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Least squares system identification class II: recursive least squares

2022-06-26 16:16:00 RuiH. AI

Preface

This paper records the principle of recursive least squares in the course of system identification .

Least squares estimate

The following system identification model :
{ y ( 1 ) = φ T ( 1 ) θ + v ( 1 ) y ( 2 ) = φ T ( 2 ) θ + v ( 2 ) . . . y ( t ) = φ T ( t ) θ + v ( t ) set up : Y t = [ y ( 1 ) y ( 2 ) . . . y ( t ) ] T H t = [ φ ( 1 ) φ ( 2 ) . . . φ ( t ) ] T V t = [ v ( 1 ) v ( 2 ) . . . v ( n ) ] T Yes : Y t = H t θ + V t \left \{ \begin{aligned} y(1)&=\varphi^T(1)\theta+v(1) \\ y(2)&=\varphi^T(2)\theta+v(2) \\ &... \\ y(t)&=\varphi^T(t)\theta+v(t) \end{aligned} \right . \\ \quad \\ \begin{aligned} set up : Y_t&=[y(1)\quad y(2)\quad ... \quad y(t)]^T \\ H_t&=[\varphi(1) \quad \varphi(2) \quad ... \quad \varphi(t)]^T \\ V_t&=[v(1) \quad v(2) \quad ... \quad v(n)]^T \end{aligned} \\ \quad \\ Yes : Y_t=H_t\theta +V_t y(1)y(2)y(t)=φT(1)θ+v(1)=φT(2)θ+v(2)...=φT(t)θ+v(t) set up YtHtVt=[y(1)y(2)...y(t)]T=[φ(1)φ(2)...φ(t)]T=[v(1)v(2)...v(n)]T Yes Yt=Htθ+Vt

The least square estimation of parameters can be expressed as :
θ ^ L S = ( H t T H t ) − 1 H t T Y t \hat\theta_{LS}=(H_t^TH_t)^{-1}H_t^TY_t θ^LS=(HtTHt)1HtTYt

The above method updates the input and output data at each new time , Must be re estimated , The computational load is heavy .

Recursive least squares estimation

principle

Recursive least square estimation uses incremental method , The following is derived :

P − 1 ( t ) = H t T H t = ∑ i = 1 t φ ( i ) φ T ( i ) = ∑ i = 1 t − 1 φ ( i ) φ T ( i ) + φ ( t ) φ T ( t ) = P − 1 ( t − 1 ) + φ ( t ) φ T ( t ) θ ^ ( t ) = ( H t T H t ) − 1 H t T Y t = P ( t ) H t T Y t = P ( t ) [ H t − 1 T φ ( t ) ] [ Y t − 1 T y ( t ) ] T = P ( t ) H t − 1 T Y t − 1 T + P ( t ) φ ( t ) y ( t ) = P ( t ) P − 1 ( t − 1 ) P ( t − 1 ) H t − 1 T Y t − 1 T + P ( t ) φ ( t ) y ( t ) = P ( t ) P − 1 ( t − 1 ) θ ^ ( t − 1 ) + P ( t ) φ ( t ) y ( t ) = P ( t ) [ P − 1 ( t ) − φ ( t ) φ T ( t ) ] θ ^ ( t − 1 ) + P ( t ) φ ( t ) y ( t ) = θ ^ ( t − 1 ) − P ( t ) φ ( t ) φ T ( t ) θ ^ ( t − 1 ) + P ( t ) φ ( t ) y ( t ) = θ ^ ( t − 1 ) + P ( t ) φ ( t ) [ y ( t ) − φ T ( t ) θ ^ ( t − 1 ) ] lead The reason is : Moment front A , B , C , if ( I + C A − 1 B ) And A all can The inverse , be : ( A + B C ) − 1 = A − 1 − A − 1 B ( I + C A − 1 B ) − 1 C A − 1 P − 1 ( t ) = P − 1 ( t − 1 ) + φ ( t ) φ T ( t ) hold after Noodles Of type Son take The inverse , root According to the lead The reason is Yes : P ( t ) = P ( t − 1 ) − P ( t − 1 ) φ ( t ) φ T ( t ) P ( t − 1 ) 1 + φ T ( t ) P ( t − 1 ) φ ( t ) P ( t ) φ ( t ) = P ( t − 1 ) φ ( t ) 1 + φ T ( t ) P ( t − 1 ) φ ( t ) L ( t ) = P ( t ) φ ( t ) P^{-1}(t)=H_t^TH_t=\sum_{i=1}^t \varphi(i)\varphi^T(i)=\sum_{i=1}^{t-1} \varphi(i)\varphi^T(i) + \varphi(t)\varphi^T(t) = P^{-1}(t-1)+ \varphi(t)\varphi^T(t) \\ \quad \\ \begin{aligned} \hat\theta(t)&=(H_t^TH_t)^{-1}H_t^TY_t \\ &=P(t)H_t^TY_t \\ &=P(t)\begin{bmatrix} H_{t-1}^T & \varphi(t) \end{bmatrix} \begin{bmatrix} Y_{t-1}^T & y(t) \end{bmatrix}^T \\ &=P(t)H_{t-1}^TY_{t-1}^T + P(t)\varphi(t)y(t) \\ &=P(t)P^{-1}(t-1)P(t-1)H_{t-1}^TY_{t-1}^T + P(t)\varphi(t)y(t) \\ &=P(t)P^{-1}(t-1)\hat \theta(t-1)+P(t)\varphi(t)y(t) \\ &=P(t)[P^{-1}(t)- \varphi(t)\varphi^T(t) ]\hat\theta(t-1)+P(t)\varphi(t)y(t) \\ &=\hat\theta(t-1)-P(t)\varphi(t)\varphi^T(t)\hat\theta(t-1)+P(t)\varphi(t)y(t) \\ &=\hat\theta(t-1)+P(t)\varphi(t)[y(t)-\varphi^T(t)\hat\theta(t-1)] \end{aligned} \\ \quad \\ lemma : matrix A,B,C, if (I+CA^{-1}B) And A It's reversible , be :\\ (A+BC)^{-1}=A^{-1}-A^{-1}B(I+CA^{-1}B)^{-1}CA^{-1} \\ \quad \\ P^{-1}(t)= P^{-1}(t-1)+ \varphi(t)\varphi^T(t) \\ \quad \\ Take the inverse of the following formula , According to lemma, there are :\\ P(t)=P(t-1)-\frac{P(t-1)\varphi(t)\varphi^T(t)P(t-1)}{1+\varphi^T(t)P(t-1)\varphi(t)} \\ \quad \\ P(t)\varphi(t)= \frac{P(t-1)\varphi(t)}{1+\varphi^T(t)P(t-1)\varphi(t)} \\ \quad \\ L(t)=P(t)\varphi(t) P1(t)=HtTHt=i=1tφ(i)φT(i)=i=1t1φ(i)φT(i)+φ(t)φT(t)=P1(t1)+φ(t)φT(t)θ^(t)=(HtTHt)1HtTYt=P(t)HtTYt=P(t)[Ht1Tφ(t)][Yt1Ty(t)]T=P(t)Ht1TYt1T+P(t)φ(t)y(t)=P(t)P1(t1)P(t1)Ht1TYt1T+P(t)φ(t)y(t)=P(t)P1(t1)θ^(t1)+P(t)φ(t)y(t)=P(t)[P1(t)φ(t)φT(t)]θ^(t1)+P(t)φ(t)y(t)=θ^(t1)P(t)φ(t)φT(t)θ^(t1)+P(t)φ(t)y(t)=θ^(t1)+P(t)φ(t)[y(t)φT(t)θ^(t1)] lead The reason is Moment front A,B,C, if (I+CA1B) And A all can The inverse , be (A+BC)1=A1A1B(I+CA1B)1CA1P1(t)=P1(t1)+φ(t)φT(t) hold after Noodles Of type Son take The inverse , root According to the lead The reason is Yes P(t)=P(t1)1+φT(t)P(t1)φ(t)P(t1)φ(t)φT(t)P(t1)P(t)φ(t)=1+φT(t)P(t1)φ(t)P(t1)φ(t)L(t)=P(t)φ(t)

therefore , The recursive least squares of the system parameters can be expressed as :
θ ^ ( t ) = θ ^ ( t − 1 ) + L ( t ) [ y ( t ) − φ T ( t ) θ ^ ( t − 1 ) ] ] L ( t ) = P ( t − 1 ) φ ( t ) 1 + φ T ( t ) P ( t − 1 ) φ ( t ) P ( t ) = [ I − L ( t ) φ T ( t ) ] P ( t − 1 ) , P ( 0 ) = p 0 I , p 0 = inf ⁡ φ ( t ) = [ . . . ] \hat\theta(t)=\hat\theta(t-1)+L(t)[y(t)-\varphi^T(t)\hat\theta(t-1)]] \\ \quad \\ L(t)=\frac{P(t-1)\varphi(t)}{1+\varphi^T(t)P(t-1)\varphi(t)} \\ \quad \\ P(t)=[I-L(t)\varphi^T(t)]P(t-1), P(0)=p_0I, p_0=\inf \\ \quad \\ \varphi(t)=[...] θ^(t)=θ^(t1)+L(t)[y(t)φT(t)θ^(t1)]]L(t)=1+φT(t)P(t1)φ(t)P(t1)φ(t)P(t)=[IL(t)φT(t)]P(t1),P(0)=p0I,p0=infφ(t)=[...]

New interest , residual

New interest e ( t ) = y ( t ) − φ T ( t ) θ ^ ( t − 1 ) e(t)=y(t)-\varphi^T(t)\hat\theta(t-1) e(t)=y(t)φT(t)θ^(t1)
residual ε ( t ) = y ( t ) − φ T ( t ) θ ^ ( t ) \varepsilon(t)=y(t)-\varphi^T(t)\hat\theta(t) ε(t)=y(t)φT(t)θ^(t)

Data saturation

The phenomenon

When the input-output time series is long enough , The new input and output do not contribute to the result of recursive least squares estimation , It is called data saturation .

reason

P ( t ) = [ P − 1 ( t − 1 ) + φ ( t ) φ T ( t ) ] − 1 = > P ( t ) ≤ P ( t − 1 ) ( α I + 1 / p 0 ) t ≤ P − 1 ( t ) = P − 1 ( 0 ) + ∑ i = 1 t φ ( i ) φ T ( i ) ≤ ( β I + 1 / p 0 ) t lim ⁡ t − > inf ⁡ P − 1 ( t ) = 0 P(t)=[P^{-1}(t-1)+\varphi(t)\varphi^T(t)]^{-1} => P(t)\le P(t-1) \\ (\alpha I+1/p_0)t\le P^{-1}(t) = P^{-1}(0)+ \sum_{i=1}^t\varphi(i)\varphi^T(i) \le (\beta I+1/p_0)t \\ \lim_{t->\inf}P^{-1}(t) = 0 P(t)=[P1(t1)+φ(t)φT(t)]1=>P(t)P(t1)(αI+1/p0)tP1(t)=P1(0)+i=1tφ(i)φT(i)(βI+1/p0)tt>inflimP1(t)=0

resolvent : Recursive least squares with forgetting factor

θ ^ ( t ) = θ ^ ( t − 1 ) + L ( t ) [ y ( t ) − φ T ( t ) θ ^ ( t − 1 ) ] ] L ( t ) = P ( t − 1 ) φ ( t ) λ + φ T ( t ) P ( t − 1 ) φ ( t ) P ( t ) = 1 λ [ I − L ( t ) φ T ( t ) ] P ( t − 1 ) , P ( 0 ) = p 0 I , p 0 = inf ⁡ φ ( t ) = [ . . . ] \hat\theta(t)=\hat\theta(t-1)+L(t)[y(t)-\varphi^T(t)\hat\theta(t-1)]] \\ \quad \\ L(t)=\frac{P(t-1)\varphi(t)}{\lambda+\varphi^T(t)P(t-1)\varphi(t)} \\ \quad \\ P(t)=\frac{1}{\lambda}[I-L(t)\varphi^T(t)]P(t-1), P(0)=p_0I, p_0=\inf \\ \quad \\ \varphi(t)=[...] θ^(t)=θ^(t1)+L(t)[y(t)φT(t)θ^(t1)]]L(t)=λ+φT(t)P(t1)φ(t)P(t1)φ(t)P(t)=λ1[IL(t)φT(t)]P(t1),P(0)=p0I,p0=infφ(t)=[...]

Postscript

This paper records the derivation of recursive least squares , Data saturation and its solution . The next chapter will record the least squares , φ T ( t ) \varphi^T(t) φT(t) How to get .

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本文为[RuiH. AI]所创,转载请带上原文链接,感谢
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