1
$\begingroup$

I have a question about determining time-invariance of a linear system. We are given this system and we need to determine if it is time-invariant or not:

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

Now after doing the shifting $t-t_o$ you get two different outputs. Now that makes sense, but what confuses me is the variable substition that happens in order to check. I'll post a link to the picture on imgur,sorry for this but I don't know how to do it otherwise.

https://imgur.com/a/CBbe4Sp

Basically my question is why is the substittution $-3τ -t0$ and not just $τ -t0$. Also why are we substituing both the Z1 as well as Z2, isn't the goal to make try to make them equal? Like try to modify one to get the same result as the other?.Maybe someone can give me some insight.Sorry for the bad post I need instructions on how to post. Cheers!

$\endgroup$
  • $\begingroup$ I've added the integral defining the system. Check how it's done and you can do the rest yourself. Also note that I've changed the input to $x(t)$ and the output to $y(t)$, because the variable $z$ is usually reserved for the argument of the Z-transform. $\endgroup$ – Matt L. May 20 at 7:29
2
$\begingroup$

You have a system with the following input-output relation:

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

In order to check whether the system is time-invariant or not, we need to compare the shifted output with the output resulting from a shifted input. The shifted output is

$$y(t-T)=\int_{-(t-T)}^{\infty}x(-3\tau)d\tau\tag{2}$$

Shifting the input means applying an input signal $x_T(t)=x(t-T)$. Note that $x_T(-3\tau)=x(-3\tau-T)$, so the response to the shifted input is

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

The system is time-invariant if $(3)$ equals $(2)$. So we try to make the integrand in $(3)$ look like the one in $(2)$ by substituting $-3\tau-T$ by $-3\zeta$, which gives

$$y_T(t)=\int_{-(t-T/3)}^{\infty}x(-3\zeta)d\zeta\tag{4}$$

Now $(2)$ and $(4)$ have the same integrand, but we see that the lower integration limits are different. Consequently, $y_T(t)\neq y(t-T)$, hence the system is not time-invariant.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Oooooooh okay okay, I see now. Thank you a lot for this! $\endgroup$ – cody1 May 20 at 8:00
2
$\begingroup$

In complement to Matt L.'s as-usually-excellent-answer, some additional bits on the intuition, a simplification of the problem (to ease resolution) and the construction of a counter-example. They could be useful to understand and solve similar time-invariant/shift-invariant questions.

First, on the intuition: the system contains a dilation on the time variable ($x(\tau) \to x(-3\tau) $). This is a strong suspicion that the system could be time-variant, because dilations vary as a multiplying factor, and not as shifts. It is not a proof though, but sometimes it is easier to find a single counter-example than to disprove the claims.

Second, on simplification. Some exercises are full of traps. So sometimes, it can be worth rewriting it in a simpler manner (and limit subsequent computation errors). Here you have a $-3\tau$ in the integrand, and a $-t$ on the integral bounds. This can be a cause of sign mistakes.

By a change of variable $u\mapsto -3\tau$, you can convert

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

into a simpler form:

$$y(t)=\int^{3t}_{-\infty}x(u)du\tag{2}$$

where the potential time-variance appear in the bounds, which provides us with an intuition for a counter-example: a function whose integral will behave in a non-invariant manner with $t$.

Third, on counterexamples. From here, we see that the integral will somehow truncate functions. Let us test this hypothesis against some simple shiftable family of functions: let $\mathbf{1_{T}}(t)$ denote the unit window, on the interval $[T,T+1]$.

The corresponding function family $y_T(t)$ is:

  • zero when $3t<T$
  • $3t-T$ when $T\le 3t< T+1$
  • $1$ when $T+1 \le 3t$

If you look at the interval $[T,T+1]$, $y_T(t) = 3t-T$, and the family of solutions is not time-invariant there, so not time-invariant globally.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Great example thanks! $\endgroup$ – cody1 May 21 at 23:04
  • 2
    $\begingroup$ +1, and not only for the "as-usually-excellent" :) $\endgroup$ – Matt L. May 24 at 15:52
  • 1
    $\begingroup$ I know that your rank is far above your ego:) $\endgroup$ – Laurent Duval May 24 at 15:58

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.