2
$\begingroup$

It is well known that ideal lowpass filter, i.e. the lowpass filter whose impulse response is $h(t) = \text{sinc}(t)$, is not BIBO-stable because $h(t)$ is not absolutely integrable. However, think about the ideal reconstruction process as a process that takes discrete signals to continuous time band-limited signals. In formulas, if $x_n$ is a discrete signal, then the corresponding output is \begin{equation} x(t) = \sum_{k = -\infty}^{+\infty} x_k \text{sinc}\left(\frac{t - kT}{T}\right). \end{equation} This process can be seen as the composition of a DAC, that transforms a discrete signal into the corresponding impulse-train continuous signal, and the ideal lowpass filter. Is this process BIBO?

$\endgroup$

1 Answer 1

2
$\begingroup$

Is this process BIBO?

Not really. First, let's write it correctly

\begin{equation} x(t) = \sum_{k = -\infty}^{+\infty} x_k \text{sinc}(\frac{t - kT}{T}). \end{equation}

We can choose a bounded input sequence $x_k$ that maximizes the output as

$$x_k = \text{sign}( \text{sinc}(\frac{t - kT}{T}))$$ where $\text{sign}()$ is the signum function.

Let's evaluate that at $t = T/2$

$$x(\frac{T}{2}) = \sum_{k = -\infty}^{+\infty} |\text{sinc}(\frac{1}{2}-k)| = \sum_{k = -\infty}^{+\infty} |\frac{\sin(\pi/2- k\pi)}{(k-1/2)\pi} | \\= \sum_{k = -\infty}^{+\infty} \frac{1}{|k-1/2|\pi} $$ which does not converge.

In practice this makes little difference since ideal reconstruction is not possible and every real DAC has a casual (but not ideal) lowpass filter.

$\endgroup$
2
  • $\begingroup$ Thank you, your answer is definitely convincing. I was assuming $T = 1$ just for ease of notation. $\endgroup$
    – avril_14th
    Mar 3, 2022 at 15:22
  • 1
    $\begingroup$ Sorry for being a stickler but that would be $T = 1s$ A non-trivial difference between continuous and discrete signals is that continuous signals are function of a continuous time variable with units of seconds. Discrete signal are a function of a unit-less index. $\endgroup$
    – Hilmar
    Mar 3, 2022 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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