1
$\begingroup$

How can I verify if the LTI system ($T$ is a real number, $y(t)$ output, $x(t)$ input):

$$y(t)=x(t-T)$$

is with or without memory? I know that, if $y(t)$ depends only on $x(t)$, the LTI system will be without memory. In the above example, does it depend on $t-T$ or only on $t$?

$\endgroup$
0
$\begingroup$

From Signals and Systems/Time Domain Analysis -- Memory

A system is said to have memory if the output from the system is dependent on past inputs (or future inputs) to the system. A system is called memoryless if the output is only dependent on the current input.

According to some textbooks, the notion of memory is stricter, and requires causality of the system. To that respect:

  • if $T\ge 0$, the system is causal. If $T=0$, it only use the current time (memoryless). If $T>0$, the output need a memory of the past of input $x$
  • if $T< 0$, the system is non-causal. The output needed a memory of the future of input $x$. You can either say "non applicable", or still term it "memory".

References (especially the first, with many practical exemples):

$\endgroup$
  • $\begingroup$ Hi @LaurentDuval. Sorry I did a mistake, now I'm editing the question. $\endgroup$ – Gennaro Arguzzi Aug 29 '17 at 8:50
  • 1
    $\begingroup$ maybe do you want to say that if $T<0$ the system is non-causal because $y(t=0)=x(-T>0)$? $\endgroup$ – Gennaro Arguzzi Aug 29 '17 at 9:26
  • 1
    $\begingroup$ Indeed, typed too fast $\endgroup$ – Laurent Duval Aug 29 '17 at 9:30

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.