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If we have a system with an impulse defined as: $$h(t)=u(t)-u(t-2)$$ Then the Laplace Transform of h(t) would be the transfer function: $$H(s)=\frac{1}{s}-\frac{e^{-2s}}{s}, \quad Re(s)>0$$

We also know that a stable system must have a RoC region that passes through the imaginary axis. In our case for $h(t)$, $\textrm{Re}(s)>0$ so it doesn't pass the imaginary axis, indicating that the system is not stable.

However, the system is stable because $$\int_{-\infty}^{\infty}|h(t)| \leq M$$

So my question is how is this possible? Am I missing something in understanding the stability condition in Laplace Transform?

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Note that for $s\to 0$, the transfer function $H(s)$ doesn't have a pole because also the numerator approaches zero. In fact, you have

$$\lim_{s\to 0}H(s)=2\tag{1}$$

The ROC is the whole $s$-plane, except for $\textrm{Re}(s)\to -\infty$.

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  • $\begingroup$ Thank you! This clears up a lot of stuff. It also explains why we can take the inverse Laplace as u(t) or -u(-t) and get back the same result. $\endgroup$ – Muaath Asali Nov 23 '20 at 13:22

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