On the meaning of s-plane and it's link to a transfer function Considering Fourier analysis and let's say I'm walking on the blue frequency axis in the below 3D plot from zero towards infinity:



So each time I encounter a non zero blue bar, I check the frequency at that point and measure the bar's height and say: Oh I found another sinusoidal component of the signal at this particular frequency and at this particular amplitude. This is how I grasp the idea of Fourier analysis.

But when I read about the s-plane I'm extremely confused to make it visual and link it to the actual signal as in the above fashion. I just cannot give a similar meaning what it is that really about. 

Now imagine a system's output and input in time domain is given. Vout(t)/Vin(t) is transformed into s-domain and we call this ratio as H(s) which is the transfer function. 

Here is a transfer function which is a surface on the s-plane:



Imagine I am walking on the s-plane. And imagine at a point on the s-plane I stop and the point s is a complex number and it is s = ωj + σ. Now if I look upwards towards the surface above me, obviously there will be a point on the surface. Lets call this point x. So what should it mean to me in terms of the signal or the transfer function?

Ok so at this particular point I found "what"? What is "x"? Is it a component of something? And my second question is if H(s) were a signal's transform not a transfer function what would x be in that case?
In the Laplace transform, being on the line $s = 0 \pm j\omega$ means you are looking at the pure sinusoids contained in the signal. This is exactly as you mentioned in your understanding of Fourier analysis. So this means that you are looking at the time-domain terms:

$$e^{j\omega{t}}$$

Which again represent the pure sinusoids composing your signal.

However once you are off this line and have some $s$ such that: $$s = \sigma \pm j\omega$$

Not only are you looking at the $e^{j\omega{t}}$ sinusoids terms, you are now looking at time domain terms:
$$e^{\sigma{t}}e^{j\omega{t}}$$

When talking about a stable system, poles being on the left-hand-plane of the $s$-domain means that $\sigma &lt; 0$ so the sinusoids you are considering contained in the signal now have a damping envelope. If $\sigma$ was positive, the poles would be placed in the right-hand-plane and the signal would explode to infinity as time went on. This is why we want poles to stay out of the right-hand-plane if we want a stable system.

The takeaway is that in Fourier analysis you are considering the pure sinusoids composing your signal. With the Laplace transform you are considering sinusoids with that exponential term attached. In the case for a stable or oscillating system, you are considering the damped sinusoids ($s = -\sigma \pm j\omega$), decaying exponentials ($s = -\sigma$), pure sinusoids ($s = \pm j\omega$), and constant amplitude ($s = 0$) components composing your signal.

This is what makes Fourier analysis appropriate for identifying frequency content and the Laplace transform appropriate for analyzing stability and performance parameters.