### Fourier Series Classically

**Chapter 4. Visualization of the Complex Fourier Series****Chapter 4.1 Visualization tool**

It is generally available and comes from **https://betterxplained.com/articles/an-interactive-guide-to-the-fourier-transform**. 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.

Click

**https://betterexplained.com/examples/fourier/**

**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

**3ω**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) **Click

**https://betterexplained.com/examples/fourier/**

**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)

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

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

Click

**https://betterexplained.com/examples/fourier/**

**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 a

**rotating 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 char**t 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:

**c1=1-0.2j**

c2=0.6-0.4j

c3=0.4-0.4j

Will we get the same waveform, i.e. from

c2=0.6-0.4j

c3=0.4-0.4j

**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

c2=0.6-0.4j–>r=0.721 φ=-33.7º

c3=0.4-0.4j–>r=0.567 φ=-45º

**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

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 o

**ppositely**rotating

**2**times smaller

**conjugate vectors**. Your imagination will surely cope with it!