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

Question

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 $x$.

Answer 1

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.

Answer 2

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.

Answer 3

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.