最小二乘法

image-20210418193326202

由

G(z)=y(k)u(k)=b1z−1+b2z−2+⋯+bnz−n1+a1z−1+a2z−2+⋯+anz−nG(z)=\frac{y(k)}{u(k)}=\frac{b_1z^{-1}+b_2z^{-2}+\dots+b_nz^{-n}}{1+a_1z^{-1}+a_2z^{-2}+\dots+a_nz^{-n}}

得

y(k)=−∑i=1naiy(k−i)+∑i=1nbiu(k−i)y(k)=-\sum_{i=1}^{n}a_iy(k-i)+\sum_{i=1}^{n}b_iu(k-i)

加上噪声得

z(k)=−∑i=1naiy(k−i)+∑i=1nbiu(k−i)+v(k)z(k)=-\sum_{i=1}^{n}a_i y(k-i)+\sum_{i=1}^{n}b_iu(k-i)+v(k)

令

θ=[a1 a2 … an b1 b2 … bn]T\theta=[a_1 \ a_2 \ \dots \ a_n \ b_1 \ b_2 \ \dots \ b_n]^T

h(k)=[−y(k−1) −y(k−2) … −y(k−n) u(k−1) u(k−2) …u(k−n)]h(k)=[-y(k-1) \ -y(k-2) \ \dots \ -y(k-n) \ u(k-1) \ u(k-2) \ \dots u(k-n)]

可得

z(k)=h(k)θ+v(k)z(k)=h(k)\theta+v(k)

令K=1,2,3,mK=1,2,3,m得

Zm=Hmθ+VmZ_m=H_m\theta+V_m

令实际值和估计值只差平方和最小

J(θ^)=(Zm−Hmθ^)T(Zm−Hmθ^)J(\hat{\theta})=(Z_m-H_m\hat{\theta})^T(Z_m-H_m\hat{\theta})

∂J∂θ∣θ=θ^=−2HmT(Zm−Hmθ^)=0\frac{\partial{J}}{\partial{\theta}}|_{\theta=\hat{\theta}}=-2H_m^T(Z_m-H_m\hat{\theta})=0

HmTHmθ^=HmTZmH_m^TH_m\hat{\theta}=H_m^TZ_m

所以

θ^=(HmTHm)−1HmTZm\hat{\theta}=(H_m^TH_m)^{-1}H_m^TZ_m