Proving linearity and time-invariance

Question

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.

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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.

Answer 1

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 $t_0$.