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Are the output functions of a continuous-time LTI system necessarily continuous (in the calculus sense) for any given input functions?

I had this question when I saw this claim in my textbook:

for the most part we will use the condition of initial rest for systems described by differential equations. In the example, since the input was 0 for t < 0, the condition of initial rest implied the initial condition y(0) = 0.

$x(t) = 0$ if $t < 0$ -(initial rest)-> $y(t) = 0$ if $t < 0$

But to conclude that $y(0)$ itself is zero, $y$ needs to be continuous (at 0) given $x$.

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  • $\begingroup$ The thing is really in the sentence fragment: "the condition of initial rest implied the initial condition y(0) = 0." You define your system to initially rest, and that very simply is defined as x(t) = 0 for t < 0, and y(0) = 0. It's a definition! You can't generalize a definition. $\endgroup$ Mar 8 at 17:23

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The tag continuous-functions made me interpret the question a bit differently, as nice and simple counter-examples have been given.

I consider here that a system is fixed, and may have some constraints in its conception. And that the inputs are not controlled and can be much wilder. Then basically, if the impulse response on the system has a limied support, the output is continuous, only requiring mildly integrable inputs (continuity necessary neither on inputs and nor on ystem). A concrete example: with a time-bounded support LTI, and inputs being all functions having at most a finite number of discontinuities on a finite interval (fairly what we have in practice), then all outputs have continuity.

Rephrasing; suppose that a continuous LTI system has an impulse response $h$ that is regular enough (integrable, continuous, differentiable, etc.), and that $f$ is any input in a large class of functions (at least with some integrability properties, like belonging to Lebesgue $L_p$ spaces, $p\ge 1$). We note $C^k$ the space of continuous functions with $k$ continuous derivatives. Can we say something about the regularity of the convolution product (when it exists, hence the requirement about integrability) $h\ast f$?

The good news if that convolution does some "smoothing": convolution most often combines at best the regularity of $h$ and that of $f$. Here are a couple of results. I do apologize in advance for mistakes, did that quite long ago:

  1. If $h\in L_1$ and $f\in L_1$, then $h\ast f\in L_1$, and it is discontinuous in general. More generally, $h\in L_p$ and $f\in L_1$, then $h\ast f\in L_p$
  2. If $h$ has a compact support (classical for systems), and $f\in L^{loc}_1$ is locally integrable (e.g. integrable on every compact interval), then $h\ast f$ is continuous, which could be akin to your question. Note that no continuity is set on $h$: it can be made of boxcar or staircase functions
  3. If $h \in C^k$ has a compact support and $f\in L_p$, then $h\ast f\in C^k\ \cap L_p$.

There are many other results of this kind, often more technical for us poor DSP folks.

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    $\begingroup$ Let's see! 1. follows from Hölder's ineq.: $\|h(t)f(\tau-t)\|_1 \le \|h(t)\|_2\, \|f(t)\|_2\le \|h(t)\|_1 \, \|f(t)\|_1$, which is $<\infty \text {iff }h,f\in L_1$, right? $\endgroup$ Mar 8 at 21:06
  • $\begingroup$ Indeed. I remember a more elementary proof with Tonelli/Fiubini $\endgroup$ Mar 8 at 21:21
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Consider the identity system $y(t) = x(t)$. This system is LTI. If the input $x(t)$ is discontinuous, then the output $y(t)$ will be discontinuous too.

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    $\begingroup$ Indeed, silly me. :D If the inputs are continuous functions on a domain $D$, then I think we can say that $y$ will also be continuous on $D$? $\endgroup$
    – HappyFace
    Mar 8 at 18:33
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    $\begingroup$ Not in general, see Marcus' answer. $\endgroup$
    – MBaz
    Mar 8 at 18:56
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To add an even worse example to MBaz (best possible) counterexample:

The derivative $\frac{\mathrm d}{\mathrm dt}$ is an LTI system. $f(t)=|t|$ is a continuous function.

$\frac{\mathrm d}{\mathrm dt}(|t|)$ is discontinuous. The output of an LTI system can be discontinuous even when the input is continuous.

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    $\begingroup$ Nice example! .. $\endgroup$
    – MBaz
    Mar 8 at 17:29
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    $\begingroup$ A sufficiently pedantic mathematician might point out that $t\mapsto |t|$ is not even differentiable at zero, and therefore cannot be fed into that LTI system in the first place. The standard nitpick-proof example would be $t \mapsto t^2\cdot\sin\tfrac1t$, which is (strongly) differentiable everywhere including zero, but whose derivative $t\mapsto2x\cdot\sin\tfrac1t - \cos\tfrac1t$ is discontinuous there. $\endgroup$ Mar 10 at 11:42
  • $\begingroup$ @leftaroundabout good example! But: why should I not be able to feed anything non-differentiable into an LTI system in the first place? Especially since we typically characterize them through their impulse response. $\endgroup$ Mar 10 at 12:00
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    $\begingroup$ @MarcusMüller yes, but mathematically that makes everything a bit more complicated – you need to clarify that you're not really talking about functions on $\mathbb{R}$ but rather elements of the $L^2(\mathbb{R})$ Hilbert space (whose elements are equivalence classes of functions), and that the derivative is to be understood in a weak sense (i.e. not as a limit of $\tfrac{f(x)+f(x+h)}h$, but rather as the equivalence class of functions $f'$ that obey the integration by parts rule). And then you again can't use $|t|$ as-is because it's not integrable over all of $\mathbb{R}$... $\endgroup$ Mar 10 at 12:11

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