# Impulse invariance vs. DT representation of a CT system: Where is the inconsistency?

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, $$h[n]$$ 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?

• i don't think you're representing the result in the wikipedia article quite faithfully. and what is "$\omega$" in your bottom equation? – robert bristow-johnson Oct 22 '20 at 15:30
• Thanks, I fixed the $\omega$. What is not faithfully represented from the wikipedia article? – divB Oct 22 '20 at 15:47
• $$h[n] = T \ h_c(nT)$$ $$H(e^{j\omega}) = \sum\limits_{k=-\infty}^{\infty} H_c \left( j\tfrac{\omega}{T} + j\tfrac{2\pi k}{T}\right)$$ – robert bristow-johnson Oct 22 '20 at 16:01