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I think this system is not time-invariant, but I'm not really sure how to plug in a couple test cases to check. The system is:

$x(t)$ -->(S)--> $y(t) = \int_{-\infty}^{3t}x(\tau) d\tau$

Without actually doing a proof, could I possibly plug in an impulse followed by a shifted impulse to show that it's not time-invariant? What would this look like?

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  • $\begingroup$ This system is time invariant, so that wouldn't work. $\endgroup$ – Omnomnomnom Dec 12 '15 at 20:20
  • $\begingroup$ It is? my professor said it wasn't so I've been trying to think of a test case. $\endgroup$ – Austin Dec 12 '15 at 20:22
  • $\begingroup$ I thought it was... I'll have to take a second look at my work $\endgroup$ – Omnomnomnom Dec 12 '15 at 20:29
  • $\begingroup$ Aha! Try to come up with an example for the system $x(t) \to x(3t)$. The same example should work here. The sum of an impulse and shifted impulse does work. $\endgroup$ – Omnomnomnom Dec 12 '15 at 20:31
  • $\begingroup$ I'm not sure about how to do it though I think the tau is confusing me. If I first use x(t) = DiracDelta(t) I think I get y(t) = 1, then x(t-1) = DiracDelta(t-1), y(t-1) = 1 still? $\endgroup$ – Austin Dec 12 '15 at 20:34
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If $y(t)$ is the response to an input signal $x(t)$

$$y(t)=\int_{-\infty}^{3t}x(\tau)d\tau\tag{1}$$

then the system is time-invariant if the response to $x(t-T)$ equals $y(t-T)$. The response to $x(t-T)$ is

$$\int_{-\infty}^{3t}x(\tau-T)d\tau=\int_{-\infty}^{3t-T}x(\tau)d\tau=y(t-T/3)\neq y(t-T)$$

Consequently, the system is not time-invariant, but time varying.

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