I have the following input output equation that I have been told is not time invariant; however, I am not sure why this is the case.
$$y(t)=\int_{-5}^5x(\tau)\space{}d\tau$$
I am not sure how this would vary with time considering $t$ doesn't appear in the integral.
Thanks.
Edit:
I was given the following explanation:
$$x_a(t)=x(t-a)$$ $$y_a(t)=\int_{-5}^5x(\tau)\space{}d\tau=\int_{-5}^5x(\tau-a)\space{}d\tau$$ $$=\int_{-5-a}^{5-a}x(\tau)\space{}d\tau$$ $$y(t-a)=\int_{-5}^5x(\tau)\space{}d\tau \space{}\text{(Note: t is not a factor)}$$
$$y(t-a)\neq{}y_a(t)$$ So not time invariant
Which I am having trouble understanding.
This is a tricky question. The output of the integrator is indeed a constant value, independent of $t$. This, however, does not necessarily imply time invariance.
Note that the output of the system is the integral of the input signal over the interval $[-5, 5]$. Let's denote the output of the system with input $x(t)$ as $y_0(t)=y_0, t\in\mathbb{R}$. Now, if the input is shifted by $a$ before application to the filter, the integration will still be performed over the interval $[-5, 5]$, however, on the shifted version, $x(t-a)$, of the original input. The output of the system will again be a constant value, say, $y_a(t)=y_a, t \in \mathbb{R}$, that is, in general, different from $y_0$.
Therefore, it holds $y_a(t)=y_a\neq y_0 = y_0(t)=y_0(t-\alpha)$, which means that the time invariance property does not hold.