thank youhi avyan.
The reason why S=jω 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 σ=0.
thank youSetting s=jω allows us to evaluate the magnitude of a transfer function along the jω axis as a function of frequency alone. We could also evaluate it along another path in the left half plane which would then have an exponential decay as well as a dependence on frequency. We can of course do the same thing with phase. When you do this you get the familiar Bode plots.
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