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Note: This question may seem extremely elementary, but I am not a beginner to signal processing, linear system theory, control theory, etc. I believe my confusion is over a subtle point, so, before responding, please read the entire question so you understand the source of my confusion first.


Consider this simple autonomous system (autonomous means it has no direct dependence on $t$):

$$\dot{x}(t) = A x(t)$$

Assume the output is the state ($y(t) = x(t)$) and $A \neq 0$. My questions are:

  1. Is this system necessarily linear?
  2. Is this system necessarily time-invariant?
  3. What is the zero-input response of this system?
  4. Would the answers above change if we assumed the system was causal?

Easy, right?

  1. Obviously yes, since it is in the form $\dot{x}(t) = Ax(t)$
  2. Obviously yes, since $A$ does not depend on $t$
  3. The zero-input response is: $x(t) = x(0)\,e^{At}$
  4. No, these are unrelated properties.

But wait!

  1. Linearity means a linear combination of the inputs must give the same linear combination of the outputs. If I scale the input, $u$, then $x$ is not affected at all. So how can it be linear?!
  2. Time-invariance means that if I delay the input, the output is delayed by the same amount of time. But if I delay the input, $u$, then $x$ is not affected at all. So how can it be time-invariant?!
  3. If the initial state is an "input": Doesn't linearity imply $x(0) = 0 \implies x(t) = 0$?
    If the initial state isn't an "input": Isn't the system nonlinear unless $x(0) = 0$? (see #1)
  4. Wouldn't zero-input give zero-output for a causal LTI system (the "ZIZO" property)?

Hopefully it's clear why I'm confused.
Where am I going wrong?

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    $\begingroup$ You don't have a linear system. Any otherwise LTI system which exhibits an output component caused by the homogeneous solution (e.g. due to non-zero initial conditions) is no longer linear (or time-invariant). You've just identified a special case of this fact. $\endgroup$ – Peter K. Mar 31 '16 at 14:04
  • $\begingroup$ @PeterK.: I would agree but if you click on my link, Boyd's PDF does call $\dot{x} = Ax$ a "continuous-time autonomous linear dynamical system" and also calls it time-invariant, so that is directly at odds with your statement... $\endgroup$ – Mehrdad Mar 31 '16 at 18:24
  • $\begingroup$ Done! I hope it's of some help. $\endgroup$ – Peter K. Mar 31 '16 at 18:39
  • $\begingroup$ yeah, @PeterK. i remember MattL and i having the same issue. depends on how one expresses the convolution integral (i like it from $-\infty$ to $+\infty$ and then identifying $x(0)$ as the result of the integral from $-\infty$ to $0$. i would say it is LTI, but if $x(0) \ne 0$, then you didn't have "ZIZO". MattL disagrees. $\endgroup$ – robert bristow-johnson Mar 31 '16 at 19:12
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LDSs are systems of differential equations. For such things the terms "linear" and "time-invariant" have different definitions when compared with control theory or signal processing. For example linearity means having solutions which can be added together in particular linear combinations to form further solutions. For example time-invariant means when the [independent] variable is time, they are also called time-invariant systems.

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The state-space model of a causal continuous-time LTI system is

$$\begin{align}\dot{x}(t)&=Ax(t)+bu(t)\\y(t)&=cx(t)+du(t)\tag{1}\end{align}$$

where $x(t)$ is the state (vector), $u(t)$ is the input, and $y(t)$ is the output. If the state $x(t)$ is a length $N$ vector, then $A$ is an $N\times N$ matrix, and $c$ is a $1\times N$ matrix (or row-vector, if you like).

In your example, $b=0$, $c=0$, and $d=0$. So there is basically no input and no output. We can only observe the state, i.e., the "internal behavior" of the system. We can only talk about linearity and time-invariance concerning the states. If $A$ is constant, the system is linear and time-invariant in that sense. But since we can't observe a response to an input signal it is impossible to talk about linearity and time-invariance of the system in terms of its input/output behavior. And, consequently, the corresponding tests become pointless.

And concerning point 4. in your question: yes, an LTI system must have zero output for zero input. It is a non-zero initial state that can produce a non-zero output for zero-input, but with a non-zero initial state, a system is not LTI anymore. If it is LTI otherwise (apart from the non-zero initial state), then it can be modeled as LTI by representing the non-zero initial state as an additional input to the LTI system (which must now have a zero initial state, otherwise it couldn't be LTI).

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  • $\begingroup$ I'm assuming $y(t) = x(t)$, sorry if that was not clear. $\endgroup$ – Mehrdad Mar 31 '16 at 9:43
  • $\begingroup$ @Mehrdad: The argument remains the same: if there is no input you can't talk about linearity and time-invariance of the system in the sense of responses to input signals. So those tests (scaling and shifting the input signal) are not relevant and pointless. If you like, it is not a system in the conventional signal processing sense of the word. $\endgroup$ – Matt L. Mar 31 '16 at 9:47
  • $\begingroup$ To be clear and explicit, are you saying there are 2 different definitions of linearity and time-invariance, neither of which entirely encompasses the other? Because I fail to see how a single definition can cover both the mathematical sense and the signal-processing sense, yet I have never seen both definitions compared and contrasted side-by-side either, leading me to believe they are not distinct. $\endgroup$ – Mehrdad Mar 31 '16 at 9:56
  • $\begingroup$ Also, if you truly mean that these are not "systems" in the signal-processing sense, would it be possible for you to provide the formal definition of the word "system" in that sense? (A simple link is fine, but I'm looking for a formal definition, not something hand-wavy.) $\endgroup$ – Mehrdad Mar 31 '16 at 10:02
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    $\begingroup$ The essence of linearity and time-invariance is always the same. You just need to apply them in a way that makes sense for the given system. If your system has no input, don't try to use the input to define linearity or time-invariance. $\endgroup$ – Matt L. Mar 31 '16 at 11:47

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