My question is whether the systems below are memoryless or not:

$1.) \ y(t)=K$ where $K$ is a constant

$2.) \ y(t) = x(t_0) $ where $t_0$ is a constant

So, from the definition I have been using so far (A system is memoryless if its output at a given time is dependent only on the input at that same time), it seems like the first system is memoryless since the output at any time is fixed and can be said to depend only on the input at the same time.

For the second system, the output at $t=t_0$ depends only on $t_0$ but for any other time $t$, it requires knowledge of some other time i.e $t_0$ so I should classify it as having memory.

My doubt is if my reasoning is correct and if yes, isn't the second system a constant also which kind of confuses me.

I would really appreciate if someone could resolve this.

  • $\begingroup$ I think you're on the right track; there is no fundamental difference between $x(t_0)$ and $K$ (as long as $x(t_0)$ is actually a well-defined value!). $\endgroup$
    – MBaz
    Oct 15, 2021 at 12:56
  • 2
    $\begingroup$ Not only does the second system have memory but it also has foresight in that at times $t$ that are smaller than $t_0$, the system already knows that at a future time $t_0$, its input is going to be $x(t_0)$ which value the system is already outputting at time $t<t_0$. $\endgroup$ Oct 15, 2021 at 13:21
  • $\begingroup$ Yeah, so what kind of confuses me and seems a little contradicting is that the second system is also a constant value (it is well-defined). So, what actually is it that leads to both systems being in different categories i.e one having memory and the other memoryless $\endgroup$
    – user35508
    Oct 15, 2021 at 15:27
  • $\begingroup$ Is it just the fact that the first system can take any input signal and the output would be the same but the second system does depend on our input signal (albeit only at a single point)? I guess, this is what differentiates the two and leads to one having memory and other being memoryless? I would appreciate if you can confirm my reasoning $\endgroup$
    – user35508
    Oct 15, 2021 at 16:24

1 Answer 1


The common definition of as system with memory (also called dynamic system) is that its output at any time instant $t$ depends on the input signal at times that are generally different from $t$. In that sense, a system described by $y(t)=x(t_0)$ clearly has memory, because the system needs to observe the input signal and remember its value at $t=t_0$.

On the other hand, a system described by $y(t)=K$ with some constant $K$ that is independent of the input signal, doesn't need to know the input signal at all, let alone remember parts of it. For this reason such a system would be referred to as being memoryless.

In a digital implementation one could argue that the second system cannot actually be implemented without memory, because one would need to store the value of $K$, but that's usually not what is meant when one classifies systems into having memory (being dynamic) or being memoryless (instantaneous).

  • $\begingroup$ The second system doesn't just have to have memory to store the value of $x(t_0)$, but since the system output is $y(t) = x(t_0)$ for all $t$, including all $t \in (-\infty,t_0)$, the system needs prescience, knowledge of the value of $x(t_0)$ at all time instances before the input $x(t)$ ever takes on value $x(t_0)$ at time $t_0$ which the system could then remember for all times $t>t_0$. $\endgroup$ Oct 16, 2021 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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