Fourier Series Classically

Chapter 4. Visualization of the Complex Fourier Series
Chapter 4.1 Visualization tool
It is generally available and comes from It’s easier to understand the Complex Fourier Series equation with real part in Fig. 3-12
in Chapter 3. And from there it is very close to the general equation for the Complex Fourier Series from
Fig. 3-21 Chapter 3.
Chapter 4.2 Constant function f(t)=c0=1
Before clicking, remember that you will return to the article by clicking the windows return arrow.

Fig. 4-1
Animation of the function f(t)=c0=1
So we start with a constant as the simplest function. Do what the picture says. That is, enter Cycles 1 0 0 0 in the window and check the boxes at the bottom. Do not enter anything into the Time window, because it will “enter itself”. The result is a stationary yellow point on a circle of radius 1 on the left and a moving blue point on the right.
In the Cycles window, you can, for example, program 3 rotating vectors c1, c2 and c3 with velocities of 1ω, 2ω and and a constant yellow  c0. In the next chapter, you will measure the period T=3sec of the slowest vector. This means that its rotational speed, the so-called pulsation, is 1ω=2.1/sec
So you can study the complex function from Fig. 3-8 Chapter 3 in a truncated version.

Fig. 4-2
Complex function after entering “1 0 0 0”
Since you entered Cycles 1 0 0 0, i.e. c0=1 and c1=0 c2=0 and c3=0, nothing rotates! Or in other words – formally, the vector c0 rotates with a speed of ω=0 and the vectors c1, c2, c3 have a length of 0, i.e. they do not exist. On the right you have a time plot f(t)=c0=1.

Chapter 4.3 Function f(t)=1*exp(j1ωt) 

Fig. 4-3
Animation of f(t)=1*exp(jωt)
We get it after typing 0 1 0 0 into the Cycles window. The value of the function changes in time t and it is the position of the blue point on the complex plane. So you will see a blue point rotating at ω speed. Its position on the plane of complex numbers Re z, Im z is precisely the complex function f(t)=1*exp(jωt).
The projection of a point on the axis Re z moves according to of the formula f(t)=1*cos(ωt) and is shown on the right. The four phases of this movement, i.e. ωt=0º,90º,180º,270º, are marked by yellow points on the left and vertical yellow lines on the right.
The Time window is the values ​​1, 0, -1, 0 of the function in these phases.
By clicking anywhere you will stop the animation. Maybe you can do it in phases ωt=0º,90º,180º,270º.
Clicking again starts the animation again. Try a few times and you’ll master it. It will be useful in further analysis.
After stopping, it is possible to move the point with the slider.
One more thing. Measure the time t, e.g. 10 revolutions. It should be t=30 sec so period T=3sec. So ω=2π/3ek≅2.1/sec.

Chapter 4.4 Function f(t)=1*exp(j1ωt-30º)

Click Chapter 4.4 Function f(t)=1*exp(j1ωt-30º)

Fig. 4-4
Animation of the function f(t)=1*exp(jωt-30°)
After entering the Cycles 0 1:-30 0 0 window, we will get the above-mentioned complex function f(t). Its value changes in time t and it is the location of the blue point on the complex plane. This point moves in a circle and its projection on the Re z axis moves according to equations f(t)=1*cos(ωt-30º).
Notice the: 
– complex function f(t)=1*exp(jωt-30°) as arotating blue point
– time chart f(t)=1*cos(ωt-30º)
contains the same information. It can describe, for example, the time chart of the current, and then the rotating point is called the current phasor by electricians. Usually, a phasor is more convenient than a time chart because you can see the amplitude and phase better.

Chapter 4.5 Function f(t)=c1*exp(j1ωt)+c2*exp(j2ωt)+c3*exp(j3ωt)
The further into the forest, the more trees there are. So let’s examine the function f(t)=c1*exp(j1ωt)+c2*exp(j2ωt)+c3*exp(j3ωt) in the Fig.3-6a Chapter 3 where:
Will we get the same waveform, i.e. from Fig. 3-6b Chapter 3?
Unfortunately, the Cycles window accepts complex parameters c1, c2 and c3 only in the module-phase version So we replace
c1=1-0.2i–>r=1.02 φ=-11.301º
c2=0.6-0.4j–>r=0.721 φ=-33.7º
c3=0.4-0.4j–>r=0.567 φ=-45º
Here I suggest using the reliable WolframAlpha.
If you want to transform e.g. c1=1-0.2i then click WolframAlpha.
Just remember that the imaginary number in WolframAlpha is i, not j.

Fig. 4-5

How to transform c1=1-0.2i to c1 in the form r=1.02 φ=-11.301º
Do what the picture says.
After transforming c1, c2 and c3, you can now study the Fourier Series.
Click https://betterexplained/com/examples/fourier

Fig. 4-6
Function animation 1.02*exp(1ωt-11.301º)+0.721*exp(2ωt-33.7º)+0.567*(3ωt-45°)
Enter in the Cycles window “0 1.02:-11.301 0.721:-33.7 0.567:-45
The first in the Cycles window is the constant component c0=0
The second one is c1=1-0.2i in version r=1.02 φ=-11.301º
The third is c2=0.6-0.4j in version 0.721*exp(2ωt-33.7º)
The fourth is c3=0.4-0.4j in the version r=0.567 φ=-45º
In the Cycles window in Fig. 4-6, you only see 0.567 as bit c3
The upper part of the figure shows the 3 harmonics and their sum (Total and Parts boxes are ticked)
The lower part of the figure shows only their sum (only Total is selected)
The waveforms are similar to Fig. 3-6b Chapter 3. And why not identical? Because there are different scaling.
Associate the rotating vectors with the corresponding harmonics, compare their amplitudes and pulsations. Try to freeze the waveform at 0º, 90º, 180º and 270º. At least check their vector sums by eye and on a time plot. It should match. In this way, you can test the sum of more harmonics, i.e. more than c0, c1, c2, and c3. How much more? I don’t know. You can now imagine the Complex Fourier Series as an infinite sum of smaller and faster rotating vectors. More precisely – the projection of this sum onto the real axis Re z from Fig. 3-13 Chapter 3.
And the classic version, i.e. Fig. 3-20 and Fig. 3-21 Chapter 3? This is the sum of oppositely rotating 2 times smaller conjugate vectors. Your imagination will surely cope with it!



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