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?


1 Answer 1


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.

  • $\begingroup$ 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

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