Consider the following system:
$$y(t-1)=\int_{-\infty}^\infty x(π)u(π-t) dπ $$
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 $u(Οβt)=0$ for all $Ο<t$ or, equivalently, for all $t>Ο$ and the integrand is zero in range $({-\infty}, t)$.
Therefore, we can show that: $$\int_{-\infty}^\infty x(π)u(π-t) dπ = \int_{t}^\infty x(π)u(π-t) dπ = \int_{t}^\infty x(π) dπ$$ So, the system is not causal! Am I right?!
$(2)$ Is the system time-invariant? Why or why not?
Let $ s=t-1$
$$y(s)=\int_{-\infty}^\infty x(π)u(π-s+1) dπ $$
For invariance, we need to show output resulted from $x(sβs_o)$ = $y(sβs_o)$.
$$y_1(s)=\int_{-\infty}^\infty x(π-s_0)u(π-s+1) dπ $$ Now when I change of variable to $z=Οβs_o$, $Ο=z+s_0$ , $dz=dΟ$ 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: $$y_2(s-s_0)=\int_{-\infty}^\infty x(π)u(π-s+1-s_0) dπ $$
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.