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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?

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  • $\begingroup$ i don't think you're representing the result in the wikipedia article quite faithfully. and what is "$\omega$" in your bottom equation? $\endgroup$ Commented Oct 22, 2020 at 15:30
  • $\begingroup$ Thanks, I fixed the $\omega$. What is not faithfully represented from the wikipedia article? $\endgroup$
    – divB
    Commented Oct 22, 2020 at 15:47
  • $\begingroup$ $$ h[n] = T \ h_c(nT) $$ $$ H(e^{j\omega}) = \sum\limits_{k=-\infty}^{\infty} H_c \left( j\tfrac{\omega}{T} + j\tfrac{2\pi k}{T}\right) $$ $\endgroup$ Commented Oct 22, 2020 at 16:01

2 Answers 2

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The Impulse Invariant method does not promise to represent the frequency response of the continuous-time system as a lowpass version. What the Impulse Invariant method does is frequency-alias the frequency response by sliding it by multiples of the sampling frequency and adding up all of the translated copies.

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There's nothing wrong, per se. There's two different methods, each of which represents an approximation of the continuous-time system in discrete time.

Each one is an approximation. You can't get around that.

In a real system design that starts with some continuous-time prototype, it's up to the system designer to decide what is the most appropriate approximation* to use.

* There is exactly one case that I know of where you can make an exact discrete-time model of a continuous-time system. But that's in the restricted case where you're exciting the continuous-time system with a ZOH (i.e., a DAC) driven at the sampling rate, then sampling the continuous-time system's output. This is very handy for control system design, because you can exactly model the behavior of a continuous-time plant from the perspective of a discrete-time controller. But it otherwise doesn't have a lot of application.

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