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I would like to find a reference for two notations of an LTI system consisting of LTI filters. In z-domain, the LTI system is given by

$$ \mathbf{y}(z) = \mathbf{C}(z) \mathbf{s}(z) + \mathbf{D}(z) \mathbf{x}(z) \\ \mathbf{s}(z) = \mathbf{A}(z) \mathbf{s}(z) + \mathbf{B}(z) \mathbf{x}(z) , $$ where $\mathbf{x}$ and $\mathbf{y}$ are the input and output of the system, respectively and $\mathbf{s}$ is the system state. The matrices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$ constist of LTI filters, e.g., $$ A_{ij}(z) = \frac{b_0 + b_1 z^{-1} + \dots + b_k z^{-k}}{a_0 + a_1 z^{-1} + \dots + a_k z^{-k}}. $$ Now, each LTI filter can be represented by its impulse response in time domain. Let the matrices $\mathbf{A'}(n)$, $\mathbf{B'}(n)$, $\mathbf{C'}(n)$ and $\mathbf{D'}(n)$ with $n$ indicating time such that e.g. $$ A'_{ij} = [a'_0, a'_1, a'_2, \dots ] $$ is the impulse response of the LTI filter $A_{ij}(z)$.

Is it then possible to rewrite the whole LTI system in time-domain as $$ \mathbf{y}(n) = (\mathbf{C} \ast \mathbf{s})(n) + (\mathbf{D} \ast \mathbf{x})(n) \\ \mathbf{s}(n) = (\mathbf{A} \ast \mathbf{s})(n) + (\mathbf{B} \ast \mathbf{x})(n) , $$ where $\ast$ denotes the convolution operation. I would especially appreciate any reference to literature / text books where this representation is discussed.

An one-dimensional example: The state-transition matrix may be a (time-invariant) one-pole filter, e.g., $\mathbf{A}(z) = \frac{b_0}{1 - a_1 z^{-1}}$. The corresponding time-domain $\mathbf{A}'(n)$ is the impulse response of this one-pole filter.

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  • $\begingroup$ Welcome to SP.SE! It's not usual for the system matrices ($\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$) to be functions of time... because that means the system you're representing is not time-invariant. Can you give an example of where that notation is used? $\endgroup$ – Peter K. May 16 '17 at 11:54
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    $\begingroup$ @PeterK. : Thank you very much for your welcome! Please note that the system matrices are not time-variant, although the time-domain notation is somehow ambiguous. I add an example to hopefully clarify. $\endgroup$ – Sebastian Schlecht May 16 '17 at 11:58
  • $\begingroup$ Hmm. Taking the $z$-transform of $\mathbf{A}$ suggests that it is time varying... otherwise the $z$-transform of $\mathbf{A}$ is $\mathbf{A}$. See this page for example. $\endgroup$ – Peter K. May 16 '17 at 12:06
  • $\begingroup$ @PeterK. Ok, this needs some clarification. We can agree that the one-pole filter above is time-invariant (it is a classic LTI filter)? Hence, the impulse response of the one-pole filter is time-invariant as well, right? $\endgroup$ – Sebastian Schlecht May 16 '17 at 12:13
  • $\begingroup$ @PeterK. In contrast, what I would call a time-variant filter is when the values of the impulse response are changing over time as in your linked example. $\endgroup$ – Sebastian Schlecht May 16 '17 at 12:15
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As I've said in chat, I do not believe the system $$ \mathbf{y}(z) = \mathbf{C}(z) \mathbf{s}(z) + \mathbf{D}(z) \mathbf{x}(z) \\ \mathbf{s}(z) = \mathbf{A}(z) \mathbf{s}(z) + \mathbf{B}(z) \mathbf{x}(z) , $$ with the matrices time-varying is an LTI system.

That means, I don't think we can answer your question.

If the matrices are not functions of $z$ then you can write the transfer function as: $$ \mathbf{y}(z)/\mathbf{x}(z) = \mathbf{C}(zI - \mathbf{A})^{-1} \mathbf{B} + \mathbf{D} $$ but it's not clear to me if this equation is valid if the matrices are time-varying / $z$-transforms.

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  • $\begingroup$ Thank you very much for your answer. I cannot help to go on thinking that we talk about different ideas of time-varying. I need to find a more compelling example and let you know. $\endgroup$ – Sebastian Schlecht May 16 '17 at 13:56
  • $\begingroup$ @SebastianSchlecht Please do! We are obviously talking at cross-purposes, so any examples would be helpful to figure out what's happening. $\endgroup$ – Peter K. May 16 '17 at 14:01

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