Question

Find the ROC of \(x(t) = e^{-2t u(t) + e^{-3t} u(t)\).

• A. \(\sigma > 2\)
• B. \(\sigma > 3\)
• C. \(\sigma > -2\)
• D. \(\sigma > -3\)

Hint:

For any system formed by adding multiple right-sided signals (signals containing $u(t)$): - The overall ROC is always to the right of the rightmost pole (the pole with the largest/most positive real value). - Here, compare the poles at $-2$ and $-3$. Since $-2$ is greater than $-3$, the final ROC is immediately defined as $\sigma > -2$.

Answer 1

The Correct Option is C

Understanding the Concept: The Region of Convergence (ROC) defines the range of values for the complex variable $s = \sigma + j\omega$ in the Laplace transform plane for which the transform integral converges toward a finite value. For a right-sided signal containing the unit step function $u(t)$, the ROC takes the form of a half-plane situated to the right of the signal's pole location: $\text{Re}\{s\} = \sigma > p$. When dealing with a sum of multiple right-sided signals, the overall ROC is the intersection (common overlapping area) of the individual ROCs of each component term.

Step 1: Finding the Laplace Transform and ROC for the first term
Let the first term of the signal be $x_1(t) = e^{-2t} u(t)$. The standard Laplace transform pair for an exponential causal function is: \[ e^{-at}u(t) \longleftrightarrow \frac{1}{s+a}, \quad \text{with } \text{Re}\{s\} > -a \] Substituting $a = 2$: \[ X_1(s) = \frac{1}{s+2} \] The pole for this term is located at $s = -2$. Since it is a right-sided signal ($u(t)$), its region of convergence is: \[ \text{ROC}_1: \text{Re}\{s\} = \sigma > -2 \]

Step 2: Finding the Laplace Transform and ROC for the second term
Let the second term of the signal be $x_2(t) = e^{-3t} u(t)$. Substituting $a = 3$ into the standard transform pair: \[ X_2(s) = \frac{1}{s+3} \] The pole for this term is located at $s = -3$. Since this is also a right-sided signal, its region of convergence is: \[ \text{ROC}_2: \text{Re}\{s\} = \sigma > -3 \]

Step 3: Finding the Overlapping Intersection ROC
By the linearity property of the Laplace Transform, the total transform is $X(s) = X_1(s) + X_2(s)$. The overall ROC must satisfy both individual constraints simultaneously: \[ \text{ROC}_{\text{total}} = \text{ROC}_1 \cap \text{ROC}_2 \] We need to find the values of $\sigma$ that satisfy both inequalities: \[ \sigma > -2 \quad \text{and} \quad \sigma > -3 \] Any value of $\sigma$ that is greater than $-2$ is automatically greater than $-3$. However, values between $-3$ and $-2$ (like $-2.5$) satisfy the second condition but fail the first. Therefore, the strict common overlapping region is bounded by the rightmost pole: \[ \sigma > -2 \]