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It is my understanding that an LTI system is BIBO stable if and only if its impulse response $h(t)$ is absolutely integrable. This also happens to be one of the Dirichlet conditions for the convergence of the Fourier Transform, from which we can conclude that $H(j\omega)$, the frequency response of the system (i.e. the Fourier transform of the impulse response) converges if $h(t)$ is absolutely integrable.

Now think of an ideal low-pass filter. Its impulse response $h(t)$ is a $\mathrm{sinc}$ function which is not absolutely integrable. In other words, it is not BIBO stable. However, it does have a Fourier transform despite violating the condition of BIBO stability (or equivalently one of the Dirichlet conditions). Why is that? Is it because the Dirichlet conditions are a set of sufficient but not necessary conditions? Or is it because the Fourier transform of a $\mathrm{sinc}$ function doesn't "exist" in the usual sense unless we allow step functions?

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The same happens with sinusoids: they have a Fourier transform even though they are not absolutely summable in the time domain. These are cases in which the Fourier transform doesn't converge uniformly, but it does converge in some other sense.

The signal $x(n)=\mathrm{sinc}(n)$ is not absolutely summable... but it is square-summable (i.e. it has finite energy). Therefore, the Fourier transform will converge but in the mean-square sense. In these cases, the transform converges to a discontinuous function (in the case of this specific signal, to a rectangular pulse).

Similarly, sequences that involve sinusoids like $x(n)=\sin(\omega_0n)$ are neither absolutely nor square summable, but a Fourier transform can be defined using distributions (Dirac delta functions).

In both cases the Fourier transforms are not continuous. They are not infinitely differentiable either.

As a general rule we can state that if some impulse response $h(n)\in l^1(\mathbb{Z})$, then its Fourier transform will be continuous. If the Fourier transform is not continuous, then you know that its inverse is not in the sequence space $l^1$.

You mentioned BIBO stability, so I think this may be of interest. There is a theorem that states:

An LTI system with impulse response $h(n)$ is BIBO stable if and only if $h(n)\in l^1(\mathbb{Z})$.

So one can judge the stability of a system by looking at the continuity of its transfer function.

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electrical engineers sometimes play fast-and-loose with some of the mathematics, particular from the P.O.V. of mathematicians.

consider the ideal reconstruction of this continuous-time signal from its samples:

$$ x(t) = \sum\limits_{n=-\infty}^{+\infty} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) $$

it's not hard to see that $x(nT)=x[n]$. one might think that if finite samples $x[n]$ went into the reconstruction, a finite output $x(t)$ would come out. but this sequence will result in a spike of $\infty$ in height:

$$ x[n] = \begin{cases} (-1)^{n+1} \qquad & n<0 \\ (-1)^n \qquad & n \ge 0 \\ \end{cases} $$

consider the value of $x(t)$ when $t=-\tfrac{T}2$:

$$\begin{align} x(t) &= \sum\limits_{n=-\infty}^{+\infty} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\ &= \sum\limits_{n=-\infty}^{-1} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) + \sum\limits_{n=0}^{+\infty} x[n] \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\ &= \sum\limits_{n=-\infty}^{-1} (-1)^{n+1} \operatorname{sinc} \left(\tfrac1T(t-nT) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\ &= \sum\limits_{n=-\infty}^{-1} -(-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\ &= \sum\limits_{n=1}^{+\infty} -(-1)^{-n} \operatorname{sinc} \left(\tfrac1T(t-(-n)T) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\ &= -\sum\limits_{n=1}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t+nT) \right) + \sum\limits_{n=0}^{+\infty} (-1)^n \operatorname{sinc} \left(\tfrac1T(t-nT) \right) \\ &= \operatorname{sinc} \left(\tfrac{t}T \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \left( \operatorname{sinc}\left(\tfrac1T(t-nT) \right) - \operatorname{sinc}\left(\tfrac1T(t+nT) \right)\right) \\ \end{align}$$

now when $t=-\tfrac{T}2$:

$$\begin{align} x\left(-\tfrac{T}2 \right) &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \Big( \operatorname{sinc}\left(-\tfrac12-n \right) - \operatorname{sinc}\left(-\tfrac12+n \right) \Big) \\ &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \Big( \operatorname{sinc}\left(n+\tfrac12\right) - \operatorname{sinc}\left(n-\tfrac12\right) \Big) \\ &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \left( \frac{\sin\left(\pi n+\tfrac\pi 2\right)}{\pi n+\tfrac\pi 2} - \frac{\sin\left(\pi n-\tfrac\pi 2\right)}{\pi n-\tfrac\pi 2} \right) \\ &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \left( \frac{\cos\left(\pi n\right)}{\pi n+\tfrac\pi 2} - \frac{-\cos\left(\pi n\right)}{\pi n-\tfrac\pi 2} \right) \\ &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n \cos\left(\pi n\right) \left( \frac{1}{\pi n+\tfrac\pi 2} + \frac{1}{\pi n-\tfrac\pi 2} \right) \\ &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} (-1)^n (-1)^n \left( \frac{1}{\pi n+\tfrac\pi 2} + \frac{1}{\pi n-\tfrac\pi 2} \right) \\ &= \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} \frac{2 \pi n}{(\pi n)^2 - (\tfrac\pi 2)^2} \\ & > \operatorname{sinc} \left(\tfrac12 \right) + \sum\limits_{n=1}^{+\infty} \frac{2 \pi n}{(\pi n)^2 } \\ & = \operatorname{sinc} \left(\tfrac12 \right) + \frac{2}{\pi}\sum\limits_{n=1}^{+\infty} \frac{1}{n} \to \infty \\ \end{align}$$

this is the output of a filter with impulse response

$$ h(t) = \tfrac1{T} \operatorname{sinc} \left( \tfrac{t}{T} \right) $$

and input

$$ x_\mathrm{s}(t) = T \sum\limits_{n=-\infty}^{+\infty} x[n] \delta(t-nT) $$

now you might rightfully say that the input (with those dirac impulse functions) is not bounded. that is correct.

so now consider this other input that is bounded:

$$ x_\mathrm{r}(t) = \sum\limits_{n=-\infty}^{+\infty} x[n] \operatorname{rect}\left(\tfrac1{T}(t-nT)\right) $$

where $\operatorname{rect}(u) \triangleq \begin{cases} 1 \qquad & |u|<\tfrac12 \\ \tfrac12 \qquad & |u|=\tfrac12 \\ 0 \qquad & |u|>\tfrac12 \\ \end{cases} $

with $ x[n] = \begin{cases} (-1)^{n+1} \qquad & n<0 \\ (-1)^n \qquad & n \ge 0 \\ \end{cases} $,

then $x_\mathrm{r}(t)$ is bounded

$$ |x_\mathrm{r}(t)| \le 1 \qquad \forall t \in \mathbb{R} $$

now do you thing that infinite sum will come out differently? (consider the commutative and associative properties of the multiplication operation. then consider of those properties should apply to convolution.)

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  • $\begingroup$ i have to admit that i haven't closed the deal on the last point yet. i gotta think about it and i wouldn't mind help or suggestions from PeterK. MattL. or most certainly Jazz. or anyone else. $\endgroup$ – robert bristow-johnson Feb 24 '18 at 7:48

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