In AC analysis, \$s=j\omega\$ when we deal with \$sL\$ or \$1/sC\$. But for a Laplace transform, \$s=\sigma+j\omega\$.
Sorry for being ambiguous but I would like to connect the questions below:
Of course, \$s = \sigma + j\omega\$, by definition. What's happening is that \$\sigma\$ is being ignored because it is assumed to be zero. The reason for it is that we are looking at the response of the system to periodic (and thus non-decaying) sinusoidal signals, whereby Laplace conveniently reduces to Fourier along the imaginary axis. The real axis in the Laplace domain represents exponential decay/growth factors that pure signals do not have, and which Fourier does not model.
For AC analysis, it is assumed that the circuit has sinusoidal sources (with the same angular frequency \$\omega \$) and that all transients have decayed. This condition is known as sinusoidal steady state or AC steady state.
This allows the circuit to be analyzed in the phasor domain.
Using Euler's formula we have:
\$v_A(t) = A \cos(\omega t + \phi) = \Re(Ae^{j\phi}e^{j\omega t}) \$
The phasor associated with \$v(t)\$ is then \$\vec V_a = Ae^{j\phi}\$ which is just a complex constant that contains the magnitude and phase information of the time domain signal.
It follows that, under these conditions, we can analyze the circuit by keeping track of the phasor voltages and currents and using the following relations:
\$\dfrac{\vec V_l}{\vec I_l} = j\omega L \$
\$\dfrac{\vec V_c}{\vec I_c} = \dfrac{1}{j\omega C} \$
\$\dfrac{\vec V_r}{\vec I_r} = R \$
We then recover the time domain solution via Euler's formula.
Now, there is a deep connection between phasor analysis and Laplace analysis but it is important to keep in mind the full context of AC analysis which is, again:
(1) the circuit has sinusoidal sources (with the same frequency \$\omega \$)
(2) all transients have decayed
Laplace transform transfer function (TF) analysis gives the complete response to a sinusoidal input signal from t=0. The solution generally contains transient terms, which decay to zero exponentially, and steady-state terms which remain after the exponentials have disappeared. When we have the poles and zeros of a TF, eg s=-a+jw, the '-a' part gives the exponential (e^-at) response, and the jw part gives the sinusoidal steady-state response: (e^jwt) = cos(wt) + jsin(wt). If we are only interested in the steady-state part of the response (as is the case in frequency response analysis) then we can just use the substitution s=jw in the TF.
Note that e^jx = cos(x) + jsin(x) is 'Euler's Identity' and is one of the most important and useful relationships in science and engineering.
The reason why \$S=j\omega\$ is chosen to evaluate AC signals is that it allows to convert the Laplace transform into Fourier transform.
The reason is that while S is a complex variable, what's used in the Fourier representation is just the rotational (imaginary) component, hence \$\sigma=0\$.
You may find some more at this Stanford page.
This is only used for "Sin" and "Cos" which is the case of AC signal. Note: The laplace trasnform of sin(at) or cos(at) "1/jw+ a" or "jw/jw+a" that This can be proven using the identiy of the sin and cos using Euler's identity which is basically just 2 exponentials, and the laplace of the exponential has only the imaginary part "jw".
I will write down the proof and post it here. :)
If you look at the formula of Fourier and Laplace transform, you will see that 's' is Laplace transform is replaced by 'jw' in Fourier transform. That is why you can get the Fourier transform from Laplace transform by replacing 's' with 'jw'.
