2
$\begingroup$

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 $$\int_{-\infty}^{\infty}|h(t)|dt\lt \infty \nRightarrow \int_{-\infty}^{\infty}|h(t)|^2dt\lt \infty$$

$\endgroup$
1
$\begingroup$

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 $$E_y \le M^2 E_g$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 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 at 17:09
  • 1
    $\begingroup$ Looks good to me -- at least using "engineering" math. I think this is a nice result. $\endgroup$ – MBaz Oct 23 at 0:20
  • 1
    $\begingroup$ @MBaz Thank you so much for reading my answer. $\endgroup$ – S.H.W Oct 23 at 20:51
0
$\begingroup$

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
| improve this answer | |
$\endgroup$
  • $\begingroup$ Could you elaborate more, please? $\endgroup$ – S.H.W Oct 15 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 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 at 18:05
  • 3
    $\begingroup$ @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 at 18:17
  • $\begingroup$ @MBaz Do you think that the statement is true? $\endgroup$ – S.H.W Oct 15 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.