Laplace Transform ROC Properties Analysis

This solution analyzes the correctness of the given statements regarding the Region of Convergence (ROC) of the Laplace transform.

Statement 1 Analysis: ROC Shape

Statement: The ROC of $X(s)$ consists of strips parallel to the $j\omega$-axis in the s-plane.

Reasoning: The ROC is determined by the convergence of the Laplace integral $\int_{-\infty}^{\infty} x(t) e^{-st} dt$. For various types of signals (causal, anti-causal, or mixed), the ROC typically defines a region bounded by vertical lines in the complex s-plane, which are parallel to the imaginary axis ($j\omega$-axis). This is because convergence often depends on the real part of $s$ ($\sigma = \text{Re}\{s\}$).

Conclusion: Statement 1 is correct.

Statement 2 Analysis: ROC and Poles

Statement: For rational Laplace transforms, the ROC does not contain any poles.

Reasoning: Poles are the values of $s$ where the Laplace transform $X(s)$ becomes infinite. The ROC is defined as the region where $X(s)$ converges (is finite). Therefore, by definition, the ROC must exclude the locations of all poles.

Conclusion: Statement 2 is correct.

Statement 3 Analysis: Finite Duration & Integrable Signals

Statement: If $x(t)$ is of finite duration and is absolutely integrable, then the ROC is the entire s-plane.

Reasoning: An absolutely integrable signal satisfies $\int_{-\infty}^{\infty} |x(t)| dt < \infty$. A finite duration signal $x(t)$ is non-zero only over a finite interval $[t_1, t_2]$. The integral $\int_{t_1}^{t_2} |x(t)| dt$ is finite if $x(t)$ is bounded within this interval. The Laplace transform integral $\int_{t_1}^{t_2} x(t) e^{-st} dt$ converges for all values of $s$ because $e^{-st}$ is bounded for finite $t$ and finite $\text{Re}\{s\}$. Thus, the ROC covers the entire s-plane.

Conclusion: Statement 3 is correct.

Final Conclusion

Since statements 1, 2, and 3 are all correct, the option including all three is the right choice.

• Statement 1: Correct
• Statement 2: Correct
• Statement 3: Correct