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.

  • 1
    $\begingroup$ 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 $x(-t)$. $\endgroup$ May 5, 2015 at 19:23
  • $\begingroup$ @DilipSarwate: edited ... $\endgroup$
    – Matt L.
    May 6, 2015 at 16:11

1 Answer 1


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


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


The response to $x_2(t)$ is


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


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.

  • 1
    $\begingroup$ 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. $\endgroup$ May 16, 2015 at 9:15

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.