# Problem understanding passage from explicit representation to implicit representation

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

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.