Is the ideal reconstruction process BIBO stable?

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 $$$$x(t) = \sum_{k = -\infty}^{+\infty} x_k \text{sinc}\left(\frac{t - kT}{T}\right).$$$$ 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?

Is this process BIBO?

Not really. First, let's write it correctly

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

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.

• Thank you, your answer is definitely convincing. I was assuming $T = 1$ just for ease of notation. Mar 3, 2022 at 15:22
• 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. Mar 3, 2022 at 16:17