# What does $σ$ equal to zero mean?

Consider the Laplace transform of an RC filter. For those who can't immediately summon it, refer equation (46) at this link: http://web.mit.edu/2.151/www/Handouts/FreqDomain.pdf for a refresher.

In the same file, equations (47) and (48) represent the frequency response function of the network. Now I am sure that I, like many others who will read this question, could have obtained the frequency response just by using time domain analysis. Inserting $σ = 0$ is just one of the many conveniences Laplace transforms affords us.

But I would like to know why that is so. Why replacing $s$ with $jω$ yields the frequency response? More than the answer to this question I would like to have an intuitive understanding of Laplace transforms and the s-domain.

I could figure out many properties of addition and the numbers domain before I was told about those properties. For instance, I knew about associative and commutative properties of addition before they were formally introduced; i.e. intuition preceded formal definition. But I could have never figured out, just by learning Laplace transforms, as to why replacing $s$ with $jω$ yields what it yields.

I would be grateful if anyone has a deeply intuitive understanding of the Laplace transform and could answer my question through that understanding. I just watched this video http://www.youtube.com/watch?v=hqOboV2jgVo which explains how we arrive at Laplace transforms. Brian Douglas has posted some really cool videos on YouTube regarding the same topic. But I still feel like I am grappling in the dark. I guess an intuition, if indeed there's one for humans, will automatically answer many basic questions about Laplace transforms including this one.

It is because, the continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform with imaginary argument $s = iω$. This relationship between the Laplace and Fourier transforms is often used to determine the frequency spectrum of a signal or dynamical system. I assume you do not have any intuition problem with the Fourier transform. But let me know if you do.
• I don't think I have an intuition for Fourier transform either, but somehow it's much easier to keep in mind: Fourier transform finds and highlights the constituent frequencies of a time varying signal. And Fourier Transform is arrived at from Fourier series, which again is quite straight forward. I don't suppose I can truly understand these concepts — perhaps it is ludicrous for mortals to expect that they will arrive at Parseval's theorem intuitively. Even then, I rarely fumble around with FTs and FSsand hence I am fine with my limited perspective. – Shashank Sawant Nov 25 '14 at 1:55
• But FTs are mere sticks compared to the howitzers that LTs are. They obliterate the differentials for us to play with the remains. It just somehow makes me wish I could understand LTs a lot better than I do right now. (EDIT: not that FTs are any less important; on the contrary they are perhaps more widely used). – Shashank Sawant Nov 25 '14 at 1:56