Bilateral Laplace Transform: Derivation and ROC

From Fourier to Laplace: A Generalization

Let's start with the familiar continuous-time Fourier transform:

As we've discussed, a key condition for the convergence of this integral is that the signal $x(t)$ must be absolutely integrable, i.e., ${\int }_{-\infty }^{\infty }|x(t)|dt<\infty$. This condition excludes many important signals, such as a growing exponential $x(t)=e^{at}u(t)$ for $a>0$.

The conceptual leap that leads to the Laplace transform is to ask: can we force a signal to be absolutely integrable by multiplying it by a weighting function? Let's consider multiplying our signal $x(t)$ by a real exponential function, $e^{-\sigma t}$, where $\sigma$ is a real number. Our new, modified signal is $x(t)e^{-\sigma t}$.

If we choose $\sigma$ carefully, we can make this new signal absolutely integrable and thus guarantee that its Fourier transform exists. The Fourier transform of this weighted signal is:

Combining the exponential terms, we get:

This resulting transform depends on both $\omega$ (from the Fourier kernel) and $\sigma$ (from our weighting function). It's natural to combine these two real variables into a single complex variable. Let's define:

Substituting $s$ into our integral gives us the definition of the bilateral Laplace transform:

So, the Laplace transform can be viewed as the Fourier transform of the original signal multiplied by an exponential weighting factor. This insight is the key to understanding its power and its relationship to the Fourier transform.

The following video provides an excellent visual and conceptual walkthrough of this exact derivation and introduces the idea of the "s-plane."

The Region of Convergence (ROC)

The video you just watched introduced a critical idea: the Fourier transform of the weighted signal $x(t)e^{-\sigma t}$ only exists for certain values of $\sigma$. The range of $s=\sigma +j\omega$ values for which the Laplace transform integral converges is called the Region of Convergence (ROC).

Let's make this formal. The condition for the Laplace transform to converge is:

Since $s=\sigma +j\omega$, we can write $e^{-st}=e^{-\sigma t}{e}^{-j\omega t}$. The magnitude is:

Because $|e^{-j\omega t}|=|\cos (\omega t)-j\sin (\omega t)|=\sqrt{{\cos }^2(\omega t)+{\sin }^2(\omega t)}=1$, the convergence condition simplifies to:

Notice that this condition depends only on $\sigma$, the real part of $s$. This is a profound result: if the transform converges for a particular $s_0={\sigma }_0+j{\omega }_0$, it must converge for all values of $s$ with the same real part, i.e., for all $s={\sigma }_0+j\omega$. This means the ROC always consists of vertical strips parallel to the $j\omega$-axis in the s-plane.

The next part of the video you started watching demonstrates this beautifully.

The ROC is Not Optional: A Tale of Two Signals

You cannot fully specify a Laplace transform without its ROC. To see why, let's analyze two different signals. We will use a standard reference text for this, which aligns with your preference for deeper mathematical sources.

As you saw in the reading, the signals are:

1. Right-sided signal: $x_1(t)=e^{-at}u(t)$
2. Left-sided signal: $x_2(t)=-e^{-at}u(-t)$

Both of these signals have the exact same algebraic expression for their Laplace transform:

However, they are completely different signals, and their transforms are distinguished only by their ROC:

• For $x_1(t)$, the ROC is a right-half plane: $\text{Re}\{s\}>-a$.
• For $x_2(t)$, the ROC is a left-half plane: $\text{Re}\{s\}<-a$.

This is a fundamental principle: The Laplace transform is defined by both its algebraic expression and its region of convergence. Without the ROC, the transform is ambiguous.

The Relationship Solidified

We can now state the relationship between the two transforms with precision. As we derived, the Laplace transform evaluated for $s=j\omega$ (i.e., for $\sigma =0$) is identical to the Fourier transform.

This leads to a critical conclusion:
A continuous-time signal $x(t)$ has a Fourier transform if and only if the Region of Convergence of its Laplace transform includes the imaginary axis ($\text{Re}\{s\}=0$).

This is why the growing exponential $x(t)=e^{2t}u(t)$ from our exercise does not have a Fourier transform: its ROC is $\text{Re}\{s\}>2$, which is a half-plane that lies entirely to the right of the $j\omega$-axis and does not include it.

The following short video summarizes this relationship concisely.

Conclusion

In this lesson, we have formally defined the Laplace transform and its Region of Convergence, building upon the foundations of the Fourier transform. This new tool allows us to analyze a much broader class of signals.

Key Takeaways:

• The bilateral Laplace transform $X(s)={\int }_{-\infty }^{\infty }x(t)e^{-st}dt$ can be seen as the Fourier transform of a signal weighted by an exponential $e^{-\sigma t}$, where $s=\sigma +j\omega$.
• The Region of Convergence (ROC) is the set of all values of $s$ in the complex s-plane for which the transform integral converges.
• The ROC is a crucial part of the transform. Two different signals can have identical algebraic expressions for $X(s)$, distinguished only by their ROCs.
• The Fourier transform of a signal exists if and only if the ROC of its Laplace transform includes the imaginary axis ($\text{Re}\{s\}=0$). If it does, the Fourier transform is simply $X(s)$ evaluated at $s=j\omega$.

Preview of the Next Lesson:
We now have the tools to represent signals in the s-plane. The true power of this representation comes from analyzing systems. In the next lesson, we will see how the characteristics of a system's Laplace transform, specifically the locations of its poles relative to the ROC, allow us to determine fundamental system properties like stability and causality.