Region of convergence of a Laplace transform

In summary, the Laplace transform of a function is defined as the integral of the function multiplied by the exponential of a complex variable. The region of convergence for a Laplace transform is a right half-plane where the real part of the complex variable is greater than a certain value. In some cases, the Laplace transform can be extended to a larger region through analytic continuation. In the case of the Laplace transform of 1, it exists at s=i since it is analytic everywhere in the complex plane except at s=0. This is due to the fact that 1/s has a Taylor series that converges for |s-i|<1. In basic electronics problems, the Laplace transform is often a rational function, making
  • #1
mjtsquared
3
3
If a Laplace transform has a region of convergence starting at Re(s)=0, does the Laplace transform evaluated at the imaginary axis exist? I.e. say that the Laplace transform of 1 is 1/s. Does this Laplace transform exist at say s=i?
 
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  • #2
I know this thread is old, but in case you still interested here is an answer.

I am assuming you are using the single-sided Laplace transform, with the standard definition
$$F(s) = \int_0^\infty f(t) e^{-s t} dt$$
so the region of convergence a right half-plane ##\Re(s)>s_0## (where the notation ##\Re(s)## indicates the real part of ##s##). This means that ##F(s)## is analytic in that half-plane, and in general is not defined outside of that half-plane. However, in many cases ##F(s)## can be extended so that it is analytic in a larger region of the complex plane (this is called analytic continuation). In your example you have ##F(s) = 1/s## with ##\Re(s)>0##. Since ##1/s## is analytic everywhere in the complex plane except at ##s=0##, ##1/s## is (trivially) the analytic continuation of ##F(s)## into the entire plane except the origin. So yes, in your example ##1/s## is defined at ##s=i##.

If the term analytic is new to you, a function ##F(s)## is called analytic at a point if there is a converging Taylor series about that point. In your example, ##1/s## has a Taylor series about ##i##: ##F(s) = -i + (s-i) + i (s-i)^2 + \ldots##, which converges for ##|s-i|<1## .

Note that in some cases the process of analytic continuation is much more complicatd. However, in the types of problems you are likely to find in basic electronics, ##F(s)## is often a rational function, so is trivially its own analytic continuation into the entire complex plane except at the poles of ##F(s)##.

Jason
 

1. What is the region of convergence (ROC) of a Laplace transform?

The region of convergence is the set of values in the complex plane for which the Laplace transform converges. In other words, it is the set of values of the complex variable s for which the integral defining the Laplace transform exists.

2. Why is the region of convergence important?

The region of convergence is important because it determines the set of values of s for which the inverse Laplace transform exists. This means that the ROC is directly related to the existence and stability of the solution to a differential equation represented by the Laplace transform.

3. How do you determine the region of convergence of a Laplace transform?

The region of convergence can be determined by examining the poles (singularities) of the Laplace transform. The ROC will be the set of values of s that lie to the right of all poles in the complex plane. In other words, the ROC is the complement of the set of poles in the complex plane.

4. Can the region of convergence be a single point?

Yes, the region of convergence can be a single point in the complex plane. This happens when the Laplace transform has only one pole, and the ROC is the point to the right of that pole. However, in most cases, the ROC is a larger region in the complex plane.

5. Can the region of convergence change?

Yes, the region of convergence can change depending on the function being transformed. For example, if the function has a different set of poles, then the ROC will also change. Additionally, the ROC can change if the function is multiplied by a different function before taking the Laplace transform.

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