · · · + bnb u(t − nb ) + e(t) if we define φ(t) = [−y (t − 1) · · · − y (t − na ) u(t − 2) θ = [a 1 · · · a na b 2 · · · b nb ] then we get the simple result(predictor): y ˆ(t|θ) = φT (t)θ + 0.5u(t − 1) ··· − u(t − nb )]
22 11 )
)
q 2 +(k1 −a22 −a11 ) q +(a11 a22 −a12 a21 +a12 k2 −a22 k1 ) q 2 +(−a11 −a22 ) q +(a11 a22 −a12 a21 ) q 3 +(k3 −a22 −a11 ) q 2 +(k2 +a11 a22 −a12 a21 −a11 k3 −a22 k3 ) q +(a21 k1 −a11 k2 +a11 a22 k3 −a12 a21 k3 ) q 3 +(−a11 −a22 ) q 2 +(a11 a22 −a12 a21 ) q