1
$\begingroup$

I am studying control theory and I am having difficulties understanding a concept. Consider the following relationships, which represent the input to state behavior and the input output behavior respectively:

enter image description here

where this is the explicit representation of the evolution of the system. I have in my notes that the passage from the explicit representation to the implicit representation is given by:

enter image description here

and it is written that it is easy to verify that:

enter image description here

I really cannot understand what it is doing here, and I feel that it is a crucial passage. I know that the explicit form of a system is defined by the first equations I have written, so from a representation with $w_0, w_1, \gamma_0, \gamma_1$, which in this case my professor in the notes called kernels, and the implicit representation if represented by differential equations.

But I cannot understand what is done here. Can somebody please help me?

$\endgroup$
3
$\begingroup$

Don't know about you, to me this looks like a notational mess. I'll try to give a general formulation then turn to the time invariant case that you've got here. Consider the set of dynamical systems $$\begin{align}\dot{x}(t)&=A(t)x(t)+B(t)u(t) \tag{1}\\y(t)&=C(t)x(t)+D(t)u(t) \tag{2}\end{align}$$ where $\mathbb{R} \ni t \mapsto x(t) \in \mathbb{R}^n$ is the state trajectory, $\mathbb{R} \ni t \mapsto u(t) \in \mathbb{R}^m$ is the control action and $\mathbb{R} \ni t \mapsto y(t) \in \mathbb{R}^p$ is the output. Well more generally you can consider the linear spaces (Normed) $(X,\mathbb{R}),(U,\mathbb{R}),(Y,\mathbb{R})$ and consider state, control and output as represrentations of these space with respect to bases $\{e_i\}_{i=1}^n,\{f_i\}_{i=1}^m$ and $\{g_i\}_{i=1}^p$. And the system matrices are maps defined as $$\mathbb{R} \ni t \mapsto A(t) \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n) \\\mathbb{R} \ni t \mapsto B(t) \in \mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)\\\mathbb{R} \ni t \mapsto C(t) \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^p)$$

Then the unique solution of (1) and (2) are two functions such that $$\begin{align}x(t)&:=s(t,t_0,x_0,u) \tag{3} \\y(t)&:=\rho(t,t_0,x_0,u) \tag{4} \end{align}$$

If you consider $D_x$ as the union of discontinuity sets of $A(\cdot),B(\cdot)$ and $u(\cdot)$ and $D_y$ as union of discontinuity sets of $C(\cdot),D(\cdot)$ and $u(\cdot)$ then for all $(t_0,x_0) \in \mathbb{R} \times \mathbb{R}^n$ and $u \in \mathcal{PC}(\mathbb{R},\mathbb{R}^m)$ where $\mathcal{PC}$ signifies piecewise-continuous function from $\mathbb{R}$ to $\mathbb{R}^m$

$$\begin{align} \star ~~x(\cdot)=s(\cdot,t_0,x_0,u):\mathbb{R} \to \mathbb{R}^n~~ \text{is continuous and differentiable}~~ \forall t \in \mathbb{R}\setminus D_x \\ \star ~~y(\cdot)=\rho(\cdot,t_0,x_0,u):\mathbb{R} \to \mathbb{R}^p~~ \text{is continuous and differentiable}~~ \forall t \in \mathbb{R}\setminus D_y\end{align}$$ i.e i'm just ignoring all the sets with measure zero. Then the solution can be written as $$ \begin{align}x(t):=s(t,t_0,x_0,u)=\Phi(t,t_0)x_0+\int_{t_0}^{t}\Phi(t,\tau)B(\tau) u(\tau)\mathrm{d}\tau\end{align} \tag{5} $$ Where the mapping $\mathbb{R}_{\ge 0}\times \mathbb{R}_{\ge 0} \ni (t,t_0) \mapsto \Phi(t,t_0) \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$ is called a state-transition-matrix (STM), and this formulation is known as variation of constant formulation. You can easily work out an expression of $y(t)$ like this in terms of STM. Now for a time-invariant system of ODEs this formulation reduces to a much simpler form i.e if you have $$\begin{align}\dot{x}(t)&=Ax(t)+Bu(t) \tag{6}\\y(t)&=Cx(t)+Du(t) \tag{7}\end{align}$$ Your STM reduces to $\mathbb{R}_{\ge 0}\times \mathbb{R}_{\ge 0} \ni (t,t_0) \mapsto \Phi(t,t_0) :=e^{A(t-t_0)} \in \mathbb{R}^{n \times n}$ and state, output pair can be written as $$ \begin{align}x(t):=s(t,t_0,x_0,u)=e^{A(t-t_0)}x_0+\int_{t_0}^{t}e^{A(t-\tau)}B(\tau) u(\tau)\mathrm{d}\tau \tag{8} \\ y(t):=\rho(t,t_0,x_0,u)=Ce^{A(t-t_0)}x_0+C\int_{t_0}^{t}e^{A(t-\tau)}B(\tau) u(\tau)\mathrm{d}\tau+Du(t) \tag{9}\end{align} $$ And if you define $K$ and $H$ such that $$\begin{align}K(t,\sigma)&=K(t-\sigma,0):=\left\{ \begin{aligned}e^{A(t-\sigma)}B ~~~\text{if}~t\ge \sigma\\0 ~~~\text{if} ~t<\sigma\end{aligned}\right. \end{align}$$ $$\begin{align}H(t,\sigma)&=H(t-\sigma,0):=\left\{ \begin{aligned}Ce^{A(t-\sigma)}B+D\delta_0(t-\sigma) ~~~\text{if}~t\ge \sigma\\0 ~~~\text{if} ~t<\sigma\end{aligned}\right. \end{align}$$ It's clear from here that the solution of an LTI system does not depend on the initial time $t_0 \in \mathbb{R}_{\ge 0}$, it only cares about how much time has been elapsed i.e $t-t_0$. So wlog you take $t_0=0$ and you get

$$ \begin{align}x(t):=s(t,0,x_0,u)=e^{At}x_0+\int_{0}^{t}e^{A(t-\tau)}B(\tau) u(\tau)\mathrm{d}\tau \tag{10} \\ y(t):=\rho(t,0,x_0,u)=Ce^{At}x_0+C\int_{0}^{t}e^{A(t-\tau)}B(\tau) u(\tau)\mathrm{d}\tau+Du(t) \tag{11}\end{align} $$ And it follows immediately

$$\begin{align}K(t,\sigma)&=K(t,0):=\left\{ \begin{aligned}e^{At}B ~~~\text{if}~t\ge \sigma\\0 ~~~\text{if} ~t<\sigma\end{aligned}\right. \end{align}$$ $$\begin{align}H(t,\sigma)&=H(t,0):=\left\{ \begin{aligned}Ce^{At}B+D\delta_0(t) ~~~\text{if}~t\ge \sigma\\ 0 ~~~\text{if} ~t<\sigma\end{aligned}\right. \end{align}$$

All the calculations in your notes follows from this. And, yes you can call them kernels, it's more like a functional operator, as you can notice that with the kernels $K$ and $H$ we may write $$\begin{align} x(t)=e^{At}x_0+\left(K *u \right)(t)\end{align} \\ y(t)=Ce^{At}x_0+ \left(H*u \right)(t)$$ where $*$ : is the continuous convolution operation.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.