Regarding Bode plots; $H(s)$ and $H(j\omega)$

Question

In circuit analysis, I understand the use of Laplace Transforms to obtain the impedance of a linear RLC circuit, ie transforming from the time domain to the frequency domain. In most texts I have seen and classes I have taken, the Bode plot looks at the rational transfer function of $H(j\omega)$, which is effectively the Fourier Transform, and not $H(s)$.

Since $\omega$ is considered to be the variable that represents frequency, what is $\sigma$ if $s = \sigma + j\omega$? Does $\sigma$ have a physical interpretation?

Answer 1

Note that each complex pole $s_{\infty}=\sigma + j\omega$ of the transfer function $H(s)$ contributes to the system's impulse response a complex exponential of the form

$$e^{s_{\infty}t}=e^{\sigma t}e^{j\omega t}$$

The term $e^{\sigma t}$ is real-valued and represents exponential damping (assuming that the system is causal and stable, i.e. $\sigma<0$), and the complex term $e^{j\omega t}$ represents an oscillation at frequency $\omega$. So $\sigma$ is a damping constant.

Answer 2

$\sigma$ is associated with attenuation in "Control Theory" and there Laplace is more suitable than Fourier. $\sigma=0$ the transformation becomes Fourier.

• There is unilateral and bilateral Laplace transformation & periodic signals and Laplace transform. @Matt L have explained somewhere here the strong relationship between Laplace and Fourier.