Why do we use $s=jomega$ in AC analysis instead of $s=sigma+jomega$?

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:

• Why is sigma equal to zero?


• Is neper frequency connected to this?


• Is sigma equal to zero as the input signal is a sinusoid of constant \$\pm V_{max}\$?

Answer

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.