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.

Let us analyze the bilateral (two-sided) Laplace transform and verify why this statement is true.

The bilateral Laplace transform of a continuous-time signal x(t) is defined as:

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

Where s = σ + jω is a complex frequency variable.

Step-by-Step Proof/Analysis:
Consider a finite-duration signal x(t) that is bounded (i.e., |x(t)| ≤ M < ∞) and non-zero only within a finite time interval [t₁, t₂].

The Laplace transform for such a signal is given by:

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

To determine if X(s) has any poles, we check if the integral converges for finite values of s by examining its absolute value:

Since |e^-jωt| = 1 and |x(t)| ≤ M, we have:

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

Since the integration limits [t₁, t₂] are finite, the integral evaluates to a finite value for every finite choice of s (i.e., for all finite σ and ω).

Therefore, the Region of Convergence (ROC) for any bounded, finite-duration signal is the entire s-plane (with the possible exception of Re(s) = ∞ or Re(s) = -∞). As a result, X(s) has no finite poles anywhere in the complex s-plane.