Suppose you have a continuous-time (CT) system $h\_c(t)$, bandlimited to $B$. Your goal is to represent the system as a discrete-time (DT) system $h[n]$, sampled at $f\_s \leq 2 B$. Clearly $h[n]$ won't be a faithful representation of $h\_c(t)$ because we do not satisfy the sampling theorem. But I am interested in the exact relationship between $h\_c(t)$ and $h[n]$ in this case. There are two ways to think about this:

### Representing the CT system as DT one

Based on this question (and answer): [Representing a continuous LTI system as a discrete one](https://dsp.stackexchange.com/questions/60677/representing-a-continuous-lti-system-as-a-discrete-one), $h[n]$ represents a **low-pass filtered** version of $h\_c(t)$:

$$
h[n] = \tilde{h}\_c(n T\_s) ,\\
\tilde{h}\_c(t) = \int h\_c(\tau) \operatorname{sinc}\left(t-\tau\right) \operatorname{d}\tau
$$

### Impulse Invariance

According to <https://en.wikipedia.org/wiki/Impulse_invariance>, the sampled $h[n]$ corresponds to an **aliased** $h\_c(t)$:

$$
H(e^{-j\omega}) = \frac{1}{T\_s} \sum\_{k=-\infty}^{\infty} H\_c\left(j\frac{\omega}{T\_s} + j\frac{2\pi}{T\_s}k \right)
$$

I could take the IDTFT:

$$
h[n] = \sum\_{k=-\infty}^{\infty} T\_s e^{-j2\pi k n} h\_c(n T\_s)
$$

which is not the same as the low-pass filtered version above.
There no low-pass filtering.

These two approaches seem to give me two different answers on how to relate $h[n]$ and $h\_c(t)$ (when sampling theorem is not obeyed).
What is wrong?