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I stumbled upon a false/true statement which goes:

A time-continuous linear system, whose impulse response $c(t) = \frac{1}{\pi t}$ has a pole at the origin, always produces an output signal $y(t)$ with $||y(t)||^2 = \infty$ for an input signal $x(t)$ satisfying $0 < ||x(t)||^2 < \infty$.

It is claimed as false and this fits with my initial assessment, but I don’t know if my explanation is actually right.

With a pole in the origin the system must be stable. The $||\cdot||^2$ is equal to the Energy of the Input $x(t)$ and if it is $\lt \infty$, this means we have a converge Signal. Therefore the energy of the output Signal also has to be $< \infty$.

Would be nice if someone could tell me if I’m in the right direction or could correct me on my assumption.

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2 Answers 2

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The given impulse response, $c(t) = \frac{1}{\pi t}$, is the impulse response of the Hilbert transform. However, there are some issues with the statement you encountered.

A pole at the origin in the frequency domain implies that the system's response increases indefinitely at DC (0 Hz). However, this does not necessarily imply instability in the time domain. Stability in the time domain is typically assessed based on whether the system's impulse response is absolutely integrable, not just based on the presence of a pole. Traditionally, a system is stable if the output is bounded for every bounded input. For a system to be $\ell_2$ stable, the impulse response must be absolutely square integrable. In DSP, this is often referred to as energy stability, meaning the total energy of the output is finite for any input with finite energy.

The Hilbert transform itself is not a causal system, and it does not have a pole-zero plot in the traditional sense used in control theory or DSP. Moreover, the Hilbert transform is not $\ell_2$ stable because its impulse response $c(t) = \frac{1}{\pi t}$ is not absolutely square integrable.

Therefore, the statement that the output signal $y(t)$ has infinite $||y(t)||^2$ for an input signal $x(t)$ satisfying $0 < ||x(t)||^2 < \infty$ could be true for a system with the given impulse response, since the Hilbert transform is not $\ell_2$ stable. However, the conclusion about the system's stability based solely on the pole at the origin is not sufficient. The key point here is the nature of the Hilbert transform and its impulse response in terms of square integrability.

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  • $\begingroup$ "A pole at the origin in the frequency domain implies that the system's response increases indefinitely at DC (0 Hz)." I think this is spot on to the fallacy since the Hilbert Transform is an all-pass in frequency EXCEPT for DC. The Hilbert transform of a constant is 0. $\endgroup$ Jan 14 at 3:02
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I came across an explination, it says:

The Hilbert-Transformation represents an All-pass filter, therefor a gain of 1 is present if f≠0. This means that every average-free input signal of the Hilbert transformer has exactly the same energy as its output Signal. If 0 < ||x(t)||^2 < ∞, then 0 < ||y(t)||^2 = ||x(t)||^2 < ∞. The statement is therefore incorrect.

But I’m not quite sure if I understood it correctly, maybe someone could explain it further.

I came a to the same solution trying to show it mathematical, nevertheless is just assumed it’s a LTI System, therefor this surely isn’t the intended thought behind it. calculation way

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