ROC and its relation to the inverse Laplace transform

In summary, the conversation discusses the concept of region of convergence and inverse Laplace transform. It explains the formulas for the bilateral Laplace transform and the inverse Laplace transform and how they are used. The conversation also touches on the Cauchy-Residue theorem and how it is used in taking the inverse Laplace transform. The main question is whether it is sensible to talk about the ILT outside of the region of convergence, and the answer may be related to the concept of analytic continuation.
  • #1
elgen
64
5
This is a conceptual question on the region of convergence (ROC) and the inverse Laplace transform (ILT).

Here the bilateral laplace transform (LT) and the ILT are given by
[itex]
F(s)=L\{f(t)\}=\int_{-\infty}^{+\infty} f(t) e^{-st} dt
[/itex]
and
[itex]
f(t)=L^{-1}\{F(s)\}=\frac{1}{i 2\pi}\int_{a-i\infty}^{a+i\infty} F(s)e^{st}ds
[/itex]
where [itex]a[/itex] denotes the abssisa of absolute convergence.

If [itex]f(t)=u(t)e^{-at}[/itex] where [itex]u(t)[/itex] denotes the Heaviside step function, then [itex]F(s)=\frac{1}{s+a}[/itex] and the ROC is given by [itex]\Re\{s\}+a > 0[/itex].
When we take the inverse Laplace transform of [itex]F(s)[/itex], we use the Cauchy-Residue theorem.

Commonly, the textbook argues that, for ILT, when [itex]t<0[/itex], the integration contour in the form of the semi-circle is to the right of the vertical line passing [itex]a[/itex] and in parallel to the imaginary axis. Since there is no pole included, the contour integration is 0 and the ILT is 0.
When [itex]t>0[/itex], the integration contour is the semi-circle to the left. Since the contour encloses the pole at [itex]s=-a[/itex], the ILT is not 0.

My question is that since the ROC of [itex]F(s)[/itex]is [itex]\Re\{s\}+a > 0[/itex], when taking the ILT for [itex]t>0[/itex], shouldn't the semi-circle still be in the ROC of the [itex]F(s)[/itex]. Since the LT is not convergent to the left of the ROC, why is it sensible to talk about the ILT outside of the ROC?

My vague feeling is that the answer may be related to the concept of analytic continuation. A more lucid explanation is appreciated.

Thank you.
 
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  • #2
elgen said:
This is a conceptual question on the ROC and the inverse Laplace transform.

Here the bilateral laplace transform and the inverse laplace transform are given by
[itex]
F(s)=L\{f(t)\}=\int_{-\infty}^{+\infty} f(t) e^{-st} dt
[/itex]
and
[itex]
f(t)=L^{-1}\{F(s)\}=\frac{1}{i 2\pi}\int_{a-i\infty}^{a+i\infty} F(s)e^{st}ds
[/itex]
where [itex]a[/itex] denotes the abssisa of absolute convergence.

If [itex]f(t)=u(t)e^{-at}[/itex] where [itex]u(t)[/itex] denotes the Heaviside step function, then [itex]F(s)=\frac{1}{s+a}[/itex] and the ROC is given by [itex]\Re\{s\}+a > 0[/itex].
When we take the inverse Laplace transform of [itex]F(s)[/itex], we use the Cauchy-Residue theorem.

Commonly, the textbook argues that, for ILT, when [itex]t<0[/itex], the integration contour in the form of the semi-circle is to the right of the vertical line passing [itex]a[/itex] and in parallel to the imaginary axis. Since there is no pole included, the contour integration is 0 and the ILT is 0.
When [itex]t>0[/itex], the integration contour is the semi-circle to the left. Since the contour encloses the pole at [itex]s=-a[/itex], the ILT is not 0.

My question is that since the ROC of [itex]F(s)[/itex]is [itex]\Re\{s\}+a > 0[/itex], when taking the ILT for [itex]t>0[/itex], shouldn't the semi-circle still be in the ROC of the [itex]F(s)[/itex]. Since the LT is not convergent to the left of the ROC, why is it sensible to talk about the ILT outside of the ROC?

My vague feeling is that the answer may be related to the concept of analytic continuation. A more lucid explanation is appreciated.

Thank you.

Please explain what are 'ROC' and 'ILT'.
 
  • #3
I think he probably means "radius of convergence" and "inverse Laplace transform".
 
  • #4
I didn't know what they were, either.

elgen, please confirm that ROC = radius of convergence and ILT = Inverse Laplace Transform.
 
  • #5
Mark44 said:
I didn't know what they were, either.

elgen, please confirm that ROC = radius of convergence and ILT = Inverse Laplace Transform.

Yes. ROC = Region of convergence and ILT = Inverse Laplace Transform. Thank you.
 

1. What is the ROC in relation to the inverse Laplace transform?

The Region of Convergence (ROC) is a set of complex numbers for which the Laplace transform converges. It is an essential aspect of the inverse Laplace transform, as it determines the range of values for which the inverse transform can be calculated.

2. How does the ROC affect the inverse Laplace transform?

The ROC dictates the range of values for which the inverse Laplace transform can be calculated. If a function's ROC is not specified, the inverse Laplace transform may not exist or may be difficult to calculate. Therefore, understanding the ROC is crucial in solving problems involving the inverse Laplace transform.

3. Can the ROC change for a given function?

Yes, the ROC can change for a given function. It depends on the specific function and its properties. In some cases, the ROC may be a single point, while in other cases, it may be a range of values.

4. How is the ROC related to the convergence of the inverse Laplace transform?

The ROC is directly related to the convergence of the inverse Laplace transform. If a function has a specific ROC, it means that the inverse Laplace transform will converge for all values within that ROC. If the function's ROC is not specified, it may still converge, but it is not guaranteed.

5. What happens if the given function has no ROC?

If a function has no ROC, it means that the inverse Laplace transform does not exist. This could be due to various reasons, such as the function being non-causal or having an infinite number of poles. In such cases, the inverse Laplace transform cannot be calculated.

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