Fourier vs. Laplace

Suppose I have an RLC network in a black box, and I bang it hard in the lab to get the impulse response. I have two options now, I can take the Fourier transform or I can take the Laplace transform to get the frequency response. How do I know which one to choose and what is the physical difference between each?

I have been told that the Laplace transform also gives you the transient response or the decay whereas the Fourier transform does not. Is this true? If I suddenly apply a sinusoidal signal at the input, then there should be a transient response for a brief period of time where the output is not a sinusoid until the system settles. Can someone give me a practical example in terms of an RLC network to show how this is true?

Also, often in circuits class, we take the Laplace transform of a circuit where the real part of \$s = \sigma + j\omega\$ is assumed to be zero anyway, so when we use \$\frac{1}{Cs}\$ to denote the Laplace transform of the capacitor, it is assumed that this is equivalent to \$\frac{1}{j\omega C}\$. I believe the real part is zero since the current through the capacitor is 90 degrees out of phase with the voltage across - is this correct? I thought Fourier transform was the same as Laplace transform with \$\sigma = 0\$. However, that does not seem to be true - consider \$x(t) = u(t)\$:

$$\mathcal{F}\{x(t)\} = \int_{-\infty}^\infty{u(t)e^{-j\omega t}}dt = \pi\delta(\omega) + \frac{1}{j\omega} \neq \mathcal{L}\{x(t)\} = \int_0^\infty{e^{-st}dt} = \frac{1}{s}$$

We can see that even if I substitute \$s = j \omega\$ with no real part at the output of the Laplace transform, they are still not equal. How come the Fourier transform has an extra impulse component but Laplace does not? When can I substitute \$s = j\omega\$ and expect the Fourier transform to equal the Laplace transform?

Edit: the latter part of my question has answers here and here.

Answer

The Fourier and the Laplace transform are not the same. First of all, note that when we talk about the Laplace transform, we very often mean the unilateral Laplace transform, where the transformation integrals starts at \$t=0\$ (and not at \$t=-\infty\$), i.e. with the Laplace transform we usually analyze causal signals and systems. With the Fourier transform this is not always the case.

In order to understand the differences between the two, it is important to look at the region of convergence (ROC) of the Laplace transform. For causal signals, the ROC is always a right-half plane, i.e. there are no poles (of a rational function in \$s\$) to the right of some value \$\sigma_0\$ (where \$\sigma\$ denotes the real part of the complex variable \$s\$). Now if \$\sigma_0<0\$, i.e. if the \$j\omega\$ axis is inside the ROC, then you simply obtain the Fourier transform by setting \$s=j\omega\$. If \$\sigma_0>0\$ then the Fourier transform does not exist (because the corresponding system is unstable). The third case (\$\sigma_0=0\$) is interesting because here the Fourier transform does exist but it cannot be obtained from the Laplace transform by setting \$s=j\omega\$. Your example is of this type. The Laplace transform of the step function has a pole at \$s=0\$, which lies on the \$j\omega\$ axis. In all such cases the Fourier transform has additional \$\delta\$ impulses at the pole locations on the \$j\omega\$ axis.

Note that it is not true that the Fourier transform cannot deal with transients. This is just a misunderstanding which probably comes from the fact that we often use the Fourier transform to analyze the steady-state behavior of systems by applying sinusoidal input signals that are defined for \$-\inftythis answer to a similar question.