Determining Causality and Time-Invariance of a system

Question

Consider the following system: $$y(t-1)=\int_{-\infty}^\infty x(𝜏)u(𝜏-t) d𝜏 $$ where $u(t)$ is the unit step function, which is zero for $t<0$ and equals $1$ for $t>0$.

$(1)$ Is the system causal? Why or why not?

I think if $u(t)=0$ for all $t<0$. This means that $u(τ−t)=0$ for all $τ<t$ or, equivalently, for all $t>τ$ and the integrand is zero in range $({-\infty}, t)$.
Therefore, we can show that: $$\int_{-\infty}^\infty x(𝜏)u(𝜏-t) d𝜏 = \int_{t}^\infty x(𝜏)u(𝜏-t) d𝜏 = \int_{t}^\infty x(𝜏) d𝜏$$ So, the system is not causal! Am I right?!

$(2)$ Is the system time-invariant? Why or why not?

Let $ s=t-1$
$$y(s)=\int_{-\infty}^\infty x(𝜏)u(𝜏-s+1) d𝜏 $$
For invariance, we need to show output resulted from $x(s−s_o)$ = $y(s−s_o)$.
$$y_1(s)=\int_{-\infty}^\infty x(𝜏-s_0)u(𝜏-s+1) d𝜏 $$ Now when I change of variable to $z=τ−s_o$, $τ=z+s_0$ , $dz=dτ$ it leads to: $$y_1(s)=\int_{-\infty}^\infty x(z)u(z+s_0-s+1) dz $$
For $y(s-s_o)$ integral becomes: $$y_2(s-s_0)=\int_{-\infty}^\infty x(𝜏)u(𝜏-s+1-s_0) d𝜏 $$
So it seems to time-variant system.
Update:I would appreciate if anyone can answer my question because it's not a homework anymore and yet I didn't find the answer.

Answer 1

Hint

Substitute $s = t-1$. That gets you the equations in a more standard form $y(s) = ...$ Then go through the same excercise.

Answer 2

Concerning causality, your conclusion is correct. When you tried to investigate time-invariance you failed because you're not sufficiently careful when substituting variables. The first error occurred already when formulating the input-output equation for $y(t)$. Using $s=t-1$ means that $\tau-t$ becomes $\tau-(s+1)=\tau-s-1$, and hence you should obtain

$$y(s)=\int_{-\infty}^{\infty}x(\tau)u(\tau-s-1)d\tau\tag{1}$$

Now you're making the same mistake when you try to evaluate $y_2(s-s_0)$. Replacing $s$ by $s-s_0$ doesn't result in what you wrote down. If you do things right, you should see that the system is in fact time-invariant.

Time-invariance is also directly obvious from the original input-output equation, because it can be seen that the output signal is given by a convolution integral, and only LTI systems can be represented by convolution.