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- #1

\[

x(t) = e^{-5t}\mathcal{U}(t) + e^{-\beta t}\mathcal{U}(t)

\]

and denote the its Laplace transform by \(X(s)\).

What are the constraints placed on the real and imagainary parts of \(\beta\)

if the region of convergence of \(X(s)\) is \(\text{Re} \ \{s\} > -3\)?

Let's start by taking the Laplace transform.

\begin{align*}

X(s) &= \int_0^{\infty}e^{-5t}\mathcal{U}(t)e^{-st}dt +

\int_0^{\infty}e^{-\beta t}\mathcal{U}(t)e^{-st}dt\\

&= \int_0^{\infty}e^{-5t}e^{-st}dt +

\int_0^{\infty}e^{-\beta t}e^{-st}dt\\

&= \frac{1}{s + 5} + \frac{1}{s + \text{Re} \ \{\beta\}}

\end{align*}

The first integral gives us a region of convergence of \(\sigma > -5\). For \(\beta\), we need the real part to be \(3\). Then the second integral would have a region of convergence of \(\sigma_1 > -3\). With the addition of the integrals, how does that affect the overall region of convergence? Do we just take the lesser of the two or is there something else?