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.