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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?

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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.

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  • \$\begingroup\$ So basically the fact, that my signal is the electrical current, which is continous, is enough as explication? But what if I got the solution for my differential equation with the assumption of a sinusodially excitation, but it is supposed to be true also for a voltage unitary step? \$\endgroup\$ – thewaywewalk Apr 7 '16 at 14:35
  • \$\begingroup\$ Steady-state is the key here. When working with differential equations and initial conditions, you are modeling a transient behavior as well and Laplace's is not equivalent to Fourier's. Once the transient settles and steady-state regime is reached, then Fourier's results should be equal to Laplace's. \$\endgroup\$ – Vicente Cunha Apr 7 '16 at 14:42
  • \$\begingroup\$ More generically, I think you can swap jw and s when you're in the Region of Convergence of the Laplace Transform. \$\endgroup\$ – Scott Seidman Apr 7 '16 at 14:49
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    \$\begingroup\$ @ScottSeidman Actually, I believe the correct way to put it is if the Region of Convergence of Laplace's include the possibility of \$\sigma=0\$ (i.e. the \$j\omega\$ axis). I nearly forgot about this, thank you, I should add it to the answer. \$\endgroup\$ – Vicente Cunha Apr 7 '16 at 14:56
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    \$\begingroup\$ (+1) The whole thing boils down to system stability. If the system is stable it will have a steady state so the Fourier transform exists (since it relates the input and output for a sinusoidal input in steady state conditions) and the relation with the Laplace transform is valid. If the system is not stable, the Fourier transform of its impulse response won't exist. \$\endgroup\$ – Lorenzo Donati Apr 7 '16 at 22:30
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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.

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  • \$\begingroup\$ I know all that. You're describing the case that you have a transfer function dependent on s and you'd like to analyze it's frequency behavior. But in my case I know the frequency behavior and want to know under what conditions I can claim, that the mathematical formula I have, dependent of jw, is also a valid transfer function. \$\endgroup\$ – thewaywewalk Apr 7 '16 at 16:12
  • \$\begingroup\$ I think, you can switch from the complex frequency variable to jw and vice versa. Both directions are allowed. Was THIS your question? \$\endgroup\$ – LvW Apr 7 '16 at 17:01
  • \$\begingroup\$ Well, I can, the results prove that, but my supervisor told me that certain conditions must be met and I try to find out which. Basically I was told, that I can't write jw in one equation and s in the next, stating it would be equivalent, without further explanation. I'm basically lacking one single sentence. \$\endgroup\$ – thewaywewalk Apr 7 '16 at 21:02
  • \$\begingroup\$ In your comment to the other answer you have mentioned steady-state conditions. Note that any transfer function (in w or in s) is valid for steady-stae conditions only. The term "frequency" is defined for periodic signals only. \$\endgroup\$ – LvW Apr 8 '16 at 7:15
  • \$\begingroup\$ but a step response is not steady-state over the whole time, it is transient, isn't it? Do I have a wrong definition of "steady-state" in my mind? \$\endgroup\$ – thewaywewalk Apr 8 '16 at 7:40

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