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This question was asked in one of our quiz. Question is

Determine if the system with the input-output relationship given by

$$y(t) = \int_{t-10}^{t} \cos(\tau)x(\tau)d\tau$$

is linear and time invariant.

I was able to prove linearity but was not able to prove the time invariance of the system.


My Try

For invariance we need to show output resulted from $x(t-t_o)$ is same as $y(t-t_o)$. Let $$y_1(t) = \int^{t}_{t-10} \cos(\tau)x(\tau-t_o)d\tau.$$ Now when I change of variable to $z=\tau - t_o$, $dz = d\tau$ it leads to $$y_1(t) = \int^{t-t_o}_{t-10-t_o} \cos(z+t_o)x(z)dz$$ where I get stuck (using $\cos(A+B)$ identity doesn't seem to work well) and not able to proceed further.

For $y(t-t_o)$ integral becomes $$y_2(t-t_o) = \int^{t-t_o}_{t-10-t_o} \cos(\tau)x(\tau)d\tau$$

From here it seems to time variant system. But when $t_o=2\pi n$ it becomes time invariant. But system cannot be time invariant and time variant. That's the doubt I have with this question whether the system is time invariant or time variant and why.

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  • $\begingroup$ If you show us how you tried proving time invariance, we can tell if you made a mistake or not. Right now, the only option we have is solve the problem for you, which is not what this site is about. $\endgroup$ – Matt L. Feb 4 at 12:23
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    $\begingroup$ @MattL. Edited the question with my try. $\endgroup$ – A Q Feb 4 at 12:50
  • $\begingroup$ $z=\tau-t_o$ is what you meant instead of $\tau-t$ $\endgroup$ – jomegaA Feb 4 at 14:26
  • $\begingroup$ @jomegaA Yes. You are right. $\endgroup$ – A Q Feb 5 at 16:59
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Your proof is correct, and the result is that the system is time varying because the response to $x(t-t_0)$ does not in general equal a delayed response to the input $x(t)$. Of course, for the special case $t_0=2\pi k$ the delayed response to $x(t)$ equals the response to $x(t-t_0)$, but for time-invariance this equality must hold for any $t_0$.

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