# Fourier Series = Re(Power Series)

## Main Question or Discussion Point

Somebody posted a question about Fourier series yesterday that got me thinking about an argument I heard some time before.

If we have a (complex-valued) analytic function f, then any closed loop in the complex plane will be mapped by f to another closed loop. (If the loop doesn't enclose any singularities or branch points.) In fact, if start = finish of the original loop, then start = finish of the image curve. Thus, the function is periodic on this loop. In particular, its real- and imaginary parts (call them u and v) are also periodic on this loop. If we write $$f(r,\theta )=re^{i\theta}=u(re^{i\theta})+iv(re^{i\theta})$$, and set r= const., we obtain real and imaginary parts that have period 2π. (Since the loop is an origin-centered circle.)

Thus, since f is analytic, we have [Taylor series of f] = [Fourier series of u]+i [Fourier series of v]. Then, if we are given a harmonic function u of period 2π, then we can find an harmonic conjugate v, and find the Fourier coefficients by simply reading off the coefficients of the real-part of the Taylor series. If the period is different, the approach can be easily modified. The claimed benefit is that it is easier to find the coefficients. (Though in all likelihood, you'll be evaluating one integral while finding a conjugate.)

Question 1: do you like this idea?

Question 2: I do not work with Fourier series often, and I never gained an appreciation of when they'd ever be used. Every article I've read merely asserts that the decomposition of a function into its Fourier series is important. Thus, while I understand the existence of Fourier Series, I do not understand the need for them. In practice, do the coeffecients come first and we build the function? (If this is the case, the above method is probably useless.) Or is it the other way around?

Last edited:

Related Topology and Analysis News on Phys.org
fresh_42
Mentor
The play a central role in signal theory.

EEs use them all the time.
For example, anyone that designs a radio receiver or transmitter uses them. Anytime someone designs a circuit with feedback like an amplifier in your stereo or a robot arm, they use them. Audiologists (think of hearing aids) and (some) architects that design concert halls use them. Anytime you hear someone talk about the frequency domain or the frequency of a signal, someone designed that with FTs.
They come in different flavors, often Laplace transforms (a more general form), or digital versions, FFTs, DFTs, etc. Sometimes other orthogonal basis, like wavelets, which are often used in image processing, like mp3, mp4 etc. Youtube wouldn't work without them. Companies build and sell instruments (spectrum analyzers, etc.) to analyze real world signals. I would venture to say that 99.9% of EE labs around the world have one of these instruments.
The FT is the most common version of a more general set of tools to decompose a signal into the sum of a bunch of simpler waveforms. Because they are so common, they are used in both directions. Sometimes to analyze a waveform by doing a FT, other times by building a waveform with the inverse FT. Often it is both, take a waveform and do a FT, perform some sort of operation on the result in the frequency domain, and then do the inverse transform to create the modified time domain signal.

Edit:
Having said that, it is pretty rare for a practicing engineer to actually calculate the FT coefficients like they did in school. They buy software and lab instruments to do that for them. But, they do have to understand them to use those tools.

Last edited:
Hi Joe

1. There exist algebraic methods for DC (constant ) and sinusoidal functions (sinusoidal steady state analysis) in circuits, to find voltage and current and power.

There does not exist any algebraic method for non sinusoidal, non DC functions, like triangles, square waves, bipolar waves etc.

Circuits that are Linear and time invariant, that is they obey superposition, and are being acted upon by non sinusoidal non DC sources, how do you solve for them?

The solution is the fourier series.

Since the impedence, DC and phasor transform methods apply only for pure sinusoids and DC, the fourier series is useful to analyse circuits which are affected upon by non sinusoidal functions, which can be transformed to a fourier series representation, we prefer to use the amplitude phase format of the fourier series.
1. The first step is to express the excitation f(t) as a fourier series.
2. Transform the circuit from the time domain to the frequency (phasor) domain.
3. Find the response to DC (zero frequency or mean value of your fourier series) and then find the responses to all the AC components.
4. Use superposition to sum up all DC and AC responses, adding them up.

Now, you can then analyse the forcing functions effects individually on a linear circuit (using superposition) and find the effect, using your DC and sinusoidal steady state analysis.

Hello.

$$\textbf{Example: RL circuit forced upon by a unipolar square wave with half duty cycle}$$
I just derived this easy example myself using this setup:
$$R = 10 \,\,\,\, \Omega \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L = 2 \,\, \text{H}$$
The wave here is:
$$v(t) = \begin{cases} U_{0} & \text{ t \in [ 0 , \frac{T_{0}}{2} ]} \\ 0 & \text{ t \in [\frac{T_{0}}{2}, T_{0} ] } \\ \end{cases}$$
Periodic with $T_{0}$. and thus the fundamental frequency is: $\omega_{0} = \dfrac{2 \pi}{T_{0}}$
We first derive the expression for the voltage across the inductance, using phasors:
$$V_{L} = V_{s} \cdot \dfrac{ j \omega L }{ R + j \omega L } = V_{s} \cdot \dfrac{j 2 \omega}{10 + j 2 \omega}$$

$$V_{L} = V_{s} \cdot \dfrac{j 2 \omega}{10 + j 2 \omega}$$
Now after deriving the fourier transform I got:
$$v(t) = \dfrac{U_{0}}{2} + \displaystyle \sum_{n = 1, 3, 5, 7, 9..}^{n \to \infty} \dfrac{2 U_{0}}{n \pi} \sin(n \frac{2 \pi}{T_{0}} t)$$
Now finding the responses for all the harmonics, with $n \omega_{0}$ and converting the sine in the fourier transform to its phasor:
$$\dfrac{2 U_{0}}{n \pi} \sin(n \frac{2 \pi}{T_{0}} t) \rightarrow - \dfrac{2 U_{0}}{n \pi}j$$
Plugging this into the impedence voltage divider for the inductor and multiplying by the admittance of the inductor, to find the current, the impedence frequency expression for all harmonics becomes:
$$- \dfrac{2 U_{0}}{n \pi}j \cdot \dfrac{j 2 n \omega_{0} }{10 + j 2 n \omega_{0} } \cdot \dfrac{1}{j 2 n \omega_{0} } = - \dfrac{2 U_{0}}{n \pi}j \cdot \dfrac{1}{ 10 + j 2 n \omega_{0} }$$
We know the phasor current for the inductor at all odd number multiples of the fundamental frequency now:

$$I_{L (n \omega_{0})} = - \dfrac{2 U_{0}}{n \pi}j \cdot \dfrac{1}{ 10 + j 2 n \omega_{0} }$$
Expressing the above as a complex number with magnitude and phase:
$$I_{L (n \omega_{0})} = \dfrac{ 2 U_{0} }{ n \pi \sqrt{100 + 4n^{2} {\omega_{0}}^{2} } } \angle{ - \dfrac{\pi}{2} - \arctan{\Big[ \dfrac{n \omega_{0}}{5} \Big]} }$$
Converting this to the time domain:

$$i(t, n \omega_{0} ) = \dfrac{ 2 U_{0} }{ n \pi \sqrt{100 + 4n^{2} {\frac{2 \pi}{T_{0}} }^{2} } } \cdot \sin \Bigg[n \frac{2 \pi}{T_{0}} t - \arctan{\Big[ \dfrac{n \omega_{0}}{5} \Big]} \Bigg]$$
The fourier series for the current is:
$$i(t) = \dfrac{U_{0}}{20} + \large \displaystyle \sum_{n = 1, 3, 5,.. }^{n \to \infty} \dfrac{ 2 U_{0} }{ n \pi \sqrt{100 + 4n^{2} {\left(\frac{2 \pi}{T_{0}} \right)}^{2} } } \cdot \sin \Bigg(n \frac{2 \pi}{T_{0}} t - \arctan{\Big[ \dfrac{n \omega_{0}}{5} \Big]} \Bigg) \,\,\,\,\, \text{A}$$
So its not that complicated.
The input waveform is expressed as a fourier series.
The output relation with the input is found using physical laws or circuit theorems in the frequency phasor domain.
The input is expressed as a phasor, and then plugged into the relation of the frequency domain, the dependence on the harmonics is also considered.
This result is then converted back to the time domain, and expressed as another fourier series.
So it is used to convert signals which we cannot really insert and solve into our algebraic relations, into a sum of DC and an infinite series of sinusoids, whom we can individually insert and solve into our algebraic relation, sum up these responses, and then have our output.