Question

The Laplace transform of \(x_{1}(t)=e^{-t}u(t)\) is \(X_{1}(s)\), where \(u(t)\) is the unit step function. The Laplace transform of \(x_{2}(t)=e^{t}u(-t)\) is \(X_{2}(s)\). Which one of the following statements is TRUE?

• A. The region of convergence of \(X_{1}(s)\) is \(\mathrm{Re}(s)>0\).
• B. The region of convergence of \(X_{2}(s)\) is confined to the left half–plane of \(s\).
• C. The region of convergence of \(X_{1}(s)\) is confined to the right half–plane of \(s\).
• D. The imaginary axis in the \(s\)-plane is included in both the region of convergence of \(X_{1}(s)\) and the region of convergence of \(X_{2}(s)\).

Hint:

- For \(e^{at}u(t)\), ROC: \(\mathrm{Re}(s)>-a\).
- For \(e^{at}u(-t)\), ROC: \(\mathrm{Re}(s)<-a\).
- Always check whether the imaginary axis (\(\mathrm{Re}(s)=0\)) lies inside the ROC.

Answer 1

The Correct Option is D

Step 1: For \(x_{1}(t) = e^{-t}u(t)\), \[ X_{1}(s) = \int_{0}^{\infty} e^{-t} e^{-st}\, dt = \int_{0}^{\infty} e^{-(s+1)t}\, dt. \] Convergence requires: \[ \mathrm{Re}(s) + 1 > 0 \;\;\Rightarrow\;\; \mathrm{Re}(s) > -1. \] Hence the ROC is the half-plane to the right of the vertical line \(\mathrm{Re}(s) = -1\).

Step 2: For \(x_{2}(t) = e^{t}u(-t)\), \[ X_{2}(s) = \int_{-\infty}^{0} e^{t} e^{-st}\, dt = \int_{-\infty}^{0} e^{(1-s)t}\, dt. \] Convergence requires: \[ \mathrm{Re}(1-s) > 0 \;\;\Rightarrow\;\; \mathrm{Re}(s) < 1. \] Thus the ROC is the half-plane to the left of the vertical line \(\mathrm{Re}(s) = 1\).

Step 3: The imaginary axis (\(\mathrm{Re}(s)=0\)) satisfies both: \[ \mathrm{Re}(s) > -1 \quad \text{and} \quad \mathrm{Re}(s) < 1. \] Hence, it lies within both ROCs.

Therefore, the only correct statement is: \[ \boxed{\text{(D)}} \]