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LTI systems in state space representation are systems of the form:

\begin{eqnarray} \dot{\mathbf{x}}(t)=\mathbf{Ax}(t)+\mathbf{Bu}(t) \end{eqnarray} \begin{eqnarray} \mathbf{y}(t)&=&\mathbf{Cx}(t)+\mathbf{Du}(t) \end{eqnarray}

where $\mathbf{x}$ is the state vector, $\mathbf{u}$ is the input and $\mathbf{y}$ is the output. These systems satisfy the superposition principle only if the initial condition $\mathbf{x}=\mathbf{0}$ holds. Let us refer to the class of these systems with the aforementioned initial condition as $\Sigma_{LTI}^0$.

Now a different class of systems, namely $\Sigma_{conv}$ are the ones that are represented by a convolution integral, i.e. the input-output behaviour of the system is described by:

\begin{equation} \mathbf{y}(t)=\int_{-\infty}^\infty h(\tau) \mathbf{u}(t-\tau)d\tau \end{equation}

My question is whether it is true that every system in $\Sigma_{LTI}^0$ admits a representation in $\Sigma_{conv}$? If yes, how?

Note: I understand that my question boils down to finding a function $h$ such that (assume for simplicity that $\mathbf{C}=\mathbf{I}$ and $\mathbf{D}=\mathbf{0}$):

\begin{equation} \int_0^t e^{\mathbf{A}(t-\tau)}\mathbf{Bu}(\tau)d\tau= \int_{-\infty}^\infty h(\tau) \mathbf{u}(t-\tau)d\tau \end{equation}

Update: Let us assume that there exists a $h\in\mathcal{L}^2(\Re;\Re^n)$ such that an LTI system is equivalent to a convolution system and assume for simplicity that the input is one-dimensional. Then, these have identical impulse responses, hence for $u(t)=\delta(t)$ we have:

\begin{equation} \int_0^t e^{\mathbf{A}(t-\tau)}\mathbf{B}\delta(\tau)d\tau= \int_{-\infty}^\infty h(\tau) \delta(t-\tau)d\tau \end{equation}

Therefore:

\begin{equation} \int_0^t e^{\mathbf{A}(t-\tau)}\mathbf{B}\delta(\tau)d\tau= h(t) \implies h(t)=e^{\mathbf{A}t}\mathbf{B} \end{equation}

But then:

\begin{equation} \int_0^t e^{\mathbf{A}(t-\tau)}\mathbf{Bu}(\tau)d\tau= \int_{-\infty}^\infty e^{\mathbf{A}(t-\tau)}\mathbf{Bu}(\tau)d\tau \end{equation}

Doesn't look good! So, if there is no mistake in the above procedure, there is no $h$ that fulfills my requirements. Then, what is the connection between the aforementioned classes of systems.

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3 Answers 3

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The transfer function of the given system is $$ \mathbf{Y}(s)=[\mathbf{D}+\mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B}]\mathbf{U}(s) $$ and it establishes an input-output relationship written compactly as $$ \mathbf{Y}(s)=\mathbf{G}(s)\mathbf{U}(s) $$ Using the convolution property of the Laplace transform, we have: $$ \mathbf{y}(t)=\mathbf{g}(t)*\mathbf{u}(t) $$ where $g(t)=(\mathscr{L}^{-1}\mathbf{G})(t)$. Therefore, every system in state-space form can be written in an equivalent input-output representation using the convolution operation.

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I think your problem is that your forgot to add that your solution for $h(t)=e^{\mathbf{A}t}\mathbf{B}$ is valid for $t>0$

And that makes sense: assuming for one moment ${\bf D}=0$, we have a system in which the output is given by a first-order linear differential equation, which indeed corresponds to an exponential $h(t)$ (causal, i.e., $h(t) =0$ for $t<0$). Adding ${\bf D}\ne0$ you're just adding a ${\bf D} \; \delta(t)$ term, so the conclusion would be that $\Sigma_{LTI}^0$ is a (small) subset of $\Sigma_{conv}$, namely, that of the LTI which $h(t)$ is given by the sum of an (casual) exponential and a Dirac delta.

This reasoning should be extended to multidimensional systems, though. Now, assume ${\bf u}$ and ${\bf x}$ are one-dimension, but we are allowed to increase the dimension of ${\bf x}$ arbritrarily (which is equivalent to extend arbitrarily the order of the ODE). Then, it's another story.

BTW, if you are familiar with discrete LTI systems, the analogy would be $\Sigma_{LTI}^0 \leftrightarrow AR(1)$ (autoregresive processes of order 1, i.e., one pole), which indeed corresponds to a subset of all LTI casual systems.

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Can you elaborate a bit more one that and actually show how you justify this equivalence. Start assuming that $\mathbf{D}=\mathbf{0}$ and then generalize if possible. –  Pantelis Sopasakis Oct 5 '11 at 11:17

Initial conditions can be represented as impulse responses applied to the system at time $t=0$. In other words, you can add a new column $x(0)$ to the $B$ matrix and extend your input function to $\pmatrix{u(t) \\ \delta(t)}$. In a way, you treat the initial condition as a perturbation to the zero initial condition system since

$$ x(t) = Ce^{At}x_0 + \int_0^t Ce^{\mathbf{A}(t-\tau)}\mathbf{Bu}(\tau)d\tau $$

Note that, $\delta(t)$ function only makes sense under the integral sign.

Therefore, every LTI system in $\Sigma^0_{LTI}$ is a member of $\Sigma^{x_0}_{LTI}$ with the special case $x_0=0$.

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I don't see how this answers the question. Where does $\Sigma_\text{conv}$ appear? –  joriki Oct 3 '11 at 9:30
    
Causality makes it possible to extend the lower limit of the integral to $-\infty$ and you can add an indicator function to cut off the response and extend the integral limit to $\infty$. –  user13838 Oct 3 '11 at 9:58

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