# Electronic – Under what conditions does jw equal the laplace variable s in an electrical circuit

fourierfrequencylaplace transformmathtransfer function

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.

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?

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.