Adaptive Control - Least-Squares Algorithm in Parameters Estimation
Background
I take adaptive control course this semester, and in the homework after the second lecture, we are going to derive the formular for the estimated parameter vector $\theta(t)$ at time $t$.
Some terms we need to demonstrate the problem:
Parametric model (linear): $y(t)=\theta^{*T}\phi(t)$.
$\theta^*$ is the truth value of the parameters in the system, which is what we want to reach or approach. However we only have the knowledge of $y(t)$ and $\theta(t)$. So we are going to estimate $\theta^*$ based on what are known.
$y(t)$ is the output of the system, $1\times1$ scaler, which could be measured and it is known.
$\phi(t)$ is the input or reference of the system, which is also known.
$\theta(t)$ is the estimated system parameter at time $t$, which is what we want.
Cost function $J(\theta(t))$ is what we would like to minimized, so that the error between $\theta\phi$ and $y$ is minimized.
There are two algorithms that can solve this problem. The first one is Gradient (Descent) Algorithm, another one is what we are going to say here, Least-squares algorithm. Basically, the differences between these two algorithms are the cost function they employed and the method to minimize the cost function.
Problem
So here is our setting:
Given the cost function in Least-Squares algorithm,
(The right hand side is a column vector, so when taking $\theta^T(t)$ out of the integral, it should transpose to become a column vector. And $\phi(\tau)\phi^T(\tau)=[\phi(\tau)\phi^T(\tau)]^T$, because it is a symmetric matrix.)
Revision and Updates
Since the solution above required $\int_0^t \phi(\tau)\phi^T(\tau)d\tau$ to be invertible, in some scenario where t is very small, usually the matrix is not full ranked and is not invertible. To deal with this situation, we revise the cost function as
(Notice that: $\frac{\partial x^TAx}{\partial x}=(Ax)^T+x^TA=x^TA^T+x^TA$, where $x$ and $A$ are matrices, $A$ is not a function of $x$.If A is a symmetric matrix, $A^T=A$, then the result would be $2x^TA$ or $2x^T A^T$.)
Positive definite matrix is always invertible, so this formula can suit for more situations.
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