Which of the following statements is true about the two sided Laplace transform?

The correct option is: It has no poles for any bounded signal that is non-zero only inside a finite time interval.

To understand why this statement is true, let us analyze the properties of the two-sided Laplace transform for a finite-duration signal.

Let $x\left(t\right)$ be a bounded signal that is non-zero only within a finite time interval $\left[t_1,t_2\right]$, where both $t_1$ and $t_2$ are finite. Since the signal is bounded, there exists some finite positive constant $M$ such that $\left|x\left(t\right)\right|\le M$ for all $t$.

The two-sided Laplace transform of $x\left(t\right)$ is defined as:

$X\left(s\right)={\int }_{-\infty }^{\infty }x\left(t\right)e^{-st}dt$

Since the signal is non-zero only within the interval $\left[t_1,t_2\right]$, the limits of integration reduce to this finite range:

$X\left(s\right)={\int }_{t_1}^{t_2}x\left(t\right)e^{-st}dt$

Here, the complex variable is $s=\sigma +j\omega$. Let us evaluate the absolute value of $X\left(s\right)$ to determine its convergence:

$\left|X\left(s\right)\right|=\left|{\int }_{t_1}^{t_2}x\left(t\right)e^{-\left(\sigma +j\omega \right)t}dt\right|$

Using the property that the absolute value of an integral is less than or equal to the integral of the absolute value, we get:

$\left|X\left(s\right)\right|\le {\int }_{t_1}^{t_2}\left|x\left(t\right)\right|\left|e^{-\sigma t}\right|\left|e^{-j\omega t}\right|dt$

Since $\left|e^{-j\omega t}\right|=1$ and $\left|x\left(t\right)\right|\le M$, this simplifies to:

$\left|X\left(s\right)\right|\le M{\int }_{t_1}^{t_2}{e}^{-\sigma t}dt$

Evaluating this integral yields:

$\left|X\left(s\right)\right|\le M{\left[\frac{e^{-\sigma t}}{-\sigma }\right]}_{t_1}^{t_2}=\frac{M}{\sigma }\left(e^{-\sigma t_1}-e^{-\sigma t_2}\right)$

Since $t_1$ and $t_2$ are finite, the term $e^{-\sigma t_1}-e^{-\sigma t_2}$ is finite for all finite values of $\sigma$ (the real part of $s$). Even if $\sigma =0$, the limit of the integral exists and equals $M\left(t_2-t_1\right)$, which is finite.

Consequently, the Region of Convergence (ROC) for a finite-duration, bounded signal is the entire $s$-plane (except possibly for $s=\infty$ and/or $s=-\infty$ depending on the signs of $t_1$ and $t_2$). Since the ROC includes the entire finite complex plane, the Laplace transform $X\left(s\right)$ cannot have any poles in the finite $s$-plane.