干扰观测器是控制中非常常见的一种方法,本节推导的是一种简单的非线性干扰观测器NDOB
朴素NDOB观测器
假设非线性系统方程为
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\begin{align} \dot x &= f(x) + g_1(x)u + g_2(x)d \nonumber\\ y &= h(x) \nonumber \end{align}
x˙y=f(x)+g1(x)u+g2(x)d=h(x)
其中
d
d
d 为扰动,
u
u
u 为控制量,我们期望对扰动进行估计,不妨假设扰动为
d
^
\hat d
d^ ,那么扰动的估计误差为
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~
=
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−
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\tilde d = d - \hat d
d~=d−d^
对扰动误差
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~
\tilde d
d~ 求导,可得(这里注意,我们对真实扰动没有一点先验知识,这里最不失一般性的假设就是,她的导数为0):
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~
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\dot {\tilde d} = \dot d - \dot {\hat d} = - \dot {\hat d}
d~˙=d˙−d^˙=−d^˙
为了让
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~
\tilde d
d~ 尽快收敛,我们可以命令其等于如下形式
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\dot {\tilde d} = - \dot {\hat d} =-l(x)g_2(x){\tilde d}
d~˙=−d^˙=−l(x)g2(x)d~
上面这个等式是我们人为设计的一种
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\tilde d
d~ 的趋近律,也就是我们希望扰动的估计误差如何趋近于0,我们可以任意选择
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l(x)
l(x) ,只要保证上面的误差估计趋近于0即可。
基于上述的趋近律,我们带入到非线性系统方程可以得到:
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\begin{split} \dot {\hat d} &= l(x)g_2(x){\tilde d} \\ &= l(x) g_2(x) (d - {\hat d}) \\ &= l(x) (g_2(x)d - g_2(x){\hat d}) \\ &= l(x) (\dot x - f(x) - g_1(x)u - g_2(x){\hat d}) \\ &= l(x) (\dot x - f(x) - g_1(x)u) - l(x)g_2(x){\hat d} \end{split}
d^˙=l(x)g2(x)d~=l(x)g2(x)(d−d^)=l(x)(g2(x)d−g2(x)d^)=l(x)(x˙−f(x)−g1(x)u−g2(x)d^)=l(x)(x˙−f(x)−g1(x)u)−l(x)g2(x)d^
上面我们就推导了最为朴素的非线性扰动观测器,但是我们发现,这个扰动观测器需要用到
x
˙
\dot x
x˙ 这一项,很明显,大多数系统都不可能得到这项的观测,所以这个观测器需要做一些改进。
改进NDOB观测器
最朴素的把
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\dot x
x˙ 消掉的思路就是,在观测器的左右两遍同时减去
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l(x)\dot x
l(x)x˙:
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\dot {\hat d} - l(x)\dot x = l(x) (- f(x) - g_1(x)u) - l(x)g_2(x){\hat d}
d^˙−l(x)x˙=l(x)(−f(x)−g1(x)u)−l(x)g2(x)d^
这样公式的右侧就没有
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\dot x
x˙ 这一项了,我们对等式的左右两边进行积分:
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\begin{align} p(x) &= \int l(x)\dot x \ \text d x\nonumber\\ \hat d - p(x) &= \int {l(x) (- f(x) - g_1(x)u) - l(x)g_2(x){\hat d}} \ \text d x\nonumber \end{align}
p(x)d^−p(x)=∫l(x)x˙ dx=∫l(x)(−f(x)−g1(x)u)−l(x)g2(x)d^ dx
设置一个中间变量
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z=\hat d - p(x)
z=d^−p(x) ,那么对
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z 进行求导可以得到:
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\begin{align} \dot z &= l(x) (- f(x) - g_1(x)u) - l(x)g_2(x){\hat d}\nonumber \\ &= l(x) (- f(x) - g_1(x)u) - l(x)g_2(x)(z + p(x))\nonumber \\ &= l(x) [- f(x) - g_1(x)u + g_2(x)p(x)] - l(x)g_2(x)z\nonumber \\ \end{align}
z˙=l(x)(−f(x)−g1(x)u)−l(x)g2(x)d^=l(x)(−f(x)−g1(x)u)−l(x)g2(x)(z+p(x))=l(x)[−f(x)−g1(x)u+g2(x)p(x)]−l(x)g2(x)z
上面是一个针对中间变量
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z 的观测器,有了
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z 以后,扰动估计量
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\hat d
d^ 也可以很容易得到:
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\hat d = z + p(x)
d^=z+p(x)
这里我们实现了扰动的观测,还遗留了一个最重要的问题,
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p(x)
p(x) 是什么,刚才我们假设了
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p(x) = \int l(x)\dot x \ \text d t
p(x)=∫l(x)x˙ dt,所以也存在了如下的关系:
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∂
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l(x) = {\frac {\partial p(x)} {\partial x}} \to {\frac {\text d p(x)} {\text d t}} = l(x)\dot x
l(x)=∂x∂p(x)→dtdp(x)=l(x)x˙
如何确定
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p(x)
p(x) 呢?有两种方式,一种是先选定
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l(x)
l(x) ,然后对
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x 积分,求
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p(x)
p(x) 。另一种方式则是,先自由选定
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p(x)
p(x) ,但是
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l(x)
l(x) 则必须满足上述关系。
李雅普诺夫证明稳定性
假设李雅普诺夫函数
V
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2
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V=\frac {1} {2} \tilde d ^ 2
V=21d~2
对其求导,之前我们就假设了扰动的导数为0,即这个扰动是一个常数,反正波动不会很大
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\dot V = \tilde d \dot {\tilde d} = \tilde d (\dot d - \dot {\hat d}) = - \tilde d \dot {\hat d}
V˙=d~d~˙=d~(d˙−d^˙)=−d~d^˙
其中
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\begin {align} \dot {\hat d} &= \dot z + \dot p(x) \nonumber\\ &= \dot z + l(x){\dot x} \nonumber\\ &= l(x) [- f(x) - g_1(x)u + g_2(x)p(x)] - l(x)g_2(x)z + l(x){\dot x} \nonumber\\ &= l(x) [- f(x) - g_1(x)u + g_2(x)p(x)] - l(x)g_2(x)z + l(x)(f(x) + g_1(x)u + g_2(x)d )\nonumber \\ &= l(x)g_2(x)p(x) - l(x)g_2(x)z + l(x)g_2(x)d \nonumber\\ &= l(x)g_2(x)(p(x) - z + d) \nonumber\\ &= l(x)g_2(x){\tilde d}\nonumber \end {align}
d^˙=z˙+p˙(x)=z˙+l(x)x˙=l(x)[−f(x)−g1(x)u+g2(x)p(x)]−l(x)g2(x)z+l(x)x˙=l(x)[−f(x)−g1(x)u+g2(x)p(x)]−l(x)g2(x)z+l(x)(f(x)+g1(x)u+g2(x)d)=l(x)g2(x)p(x)−l(x)g2(x)z+l(x)g2(x)d=l(x)g2(x)(p(x)−z+d)=l(x)g2(x)d~
所以
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\begin {align} \dot V &= - \tilde d \dot {\hat d}\nonumber \\ &= -{\tilde d} l(x)g_2(x){\tilde d}\nonumber \\ &= -l(x)g_2(x){\tilde d}^2\nonumber \\ &= -2l(x)g_2(x){\tilde d}^2\nonumber \\ \end {align}
V˙=−d~d^˙=−d~l(x)g2(x)d~=−l(x)g2(x)d~2=−2l(x)g2(x)d~2
**定理1:**若李雅普诺夫满足以下等式,则系统稳定
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\dot V \le -KV^{\lambda} \ ;K,\lambda > 0
V˙≤−KVλ ;K,λ>0
所以系统稳定
实际的控制例子
一个二阶系统如下
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\begin{align} \dot x_1 &= x_2\nonumber \\ \dot x_2 &= -k|x_2|x_2+bu+d\nonumber \\ \end{align}
x˙1x˙2=x2=−k∣x2∣x2+bu+d
其中
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x_1
x1 代表位置,
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x_2
x2 代表速度,
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u 是控制量,
d
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d 代表扰动。
{
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\begin{cases} f(x) = -k|x_2|x_2 \nonumber\\ g_1(x) = b\nonumber \\ g_2(x) = 1 \end{cases}
⎩
⎨
⎧f(x)=−k∣x2∣x2g1(x)=bg2(x)=1
设计
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l(x) = c
l(x)=c ,则有
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p(x) = cx_2
p(x)=cx2
定义中间变量为
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z=\hat d-p(x)
z=d^−p(x) ,于是扰动观测器的形式为
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\begin{align} \dot z &= -l(x)g_2(x)z-l(x)[f(x)+g_1(x)u+g_2(x)p(x)]\nonumber\\ &=-cz-c(-k|x_2|x_2 + bu + cx_2) \nonumber\\ \hat d &= z+p(x)=z+cx_2\nonumber \end{align}
z˙d^=−l(x)g2(x)z−l(x)[f(x)+g1(x)u+g2(x)p(x)]=−cz−c(−k∣x2∣x2+bu+cx2)=z+p(x)=z+cx2文章来源:https://www.toymoban.com/news/detail-404552.html
注意
干扰必须是有界的,否则定理1将不成立,如果只参考结论的话,可以只看最后一个结论文章来源地址https://www.toymoban.com/news/detail-404552.html
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