Electrical Engineering ETR › Complex AC Analysis

Laplace Transform

There is a general approach when you get stuck with the mathematics: the transformation. A transformation converts a problem into a different “kind” of mathematics.

The Laplace transform makes it possible to simplify the mathematics so much that networks with AC quantities can be calculated with similarly little mathematical effort as for DC quantities. You will need this kind of mathematics later in your studies in measurement technology, control engineering, power engineering, digital filters and many other disciplines. It is worth understanding this chapter properly now.

The general formula of the Laplace transform is:

\[ \begin{gathered} \text{Formula in the time domain that is to be transformed: } x(t) \\[6pt] \text{Laplace transform: } x(\omega) = \int_0^{\infty} x(t) \cdot e^{-j\omega t}\,dt \\[6pt] \text{The only formula in the time domain relevant to you (in Fundamentals of EE):} \\[4pt] x(t) = \hat{x} \cdot \sin(\omega t + \varphi) \\[6pt] \text{Laplace transform: } x(\omega) = \int_0^{\infty} \left[\hat{x} \cdot \sin(\omega t + \varphi)\right] \cdot e^{-j\omega t}\,dt \;\rightarrow\; \hat{x} \cdot e^{j\varphi} \end{gathered} \]

The Laplace transform itself is not mathematically pretty. Fortunately, we only ever have to transform one function, because with alternating voltage we always calculate with the sine function. And this one solution is given above in the bottom equation. As far as the transformation mathematics with the integral is concerned, we are therefore already done.

All AC quantities (alternating voltage, alternating current, alternating power) always consist of peak value, sine, angular frequency, time and phase angle. For the example of an alternating voltage, we have the following parameters:

\[ \begin{gathered} u(t) = \hat{u} \cdot \sin(\omega t + \varphi) \\[4pt] \text{Parameters: } \hat{u}, \omega, t \text{ and } \varphi \\[4pt] \text{Function: sine} \end{gathered} \]

The Laplace transform of this function is known. So we can simply take the result and work with it.

\[ \begin{gathered} \text{Time domain: } x(t) = \hat{x} \cdot \sin(\omega t + \varphi) \\[6pt] \text{Laplace transform into the “frequency domain”: } x(\omega) = \hat{x} \cdot e^{j\varphi} \end{gathered} \]

For the Laplace transform, we only need two of the four parameters of an AC quantity: the peak value and the phase shift. The Laplace transform massively reduces the amount of information about the signal. In the time domain, all 4 parameters of the AC quantity are taken into account.

Let us look at two examples:

\[ \begin{gathered} \text{General formula for an alternating voltage: } u(t) = \hat{u} \cdot \sin(\omega t + \varphi) \\[4pt] \text{Laplace transform into the frequency domain: } u(\omega) = \hat{u} \cdot e^{j\varphi} \\[4pt] \text{Example of an alternating voltage:} \\[2pt] u(t) = 3\,\mathrm{V} \cdot \sin\left(\omega t + \frac{\pi}{2}\right) \rightarrow u(\omega) = 3\,\mathrm{V} \cdot e^{j\frac{\pi}{2}} \\[8pt] \text{General formula for an alternating current: } i(t) = \hat{\imath} \cdot \sin(\omega t + \varphi) \\[4pt] \text{Laplace transform into the frequency domain: } i(\omega) = \hat{\imath} \cdot e^{j\varphi} \\[4pt] \text{Example formula for an alternating current:} \\[2pt] i(t) = 5\,\mathrm{mA} \cdot \sin\left(\omega t - \frac{\pi}{4}\right) \rightarrow i(\omega) = 5\,\mathrm{mA} \cdot e^{-j\frac{\pi}{4}} \end{gathered} \]

Now we still have to clarify what “frequency domain” means, what the “j” in the formulas is for and how to calculate with these exponential functions. I can promise you here: it is much easier than it seems at first glance.