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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

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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

If the term

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

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