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

Laplace Transform Fundamentals

The two-sided Laplace transform, also known as the bilateral Laplace transform, is a powerful mathematical tool used in signal processing and system analysis to transform a time-domain signal, $x(t)$, into a complex frequency-domain representation, $X(s)$. It is defined by the integral:

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

where $s=\sigma +j\omega$ is a complex frequency variable. The existence of the Laplace transform depends on the Region of Convergence (ROC) of the integral.

Analyzing Laplace Transform Statements

Let's evaluate each statement regarding the two-sided Laplace transform:

Laplace Transform Existence

- Statement 1: "It exists for every signal that may or may not have a Fourier transform"
- Analysis: This statement is generally false. While the Laplace transform is a more general transform than the Fourier transform (the Fourier transform is a special case of the Laplace transform where the ROC includes the $j\omega$ -axis, i.e., when $s=j\omega$), it does not exist for every signal. For instance, signals that grow faster than any exponential (e.g., $e^{t^2}$) do not have a Laplace transform. Many signals that do not have a Fourier transform (because they are not absolutely integrable, like $e^{at}u(t)$ for $a>0$) can still have a Laplace transform. However, the use of "every signal" makes this statement incorrect.

Laplace Transform Poles and Finite Duration Signals

- Statement 2: "It has no poles for any bounded signal that is non-zero only inside a finite time interval"
- Analysis: This statement is true. - A signal that is non-zero only inside a finite time interval, say $t_1\le t\le t_2$, is called a finite duration signal. - If such a signal $x(t)$ is also bounded, it means that $|x(t)|<M$ for some finite constant $M$ over its non-zero interval. - For such a signal, the integral for the Laplace transform becomes: $X(s)={\int }_{t_1}^{t_2}x(t)e^{-st}dt$ - This integral will converge for all finite values of $s$ (i.e., for all $\sigma$ and $\omega$). This means the Region of Convergence (ROC) for a bounded, finite duration signal is the entire $s$ -plane. - If the ROC is the entire $s$ -plane, it implies that there are no finite poles in the $s$ -plane, because poles are points where the transform $X(s)$ becomes infinite and thus the ROC cannot include them. - Therefore, a bounded, finite duration signal has a Laplace transform with no finite poles.

Poles and Zeros of Laplace Transform

- Statement 3: "The number of finite poles and finite zeroes must be equal"
- Analysis: This statement is generally false. For a rational Laplace transform $X(s)=\frac{N(s)}{D(s)}$, where $N(s)$ is the numerator polynomial and $D(s)$ is the denominator polynomial, the number of finite zeros is the degree of $N(s)$ and the number of finite poles is the degree of $D(s)$. These degrees are not necessarily equal. For example, $X(s)=\frac{1}{s+1}$ has one pole and zero finite zeros. If we consider poles and zeros at infinity, then the total number of poles and zeros might be equal, but the statement specifies "finite poles and finite zeroes".

Laplace Transform of Exponential Sums

- Statement 4: "If a signal can be expressed as a weighted sum of shifted one sided exponential, then its Laplace Transform will have no poles"
- Analysis: This statement is false. A one-sided exponential, such as $e^{at}u(t)$, has a Laplace transform $X(s)=\frac{1}{s-a}$, which clearly has a pole at $s=a$. If a signal is a weighted sum of such exponentials, for example, $x(t)=A_1{e}^{a_{1}t}u(t)+A_2{e}^{a_{2}t}u(t)$, its Laplace transform would be $X(s)=\frac{A_1}{s-a_1}+\frac{A_2}{s-a_2}$, which would have poles at $s=a_1$ and $s=a_2$. Therefore, a sum of exponentials will typically have poles.

Conclusion on Laplace Transform Properties

Based on the detailed analysis of each statement, the only true statement about the two-sided Laplace transform is that for any bounded signal that is non-zero only inside a finite time interval, its Laplace transform will have no finite poles because its Region of Convergence covers the entire s-plane.