# Can $\delta(t+\infty)$ be a legitimate signal?

Mathematically speaking, when I try to use some signal to disprove a system is invertible, can I use the signal like $$\delta(t+\infty)$$ ($$\delta$$ representing the Dirac distribution)? For example, the two inputs $$0$$ and $$\delta(t+\infty)$$ are distinct, but their outputs from the system $$\int_{-\infty}^{t}e^{\tau}x(\tau)d\tau$$ will then both be $$0$$. In this way, can I say the system is invertible?

• Personally, I'd say that you can't do that. In any case, it wouldn't be useful, since the system might be invertible for all signals that one actually cares about. – MBaz Feb 9 at 0:49

It's possible, in mathematics, to complete real numbers with "infinite" values, with sound topological properties; for instance non-standard analysis or the Extended real number line (discussion at Math StackExchange).

However, in this context, for any standard real number $$t$$, the rule is to set $$t\pm \infty = \pm \infty$$ (see Arithmetic operations). So, using your notations, for every $$t$$, you would have [CAUTION ADVISED] $$\delta(t+\infty)=\delta(\infty)$$, to which I am not able to give a meaning, else than something being a constant evererywhere equal to $$\delta(\infty)$$. Of course, $$\delta$$ is a distribution, and should not be treated as a function, but defined as a functional operator, or via limits of functions.

However, so far, I have never seen

• a (serious and useful) use of the extended real system in signal processing,
• a reference trying to define such a Dirac at infinity.

For the latter, I fear (just intuition) that constructing this with test functions can be troublesome, since two different limits are required (location and amplitude). This is just an analogy, but $$\delta(0)\times \delta(0)$$ is, as far as I know, not defined.

• The only sensible use I could see of this would be to use it as a test function, e.g. to have something to convolve a system with to get the value it converges to at $\infty$; but that would essentially require us to define that the convolution with $\delta(t+\infty)$ does that; it would feel consistent, but exactly as you state, there's nothing inherently meaningful about $\delta(\infty)$ – Marcus Müller Feb 9 at 13:30
• Yes, I feel that one would need a double limit, one classical on the amplitude, and one on the ordinal variable, which could lead to ambiguities – Laurent Duval Feb 9 at 14:43
• As far as I can remember when I used then, that mostly simplifies proofs and notationt – Laurent Duval Feb 9 at 20:14

The Dirac delta generalized function is only defined under an integral

$$Dirac(t-a) = \int_{-\infty}^\infty f(t)\delta(t-a)dt = f(a)\,$$

Typically there's also a function under the integral.

Hence

$$Dirac(t-\infty) = f(-\infty)\,$$

if one considers $$-\infty$$ to be a point - not to mention problems evaluating the $$Dirac$$ at -$$\infty$$.