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I need to do an absolute integral of my impulse response of my LTI system so I can find out if the system is stable or not. The general formula is:
$\int_{-\infty}^{\infty} \! |h(t)| \, \mathrm{d} t$ so after plugging in my $h(t)$ my integral looks like this:

$\int_{-\infty}^{\infty} \! |\frac{1}{2}e^{-3t}u(t)+\frac{1-e^{-3t}}{6}\delta(t)| \, \mathrm{d} t$

So since there is an absolute value does that mean we lose the step function $u(t)$ and if (or not) so, then is the correct answer:
$\frac{1}{6}(1-e^{-3t})u(t)\bigg|^{\infty}_{-\infty}$ which then evaluates to $\frac{1}{6}$.

I guess the only question I have is does the absolute value remove the step function and is my final answer, $\frac{1}{6}$ correct?

share|cite|improve this question
The hole integrand is nonnegative as $e^{-3t} \le 1$, hence you can loose the absolute value. – Pragabhava Oct 25 '12 at 2:58

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