Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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}


\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.

share|cite|improve this question

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 $x_0=0$.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.