is there any concept of state transition matrix for this discrete system?


$N(k)$ are nilpotent matrices for $k=0,1,2\dots$

I know state transition matrix for the system $x(k+1)=A(k)x(k)+B(k)u(k)$

is given by $\phi(k,k_0)=A(k-1)\times \dots \times A(k_0)$

Thanks for helping.

$ ax(k+1)=x(k)+bu(k)$ where $a$ is an nilpotent matrix, I am given that its solution is $x(k)=-\sum_{i=1}^{q-1} a^ibu(k+i)$ where $q$ is the degree of nilpotency of $a$.

  • $\begingroup$ That system doesn't determine a sequence given the initial point. Before the edit it did. $\endgroup$ – user550929 Apr 12 '18 at 6:12
  • $\begingroup$ sorry I have edited again......... $\endgroup$ – Marso Apr 12 '18 at 6:14

You can extend the definition of a discrete (and continues) state space model to

$$ \begin{array}{rl} E\,x[k+1] \!\!\!\!&= A\,x[k] + B\,u[k] \\ y[k] \!\!\!\!&= C\,x[k] + D\,u[k] \end{array} $$

which is also known as a descriptor state space model.

This will also allow you to represent improper transfer functions when $E$ is not invertible. This is the case when $E$ is a nilpotent matrix. So your equivalent transfer function and thus also your state space model is noncausal. So one can only define the state transition matrix $\Phi(k,\kappa)$ for $\kappa\geq k$ with

$$ \Phi(k,\kappa) = N(k)\,N(k+1)\,\cdots\,N(\kappa-1). $$

  • $\begingroup$ thanks for your comment but I am afraid how will I get an expression for $x(k)$? none of the $N(k)$ are invertible!, after all I need to find $x(k)$ which I am not getting from your comment. $\endgroup$ – Marso Apr 12 '18 at 10:41
  • $\begingroup$ @Wow Your question is about finding the state transition matrix and for the zero input case $x(k)$ can be found using the state transition matrix, namely $x(k) = \Phi(k,\kappa)\,x(\kappa)$. $\endgroup$ – Kwin van der Veen Apr 12 '18 at 11:27
  • $\begingroup$ are you saying solution of the system $1$ can be given by, $x(k)=\Phi(k,\kappa)x(\kappa)$? then what is the different between the system $1$ and this system: $x(k+1)= A(k)x(k)+B(k)u(k)$? $\endgroup$ – Marso Apr 13 '18 at 6:06
  • $\begingroup$ and what do you mean by zero input case? $u(k)=0\forall k$? but in my system $1$, inputs are there $\endgroup$ – Marso Apr 13 '18 at 6:10

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