Let $\mathcal{L}$ be a stable LTI system. Is it true that if input has finite energy then output also has finite energy? I'm not sure about that. We know that $$\int_{-\infty}^{+\infty}|h(t)|dt\lt\infty \tag{1}$$Where $h(t)$ is the impulse response. Also we have $$\int_{-\infty}^{+\infty}|y(t)|dt = \int_{-\infty}^{+\infty}|Y(s)|ds = \int_{-\infty}^{+\infty}|X(s)||H(s)|ds \tag{2}$$Since $y(t) = \mathcal{L}(x(t)) = x(t)\star h(t)$ which implies $Y(s) = X(s)H(s)$. Applying Cauchy–Schwarz inequality to $(2)$, $$\left(\int_{-\infty}^{+\infty}|X(s)||H(s)|ds\right)^2 \le \left(\int_{-\infty}^{+\infty}|X(s)|^2ds\right)\left(\int_{-\infty}^{+\infty}|H(s)|^2ds\right) \tag{3}$$We know that $$\int_{-\infty}^{+\infty}|X(s)|^2ds = \int_{-\infty}^{+\infty}|x(t)|^2dt<\infty$$Since input is an energy signal but $$\int_{-\infty}^{+\infty}|H(s)|^2ds$$doesn't necessarily exists. So is this indicates that we can find a counterexample to the statement or we can prove that by other methods?

Edit: Here is a counterexample which shows $$\int_{-\infty}^{\infty}|h(t)|dt\lt \infty \nRightarrow \int_{-\infty}^{\infty}|h(t)|^2dt\lt \infty$$


I think I've found the answer. Please correct me if I'm wrong. First of all, I've made a silly mistake $$\int_{-\infty}^{+\infty}|y(t)|dt = \int_{-\infty}^{+\infty}|Y(s)|ds$$which is clearly false. Let $y(t) = x(t)\star h(t)$. We have $$E_y = \int_{-\infty}^{+\infty}|y(t)|^2dt = \int_{-\infty}^{+\infty}|Y(s)|^2ds = \int_{-\infty}^{+\infty}|H(s)X(s)|^2ds = \int_{-\infty}^{+\infty}|H(s)|^2|X(s)|^2ds$$Also we have $$|H(s)| = \left|\int_{-\infty}^{+\infty}e^{-2\pi ist}h(t)dt \right | \le \int_{-\infty}^{+\infty}|e^{-2\pi ist}h(t)|dt = \int_{-\infty}^{+\infty}|h(t)|dt \lt\infty$$So $\exists M\in\mathbb{R}:\ \ |H(s)|\le M$ for all $s$. This means that $|H(s)|^2\le M^2$ and then $$\int_{-\infty}^{+\infty}|H(s)|^2|X(s)|^2ds\le M^2\int_{-\infty}^{+\infty}|X(s)|^2ds$$By assumption $$E_x = \int_{-\infty}^{+\infty}|x(t)|^2dt=\int_{-\infty}^{+\infty}|X(s)|^2ds$$The result is $$E_y \le M^2 E_g$$

  • $\begingroup$ Help me understand this. Assume $x(t) = t$. I would say that $x(t) < \infty$ for all $t$ (I would say, though, that $x(t) \rightarrow \infty$ as $t \rightarrow \infty$). However, in this case there is no $M$ such that $x(t) < M$ for all $t$. $\endgroup$ – MBaz Oct 17 '20 at 14:55
  • $\begingroup$ @MBaz Note that by a bounded function we mean there exists $M\gt 0$ such that $|x(t)|\le M$ for all $t \in \mathbb{R}$. Here $|H(s)|$ is bounded by $M = \int_{-\infty}^{+\infty}|h(t)|dt$ and so $|H(s)|\le M$ for all $s$. See this for better explanation. $\endgroup$ – S.H.W Oct 17 '20 at 17:09
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    $\begingroup$ Looks good to me -- at least using "engineering" math. I think this is a nice result. $\endgroup$ – MBaz Oct 23 '20 at 0:20
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    $\begingroup$ @MBaz Thank you so much for reading my answer. $\endgroup$ – S.H.W Oct 23 '20 at 20:51

Rough outline of proof:

  1. A (real-valued) system that's IIR has an impulse response that's $h(t) \ne 0$ for infinitely long
  2. Since $\lvert h(t)\rvert \ge h(t) > 0$ for infinity, it follows that this impulse response doesn't have finite energy
  3. there are stable LTI IIR systems
  • $\begingroup$ Could you elaborate more, please? $\endgroup$ – S.H.W Oct 15 '20 at 17:50
  • $\begingroup$ I could, but I don't know where it'd be necessary. Could you point out what specifically you're unclear about? $\endgroup$ – Marcus Müller Oct 15 '20 at 17:55
  • $\begingroup$ I couldn't follow your reasoning. First of all, why IIR system came into the solution? Second, what do you mean by "for infinity"? Honestly, I didn't understand what you proposed. Are you trying to prove the statement or disproving it? $\endgroup$ – S.H.W Oct 15 '20 at 18:05
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    $\begingroup$ @MarcusMüller It's possible for $h(t) > 0$ for all $t$ but its integral to be finite, if $h(t) \rightarrow 0$ fast enough. For example, a Gaussian pulse. $\endgroup$ – MBaz Oct 15 '20 at 18:17
  • $\begingroup$ @MBaz Do you think that the statement is true? $\endgroup$ – S.H.W Oct 15 '20 at 18:24

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