# In what sense does the Laplace Transform give components of a signal in something called the 's-domain'?

For example, I understand how this make sense for the Fourier transform. When we do the transform, we get 'how much of each frequency' in present in the signal. We get the value of the coefficients, such that when we multiply each coefficient with the corresponding constant frequency complex exponential and add those up, we get back our original signal. When engineers analyse the graph of amplitude squared in the frequency domain, they get the idea of how much of each frequency is present.

Is the same true for the Laplace transform as well? I know that Laplace transform is about getting "components" of a signal in terms of decaying complex exponentials, as opposed to unit magnitude complex exponentials in Fourier transform.

I haven't read much in detail about the inverse of Laplace transform. But its formula does not seem like we're summing up 'components times decaying exponential' over the entire s-plane. So what do we see when we look at, say, a colored graph in the 's-domain'? Is it really giving us the strength of the signal at each point in the s-plane? How is that true if we are not summing over the entire s-plane to get back our signal? Is this question clear?

• This may not answer your exact question, but this video should lend some insight into where the Laplace transform "comes from" and what it is actually doing: youtu.be/sZ2qulI6GEk?t=88 – Math1000 Nov 2 '19 at 15:24

You just have to accept the simple formulas belonging to the Laplace transform. Unlike the Fourier transform of functions or signals the Laplace transform of transient phenomena $$t\mapsto f(t)$$ $$\>(t\geq0)$$ has no intuitive physical interpretation. It is a purely formal operation applied to given or as yet unknown function terms ("expressions") resulting in other function terms hopefully listed in a catalogue. – While functions $$f$$ available only in the form of a discrete time series are Fourier transformed all the time, and interesting information about $$f$$ is revealed in this way, nobody would consider Laplace transforming such a data set, nor would anybody look at the graph of a Laplace transform.