In section 1.3 from Signals, Systems and Inference, Alan V. Oppenheim, George C. Verghese. Finite-action signals, which are also called absolutely summable signals, are defined by the condition \begin{equation} \int_{-\infty}^{\infty}|x(t)|dt<\infty \end{equation} whereas for discrete time signal, its as \begin{equation} \sum_{k = -\infty}^{\infty}|x[k]|<\infty \end{equation} The integration and sum on the left are called the action of the signal. Therefore also known as finite-action signals. So I would like to know what does finite action signal means? Or is there intuitive explanation about it.
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1$\begingroup$ The definition is straightforward, so why go into transformation theory? $\endgroup$ – Max Mar 20 '18 at 10:40
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$\begingroup$ What exactly don't you understand? Your description suggests you understand what the definition means. It's simply a signal that has a finite amount of energy. $\endgroup$ – Izzo Apr 8 '19 at 20:19
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A finite-action signal is one for which the Fourier transform exists in the conventional sense (i.e. no impulses a.k.a. Dirac deltas allowed). If the function were an impulse response of a LTI system, then the system would be considered to be a BIBO-stable system.
answered Mar 9 '19 at 18:05
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$\begingroup$ Can you please elaborate it without going into transform theory. $\endgroup$ – Meet Mar 15 '19 at 12:03
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It's something like "finite energy signals"
$$ \int_{-\infty}^{\infty}|x(t)|^2 dt<\infty $$
$$ \sum_{n=-\infty}^{\infty}|x[n]|^2 <\infty $$
except that the "finiteness" is a tougher constraint with
$$ \int_{-\infty}^{\infty}|x(t)| dt<\infty $$
$$ \sum_{n=-\infty}^{\infty}|x[n]| <\infty $$
for example, the signal
$$ x(t) = \frac{1}{\sqrt{1+t^2}} $$
is finite energy, but is not finite action.
answered Apr 8 '19 at 21:19
