**Rotating Fourier Series**

**Chapter 11. Checking the Fourier Series formulas with the Wolfram Alfa program**

**Chapter 11.1 Introduction**My main goal is to convince the reader of the following formulas, thoroughly discussed in

**chapter. 7.2**. They are not easy, but I hope that the large number of examples with animations helped a bit. It is known that the centroid

**scn**of the

**trajectory**for a given rotational speed

**nω0**is almost the

**n-th**harmonic. The formula of

**Fig. 11-1b**generally agrees with intuition, especially for

**scn=(0,0)**when there is no harmonic for

**nω0**. That’s right, it matched the intuition, but it wasn’t calculated. Now we will use the

**WolframAlfa**program from the

**Internet**. You don’t have to install anything or pay anything. You’ll get to know it along the way. It’s a pity that I couldn’t use it as a student in the

**1960**s.

**Fig. 11-1**

Formulas to remember from

**chapter 7**.

We will check them with the

**WolframAlfa**program. Then they will become more intuitive.

**Fig. 11-1a**

Trajectory

**F(njω0t)**

The point moves on the real axis

**Re Z**according to the periodic function

**f(t)**and the

**Z**plane rotates with

**ω=-nω0**. In this way, the trajectory

**F(njω0t)**is drawn.

**Fig. 11-1b**

Formula for the centroid

**scn**of the trajectory

**F(njω0t)**.

The

**integrand**function is the trajectory

**F(njω0t)**. We can treat the

**integral**(divided by

**T**) as a point

**scn**that is on average distant in the trajectory

**F(njω0t)**from the point

**(0,0)**in time

**T sec**. Just as the

**Earth**revolves around the

**Sun**, the trajectory

**F(njω0t)**revolves around the centroid

**scn**.

**Fig. 11-1c**

Formula for the

**constant**component

**c0=a0**of the

**Fourier Series**. The classical

**average**of the periodic function

**f(t)**when

**n=0**in the formula

**Fig. 11-1b**.

**Fig. 11-1d**

The formula for the remaining

**nth**coefficients of the Fourier Series

**cn**for

**ω=1ω0, ω=2ω0…ω=nω0**.

In other words, the

**nth**complex amplitude of the harmonic

**hn(t)**

**Fig. 11-1e**

**cn**as a vector (complex number) with components

**an**and

**jbn**

**Fig. 11-1f**

**cn**as a algebraic version

**cn=an-jbn**

**Fig. 11-1g**.

**cn**as a exponential version

**cn=|cn|*exp(-jϕn)**

The

**|cn|**module and phase

**ϕn**is clearly visible for pulsation

**ω=nω0**.

**Fig. 11-1h**

**nth**harmonic

**hn(t)**as the sum of the

**cosine**and

**sinusoidal**components.

**Fig. 11-1i**

**nth**harmonic

**hn(t)**as

**cosine**with phase shift

**ϕ**

Module

**|cn|**, is a “Pythagoras” of

**an**and

**bn**, and

**tan(ϕ)=bn/an**.

**Chapter 11.2 WolframAlfa Program**

We will use it to check the formulas for the centroids of the known trajectories. Most of these types of programs have one drawback. To solve a problem, you must define it precisely. If you make a mistake in placing a dot, an “error” or other message that is difficult to understand appears. **WolframAlfa** is so smart that just ask him a general question and he will give you many answers. Even more so, the more general the question was. You will choose the answer that suits you best. Thanks to this, you do not have to remember all** WolframAlfa** instructions and you can completely focus on the problem.**For example**, we want to solve the equation **2x+3=7**.**1.** Download the program from the Internet.**2.** Enter** 2x+7=3** in the box.**3.** The program doesn’t really know what you mean. Just to be on the safe side, he will draw a **graph** and give the solution **x=-2**. You cared about the latter.**Note:**

Once you invoke the **WolframAlfa**, you will exit this blog. To return to it, click the windows** return arrow –>**.

Click on https://www.wolframalpha.com and do what the image below tells you to do**Fig. 11-2**

How did **WolframAlfa** solve the equation **2x+7=3**?

The most important thing for you is the solution **x=–2**. And a time chart as an additional answer wouldn’t hurt. It was a primer. I wonder how **WolframAlfa** will handle integration? Especially with integrating **vectors** or **complex** numbers. Let us now calculate the** centroids** and the** nth** harmonics for several known periodic functions **f(t)**. Previously you had to take the **centroids** exact values on faith. Now you will calculate them.

**Chapter 11.3 Centroid sc1 of the trajectory F(jnω0t)=F(-1j1t)=1*exp(-1j1t)****Chapter 11.3.1 Introduction**

That is, for a constant function **f(t)=1**. from **chapter. 7.4.1.** It is a **periodic** function because the function repeats itself every period **T**. What’s more, any **T** period!

**Fig. 11-3**

Complex function** 1exp(-1j1t)** as a trajectory and its centroid **sc1**.**Fig. 11-3a**

Complex function **1exp(-1j1t)** as a rotating vector.

Click. During the period **T=2π/ω≈6.28sec** the vector will make **one** revolution. What if we summed up subsequent vectors during rotation?**Fig. 11-3b**

A **circle** as a trace of a rotating vector, i.e. a trajectory of **1exp(-1j1t)**.**Fig.11-3c**

Formula for the centroid **scn** of a function trajectory. When, for example, **n=1**, the formula **Fig. 11-3c** **(1/2π)*exp(-1jω0t)** for **ω0=-1/sec**.

Without any calculations, you can see that the centroid is **sc1=(0,0)**. Will the formula in **Fig.11-1b** confirm this?**Chapter 11.3.2 Checking with the WolframAlfa**Let us calculate the centroid

**sc1**of the trajectory

**https://www.wolframalpha.com**and do what the picture says. Instead of laboriously typing instructions, you can simply copy them

**integrate 1*exp(-i1t)/(2pi) from t=0 to t=2pi**and paste into the window.

In other words, you will paste the integral formula

**Fig. 11-1b**written in the

**WolframAlpha**language into the window. Please note that the imaginary number for

**WolframAlfa**is the “mathematical

**i**”, not the “electrical

**j**” used in the blog.

**Fig. 11-4**

Centroid

**sc1=(0,0)**trajectory

**F(-1j1t)=1exp(-1j1t)**

As we expected, the centroid

**sc1=0**, or more precisely,

**sc1=(0,0)**, because it is a complex number. This

**WolframAlfa**is nice. For example, the instruction from the window is converted into human, i.e. mathematical, language. This can be seen with the arrow “

**3. You will get it**.”

Coming back to the result of

**sc1=0**, what does it mean? Well, the

**amplitude**of the first harmonic

**c1**for the harmonic

**1ω0=1/sec**is

**zero**, because

**cn=2*sn**. So there is no harmonic with pulsation

**ω=1/sec**, nor for any other pulsation

**ω**. This is what we expected and the only component of the constant function

**f(t)=1**is the

**constant**component

**c0=1**.

**Chapter 11.4 Centroids scn of trajectories F(-njω0t)=0.5cos(4t)*exp(-njω0t) for n=0, 1, 4 and ω0=1/sec****Chapter 11.4.1 Introduction**

This is a shortened version of** Chapter 4**, which had** 9** versions of this trajectory for **n=0…9**. Now there are only **3** versions for** n=0.1** and only these formulas will we check for calculations.

Let’s throw the function **f(t)=0.5cos(4t)** into a centrifuge.

Let’s turn on the rotations to **n=0, n=1** and **n=4**. So at a speed **of ω=0** (the centrifuge is stationary!), **ω=-1/sec** and **ω=-4/sec**.

There will be **3** trajectories:**n=0–>ω=0**–->centrifuge standing–>**F(-0j1t)=0.5cos(4t)****n=1–>ω=-1/sec** –->centrifuge rotating–>**F(-1j1t)=0.5cos(4t)*exp(-1j1t)****n=4–>ω=-4/sec** –->centrifuge rotating–>**F(-4j1t)=0.5cos(4t)*exp(-4j1t)**

**Fig. 11-5**Three trajectories

**F(-nj1t)=0.5cos(4t)*exp(nj1t)**for

**n=0, 1**and

**4**

**ω=0**

The centrifuge is standing because

**ω=0**and

**sc0=(0,0)**

**ω=-1/sec**

Centrifugation

**ω=-1/sec**and

**sc1=(0,0)**

**ω=-4/sec**

Centrifugation

**ω=-4/sec**and

**sc4=(0.25,0)**

The above parameters

**sc0, sc1**and

**sc0**, although intuitive, were taken on my word of honor. Now we will calculate them with the

**WolframAlfa**program. The results should be the same.

**Chapter 11.4.2 Centroid sc0, i.e. for ω=0, i.e. the constant component sc0=c0=a0**.

That is, when the centrifuge is standing, because

**n=0**

Then the trajectory

**F(-0j1t)=f(t)=0.5cos(4t)**

The animation

**Fig. 11-4a ω=0**shows that the trajectory is a horizontally

**swinging**line according to. function

**f(t)=0.5cos(4t)**. Its centroid is clearly

**sc0=(0,0)**. Will

**WolframAlfa**confirm this? Let’s put f

**(t)=0.5cos(4t)**into the formula

**Fig. 11-1c**and calculate this

**integral**.

Click on

**https://www.wolframalpha.com**and do whatever the picture tells you to do.

In the box, type or paste

**integrate 0.5cos(4t)/(2pi) from t=0 to t=2pi**

**Fig. 11-6**

Calculation of the centroid

**sc0**for

**F(0j1t)=0.5cos(4t)*exp(-0j1t)**=

**0.5cos(4t)**

**sc0=0**

Note that although the fundamental period of

**f(t)**is

**T=π/2**, the result of

**sc0=0**for the integration limits of

**T=π/2**and

**T=2π**will be the same!

The

**integration limits**in the formula

**Fig. 11-1c**can be arbitrary and the result will be the same for e.g.

**T = 0.234 sec**,

**T = 5.27 sec**or

**T=2π sec**as in the above formula! The only important thing is that

**T**is the

**period**of the function. This note is more

**general**and applies to the

**Fourier Series**formulas in

**Fig. 11-1b**. Some give arbitrary limits of integration

**T**, and others, like me, specify specific

**T=2π sec**. The

**Fourier coefficients**of, for example, a

**square wave**do not depend on the period

**T**, but on the

**shape**of the wave! The

**centroid**

**sc0**for the trajectory of each function

**f(t)**is also the

**constant**component of this function and is the coefficient

**c0=a0**of the

**Fourier Series**. Unlike the other

**Fourier coefficients**i.e.

**c1, c2…cn**, the coefficient

**c0**is always a

**real**numbe

**Chapter 11.4.3 Centroid sc1, i.e. for ω=-1/sec.**

That is, **sc1** of the trajectory **F(-1j1t)=0.5cos(4t)*exp(-1j1t)** from** Fig. 11-5 ω=-1/sec**. We are looking for a** harmonic** with pulsation **ω=1/sec** from the function **f(t)=0.5cos(4t)**. It is obvious that it does not exist. After all, the only harmonic of **f(t)** is itself, i.e. **0.5cos(4t).** This also follows from **Fig. 11-4** **ω=-1/sec**, where **sc1=(0,0)**. **Will WolframAlfa** confirm this?

Click on **https://www.wolframalpha.com **and do whatever the picture tells you to do.

In the box, type or paste integrate **[0.5cos(4t)/(2pi)]*exp(-1i1t)** from** t=0** to **t=2pi**

**Fig.11-7**Centroid

**sc1=0**trajectory

**0.5cos(4t)*exp(-1j1t)**

Amplitude

**c1=0**for harmonic

**ω=1/sec**because

**c1=2sc1=0**. So there is no such

**harmonic**.

As you saw in

**Chapter 4**, the centroid for all

**ω**different from

**4/sec**(that is, from

**ω**of the function

**0.5cos(ωt)**) are

**zero**. It is obvious. After all, the

**0.5cos(4t)**function has only one

**harmonic**, and that is itself! The

**remaining**harmonics are

**zero.**

**Chapter 11.4.4 Centroid sc4, i.e. for ω=-4/sec.**

That is,** sc4** of the trajectory **F(-4j1t)=0.5cos(4t)*exp(-4j1t)** from **Fig. 11-4 ω=-4/sec**

We are looking for a **harmonic** with pulsation **ω=4/sec** from the function **f(t)=0.5cos(4t)**. I guess everyone sees this harmonic. It is itself **0.5cos(4t)**.

The function** f(t)=0.5cos(4t)** rotates clockwise at a speed of** ω=-4/sec**. Note that **f(t)** and the centrifuge have** ω=4/sec** (though opposite signs)! This caused the centroid to shift from **sc4=(0,0) **to **sc4=(+0.25,0)**.

To check the calculation, click** https://www.wolframalpha.com** and do what the picture tells you.

In the box, enter or paste integrate **[0.5cos(4t)]*exp(-i4t)/(2pi)** **from t=0 to t=2pi**.

**Fig. 11-8**Centroid

**sc4**or more precisely

**sc4=(0.25,0)**of the trajectory

**0.5cos(4t)*exp(-4j1t).**It will turn out that for each spin speed

**ω**of the function

**0.5cos (4t)/(2pi)**different from

**ω=4/sec**the center of gravity

**scn=(0,0)**, and only for

**ω=4/sec**there is a non-zero

**sc4=(+ 0.25.0)**!

Acc. to

**Fig. 11-1d c4=2sc4=(+ 0.5,0)**that is

**a4=0.5**and

**b4=0**

Acc. to

**Fig. 11-1h h4(t)=0.5cos (4t)**.

**C**

**hapter 11.5 Centroids scn of trajectories F (-njω0t) = [1.3+0.7cos(2t)+0.5cos (4t)] exp (-njω0t)**

**that is F (-njω0t) for n = 0,2,3,4 and ω0=1/sec**

**Chapter 11.5.1 Introduction**

It is an abbreviated version of

**Chapter 5**, in which there were

**9**trajectories

**for n = 0… 9**. Now there will only be

**4**for

**n=0,2,3**and

**4**.

Let’s put the function

**f (t)=1.3+0.7cos (2t)+0.5cos (4t)**into the centrifuge

Let the rotation be at

**ω=0, ω=-2 sec, ω=-3/sec, and ω=-4/sec**.

**4**trajectories will be created:

**n=0**–>

**ω=0**–> centrifuge is standing–>

**F(0j1t)=1.3+0.7cos(2t)+0.5cos (4t)**

**n=1**–>

**ω=-2/sec**–> centrifuge is rotating–>

**F(-1j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp (-2j1t)**

**n=3–>ω=-3/sec**–> centrifuge is rotating–>

**F(-3j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp(-3j1t)**

**n=4–>ω=-4/sec**–> centrifuge is rotating–>

**F(-4j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp (-4j1t)**

**Fig. 11-9**

**Four**trajectories

**F(-nj1t)=[1.3+0.7cos (2t)+0.5cos(4t)]exp(-nj1t)**for

**n=0, 1, 2**and

**4**and their centroids

**scn**.

The waveforms during

**one**period

**T=2π sec**. The waveforms

**ω=-2/sec**and

**ω=-4/sec**are “drawn on top of each other” and therefore seemingly stopped after

**1π sec**.

**ω=0**The centrifuge is standing–>

**sc0=(1.3,0)**

**ω=-2/sec**The centrifuge is rotating–>

**sc2=(0.35.0)**

**ω=-3/sec**The centrifuge is rotating–>

**sc3=(0,0)**

**ω=-4/sec**The centrifuge is rotating –>

**sc4=(0.5,0)**

The centroidsy

**scn**are fairly

**intuitive,**but were included in

**Chapter 5**without justification. Now we will calculate them by of the formula

**Fig. 11-1b**with the

**WolframAlfa**program. The results should be the same.

**Chapter 11.5.2 Centroid sc0 for ω=0**

So

**sc0**of the trajectory

**F(-0j1t)=[1.3+0.7cos (2t)+0.5cos (4t)] from Fig. 11-9 ω = 0**. The centrifuge is standing. The trajectory is the horizontally swinging line acc. to

**F(-0j1t)=f (t)= 1.3+ 0.7cos (2t)+0.5cos(4t)**. Its centroid is

**sc0=1.3=(1.3,0)**.

Let us calculate

**sc0=c0=a0**that is the constant component acc. to

**Fig. 1-11c**.

Click on

**https://www.wolframalpha.com**and do what the picture tells you to do.

Enter or paste the

**WolframAlfa instruction**for the

**integral**, i.e.

**integrate[1.3+0.7*cos (2t)+0.5*cos (4t)]/(2pi) from t = 0 to t = 2pi**

**Fig. 11-10**

Calculation of the

**sc0**centroid for

**F**

**(0j1t)=[1.3+0.7cos(2t)+ 0.5cos(4t)]/(2pi)**

The centroid

**sc0=(1.3,0)**centroid

**=1.3**for the non-rotating trajectory of the function

**f(t)**

is its constant component, i.e. the coefficient

**c0=a0=1.3**of the

**Fourier Series**.

**Chapter 11.5.3 Center of gravity sc2 for ω=2/sec**

So

**sc2**trajectory

**F(-2j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-2j1t)**for

**Fig. 11-9 ω=-2/sec**

Click

**https://www.wolframalpha.com**

Enter or paste

**integrate**

**[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp (-2i1t)]/(2pi) from t = 0 to t = 2pi**

Fig. 11-11

Fig. 11-11

Calculation

**sc2**of the centroid for

**F (2j1t)=[1.3+0.7cos (2t) + 0.5 cos (4t)]*exp (-2j1t)/(2pi)**

The centroid

**sc2=0.35=(0.35.0)**, i.e. the

**2nd**harmonic amplitude is

**c2=2(0.35.0]=(0.7,0).**i.e.

**a2=0.7 b2=0**

The

**second**harmonic is

**h2(t)=0.7cos(2t)**acc.

**Fig. 11-1h**.

**Chapter 11.5.4 Centroid sc3 for ω=3/sec**

So

**sc3**trajectory

**F(-3j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-3j1t)**for

**Fig. 11-9 ω=-3/sec**

Click

**https://www.wolframalpha.com**

Enter or paste

**integrate**

**[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-3j1t)]*exp (-3i1t)]/(2pi) from t=0 to t=2pi**

**Fig. 11-12**

**Calculation**of the

**sc3**centroid of the trajectory

**[1.3+0.7cos (2t)+0.5cos(4t)]*exp (-3j1t)/(2pi)**

The centroid

**sc3=0=(0,0)**, i.e.

**3rd**harmonic amplitude

**c3=2*0=0**.There is no

**third**harmonic. There is also no

**5,6,7**… harmonics.

You can see this in the

**f(t**) function

**itself**, of course, but you can check it with

**WolframAlfa**.

**Chapter 11.5.5 Centroid sc4 for ω=4/sec**

So

**sc4**trajectory

**F(-4j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]*exp(-4j1t)**for

**Fig. 11-9 ω=-4/sec**

Click

**https://www.wolframalpha.com**

Enter or paste

**integrate**

**[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-4i1t)]/(2pi) from t = 0 to t = 2pi**

**Fig. 11-13**

Calculation of the

**sc4**centroid for

**F(4j1t)=[1.3+0.7cos (2t)+0.5 cos (4t)]*exp (-4j1t)/(2pi)**

Centroid

**sc4=0.25=(0.25.0)**, i.e. the

**4th**harmonic amplitude is

**c4=2(0.25.0]=(0.5,0).**i.e.

**a4=0.5 b4=0**

The

**fourth**harmonic is

**h4(t)=0.5cos(4t)**acc.

**Fig. 11-1h**.

**Chapter 11.6 Centroids of trajectory F(nj1t)=0.5cos(4t-30°)*exp(-4j1t)for n = 0.2,3.4 and ω0=1/sec that is with completely complex type numbers**

Until now, the centroidsy

**scn**of the

**harmonics**were

**real numbers**, eg

**Fig. 11-9 ω=-2/sec**–->

**sc2=(0.35.0)= 0.35**. They are also

**complex**numbers, but with

**zero imaginary**components. What if the

**cosine**function has a

**phase**

**shift**?

So let’s throw into the centrifuge e.g.

**f(t)=0.5cos(4t-30 °)**.

Let us turn the rotation to

**n=0**,

**n=1**and

**n = 4**.

There will be

**3**trajectories:

**n=0**–>

**ω=0**–> centrifuge is standing–>

**F(-0j1t)=0.5cos (4t-30 °)**

**n=1**–>

**ω=-1/sec**–>centrifuge is rotating–>

**F(-1j1t)=0.5cos(4t-30 °)*exp (-1j1t)**

**n=4**–>

**ω=-4/sec**–>centrifuge is rotating–>

**F(-4j1t) = 0.5cos (4t-30 °)*exp (-4j1t)**

**Fig. 11-14**

**F(nj1t)=0.5cos(4t-30°)*exp(-jnt)**for

**n=0,1 i 4**

The animation lasts

**T=2π≈6.28sec**.

**ω=0**

F(0j1t)=0.5cos(4t-30°)

The difference between the animation

F(0j1t)=0.5cos(4t-30°)

**Fig. 11-5 ω=0**is minimal, but try to notice it.

**ω=-1/sec**.

**F(4j1t)=0.5cos(4t-30°)*exp(-j1t)**

The trajectory is

**rotated**by (I suspect?)

**-30°**relative to

**Fig. 11-5 ω=-1/sec**. The centroid

**sc0=(0,0)**. This means that there is no

**harmonic**of the function

**f(t)=0.5cos(4t)**for

**ω=1/sec**. Also for any other, except for

**ω=4/sec**.

**ω=-4/sec**.

**F(4j1t)=0.5cos(4t-30°)*exp(-4j1t)**

The centroid

**sc4**is a full-fledged

**complex**number

**sc4=(a,b)**where:

**a=0.25cos(-30°)≈+0.217**

b=0.25sin(-30°)≈-0.125

sc4=(a,b)≈(+0.217,-0.125)=+0.217-j0.125

So the

b=0.25sin(-30°)≈-0.125

sc4=(a,b)≈(+0.217,-0.125)=+0.217-j0.125

**complex**fourth harmonic

**h4(t)≈2*sc≈2*(+0.217-j0.125)*exp(j4t)**. It corresponds to the function

**f(t)=0.434cos(4t)+0.25sin(4t)=0.5cos (4t-30°)**according to the formulas

**Fig. 11-1h**and

**Fig. 11-1i**What does

**WolframAlfa**say?

Click

**https://www.wolframalpha.com**.

Enter or pastej

**integrate [0.5cos(4t-pi/6)]/(2pi)*exp(-4i1t) from t=0 to t=2pi**

Note

**-30°**=-

**π/6**

**Fig. 11-15**

**sc4=0.216506-j0.125**

Such centroid

**sc4**of the trajectory

**F(4j1t)=0.5cos(4t-30°)*exp (-j4t)**was calculated by

**WolframAlfa**. A result similar to ours, only more accurate. Note that

**sc4**is a full

**complex number**. This is the case when the

**cosine/sinus**type function has a

**non-zero**

phase shift

**ϕ**. Here

**ϕ=30°=-π/6**.

**Fun fact**

Compare the animations

**Fig. 11-13 ω=-1/sec**and

**ω=-4/sec**.

**ω=-1/sec**

**Four Leaf**Clover

**ω=-4/sec**

It is also some kind of a

**leafy clover**except that each leaf is a

**circle**and is drawn “on top of each other”.

**Chapter 11.7 Centroids of trajectoriesF(-njω0t)=[0.5+1.08cos (1t-33.7 °)+0.72cos(3t+33.7°)+0.45cos(5t-26.6 °)]*exp(-njω0t)for n = 0,1,2…8 and ω0=1/sec**

**Chapter 11.7.1 Introduction**

The function

**f(t)**with the period

**T=2πsec**looks like this and you don’t see its

**3**

**cosines**with phases

**ϕ**. Otherwise

**WolframAlfa**theoretical doesn’t need to know the formula

**f(t)**. For him, the

**f(t)**diagram alone is enough.|

**Fig. 11-16**

**f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)**

In

**Chapter 6**we examined

**9**trajectories

**F(-njω0t)**of this function

**f(t)**for

**n=0,1,2,… 8**and

**ω0=1/sec**. It is interesting because due to the phase shifts

**ϕ**, the

**scn**centroids are

**completely**complex numbers. The function

**f(t)**is equivalent to the following, in which each harmonic has been decomposed into a

**cosine**and a

**sinusoidal**component–>

**f(t)=0.5+0.9cos(1t)+0.6sin(1t)+0.6cos(3t)-0.4sin(3t)+0.4cos(5t)+0.2sin(5t)**.

e.g.

**0.9*cos(1t)+0.6*sin(1t)=**

**1.08*cos(1t-33.7°)**

This is the same function, but

**complex**Fourier coefficients, or

**complex**Fourier amplitudes, are easily determined here. From them,

**harmonics**are read as

**h1(t)**,

**h3(t)**and

**h5(t)**waveforms with

**sine/cosine**components.

**c1=0.9-j0.6**—>

**h1(t)=0.9*cos(1t)+0.6*sin(1t)**

**–>**

c3=0.6+j0.4

c3=0.6+j0.4

**h3(t)=0.6*cos(3t)-0.4*sin(3t)**

c5=0.4-j0.2—>

c5=0.4-j0.2

**h5(t)=0.4*cos(5t)+0.2*sin(5t)**

**Note:**

Both forms of the function

**f(t)**are equivalent, but for the calculations we will take the version with “cosines and sines”. We will put the functions

**f(t)**into the centrifuge with different speeds

**nω0=-n/sec for n=0,1,2,3 and 5**. We will check if the calculated

**scn**centroids are the same as in

**Chapter 6**.

**Chapter 11.7.2 Centroids scn of trajectory F(-njω0t)=f (t)exp(-njω0t) for n = 0, 1,2, 3, 5 and ω0=1 sec**

Let’s put the

**function**from

**Fig. 11-15**into the centrifuge. I would like to remind you that it can also be presented in the

**sine/cosine**version, i.e:

**f(t)=0.5+0.9cos(1t)+0.6sin(1t)+0.6cos(3t)-0.4sin(3t)+0.4cos(5t)+0.2sin(5t)**.

Let’s turn on the rotation

**for n=0, 1, 2, 3 and 5**. So for the speeds

**ω=0**(the centrifuge is standing!),

**ω=-1/sec**,

**ω=-2/sec, ω=-3/sec**and

**ω=-5/sec**. The formation of

**5 trajectories:**

**ω=0**–>centrifuge is standing–>

**F(-0j1t)=f(t)**

**ω=-1/sec**–>centrifuge is rotating—>

**F(-1j1t)=f(t)exp(-1j1t)**

**ω=-2/sec**–>centrifuge is rotating—>

**F(-2j1t)=f(t)exp(-2j1t)**

**ω=-3/sec**–

**>**centrifuge is rotating—>

**F(-3j1t)=f(t)exp(-3j1t)**

**ω=-5/sec**–

**>**centrifuge is rotating—>

**F(-5j1t)=f(t)exp(-5j1t)**

**Fig. 11-17**

**5**trajectories

**F(-nj1t)=f t)exp (nj1t)**for

**n = 0,1,2,3 and 5 ω=0**and their

**scn**centroids.

**ω=0**

The centrifuge is standing and

**sc0=(+ 0.5.0)**

**ω=-1/sec**

Rotating

**ω=-1/sec**–>

**sc1=(+0.45,-0.3)**

**ω=-2/sec**

Rotating

**ω=-2/sec**–>

**sc2=(0,0)**

**ω=-3/sec**

Rotating

**ω=-3/sec**–>

**sc3=(+0.3,+0.2)**

**ω=-5/sec**

Rotating

**ω=-5/sec**–>

**sc5=(+0.2,-0.1)**

Let me remind you that the vector

**scn**is almost the

**nth**harmonic of the

**periodic**function

**f(t)**. More specifically, the

**doubled**vector i.e.

**2scn**is the

**complex**amplitude of the

**nth**harmonic. In the period

**T=2π sec**, each trajectory is drawn by a changing vector and

**scn**is the

**average**of these

**changing**vectors and therefore their sum is

**divided**by

**2π**.

**Chapter 11.7.3 Centroid sc0, i.e. for ω=0**

Trajectory

**F(0j1t)=f(t)**from

**Fig. 11-16 ω=0.**

The centrifuge is standing. The trajectory is the

**horizontally swinging line**in

**Fig. 11-17 ω=0**acc. to the function

**F(0j1t)=f(t)**. Its centroidy is

**sc0=(+0.5,0)=+0.5**.

Let’s put

**f(t)**into the formula

**Fig. 11-1c**.

Click

**https://www.wolframalpha.com**and do what the picture tells you to do. Enter or paste the

**WolframAlfa**instruction.

**integrate [0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t)]/(2pi) dt from t=0 to t=2pi**

**Fig. 11-18**

Calculation of the centroid

**sc0**for

**F(0j1t)=f(t)/(2pi)**

You see

**sc0**as a surface of the

**f(t**) ((divided by 2π). Note that they are

**plus**and

**minus**sufaces.

The centroid

**sc0=(0.5,0)=a0=0.5**for the

**non-rotating**trajectory of the function

**f(t)**is its constant component, i.e. the coefficient

**c0=a0**of the

**Fourier Series**.

**Chapter 11.7.4 Centroid sc1 for ω=1/sec**.

Trajectory

**F(-1j1t)=f(t)exp(-1j1t)**from

**Fig. 11-17 ω=-1/sec**

Click

**https://www.wolframalpha.com**. Type or paste into the box

**integrate [0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t)]*exp(-1i1t)]/(2pi) dt from t=0 to t=2pi**

**Fig. 11-19**

Trajectory

**F(-1j1t)=f(t)*exp(-1j1t)**

**Note:**

In the window, you only see the part of the expression you pasted.

The

**calculated**centroid is

**sc1=0.45-j0.3**

The

**complex amplitude**of the first harmonic is

**2**

**sc1=**

**0.9-j0.6**

So the

**first**harmonic according to

**Fig. 11-1h**and

**Fig. 11-1i**is:

**h**

**1(t)=**

**0.9cos(1t)+0.6sin(1t)=1.08*cos(1t-33.7°)**.

Success!

**WolframAlfa**perfectly filtered the

**first**harmonic from the periodic function

**f(t)**. We expect the same in the following chapters.

**Chapter 11.7.5 Centroid sc2 for ω=2/sec.**

Trajectory

**F(-2j1t)=f t)exp(-2j1t)**from

**Fig.11-17 ω=-2/sec**

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**integrate (0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t))*exp(-2i1t)/(2pi) dt from t=0 to t=2pi**

**Fig. 11-20**Trajectory

**F(-2j1t)=f(t)*exp(-2j1t)**

**sc2=0**

That is, there is no harmonic with pulsation

**ω=2/sec**

**Chapter 11.7.6 Centroid sc3 for ω=3/sec.**

Trajectory

**F(-3j1t)=f t)exp(-3j1t)**from

**Fig. 11-17 ω=-3/sec**

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**integrate**

**(0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t))*exp(-3i1t)/(2pi) dt from t=0 to t=2pi**

**Fig. 11-21**

Trajectory

**F(-3j1t)=f(t)*exp(-3j1t)**

The

**calculated**centroid is

**sc3=0.3+j0.2**

The

**complex amplitude**of the

**third**harmonic is

**2**

**sc1=**

**0.6-j0.4**

So the

**third**harmonic according to

**Fig. 11-1h**and

**Fig. 11-1i**is:

**h3(t)=**

**0.6cos(3t)-0.4sin(3t)≈0.72*cos(3t+33.7°)**..

**Chapter 11.7.7 Centroid sc5 for ω=5/sec.**

Trajectory

**F(-5j1t)=f t)exp(-5j1t)**from

**Fig. 11-17 ω=-5/sec**

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**integrate**

**(0.5+0.9*cos(1t)+0.6*sin(1t)+0.6*cos(3t)-0.4*sin(3t)+0.4*cos(5t)+0.2*sin(5t))*exp(-5i1t)/(2pi) dt from t=0 to t=2pi**

**Fig. 11-22**

Trajectory

**F(-5j1t)=f(t)*exp(-5j1t)**

The

**calculated**centroid is

**sc5=0.2-j0.1**

The

**complex amplitude**of the

**third**harmonic is

**2**

**sc1=**

**0.4-j0.2**

So the

**fifth**harmonic according to

**Fig. 11-1h**and

**Fig. 11-1i**is:

**h5(t)=**

**0.4cos(5t)+0.2sin(5t)≈0.45*cos(5t-26.6°)**

**Chapter 11.7.8 Centroids scn of the trajectories F (-njω0t)=f (t)exp(-njω0t) for n= 4,6,7,8 and ω0=1/sec**

There are still centroids

**sc4, sc6, sc7 and sc8**i.e. for

**n=4,6,7 and 8**. They are

**zero**, i.e. there are

**no harmonics**for these pulsations. I suggest you make the calculations yourself with the

**WolframAlfa**program.

**Chapter 11.8 Centroids scn of the even square wave trajectory for n=0,1,2,… 8 and ω0=1/sec**** Chapter 11.8.1 Introduction**

We will repeat **chaper 8,** but this time we will calculate the centroids of the trajectory** scn** using the **WolframAlfa** program. Previously, we took them on faith.**Fig. 11-23**

**Even**square wave

**f(t) A=1, ω=1/sec, ϕ=0**and

**ω=50%.**

We will calculate the successive harmonics with the

**WolframAlfa**program using the formulas

**Fig. 11-1**. But first we will check how

**WolframAlfa**deals with

**square waves**.

**Chapter 11.8.2 Square wave f t) and WolframAlfa**

We know what an instruction of a function looks like, e.g.

**sin(t)**in

**WolframAlfa**.

The function

**sin(t)**is just

**sin(t)**and the natural logarym

**ln(t)**is

**ln(t)**. A

**square wave**, on the other hand, is

**squarewave[t]**.

Let’s plot this function with the

**plot**instruction.

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**plot squarewave[t] from t=-2 to 2**

**Fig.11-24**

**Squarewave[t]**from

**t=-2 to 2**

It is an

**odd, periodic**function

**T = 1sec**,

**A = 2**without a

**constant**component. How to modify it to get the

**even**as in

**Fig.11-23**?

One should

**1.**Multiply by

**0.5**to reduce the amplitude from

**A=2**to

**A=1**

**2.**“Stretch” from

**T=1sec**to

**T=2πsec**

**3.**Move to the

**left**

**π/2sec**to change the

**odd**function to

**even**.

**4.**Move up

**0.5**

Let’s test it

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**plot 0.5*squarewave[(t+0.5pi)/(2pi)]+0.5 from t=-2π to 2π**

**Fig.11-25**

The square wave from

**Fig. 11-23**generated by the

**WolframAlfa**program.

The period

**T=2π**is shown, more precisely

**from -π to + π**.

In the following

**chapters**we will designate its trajectories, centroids

**scn**and harmonics

**hn(t)**.

**Chapter 11.8.3 Centroid sc0, that is for ω = 0, that is the constant component a0.**

We will use the formula

**Fig. 11-1c**

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**(1/(2pi))*integrate [[0.5+ 0.5 * SquareWave [(t + pi/2)/(2pi)]] from t=-pi to t=+pi**

**Fig.11-26**

Calculation the constant component

**sc0=c0=a0=+0.5**

As expected,

**a0=0.5**as the mean value of the square wave over the period

**-π to + π**.

It was not even necessary to integrate, because

**a0**as the mean can be seen in

**Fig. 11-25**.

**Chapter 11.8.4 Centroid sc1, that is for ω=-1/sec.**

So

**sc1**of the trajectory

**F (-1j1t)=f (t)exp(-1j1t)**where

**f(t)**is a

**square wave**from

**Fig. 11-25**.

We will repeat the

**animation**from

**Fig. 8-3 chapter. 8**.

**Fig. 11-27**

Trajectory **F (1j1t)** of an **even square** wave**a.** Vector **1exp(1j1t)** as radius **R=1** rotating at speed **ω=-1/sec****b.** Vector** F(1j1t)** as a radius **R=1** modulated by a **square wave** **f (t)** from **Fig. 11-25**.**c.** Trajectory** F(1j1t)** of an **even square** wave drawn by the vector **b**. The vector of the centroid of the trajectory

**sc1=(+1 /π, 0)**more or less agrees with the

**intuition**, But as the

**mean**of the rotating vector

**b**must lie somewhere between

**(0,0)**and

**(1,0)**.

Now let’s count it exactly with the

**WolframAlfa**program. We will use the formula

**Fig. 11-1c**.

Click

**https://www.wolframalpha.com**. Type or paste into the box

**[(1/(2pi)]*integrate[[0.5+0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-1i1t) dt from t=0 to t=+2pi**

**Fig. 1-28**

Calculation of sc1=(1/ π, 0)

Calculation of sc1=(1/ π, 0)

The program calculated the

**average**of the vectors over time

**T=2πsec**and it came out

**0.31831**.

**First**, it calculated a more precise value but rounded to

**5**decimal places.

**Second**, it is exactly the number

**1/π**

**Third**, it is a vector in the form of a complex number

**sc1=(1/π, 0)**

**Chapter 11.8.5 Centroid sc2=0, that is for ω=-2/sec.**

So

**sc2**of the trajectory

**F(-2j1t)=f(t)exp(-2j1t)**where

**f(t)**is a square wave from

**Fig. 11-23**. We will repeat the animation from

**Fig. 8-5 chapter. 8**.

**Fig. 11-29**

Trajectory **F(2j1t)** of an even square wave**a.** Vector **1exp (j2j1t)** as radius **R=1** rotating at speed **ω=-2/sec****b.** Vector **F(2j1t)** as radius** R=1** modulated by square wave** f(t)** from **Fig. 11-25**.**c.** Trajectory **F(2j1t)** of an even square wave drawn by the vector **b**. The trajectory centroid **sc2=(0,0)** agrees as the mean.

Now let’s calculate it exactly with the **WolframAlfa** program.

Click **https://www.wolframalpha.com**. Type or paste into the box**[(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-2i1t)dt from t=0 to t=+2pi**

**Calculation**

Fig. 11-30

Fig. 11-30

**sc2=0=(0,0)**

That is, the

**second**harmonic, more precisely for

**ω=2/sec**, doesn’t exist

**Chapter 11.8.6 Centroid sc3, that is for ω=-3/sec**.

So

**sc3**trajectory

**F(-3j1t)=f (t)exp(-3j1t)**where

**f (t)**is a square wave from

**Fig. 11-25**. We will repeat the animation from

**Fig. 8-6**

**chapter. 8**.

**Fig. 11-31**Trajectory

**F(3j1t)**of an even square wave

**a.**Vector

**1exp (j3j1t)**as radius

**R=1**rotating at speed

**ω=-3/sec**

**b.**Vector

**F(3j1t)**as radius

**R=1**modulated by square wave

**f(t)**from

**Fig. 11-25**.

**c.**Trajectory

**F(3j1t)**of an

**even square**wave drawn by the vector

**b**

Vector of the centroid of the trajectory

**sc3=(-1/3π, 0)**. The animation

**b**is best suited for interpretation. Notice that the vector makes

**2×3/4**turns. The

**left**direction of the vector on the

**Re z**axis is obvious when we consider

**2**empty

**quarter**turns. If they were not there (2×1 turn), then

**sc3=(0,0)**. And why is the length of the vector

**sc3**smaller than the length of

**sc1**in Figure

**11-27c**? Because there are more

**gaps**in the period

**T=2πsec**, which reduces the average. Ultimately,

**WolframAlfa**will convince you.

We will use the formula

**Fig. 11-1c**

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**[(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-3i1t)dt from t=0 to t=+2pi**

**Fig.1-32**

Calculation

**sc3=(-1/3π,0)**

**Chapter 11.8.7 Centroid sc4=0, that is for ω=-4/sec.**

So

**sc4**of the trajectory

**F(-4j1t)=f (t)exp (-4j1t)**where

**f (t)**is a square wave from

**Fig. 11-25**. We will repeat the animation from

**Fig. 8-5 chapter. 8**.

**Fig. 11-33**Trajectory

**F(4j1t)**of an

**even square**wave

**a.**Vector

**1exp(4j1t)**as radius

**R=1**rotating at speed

**ω=-4/sec**

**b.**Vector

**F(4j1t)**as radius

**R=1**modulated by

**square wave**

**f (t)**from

**Fig. 11-25**.

**c.**Trajectory

**F(4j1t)**drawn by the vector from

**b**.

The trajectory center of gravity vector

**sc4=(0,0)**agrees as the mean.

Now let’s count it exactly with the

**WolframAlfa**program. We will use the formula

**Fig. 11-1c**

We will use the formula

**Fig. 11-1c**

Click

**https://www.wolframalpha.com**.

Type or paste into the box

**[(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-4i1t)dt from t=0 to t=+2pi**

**Fig. 11-34**

Calculation

**sc4=0=(0,0)**

That is, the

**fourth**harmonic, more precisely for

**ω=4/sec**, doesn’t exist

**Chapter 11.8.8 Centroid sc5, sc6, sc7 and sc8, that is for ω = -5 / sec, -6 / sec, -7 / sec and- 8 / sec.**

As homework. I will just suggest that it is enough to paste it into the WolframAlfa window

**[(1/(2pi)]*integrate [[0.5+ 0.5*SquareWave[(t+pi/2)/(2pi)]]*exp(-ni1t)dt from t=0 to t=+2pi**

with an appropriately changed parameter

**n.**