# Proving linearity and time-invariance

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.

• 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. Feb 4 '20 at 12:23
• @MattL. Edited the question with my try.
– A Q
Feb 4 '20 at 12:50
• $z=\tau-t_o$ is what you meant instead of $\tau-t$ Feb 4 '20 at 14:26
• @jomegaA Yes. You are right.
– A Q
Feb 5 '20 at 16:59

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 $$t_0$$.