0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ i don't think you're representing the result in the wikipedia article quite faithfully. and what is "$\omega$" in your bottom equation? $\endgroup$ – robert bristow-johnson Oct 22 at 15:30
  • $\begingroup$ Thanks, I fixed the $\omega$. What is not faithfully represented from the wikipedia article? $\endgroup$ – divB Oct 22 at 15:47
  • $\begingroup$ $$ h[n] = T \ h_c(nT) $$ $$ H(e^{j\omega}) = \sum\limits_{k=-\infty}^{\infty} H_c \left( j\tfrac{\omega}{T} + j\tfrac{2\pi k}{T}\right) $$ $\endgroup$ – robert bristow-johnson Oct 22 at 16:01
0
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.