I have the following model: $$ x[k+1] = Ax[k] + Bu[k]$$ $$ y[k] = x[k] $$ where $x \in \mathbb{R} $ is the state, $A \in \mathbb{R}$, $u \in \mathbb{R}^4$ are the inputs and $B \in \mathbb{R}^4$ and $y[k]$ is the output.

From theory this can be solved using: $$ y[k] = A^k x[0] + \sum_{i=0}^{k-1} A^{k-1-i}Bu[i]$$

How can I find the time instance $k$ when $y[k]$ has a specific value?


1 Answer 1


The answer depends on your initial value and your input.

For specific initial values and inputs there are easy solutions, but in general I think the fastest way is to let the computer solve this for you. So crank up your matlab / python / mathematica / excel / other tool of choice and let the brute force of computation serve you.

  • $\begingroup$ Yeah I thought so, I wish there was a cleaner way $\endgroup$
    – Marc
    Commented Sep 19, 2016 at 15:19
  • $\begingroup$ @Marc The reason why, in general, you can't solve for $k$, is because for a given input and initial condition the output $y$ might only be able to obtain values from a fixed (finite) set of values. For example $x[0]=0$, $u[k]=\vec{0}$, can only obtain $y[k]=0$. $\endgroup$
    – fibonatic
    Commented Sep 20, 2016 at 1:42

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