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When using the Nyquist stability criterion, amplitude-frequency characteristic etc. we go from the Laplace image $G(s)$ to $G(j\omega )$. By definition of the Laplace transform, $s=\sigma + j\omega$. So why is $\sigma$ (real part of $s$) approximated as $0$ or whatever the case might be?

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Did you see Wikipedia entry section Nyquist stability criterion Mathematical Derivation? –  Américo Tavares May 4 '13 at 12:01

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If $G(s)$ is the laplace transform of $g$, then $F(\omega)=G(i\omega)$ is the fourier transform of $g$, if the fourier transform exists. You can easily see that by comparing the formulas for the two transforms - in the case of the laplace transform the kernel is $e^{-s}$ whereas for the fourier transform it's $e^{-i\omega}$. (Since you usually assume that $fg(x)=$ for $x < 0$ when using the laplace transform, the fact that the integration boundaries are different doesn't matter).

This explains (at least if the fourier transform exists) why $G(i\omega)$ gives you the amplitude-frequency-characteristic $g$ - it's simply the fourier transform of $g$.

BTW, if you laplace transform some $g$ for which the fourier transform does not exists, you often still end up with an algebraic expression which does, in fact, evaluate to finite values even for $s=i\omega$. If this happen, what you get is not the fourier transform, but rather the analytic extension of $G$ to the imaginary axis. You have to be very carefull in this case with drawing conclusions about the behaviour of $g$ from $G(i\omega)$.

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So in a way we use the given transfer formula in Laplace transform to get the frequency spectrum which is given by the Fourier Transform. Thanks a lot! –  Martin V May 4 '13 at 16:08

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