Integrating output y of continuous-time state-space representation wrt. time

control theoryintegrationlinear algebra

I have a state-space representation:

$$ \dot x(t) = A x(t) + B u(t) $$

$$ y(t) = C x(t) + D u(t) $$

and I would like to integrate $y(t)$ to get something like:

$$ \dot z(t) = \bar A z(t) + \bar B u(t) $$
$$ \int y(t) = \bar C z(t) + \bar D u(t) $$

I have not been able to find solutions to this online, so I came up with my own approach, but I am not sure that it is correct:

  1. solving for $x(t)$:
    $$ \dot x(t) = A x(t) + B u(t) \implies x(t) = A^{-1} \dot x(t) – A^{-1} B u(t) $$
  2. integrating $x(t)$:
    $$ \int x(t) = A^{-1} x(t) – A^{-1} B \int u(t) $$
  3. integrating $y(t)$:
    $$ \int y(t) = C \int x(t) + D \int u(t) $$
  4. substituting (2) into (3):
    $$ \int y(t) = C A^{-1} x(t) + (-C A^{-1} B + D) \int u(t) $$
  5. defining:
    $$ z(t) = \begin{bmatrix} x(t) \\ \int u(t) \end{bmatrix} $$
    $$ \bar A = \begin{bmatrix} A & 0 \\ 0 & 0 \end{bmatrix} $$
    $$ \bar B = \begin{bmatrix} B \\ I \end{bmatrix} $$
    $$ \bar C = \begin{bmatrix} C A^{-1} & -C A^{-1} B + D \end{bmatrix} $$
    $$ \bar D = \begin{bmatrix} 0 \end{bmatrix} $$

clearly $ \int y(t) $ is as in (4) and also:
$$ \dot z(t) = \begin{bmatrix} \dot x(t) \\ u(t) \end{bmatrix} = \begin{bmatrix} A x(t) + B u(t) \\ u(t) \end{bmatrix} $$
right?

So my question(s) are:

  1. is this solution valid or is there something that I have overlooked?
  2. is there a better / more "correct" way to do this?
  3. do you know of any references to this kind of integration of the output of a state-space representation?
  4. clearly the new system cannot be integrated again using the same approach, due to $ \bar A $ being singular, is this a result of my (possibly flawed) solution? or is there perhaps some fundamental reason why it is not possible to twice-integrate the output of a state-space representation?

I am grateful for any constructive criticism and for answers that help further my understanding.

(I know that technically I am asking more that one question, but I don't feel that making them separate SE questions would make sense)

EDIT:

  1. I suspect $ \bar D = \begin{bmatrix} 0 \end{bmatrix} $ is wrong, and $\bar D$ should instead be some unknown constant matrix (the initial-conditions) right?
  2. I corrected an error in step 3,4 and 5, pointed out to me by @KBS, there was a B which should have been D.

Best Answer

First the answer to your questions:

  1. It seems that you have forgotten the initial conditions. Also, there is a term $B$ that should be $D$ in the second step and the next ones. Note that your approach is fully correct when the initial conditions are zero.

2, 3, and 4. The solution below simplifies your problems:

There is a simpler solution that does not require the invertibility of $A$. Just define the system

$$\dot{z}=\begin{bmatrix}A & 0\\C & 0\end{bmatrix}z+\begin{bmatrix}B\\D\end{bmatrix}u$$

where $z_1=x$ and $\dot{z}_2=y$. So, obviously, you get that $z_2(t)=\int_0^t y(s)ds$.

This procedure is called state-augmentation is used for the output regulation of linear systems where an integrator is added in order to regulate the output around a desired set-point.

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