I'm confused on how to test this for time invariance. What is $x(t-t_0)$? Is it $x(2t-t_0)$ or $x(2(t-t_0))$?
If it is the former, it would be considered time varying right? Because $x(2t-t_0)$ is not equal to $y(t-t_0)=x(2(t-t_0))$
Is $y(t) = x(2t)$ time invariant?
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1 Answers
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Let $(\sigma_Tx)(t) = x(t-T)$, then an operator $F$ is time invariant iff $\sigma_T F = F \sigma_T$ for all relevant $T$.
That is, $\sigma_T (Fx) = F (\sigma_T x)$ for all $x$.
You have $(Fx) (t) = x(2t)$, hence we have $(\sigma_T (Fx))(t) = x(2 (t-T)) = x(2t -2T)$ and $(F(\sigma_T x))(t) = F(s \mapsto x(s-T))(t) = x(2t-T)$.
If we choose some $x$ such that $x(0) \neq x(T)$, then we see that $\sigma_T (Fx) \neq F (\sigma_T x)$ and so $F$ is not time invariant.