$y(t)=\mathcal{T} \{ x(t) \} = \int_{-\infty}^{x(t)} x(\tau) d\tau$

Is the above system time invariant ?

  • $\begingroup$ So have you applied the fundamental test ? $\endgroup$
    – Fat32
    Oct 4 '17 at 14:42
  • $\begingroup$ Yes i have applied the test and i think its time invariant but after changing the variable and limits in the integral i got little bit confused $\endgroup$
    – fpsshubham
    Oct 4 '17 at 14:46
  • $\begingroup$ so, how exactly did you get confused? $\endgroup$ Oct 4 '17 at 15:28
  • $\begingroup$ inaggressive @MarcusMüller is welcomed too ! :-)) You know that many beginning students (as we all were once) might have problems with manipulating such equations especially when they involve variables in a less accustomed way. The primary difficulty of a beginner is his reliance on intuition and experience and different notation immediately confuses (even right for us I belive). Why did you down vote the question anyway ? $\endgroup$
    – Fat32
    Oct 4 '17 at 15:47
  • $\begingroup$ Because I thought it to be a good question and wanted, considering op already had a good answer, to encourage a more rigid explanation of what's confusing! I think that would greatly improve the future value of the question. $\endgroup$ Oct 4 '17 at 15:59

The system is not time invariant as can be seen by the fundamental test procudere:

Given the system with an input/output relation: $$y(t)=\mathcal{T} \{ x(t) \} = \int_{-\infty}^{x(t)} x(\tau) d\tau$$

First check the new output for a shifted input $x(t-d)$ as $$y_d(t) = \mathcal{T} \{ x(t-d) \} = \int_{-\infty}^{x(t-d)} x(\tau-d) d\tau $$ $$y_d(t) = \mathcal{T} \{ x(t-d) \} = \int_{-\infty}^{x(t-d)-d} x(\tau) d\tau $$

And then also check if the output $y(t)$ shifted by $d$ yields th same formula? We know that output $y(t)$ is given in terms of input $x(t)$ with the formula $$y(t) = \int_{-\infty}^{x(t)} x(\tau) d\tau$$

Hence the shifted output $y(t-d)$ is obtained by simply replacing every $t$ variable with $t-d$ in that formula which yields: $$y(t-d) = \int_{-\infty}^{x(t-d)} x(\tau) d\tau $$

And now looking at the formulas for both $y(t-d)$ and $\mathcal{T} \{ x(t-d) \}$ given above, we see that since $x(t-d)-d \neq x(t-d)$ for all $d$ then $y(t-d) \neq y_d(t)$ and we conclude that the system is not time invariant.

  • $\begingroup$ sorry but can you please explain the "Shifted version of the output $y(t-d)$" this Line in details and in little simple language. $\endgroup$
    – fpsshubham
    Oct 4 '17 at 15:05
  • $\begingroup$ ok let me add, btw I supposed it was the most apprent line anyway :-) $\endgroup$
    – Fat32
    Oct 4 '17 at 15:18
  • $\begingroup$ Thank you very much. off-topic but what do you do ? i mean are you engineer,student or ...?? $\endgroup$
    – fpsshubham
    Oct 4 '17 at 15:31
  • $\begingroup$ your welcome! answer:both and more. $\endgroup$
    – Fat32
    Oct 4 '17 at 15:32
  • 3
    $\begingroup$ They are a file system 😜 $\endgroup$
    – Peter K.
    Oct 4 '17 at 18:16

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