# RoC and Stability of a Rectangular Signal

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?

## 1 Answer

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 $$\textrm{Re}(s)\to -\infty$$.

• 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. – Muaath Asali Nov 23 '20 at 13:22