Question

Given X(s) = $\frac{1}{s+2}$, ROC: Re(s) $>$ -2. Signal is:

• A. e$^{-2t}$u(t)
• B. -e$^{-2t}$u(-t)
• C. e$^{2t}$u(t)
• D. e$^{-2t}$u(-t)

Hint:

The algebraic expression of the Laplace transform is not unique by itself.
The ROC is mandatory to uniquely define the inverse transform:
$Re(s) > \sigma$ indicates a right-sided signal ($u(t)$).
$Re(s) < \sigma$ indicates a left-sided signal ($u(-t)$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks for the inverse Laplace transform of a first-order transfer function, given its algebraic form and its Region of Convergence (ROC).

Step 2: Key Formula or Approach:
The Laplace transform of $e^{-at}u(t)$ (a right-sided causal signal) is:
\[ \mathcal{L}\{e^{-at}u(t)\} = \frac{1}{s+a}, \quad \text{ROC: } Re(s) > -a \]
The Laplace transform of $-e^{-at}u(-t)$ (a left-sided anti-causal signal) is:
\[ \mathcal{L}\{-e^{-at}u(-t)\} = \frac{1}{s+a}, \quad \text{ROC: } Re(s) < -a \]

Step 3: Detailed Explanation:

• We are given:
\[ X(s) = \frac{1}{s+2} \]
• This function has a single pole located at $s = -2$.

• The given Region of Convergence (ROC) is:
\[ Re(s) > -2 \]
• The ROC lies to the right of the pole $s = -2$ (indicated by the "$>$" sign).

• In Laplace transform theory, an ROC that is a right-half plane (to the right of the rightmost pole) uniquely corresponds to a right-sided (causal) time-domain signal.

• Matching this with the standard causal transform pair where $a = 2$:
\[ e^{-2t} u(t) \longleftrightarrow \frac{1}{s+2}, \quad \text{ROC: } Re(s) > -2 \]
• Therefore, the inverse transform of $X(s)$ is $e^{-2t}u(t)$.

Step 4: Final Answer
Thus, the time-domain signal is $e^{-2t}u(t)$, which matches option (A).