# Notation of an LTI system consisting of LTI filters

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.

• 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?
– Peter K.
May 16 '17 at 11:54
• @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. May 16 '17 at 11:58
• 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.
– Peter K.
May 16 '17 at 12:06
• @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? May 16 '17 at 12:13
• @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. May 16 '17 at 12:15

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.
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.