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In my Electrical Engineering degree I recently learned how to analyse circuits using Laplace transforms and I know that the Laplace transform of the voltage across an inductor is \$V_L(s) =s*L*I(s) - L*i(0)\$, but I don't understand its physical meaning. I know that \$s*L\$ is the impedance so with Ohm's law we get \$V_L(s) =s*L*I(s)\$ but when it gets to subtracting \$L*i(0)\$ I don't understand where it comes from, I know it is supposed to represent the initial voltage but shouldn't it be also \$s*L*i(0)\$ ?

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  • \$\begingroup\$ All terms having an s in the laplace domain are time-dependent in the time domain. So, no, initial values cannot have an s. \$\endgroup\$
    – Janka
    Jun 22 '18 at 8:58
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The Laplace transform of a derivative is

$$\mathcal{L}\left\{ \frac{dy}{dt} \right\} = s\mathcal{L}\left\{f\right\} - y(0^+)$$

Ie. \$y(0^+)\$ is the initial condition, for time approaching zero from the right side.

Since the "law" of inductors is

$$v_L = L\cdot \frac{di_L}{dt}$$

Its Laplace transform is given by

$$\begin{align} V_L(s) &= L\left( sI_L(s) - i_L(0^+) \right) \\ &= Ls\cdot I_L(s) - L\cdot i_L(0^+) &\Downarrow \\ I_L(s) &= \frac{V_L(s)}{Ls} + \frac{i_L(0^+)}{s} \end{align}$$

So in this formula, \$i_L(0^+)\$ takes the meaning of the initial current through the inductor. From the last equation, you also can assign a physical meaning to it. This equation coincides to having a parallel constant current source with the inductor!

In other words, you can replace an inductor with

schematic

simulate this circuit – Schematic created using CircuitLab

For a capacitor, the same can be done. It will result in putting a constant voltage source in series with the capacitor.

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No, you're in the Laplace domain now. You're dealing in terms of frequency. If you did take the Laplace transform of an initial condition, the constant divided by an 'integrator' or a delta function. This delta function is also placed at zero in the Laplace world (which is DC in terms of frequency)

$${\mathcal{L}(c) = \dfrac{c}{s}} = \delta $$

This makes total sense, if you have a constant value, it's pretty much a DC value, if you want to know which frequency its at zero frequency, there is no frequency because it's constant.

enter image description here

So across an inductor this would signify that there was already something there before the system started, but it wouldn't make a lot of sense until you transform it back into the time domain, you can also think of it as a placeholder.

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  • \$\begingroup\$ Agreed on all but frequency. s doesn't represent a frequency because of the ability of the Laplace domain to image real exponents of e. \$\endgroup\$
    – Janka
    Jun 22 '18 at 9:00
  • \$\begingroup\$ Technically no, but it's a close analog to the FFT \$\endgroup\$
    – Voltage Spike
    Jun 22 '18 at 19:40
  • \$\begingroup\$ Yes, sure, I am nitpicking. But this is the case where it really matters! Because the OP wanted to know the difference between a static value and a one-time signal. The latter has an s in the laplace domain. The frequency analogy will simply say "it's all frequencies as it's a jump" but this is even more mysterious. \$\endgroup\$
    – Janka
    Jun 22 '18 at 21:26
  • \$\begingroup\$ Actually they do call s the complex number frequency parameter. \$\endgroup\$
    – Voltage Spike
    Jun 22 '18 at 21:35
  • \$\begingroup\$ Yes, but as my control theory teacher put it: "Gee, what's that supposed to mean?" It's obstructive, not helping at all. \$\endgroup\$
    – Janka
    Jun 22 '18 at 22:03

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