How do you augment a state space model with the derivative of a state? I know how to augment a state space model with the integral of a state by doing the following. Given a linear system

$$ \dot{x}=ax\tag{1}\label{1}$$

which has a state space representation given by

$${\bf\dot{x}}={\bf A}{\bf x} \tag{2}\label{2}$$

one can introduce the integrated state by augmenting the ${\bf A}$ matrix above by writing

$$\frac{d}{dt}\int xdt=x \tag{3}\label{3}$$

and taking $\int xdt$ to be the augmented state. This gives the following augmented state space model:

$$\frac{d}{dt}\left[\matrix{x \cr \int xdt} \right]= \left[ \matrix{a & 0\cr 1 & 0} \right] \left[\matrix{x \cr \int xdt} \right] \tag{4}\label{4}$$

Solution Attempt

Is there a way to add $\dot{x}$ as a state? It seems that it should be the reverse of the above. I can take the derivative of $\ref{1}$ to get

$$\ddot{x}=a\dot{x} \tag{5}\label{5}$$

This gives the following augmented state space model:

$$\frac{d}{dt}\left[\matrix{\dot{x} \cr x} \right]= \left[ \matrix{a & 0\cr 0 & a} \right] \left[\matrix{\dot{x} \cr x} \right] \tag{6}\label{6}$$

But this seems wrong because now I have essentially the same equation twice... So I tried substituting $\ref{1}$ into $\ref{5}$ yielding

$$ \ddot{x}=a^2x \tag{7}\label{7}$$

This gives the following augmented state space model:

$$\frac{d}{dt}\left[\matrix{\dot{x} \cr x} \right]= \left[ \matrix{0 & a^2\cr 0 & a} \right] \left[\matrix{\dot{x} \cr x} \right] \tag{8}\label{8}$$

I think somehow what I am doing above is incorrect, because if the initial state of $x$ is zero, the system remains static.

Any help greatly appreciated and thanks in advance!! I asked this question over on math.exchange here but had not gotten any help so I decided to try my luck on dsp.exchange :).

  • $\begingroup$ Did you forget the B matrix? $\endgroup$ – Ben Jan 13 at 18:17
  • $\begingroup$ @Ben, thanks for the comment! I did not forget the ${\bf B}$ matrix. I'm just considering the zero input (homogeneous) response to non-zero initial conditions. Is this a problem? $\endgroup$ – eball Jan 13 at 21:58
  • $\begingroup$ No, but it is possible for a system to have an empty first matrix row. For example the system F = mA. The acceleration would be provided by an external input. x would be the position and x' the speed. Thus in this case, the derivative of x' (i.e x'') does not depend on x' and x. $\endgroup$ – Ben Jan 13 at 22:16
  • $\begingroup$ Ahh interesting point. Not sure I understand how that helps answer my question tho haha $\endgroup$ – eball Jan 13 at 22:41

Your Answer

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

Browse other questions tagged or ask your own question.