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I have a system with the following input/output relation:

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

and I want to prove (not graphically/draw) that its not time invariant (TI).

I tried to write down $y(t-T)$ and compare it to the response to $x(t-T)$ but I get both terms equal which shouldn't be the case.

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    $\begingroup$ A signaL cannot be time-invariant unless it happens to have constant value for all time. What you are asking about is a system that transforms its input signal (here denoted by $x(t)$) into the output signal $y(t)$ that for all time instants $t$ has value $x(-t)$. $\endgroup$
    – Dilip Sarwate
    May 5, 2015 at 19:23
  • $\begingroup$ @DilipSarwate: edited ... $\endgroup$
    – Matt L.
    May 6, 2015 at 16:11

1 Answer 1

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Let $y_1(t)$ be the response to the signal $x_1(t)$:

$$y_1(t)=x_1(-t)\tag{1}$$

Now let $x_2(t)$ be a shifted version of $x_1(t)$:

$$x_2(t)=x_1(t-T)\tag{2}$$

The response to $x_2(t)$ is

$$y_2(t)=x_2(-t)=x_1(-t-T)\tag{3}$$

If the system were time-invariant, its response to $x_2(t)$ should be a shifted version of its response to $x_1(t)$:

$$y_2(t)=y_1(t-T)\tag{4}$$

However, from (1) we have $y_1(t-T)=x_1(-(t-T))=x_1(-t+T)$. Comparing this to (3) we see that (4) is not satisfied, and, consequently, the system is not time-invariant. Now that you've seen the proof, try drawing all the signals in order to gain a better understanding.

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    $\begingroup$ I like the clarity of your notation. Most of the tutorials I've seen used very messy notation and I got a feeling they did most of substitutions very non rigorously. $\endgroup$
    – Trismegistos
    May 16, 2015 at 9:15

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