## Rotating Fourier Series

### Chapter 11. Checking the formulas for Fourier Series with WolframAlpha program

**Chapter11.1 Introduction
**My main goal is to convince the

**Reader**of the following formulas, which are thoroughly discussed in

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

**scn**of the trajectory for a given

**rotational**speed

**nω0**is almost the

**nth harmonic**. The formula of

**Fig. 11-1b**is generally

**intuitive**, especially for

**scn=(0,0)**when the harmonic for

**nω0**does not exist. Yes, it was

**intuitive**, but not

**calculated**. Now we will use the

**WolframAlfa**program from the

**Internet**. You don’t have to install anything or pay anything. By the way, you will get acquainted with this program. It is a pity that as a student in

**the 60’s**I could not use it.

Fig. 11-1

Fig. 11-1

Formulas to remember from

**chapter**

**7**.

We will check them with the

**WolframAlfa**program. Then they will be more understandable.

**Fig. 11-1a**

Trajectory

**F(njω0t)**The point moves on the real axis

**Re Z**according to the periodic function

**f(t)**and the plane

**Z**rotates with

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

**F(njω0t)**is drawn for any

**n**.

**Fig. 11-1b**

Formula for the center of gravity

**scn**of the trajectory

**F(njω0t)**. The

**subscript**function is the trajectory

**F(njω0t)**. The

**integral**(divided by 2π) can be treated as a point

**scn**, average distant from the trajectory

**F(njω0t)**from the beginning

**z=(0,0)**at the time

**T=2π sec**.

**Fig. 11-1c**

Formula for the

**constant**component

**c0=a0**of the Fourier series. Classic mean of the

**periodic**function

**f(t)**when

**n=0**in the formula

**b**.

**Fig. 11-1d**

Formula for the remaining

**nth**coefficients of the Fourier

**cn**series

**for ω=1ω0**,

**ω=2ω0… ω=nω0**. Otherwise, the nth

**complex amplitude**of the harmonic

**hn(t)**

**Fig. 11-1e**

A complex number

**scn**as a vector with components

**an**and

**jbn**

**Fig. 11-1f**

**scn**in the

**algebraic**version

**scn=an+jbn**

**Fig. 11-1g**.

**cn**in the

**exponential**version

**cn=|cn|exp (jϕn)**The module

**|cn|**is clearly visible and the phase

**ϕn**for pulsations

**ω=nω0**.

**Fig. 11-1h**

**nth**harmonic

**hn(t)**as a

**sum**of

**cosine**and

**sinus**components.

**Fig. 11-1i**

**nth**harmonic

**hn(t)**as

**cosine**with phase

**ϕ.**Module

**|cn|**is a “pythagoras” of

**an**and

**bn**, and

**tg(ϕ)=bn an**.

**Chapter 11.2 The WolframAlfa Program**

We will use it to check the formula for the centers of gravity **scn** of the known **trajectories**. Most of these types of programs have one drawback. To solve the problem, you need to define it strictly. You make a mistake in putting the period and already receive an “error” or other difficult to understand message. **WolframAlfa** is so wise that all he needs to do is ask a general question and he will give many answers. All the more, the more general the question was. You will choose the answer that best suits you. Thanks to this, you do not have to remember all the instructions and you can completely focus on the problem.

**For example**, we want to solve the equation **2x+3=7**.

**1.** Call up the program from the Internet.

**2.** Enter **2x+3=7** in the box.

**3.** The program doesn’t really know what you mean. Just in case, it will draw a graph and give the solution **x=-2**. You cared about the** latter i.e**. **x=-2**

**Note:**

After calling the program, you will exit this blog. To return to it, click the **Windows** return arrow** –>**.

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

**Fig. 11-2
**How did

**WolframAlfa**solve the equation

**2x+7=3**?

For you, the most important solution is solution

**x=–2**. And the chart as an additional answer will not hurt.

It was a primer. I wonder how

**WolframAlfa**will cope with integration? Especially with integration of vectors or complex numbers.

Let us now calculate the centers of gravity

**scn**and by the way the

**nth**harmonics for several known periodic functions

**f(t)**. Before that, you had to take the

**center of gravity**on faith. Now you calculate them.

**Chapter 11.3 Center of gravity sc1 of the trajectory F(jnω0t)=F(-1j1t)=1exp(-1j1t) **

**Chapter 11.3.1 Introduction**

So for the constant function **f(t)=1** with **p. 7.4.1 Chapter 7**.

It is a **periodic** function, because the function repeats every period of **T**. Not only that, what **any T** period!

**Fig. 11-3**

Complex function **1exp(-1j1t)** as a trajectory and its “center of gravity” **scn**.

**Fig. 11-3a**

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

Click. During the period **T=2π/ω≈6.28sec**, the vector will perform one turn. What if we added successive vectors during the rotation?

**Fig. 7-3b**

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

**Fig. 11-3c**

The formula for the center of gravity **scn** of the trajectory of a function when **n=1 **and **f(t)=1**.

Even without calculations, it can be seen that the center of gravity is **sc1=(0,0**). Will the formula **Fig. 11-1b** confirm this?

**Chapter 11.3.2 WolframAlfa**

Let us calculate the center of gravity **sc1** of the trajectory **F(-1j1t)=f (t)exp (-1j1t)**.

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

Instead of laboriously typing an instruction, you can just copy it.

**integrate 1*exp(-i1t)/(2pi) from t=0 to t=2pi**

and paste into the window.

In other words, you will paste into the window the integral formula **Fig. 11-1b** written in **WolframAlfa **language.

Remember that the **imaginary** number for **WolframAlfa** is “**mathematical i**” and not the “**electric j**” used in the blog.

**Fig. 11-4
**Center of gravity

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

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

As we expected, the center of gravity

**sc1 = 0**or more precisely

**sc=(0,0)**, because it is a

**complex**number. Wolfram is nice! For example, he changes the instruction from the window into a human, or mathematical language. It can be seen by the arrow “

**3.You will get it**. ” Returning to the result

**sc1=0**, what is the result of it? Well, the

**amplitude**of the

**first**harmonic

**c1**for the harmonic

**1ω0=1/sec**is

**zero**, because

**cn=2scn**. 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=a0=1**.

**Chapter 11.4 Centers of gravity scn of trajectory F(-njω0t)=0.5cos(4t exp(-njω0t) for n=0,1,4 and ω0=1/sec **

**Chapter 11.4.1 Introduction**

This is an abbreviated version of **Chapter 4**, in which there were **9** versions of this trajectory for** n=0…9**. Now there are only **3** versions for** n=0,1,4** and only these formulas will be checked in the accounting

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

Let us turn the rotation to** n=0, n=1 and n=4.** So for the speed **ω=0** (the centrifuge is standing!), **ω=-1/sec** and **ω=-4/sec**.

There will be **3** trajectories:

**n=0–>ω = 0–>**centrifuge is standing**–>F(-0j1t)=0.5cos (4t) **

**n=1–>ω=-1/sec ->F(-1j1t)=0.5cos(4t)exp(-1j1t) **

**n=4–>ω=-4/sec ->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 –>

**sc0=(0,0)**

**ω=-1sec**

Rotating

**ω=-1sec**–>sc1=(0.0)

**ω=-4/sec**

Rotating

**ω=-4/sec**–>sc4=(0.25.0)

The above parameters

**sc0, sc1**and

**sc4**although intuitive, were taken as “my word of honor”. Now we will calculate them with the

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

**Chapter 11.4.2 Center of gravity sc0, that is for ω=0, that is the constant component sc0=c0=a0.**

So when the centrifuge is stopped, 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 of the function **f (t)=0.5cos (4t)**. Its center of gravity is evidently **sc0=(0,0)**. Will **WolframAlfa** confirm it? 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 as the picture tells you.

Enter or paste i**ntegrate 0.5cos(4t)/(2pi) from t=0 to t=2pi** into the box

**Fig. 11-6
**Calculation of the center of gravity

**sc0**for

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

**sc0 = 0**

Note that although the base period f(t) is

**T=π/2**, the result of

**sc0=0**for the integration limits

**T=π/2**and

**T=2π**will be the same!

The

**limits**of integration in the formula

**Fig. 11-1c**can be any 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! It is only important that

**T**is a period of the function. This remark is more general and applies to the

**Fourier Series**formulas in

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

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

**T=2π**. After all, the Fourier coefficients for example of a

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

**T,**but on the shape of this wave!

The center of gravity

**sc0**for the

**trajectory**of each function

**f (t)**is also a 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 number.**

**Chapter 11.4.3 Center of gravity sc1, that is for ω=-1/sec.**

So **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 **f (t)** is itself, i.e. **0.5cos (4t)**. It also results from **Fig. 11-4 ω=-1 sec**, where **sc1=(0,0)**. **Will WolframAlfa** confirm it?

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

Enter or paste integrate **[0.5cos (4t)/(2pi)]exp(-1i1t) from t=0 to t= pi** into the window.

**Fig. 11-7
**Center of gravity

**sc1=0**trajectory

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

Amplitude

**c1=0**for the harmonic

**ω=1/sec**because

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

As you saw in

**Chapter 4**, the centers of gravity

**scn**for all ω other than

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

**ω**of the function

**0.5cos (ωt)**) are

**zero**. It is like, as the classic used to say, obvious obviousness. After all, the function

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

**one**harmonic and it is

**itself**! The other harmonics are

**zero**.

**Chapter 11.4.4 Center of gravity sc4, that is for ω=-4/sec.**

So **sc4** 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 think 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** (albeit opposite signs)! This resulted in a **shift** of the **center of gravity** from **(0.0)** to **sc4=(+0.25.0)**.

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

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

**
Fig. 11-8
**Center of gravity

**sc4=0.25**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)**.

**Chapter 11.5 Centers of gravity scn of trajectory 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** -> **F(-1j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp (-2j1t)**

**n=3–>ω=-3/sec** -> **F(-3j1t)=[1.3+0.7cos(2t)+0.5cos(4t)]exp(-3j1t) **

**n=4–>ω=-4/sec** ->** 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 centers of gravity

**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**–>

**sc2=(0.35.0)**

**ω=-3/sec**–>

**sc3=(0,0)**

**ω=-4/sec**–>

**sc4=(0.5,0)**

The centers of gravity of the trajectories

**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 Center of gravity 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 accl to **F(-0j1t)=f (t)= 1.3+ 0.7cos (2t)+0.5cos(4t)**. Its center of gravity 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 center of gravity

**sc0**for

**F**

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

The center of gravity

**sc0=(1.3,0)=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 of the center of gravity

**sc2**for

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

Center of gravity

**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 Center of gravity 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 center of gravity

**sc3**of the trajectory

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

Center of gravity

**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**… harmonic.

You can see this in the

**f(t**) function

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

**WolframAlfa**.

**Chapter 11.5.5 Center of gravity 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 center of gravity

**sc4**for

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

Center of gravity

**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 Centers of gravity scn 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 center of gravity scn**

Until now, the centers of gravity **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**->**F(-1j1t)=0.5cos(4t-30 °)*exp (-1j1t)**

**n=4**–>**ω=-4/sec** ->**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≈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**. Center of gravity

**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 center of gravity

**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 harmonica

**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 center of gravity

**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 Centers of gravity scn trajectories
F(-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 Introduction**

The function

**f(t)**with the period

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

**3**

**cosineoids**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 centers of gravity

**scn**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°)**

It 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 centers of gravity

**scn**are the same as in

**Chapter 6**.

**Chapter 11.7.2 Centers of gravity 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 **4 trajectories:**

**ω=0**–>centrifuge is standing–>**F(-0j1t)=f(t)**

**ω=-1/sek ** –>**F(-1j1t)=f(t)exp(-1j1t) **

**ω=-2/sek **–>**F(-2j1t)=f(t)exp(-2j1t)
**

**ω=-3/sek**–>

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

**ω=-5/sek**–>

**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 centers of gravity

**scn**.

**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 Center of gravity 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 of the function

**F(0j1t)=f(t)**. Its center of gravity 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

**center of gravity 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 center of gravity

**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 Center of gravity 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 center**of gravity 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 Center of gravity 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 Center of gravity 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 center**of gravity 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 Center of gravity 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 center**of gravity 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 “Center of gravity” scn trajectory F (-njω0t)=f (t)exp(-njω0t) for n= 4,6,7,8 and ω0=1/sec**

There are still centers of gravity **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 **n*ω0**. I suggest you make the calculations yourself with the **WolframAlfa** program.

**Chapter 11.8 Centers of gravity 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 centers of gravity 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, centers of gravity

**scn**and harmonics

**hn(t)**.

**Chapter 11.8.3 Center of gravity 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 Center of gravity 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 chap. 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

**center of gravity**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 Center of gravity 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 center of gravity vector **sc2=(0,0)** agrees as the mean.

Now let’s count 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**

**
Fig. 11-30
**Calculation

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

That is, the

**second**harmonic, more precisely for

**ω=2/sec**, doesn’t exist

**Chapter 11.8.6 Center of gravity 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 center of gravity 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 11.8.5 Center of gravity 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 Centers of gravity 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.**