3
$\begingroup$

Are there any examples of "Causal" Continuous-Time Linear-Time-Invariant Systems (CT-LTI) with output given by finite-duration functions $y(t)$??

With output I am meaning that $y(t)$ is such that $y(t)=h(t)\circledast x(t)$ with $h(t)$ the impulse response function, and the input $x(t)$ such the output $y(t)$ is given by a finite-duration/time-limited continuous function (also $y(t)$ absolutely integrable so it could have a defined Laplace transform for an specific ROC). Also, explain if you can, which conditions must fulfill $h(t)$ and $x(t)$.

With finite-duration/time-limited I am meaning that there exist a starting time $t_0$ from which $y(t)=0,\,t<t_0$, and also an ending time $t_F$ from which $y(t)=0,\,t>t_F$, so, $y(t)$ is compact-supported with domain within the edges $\partial t = \{t_0,\,t_F\}$.

The whole picture is that I am trying to find if for continuous time LTI systems, the condition of causality will create a restriction for the upper bound of the rate of change $y'(t)$, since I can write $y(t) = y'(t)\circledast \theta(t)$ with $\theta(t)$ the standard unitary step function.

All the continuous-time examples of LTI Systems I have found so far are related to linear ordinary differential equations, which outputs cannot be time-limited (since its solutions are analytical, and analytical functions cannot be compact-supported), so any counter-example will be welcome. Thanks.

Added later:

This doubt rises from the book I am studying signals, there, it is said that a LIT causal system must have a ROC that is a right-side plane in $s$ with $s$ the complex plane defined when taking the Laplace transform... but the book also states on a previous subsection that every finite-duration signals must have a ROC that contains the whole $s$-plane, so I feel its kind of contradicts the requirement to be an LIT causal system, but surely I am missing something (maybe there is a mixed with properties of one-side and bilateral Laplace Transforms, but I don´t really know)... but at these extent, it looks like there is a problem for signals for being being causal LIT and of finite-duration at the same time.

Last update

Thanks you for your comments. Being true that a "naive" physical system must be of finite-duration, and also that in real systems thermal noise will made that there will be always an small amount of garbage-in / garbage-out noise being output by your system, when I made this question I was thinking in something much more "innocent": just to find examples of functions $x(t)$, $h(t)$ and $y(t)$ that will behave as continuous-time LIT system of finite duration.

But working into this through another question in the maths forum, I believe is showed through the question and the comments of this question that is impossible to have a non-constant finite-duration continuous function that is linear (I am no mathematician so I can´t prove anything related - I am just following the comments made there. I will add at the end the analysis done there.), so in principle, no continuous-time LIT system will be described by finite duration functions, meaning here, they are compact-supported - different from the standard assumption of having functions that vanishes at infinity so their outputs disappear within the noise at any positive SNR.

I left this here since I am interesting in hear your comments, for me have been really shocking that no non-zero finite-duration function could be described through analytical functions, LIT systems, or linear differential equations, so almost everything I learned in engineering, being totally smart, are "philosophically speaking" only approximation, given that "common sense" tells that every phenomena should be of finite-duration and the theory they teach me can´t model these functions - and also thinking in quantum dynamics, where Schrödinger equation is linear (so with finite extent in time and space solutions), could be "real" or an approximation? at least every electron is moving since the very beginning, isn´it?... many interesting question rises because of these issues.

If you know a theory to work with finite-duration phenomena hope you share it (I am looking if there any restriction imposed by causality on the maximum rate of change of time-limited functions). Thanks you all.

Why I believe no finite-duration function could be linear:

Let define an arbitrary finite-duration function as: $$f(t) = \begin{cases} 0,\, t<t_0 \\ x(t),\, t_0\leq t \leq t_F \\ 0,\, t> t_F \end{cases}$$ with $x(t)$ and arbitrary continuous function. Note that $f(t)$ is compact-supported, and since is continuous within its domain $[t_0,\,t_F]$, it is also bounded $\|f\|_\infty < \infty$.

Lets define $\Delta T = t_F - t_0 > 0$ the length of the support of the finite duration function, and lets choose a constant $b \geq 0$ such that the time $t = t_0 + b$ is contained into the support of $f(t)$, so $t_0 \leq t_0 + b \leq t_F$.

Then, for any arbitrary point within the support of $f(t)$ I can state that if $f(t)$ fulfill the additive property of a linear operator $f(x+y) = f(x)+f(y)$, then by adding zero it will becomes: $$\begin{array}{r c l} f(t) & = & f(t_0+b) \\ & = & f(t_0 +b +\Delta T - \Delta T) \\ & = & f(t_0 + b + t_F - t_0)-f(\Delta T) \\ & = & f(t_F+b) - f(t_F) + f(t_0) \\ & = & f(t_0) - f(t_F) \end{array}$$ because $f(t_F + b) = 0$ since $t_F+b$ is outside the support for any $b>0$, so if the operator has the additive property, every point within the support must be equal to a constant $f(t) = f(t_0)-f(t_F)$ (or $f(t) = f(t_0)$ if $b=0$), so the only operator $f(t)$ will be a constant function, not an arbitrary function as is intended. So no compact-support/finite-duration non-constant function could be described through an additive operator.

I have found this paper that analyze finite-time controllers as continuous time differential equations, and the examples look highly non-linear, but is too advanced for me to fully understand what is said (I have no clue about Lyapunov theory). Maybe someone can explain it in simple.

Improve this question
$\endgroup$

3 Answers 3

Reset to default
1
$\begingroup$

You may be overthinking this. If the input has length $T_x$ and the impulse response of the LTI system has length $T_h$ than the output will have have length $T_x + T_h$. If either input or impulse response are infinite so will be the output. That's all there is to it.

In other words: if both input and impulse response have finite support so will have the output.

Also, explain if you can, which conditions must fulfill h(t) and x(t).

Both must have finite support and h(t) must be causal.

EDIT: adding a few examples.

  1. Anything with a non recursive delay (or reflection). For example a $2pi$ anechoic chamber, a full anechoic chamber with a decent chunk of plywood, a properly terminated transmission line or wave tube.
  2. A good old fashioned tape delay
  3. A discrete FIR on a processor with an ADC and DAC wrapped around it
  4. A laser in front of mirror with a beam splitter, an optical fiber.

You need to be fairly precise about what you mean by "are there any examples?" and what specific criteria you use to accept or dismiss an example.

Any real-world physical system will only approximately behave like the underlying model. A constraint like $h(t) = 0 , t < t > t_{max}$ is not really practical since no physical quantity can be measured to be truly zero (as long as we are staying out of quantum mechanics). Best you can do is $|h(t)| = \epsilon , t < t > t_{max}$ where $\epsilon$ is a sufficiently small number determined by either your application requirements or the inevitable noise floor.

Once you take that step most IIR filters become finite as well. Let's look at a simple RC lowpass with a cutoff at 1 kHz. The impulse response is an exponential decay, but it decays fast. The half energy point is at about $50\mu s$ . At $1ms$ it's 55 dB down, at $10ms$ it's a whopping 550dB down, and at around $120ms$ even double precision math results in an underflow (at about -6500dB).

So we can say "the simple mathematical model we use to describe an RC circuit predicts an impulse response that's infinite in time". But physically that makes little sense and it's more appropriate to say "any RC circuit can be modelled with having a finite impulse response without adding additional noise".

Improve this answer
$\endgroup$
15
  • $\begingroup$ So how are you gonna make that FIR outa op-amps, resistors, and capacitors? $\endgroup$
    – robert bristow-johnson
    Dec 11, 2021 at 7:19
  • $\begingroup$ Singing in front of a single flat mountain face would seem to satisfy the criteria? $\endgroup$
    – Knut Inge
    Dec 11, 2021 at 8:14
  • 1
    $\begingroup$ i guess, so also would a length of transmission line, properly terminated. $\endgroup$
    – robert bristow-johnson
    Dec 11, 2021 at 13:57
  • 1
    $\begingroup$ @robertbristow-johnson Or you could try this way (link on ee.se), half amusement, half serious. $\endgroup$
    – a concerned citizen
    Dec 11, 2021 at 14:54
  • 1
    $\begingroup$ @Joako: I added a few to my answer. This is actually a somewhat philosophical question $\endgroup$
    – Hilmar
    Dec 12, 2021 at 13:30
1
$\begingroup$

I have found the following paper named "Finite time differential equations" by V. T. Haimo (1985), where continuous time differential equations with finite-duration solutions are studied, an it is stated the following:

"One notices immediately that finite time differential equations cannot be Lipschitz at the origin. As all solutions reach zero in finite time, there is non-uniqueness of solutions through zero in backwards time. This, of course, violates the uniqueness condition for solutions of Lipschitz differential equations."

Since, linear differential equations have solutions that are unique, and finite-duration solutions aren´t, finite-duration phenomena models must be non-linear to show the required behavior (non meaning this, that every non-linear dynamic system support finite-duration solutions).

The paper also show which conditions must fulfill the non-linear differential equation to support finite-duration solutions, at least for first and second order scalar ODEs.

So, if the paper is right (I can't prove or disprove it), no LTI system could have finite-duration solutions, since they are described through linear differential equations.

Improve this answer
$\endgroup$
0
$\begingroup$

A length of coax cable, properly terminated. For extra credit, add properly terminated taps along the length of the coax cable, and add their outputs.

Back in the 1990s, and today for all I know, there were radar processing circuits that used surface acoustic wave filters that were basically continuous time FIR filters.

Why I believe no finite-duration function could be linear:

Let $y(t) = h(x(t))$ define a system, where

$$y(t) = \int_{0}^{T} x(t - \tau) d\tau$$

  • It has a finite impulse response
  • It obeys superposition
  • It's properties are not dependent on the value of $t$; only on the values of $x$ as a function of $t$

So -- it's a FIR continuous-time linear time invariant filter.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ A modern iPhone contains SAW filters in its radio front end, K believe. $\endgroup$
    – Knut Inge
    Dec 12, 2021 at 15:03
  • $\begingroup$ AFAIK you can arrange a SAW device either as a resonator or as a delay line -- do you know if they're one or the other? $\endgroup$
    – TimWescott
    Dec 12, 2021 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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