Under what conditions does jw equal the laplace variable s in an electrical circuit?

Question

Imagine I've got an electromagnetic actuator and solved the underlying differential equations. I got an solution for sinusodially excitation, as follows (just an example, no physical backround):

$$G(\mathrm{j}\omega) = \frac{I(\mathrm{j}\omega)}{U(\mathrm{j}\omega)} = \frac{1}{1 + \mathrm{j}\omega T}$$

Now I just set

$$\mathrm{j}\omega = s, \quad s \rightarrow \text{Laplace Variable}$$

to get

$$G(s) = \frac{1}{1 + s T}$$

and I claim that this is the transfer function for the electrical circuit feeding the actuator. I'm sure the results I get are correct, but I was told I can't just set $$\mathrm{j}\omega = s$$ without further explication.

So I wonder what are the conditions to do what I've done, how can I proof that both equations are correct and explain that in a correct mathematical way? I feel that everybody (in literature) either just does it or avoids the issue.

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Further explanation:

Above system G(jw) shows a low-pass behavior in the frequency domain, means for sinusodially voltage excitation $$ i(\mathrm{j}\omega) = G(\mathrm{j}\omega) \cdot u(\mathrm{j}\omega) = \frac{1}{1+\mathrm{j}\omega T} \cdot u(\mathrm{j}\omega)$$ I get a higher damping the higher the frequency of the excitation.

But we also know that a transfer function $$ i(s) = G(s) \cdot u(s) = \frac{1}{1+s T} \cdot u(s)$$ excited by a unitary step $$u(s) = \frac{u_0}{s}$$ will also lead to a valid solution, namely the typical exponential $$ i(t) = u_0 \cdot (1- e^{-t/T})$$ PT1-behavior.

So what is the condition, that a system which is valid in the frequency domain for harmonic signals in steady-state, is also valid in the Laplace-domain by setting $$ \mathrm{j}\omega = s$$ for the non-steady-state e.g. for excitation with a unitary step?

Answer 1

The Laplace variable \$s\$ relates to Fourier's \$j\omega\$ as follows:

$$ s = \sigma + j\omega $$

Fourier transform can be seen as a Laplace transform when \$\sigma=0\$. The \$\sigma\$ allows the Laplace integral transformation to converge for signals that Fourier transform does not, e.g. a unitary step (Heaviside function).

If you are working with real signals, in a steady-state regime, chances are that waveforms will converge for both Fourier and Laplace (the signal won't suddenly become unbounded or will present a non-differentiable spot), so \$s\$ and \$j\omega\$ become interchangeable. In a mathematically rigorous fashion, the Laplace ROC (Region of Convergence) must include \$\sigma=0\$, also known as the \$j\omega\$ axis of the s-plane.

The convergence limitations of Fourier are grounds for some hardcore maths, if you are interested, Wikipedia has a great link on this.

Answer 2

Let me try a short answer without any mathematics. Let`s assume that you are not interested in the relation between time and frequency domain - that means: You are interested in the frequency-dependent properties of a system or circuit only. In this case, you do not need the Laplace Transformation at all - and you can interprete the symbol s as an abbreviation for jw only (s=jw). In this case, you can calculate and draw the frequency-dependent quantities of interest (magnitude, phase) for your circuit.

However, for discussing some special properties of the system (in particular for systems with feedback) - it is helpful to switch to the complex frequency domain (setting s=σ+jω). And in the complex frequency plane we can visualize some interesting parameters like pole/zero distribution, pole frequencies, pole-quality figures,...).

These parameters are very helpful for describing the system properties in comparison to other (similar) systems. A good example are filter circuits, which can be characterized by the corresponding pole and zero locations.

Finally, it is worth mentioning that the most important applications of the complex frequency domain are stability analyses (Nyquist diagram).

But you must realize that the complex frequency domain is a pure artificial product. Complex frequencies do not exist an cannot be produced. But they are a very efficient tool for analyzing (and also for designing) frequency-dependent circuits and systems.