Just for clarity, a system is invertible if distinct inputs lead to distinct outputs.
That said, I have two systems, a continuous time system and a discrete time system:
(1) $y(t) = x(2t)$ (Continuous time) and
(2) $y[n] = x[2n]$ (Discrete Time)
I know that the CT system is invertible, $y(t) = x(t/2)$. However, my book says that the DT system is not invertible, citing that $x_{1}[n]=\delta[n-1] \rightarrow y[n] = 0$ and $x_{2}[n]=0 \rightarrow y[n] = 0$. Seeing that these two distinct inputs yield the same output, I see that it's not invertible.
My question, then, is why doesn't the same counter-example used to show that the DT system isn't invertible work on the CT system? That is, why doesn't $x_{1}(t) =\delta(t-1) \rightarrow y(t) = 0$ and $x_{2}(t)=0 \rightarrow y(t) = 0$?
I suspect that there is a fundamental difference between CT and DT delta functions that I'm missing. From my book, it states that $\delta[n] \equiv 0$ if $n \neq 0$, $1$ if $n = 0$. For CT, $\delta(t) = \frac{du(t)}{dt}$, where $u(t)$ is the unit step function.
Any insight would be appreciated.
In the CT system, we would calculate $$ y[\delta(t-1)] = \delta(2t - 1) $$ whereas $$ y[0] = 0 $$ Note that $\delta(2t - 1) \neq 0$. In particular, this function is "non-zero" at $t = 1/2$.
In the discrete system, there is no "$n = 1/2$" that would cause distinction between the two outputs.