LTI system with impulse response:

$h(t) = e^t \sin(-3t) \text{u}(-t)$

I know the rule and formula , but I am lost in how to do this because of the absolute. Please help.

Sorry if I posted in the wrong section.

Thank you.

  • 1
    $\begingroup$ What do you want to do with this impulse response? What is "the absolute"? $\endgroup$ – msm Nov 19 '16 at 23:49
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    $\begingroup$ the system is stable, but it is also anti-causal. (it's not a real system that anyone can actually build, since it is predictive of the future input.) $\endgroup$ – robert bristow-johnson Nov 20 '16 at 2:19
  • $\begingroup$ @msm by "the absolute" meant the absolute on h(t) , [see navin's answer] $\endgroup$ – Frapto Nov 20 '16 at 6:36
  • $\begingroup$ @robert yeah it is not causal, you know exams have weird stuff :) $\endgroup$ – Frapto Nov 20 '16 at 6:37
  • $\begingroup$ @user6019827 Consider improving your post. I mean writing it in form of an actual question with the necessary details and also grammatically. $|h(t)|$ is read the absolute value of $h(t)$. $\endgroup$ – msm Nov 20 '16 at 6:56

A system is stable if $ \int_{-\infty}^{\infty} |h(t) |dt < \infty$

In your case $ \int_{-\infty}^{\infty} |h(t) |dt = \int_{-\infty}^{\infty} |e^t \sin(-3t)u(-t)| \ dt$

Since $\sin(-3t) = -\sin(3t)$,

$ = \int_{-\infty}^{\infty} | e^t \sin(3t) u(-t)| \ dt$

$ = \int_{-\infty}^{0} e^t |\sin(3t)| \ dt$ , $u(-t) = 1$ if $ -\infty < t < 0$ else $0$

$ < \int_{-\infty}^{0} e^t \ dt < \infty$

So the system is stable.

  • $\begingroup$ Thank you, but I did not understand something: after we changed the bounds of the integral by u(-t) where did the sin(3t) go? $\endgroup$ – Frapto Nov 20 '16 at 6:35
  • $\begingroup$ Integral with sin3t will be less than integral without sin3t which is less than infinity. So the integral with sin3t is also less than infinity.. $\endgroup$ – Navin Prashath Nov 20 '16 at 6:41

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