**Rotating Fourier Series**

**Chapter 7. How to calculate centroids of scn trajectories and harmonics detector.**

**Chapter 7.1 Introduction**In

**chapter 4**(4.5, 4.6), we extracted from the rotating trajectory

**F(njω0t)**its

**“centre of gravity”**or

**“centroid”**

**scn**as a

**complex number**. And this is almost the

**nth**harmonic with the pulsation

**ω=n*ω0**of the function

**f(t)**

**.**At that time, we did not know the formula for

**scn**and we accepted the result without proof. Now we will get to know it in

**Fig. 7-2b,c**as well as the relationship of the

**centroid**scn of the

**nth**trajectory with the

**nth**harmonic

**f(t)**in

**Fig. 7-2i,j**.

**Chapter 7.2 The centroids of trajectories and harmonics**

**Strange Formulas**

Fig. 7-1

Fig. 7-1

Don’t go into details for now. Let me just say that formulas are among the most important in mathematics, just like

**mc_square**in physics. They are not intuitive, because what is not here?

**Complex**functions,

**integrals**,

**infinities**,

**ω**

**pulsations**. The

**equations**correspond to some

**trajectories**in the

**complex plane**. And the so-called the “center of gravity” or “centroids” of the trajectories will make it much easier to understand these strange and difficult formulas.

**Chapter 7.3 Centroid scn of the trajectory F(njωot) of the function f(t) rotating at speed -nω0.**

The **centroid scn** (center of gravity) for the **nth F(njωot)** trajectory is a **complex number** from which we can easily read the **nth** harmonic of the** f(t)** function with **ω=n*ω0** pulsation. As if we threw a periodic function** f(t)** into a **centrifuge** rotating at a speed of **ω=n*ω0**. And what comes out of it? Of course it’s a** harmonic** with pulsation** ω=n*ω0**! So far, we have determined the **centroid **of the **scn** intuitively. In simple cases where **scn=(0,0)** this was as expected. Such as, for example, the centroid of a** circle**. Now it’s time for the exact formula for the **scn**. There will be no strict proof. However, I will try to make it as obvious as riding a bike. Although is cycling obvious? The first cyclists in the **19th** century were treated like **circus performers**. How does it move and not fall over?

**Fig. 7-2****Relation** of the centroid **scn** of the trajectory **F(njω0t)** to the **complex Fourier** coefficients **c0, cn=an-jbn*** and the harmonics **hn(t)** of the function **f(t)** of period **T**.***** It seems that the formula **7-2g **as **cn=an+jbn** would be more natural.

But then the formula **7-2i** would appear in **Fourier Series** as **hn(t)=an*cos(nω0t)-bn*sin(nω0t)**.**Fig. 7-2a****Trajectory F(njω0t)**

The point moves on the real axis **Re Z** according to the periodic function** f(t)**. When the centrifuge (**Z** plane) with the function** f(t)** rotates with the speed **-nω0** (sign -, because “clockwise”), the trajectory is drawn according to complex function** F(njω0t)**. For **n=0**, the centrifuge is stationary. The concept of a **trajectory** is already familiar to you after **Chap. 4.5** and** 6**. If something is wrong, I recommend at least **p. 4.3** from **chapter 4**.**Fig. 7-2b**

General formula for the centrroid **scn** of the trajectory **F(njω0t)**.

The integrand function is the trajectory **F(njω0t)**. The **integral** (divided by **2T**) can be treated as the point** scn**, the average distance of the trajectory **F(njω0t)** from **(0,0)** in the period **T sec***. Then **for n=1** the trajectory **F(njω0t)** will make **1** rotation in the period **T=2π/1ω0**, for** n=2**–> 2 rotations etc… But always in the same period **T**, only with a higher speed **ω**!

*If such a definition of the centroid of the trajectory** F(njω0t)** is not convincing, it may help **chapter. 7.7****Fig. 7-2c**

For functions that are “similar in time”, eg** square waves** with a duty cycle of** 50%** and amplitudes **A=1** but with different pulsations, the centroids **scn** do not depend on the pulsations **ω**. They are the same for **ω=1/sec** and **ω=1000/sec**! Therefore, in the formula **Fig. 7-2b** we can assume **ω=1/sec** and the simplified formula **Fig. 7-2c** will be obtained. More on this topic in **chapter 7.6**

**Fig. 7-2d**

The formula for the

**c0**coefficient of the

**Fourier Series**for

**n=0**.

This is a

**constant**component of

**f(t)**and is therefore always a

**real**number

**sc0=c0=a0**. Substitute

**n=0**into the formula in

**Fig. 7-2b**, and you get the classical average of the periodic function

**f(t)**over the a period

**T**.

Notes for formulas

**7-2b**and

**7-2d**

The integration interval is

**0…T**. But it can also be e.g.

**-T/2…+T/2**. It is important that the interval is period

**T**.

**Fig. 7-2e**

The

**formula**for the remaining

**nth**coefficients of the

**cn**Fourier Series for

**ω=1ω0, ω=2ω0…ω=nω0**.

These are the

**complex**harmonic amplitudes

**hn(t)**as doubled centroids

**scn**. Why is

**scn**doubled for

**n=1,2…**and

**sc0**not doubled? Good question. I will try to answer in

**chapter 12**.

**Fig. 7-2f**

**Centroid**of the trajectory as a complex number

**scn**, or a

**vector**with components

**an**and

**jbn**. It can be presented in algebraic and exponential versions

**Fig.7-2g**

**scn**version of the

**algebraic**version

**cn=an-jbn**

–

**cn**is the

**complex amplitude**of the

**nth**harmonic

**nω0**

**-an**is the

**amplitude**of the

**nth**harmonic

**cosine**component

**-bn**is the amplitude of the

**nth**harmonic

**sine**component

**Fig. 7-2h**.

**cn**exponential version

**cn=|cn|*exp(+jϕn)**

The

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

**ϕn**for pulsations

**ω=nω0**. Most often, the output

**sine**wave is delayed relative to the input

**sine**wave and therefore

**ϕn**is negative.

**Fig. 7-2i**

**-nth**harmonic

**hn(t)**as the

**sum**of the cosine and

**sine**components.

**Fig. 7-2j**

**-nth**harmonic

**hn(t)**as

**cosine**with phase shift

**ϕ**

The

**|cn|**module is

**“Pythagoras”**of

**an**and

**bn**, and

**tan(ϕ)=bn/an**.

**And now**something concrete. How to calculate the harmonics of a periodic function with a known

**time chart**and pulsation, e.g.

**ω0=10/sec –>**

**T=2π/ω0≈0.628 sec**

**Firstly.**We already know that if the shape of the function in the period is known, then the parameters

**an, bn**, i.e. the harmonics, do not depend on the period

**T**of the function. Let it be, for example, normalized

**T=2π sec**. So for

**ω0=1/sec**.

**Secondly.**From the formula

**Fig. 7-2c**we calculate

**c0=a0**as the mean value of the function in the period

**T=2π sec**.

**Suppose**, for example,

**c0=a0=0.5**

**Thirdly.**From the formula

**Fig. 7-2c**we will calculate the

**centroids**of the trajectory

**scn**for

**1ω0, 2ω0,…nω0…**

**Suppose**, for example,

**sc1=4-2j**for

**1ω0=1/sec**

**sc3=2-1j**for

**3ω0=3/sec**

**sc5=1+0.5j**for

**5ω0=5/sec**

For the remaining

**nω0,**the

**centroids**of the trajectory

**scn=0**

**Note:**

The formula in

**Fig. 7-2c**is dizzying.

**Integrals, complex numbers, infinities…**Apage Satanas!

But what do we have

**WolframAlpha**for. Let that be his concern in

**Chap. 11**

**Fourthly**

From the formula

**cn=2*scn**we will calculate the parameters

**scn**for pulsations

**ω=1/sec, ω=2/sec…, ω=n/sec…**

in the

**algebraic**version

**Fig. 7-2f**

**c1=8-4j**

**c3=4-2j**

**c5=2+1j**

or

**exponential**version

**Fig. 7-2g**

**c1≈8.94*exp(-j26.6°)**

**c3≈4.47*exp(-j26.6°)**

**c5≈2.23*exp(+j26.6°)**

Let me remind you that, for example,

**8.94**is the module

**|8-4j**| i.e. “Pythagoras of

**8**and

**4**″ and

**tan(26.6°)≈4/8**.

**Fifth**

Now we can represent the next harmonics

**h1(t),h2(t),h3(t)…**

according to

**Fig. 7-2h**

**h1(t)=8cos(1t)+4sin(1t)**

**h2(t)=4cos(2t)+2sin(2t)**

**h3(t)=2cos(3t)-1sin(3t)**

or according to

**Fig. 7-2i**

**h1(t)=≈8.94*cos(1t-26.6°)**

**h2(t)=≈4.47*cos(2t-26.6°)**

**h3(t)=≈2.23*exp(3t+26.6°)**

Okay, but we want harmonic formulas for f(t) with

**ω0=10/sec**.

The

**complex amplitudes**will be the same, that is:

**h1(t)=8cos(10t)+4sin(10t)****h2(t)=4cos(20t)+2sin(20t)****h3(t)=2cos(30t)-1sin(30t)**

or acc. to **Fig. 7-2i****h1(t)=≈8.94*cos(10t-26.6°)****h2(t)=≈4.47*cos(20t-26.6°)****h3(t)=≈2.23*exp(30t+26.6°)**

**Chapter 7.4 Centroids scn trajectories F(-njω0t) for different functions f(t),****Chapter 7.4.1 Introduction**

The formula in** Fig. 7-2a** applies to the trajectory **F(njω0t)** when the plane **Z** with the function** f(t)** rotates at different speeds** ω=n*ω0**. For **n=0** the** Z** plane does not rotate. The **trajectory** should be obvious to you by now, especially after the animations.

We will consider the formula in **Fig. 7-2b** for the centroid** scn** of the trajectory **F(njω0t)** for different functions **f(t)** and different rotating speeds **n*ω0**. Next, we will examine the relationship of the centroids **scn** with the successive harmonics **hn(t)** of the function **f(t)**.

We will explore the following functions starting with the simplest.**1. f(t)=1**–> **Chapter 7.4.2****2. f(t)=0.5cos(4t)**–> **Chapter 7.4.3****3. f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**—**>** **Chapter 7.4.4****4. f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)**–> **Chapter 7.4.5**

Main conclusions:**1.** The formula **Fig.7-2b** indicates the point **scn** in the** Z** plane, which is the mean distance between** z=(0,0)** and the trajectory **F(njω0t)** when the **Z** plane makes **n** rotations **0…2π**. In other words, the point **scn** is like the centroid of the trajectory** F(njω0t)**.

“Like”? Because the true centroid is calculated by different formulas than on the **scn**, although sometimes they overlap.**2.** From the centroid **scn** of the trajectory, we can easily compute the **nth** harmonic of** f(t)** for **n*ω0**. In our examples, **ω0=1/sec**.

In **Chapter 11**, we will check the formulas in **Fig.7-2** with **WolframAlfa**. Then you will be even more convinced that the **Fig. 7-2** formulas are as obvious as riding a bicycle.**Chapter 7.4.2 Centroids scn of the trajectory F(-njω0t)****for the function f(t)=1 ****when n=1 and ω0=1/sec**

We will start with the simplest case, i.e. the constant function **f(t)=1**.

The answer is obvious without formulas.**1.** The function has no harmonics. So the centroids **scn** of the trajectories for each pulsation **n*ω0** should be **zero**, i.e. **scn=(0,0)**.**2.** Constant component **c0=a0=1**.

We will check whether the formulas** Fig. 7-2c** and **7-2d** confirm this.

We will examine the trajectory **F(njω0t)** for **f(t)=1, n=1** and** ω0=1/sec** acc. formula **Fig. 7-2a**.

Some may turn their nose up at whether **f(t)=1** is a **periodic** function? It is, because every period **T** (even any **T**!) the function **repeats.**

**Fig. 7-3**

Complex function **1exp(-1j1t)** as a trajectory and its centroid **sc1**.**Fig. 7-3a**

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

During the period** T=2π/ω≈6.28sec**, the vector will make **one** revolution. What if during the rotation we summed up subsequent vectors?**Fig. 7-3b**

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

Without calculations, we know that the centroid is **scn=(0,0)**. Will the formula **Fig.7-3c** confirm this? You see the **successive** positions of the vectors from **z0, z1 … z11** with the angle increment **-Δωt=-30º**. Their** sum** is the **zero** vector **z=(0,0)**. Why zero? Because **each** rotating vector corresponds to a **compensating** vector (e.g. **z4** and **z10**) and their sum is **zero**. So the sum of all vectors** z=(0,0)** is sort of the centroid of the **trajectory** of the complex function **1exp(jωt)**. More precisely, in order to obtain the **average** distance, the sum of the vectors must be divided, i.e. **z=(0,0)** by **12*30º=360º**, otherwise by **2π**.**Fig. 7-3c**

A more accurate formula for the centroid **scn** of the trajectory of the complex function **1exp(-1jω0t)** for **ω0=-1/sec.**

The expression **α=-ω0*t** is the angle α rotating at the speed **ω0=1/sec**.

The** definition** of the centroid in **Fig. 7-3b** as the sum of **12** vectors was not very **precise**. The angle increment is **Δα=30º**. And yet, there are infinitely many, infinitely small increments **Δα=d(ωt)** from **α=0 to α=2π**. Summation will turn into integration from **0 to 2π**.

**Chapter 7.4.3 Centroids scn of the trajectory F(-njω0t) for the function ****f(t)=0.5cos(4t) ****when n=0,1,2,…8 and ω0=1/sec**

We will examine the trajectories **F(njω0t)=f(t)*exp(-njω0t)** for:**f(t)=0.5cos(4t)****n=0,1,2…8****ω0=1/sec**

**Fig. 7-4**Complex functions

**F(n1t)=0.5cos(4t)*exp(-n1t)**as trajectories and its “centroids”

**scn**.

**ω=0**(n=0)

The stationary complex plane

**Z**, on which the function

**f(t)=0.5cos(4t)**performs “there and back again” harmonic movements.

**ω=4/sec**(n=4)

Only at this speed

**ω**will there be a trajectory with a

**non-zero**centroid

**sc4=(0.25,0)**. Intuition suggests that it is

**medium**distant from

**z=(0,0)**and we will calculate this result with the formula

**Fig. 7-2c**. Note that you only see the movement of the trajectory at the beginning. Then it follows the same path and therefore the trajectory is seemingly

**stationary**.

Other

**ω**(n=1.2, 3, 5, 7, 8)

Centroids

**scn=(0,0)**, which also fall under the formula

**Fig. 7-2b**.

**Constant**component

**c0**.

It clearly confirms the formula

**Fig.7-2d–> sc0=(0,0)=a0=0**.

The

**fourth**harmonic

**h4(t)**i.e. for

**ω=4/sec**

acc. to

**Fig. 7-2e–>c4=2*sc4=2*(0.25,0)=(0.5,0)=0.5+j0–>a4=0.5 and b4=0**

acc. to

**Fig. 7-2i–>h4(t)=0.5*cos(4t)**

**Harmonics**for the remaining

**n*ω0**pulsations do not exist, because the centroids

**scn**of their trajectories are

**zero**.

**Conclusion**

The function

**f(t)=0.5*cos(4t)**consists of only

**one**harmonic

**h4(t)=0.5*cos(4t)**. This is not the discovery of America, but we mainly want to confirm the formulas

**Fig.7-2**.

**Note:**

Here I assumed that the function

**f(t)=0.5cos(4t)**consists of only

**one**, i.e. the

**fourth**harmonic

**h4(t)=0.5cos(4t)**where the

**first**harmonic for

**1ω0=1/sec**and the

**others**for

**2ω0=2/ sec, 3ω0=3/sec, 5ω0=5/sec, 6ω0=6/sec,**i.e.

**h1(t)=h2(t)=h3(t)=h5(t)=h6(t)=…= 0**are

**zero**.

It is equally well, and perhaps more natural, that the function

**f(t)=0.5cos(4t)**consists of only one

**first**harmonic

**h1(t)=0.5cos(4t)**, the others for

**2ω0=8/sec, 3ω0=12 /sec, 4ω0=16/sec, 5ω0=20/sec,…**are

**zero**.

**Chapter 7.4.4 Centroids scn of the trajectory F(-njω0t)****for the function f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**

when** n=0,1,2,4 and ω0=1/sec**

The **f(t)** function has** 2** harmonics, i.e. **0.7*cos(2t)** and **0.5*cos(4t)** and a **constant** component **c0=1.3**. I wonder how they will be rotated?

We will examine the trajectories **F(njω0t)=f(t)*exp(-njω0t)** for:**f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)****n=0,1,2** and **4****ω0=1/sec**

So these will be trajectories for **ω=0, ω=1/sec, ω=2/sec** and **ω=4/sec**.

The other trajectories, i.e. for **n=3,5,6,7** and **8** rotate around, similarly to **n=1**

centroids** scn=(0,0)** and we do not study them. You can see them in **Chapter 5**.

**Fig. 7-5**Complex functions

**F(nj1t)=[1.3+0.7*cos(2t)+0.5*cos(4t)]*exp(-nj1t)**as trajectories and its centroids

**scn**.

**ω=0**

The stationary complex

**Z**plane on which the function

**f(t)=1.3+0.7*cos(2t)+0.5*cos(4t)**makes „there and back again” movements around

**sc0=(1.3,0)**and which is the constant component

**c0= a0=1.3**of the function

**f(t)**.

**ω=2/sec, ω=4/sec**,

At these speeds,

**ω**, trajectories with

**non-zero**centroids will be created

**sc2=(0.35.0)**and

**sc4=(0.25.0)**. They are medium distance between the trajectory points

**F(2j1t)**and

**F(4j1t)**a

**z=(0,0)**. In other words, these are “as if” (because not entirely in terms of mechanics) the centroids of these trajectories, which we will calculate with the formula

**Fig. 7-2c**.

**ω=1/sec**and the remaining

**ω**

The centroids

**sc1=(0,0)**and the

**others**also fall under the formula

**Fig. 7-2c**

**Constant**component

**c0**.

acc. to of the formula

**Fig.7-2d**–>

**c0=a0=1.3**

It is also the centroid

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

**ω=0**

The

**second**harmonic

**h2(t)**i.e. for

**ω=2/sec**

acc. to

**Fig. 7-2e**–>

**c2=2*sc2=2*(0.35,0)=(0.7,0)=0.7+j0–>a2=0.7**and

**b2=0**

acc. to

**Fig. 7-2i**–>

**h2(t)=–>0.7*cos(4t)**

The

**fourth**harmonic

**h4(t)**i.e. for

**ω=4/sec**

acc. to

**Fig. 7-2e–>c4=2*sc4=2*(0.25,0)=(0.5,0)=0.5+j0–>a4=0.5**and

**b4=0**

acc. to

**Fig. 7-2i–>h4(t)=–>0.5*cos(4t)**

The

**other**harmonics do not exist because the centroids of their trajectories are

**zero**.

**Chapter 7.4.5 Centroids scn of trajectory F(-njω0t) ****for the functionf(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)when n=0,1,2,3,5 and ω0=1/sec**

The function

**f(t)**with period

**T=2πsec**consists of

**three cosines**with different amplitudes

**A**and phases

**ϕ**and a constant component

**c0**.

**Fig. 7-6**

**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 studied

**9**trajectories

**F(-njω0t)**of the above function

**f(t)**for

**n=0,1,2,…8 and ω0=1/sec**.

It is interesting because due to the phase shifts

**ϕ**, the

**centroids**scn are completely

**complex numbers**.

Each harmonic can be decomposed into a

**cosine**and

**sine**component, and then:

**f(t)=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).**

This is the same

**function**, but complex

**Fourier**

**coefficients**, or complex

**Fourier amplitudes**, are easily determined here.

From them,

**harmonics**are read as time waveforms

**h1(t), h3(t)**and

**h5(t)**with

**sine/cosine**components.

**c1=0.9-j0.6—>h1(t)=0.9*cos(1t)+0.6*sin(1t)≈1.08*cos(1t-33.7°)**

**c3=0.6+j0.4–>h3(t)=0.6*cos(3t)-0.4*sin(3t)≈0.72*cos(3t+33.7°)**

**c5=0.4-j0.2—>h5(t)=0.4*cos(5t)+0.2*sin(5t)≈0.45*cos(5t-26.6°)**

We will put the function

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

**nω0=n*1/sec**.

Let’s turn on the rotations at

**n=0, 1,2,3**and

**5**, i.e. at the speed

**ω=0**(the centrifuge is stationary!)

**, ω=-1/sec, ω=-2/sec, ω=-3/sec**and

**ω= -5/sec**. The trajectories F(nj1t) of those n corresponding to the centroids

**scn**will be created.

**Fig. 7-7Trajectories F(nj1t)=[0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)]*exp(-nj1t)**

**for n=0—>ω=0—>F(0j1t)**

The function

**f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)**moves back and forth around

**sc0=(0.5 ,0)**on the stationary

**Z**plane. The point

**sc0**is the constant component

**c0=0.5**of the function

**f(t)**.

**for n=1,3,5, i.e. for ω=1/sec, ω=3/sec**and

**ω=5/sec–>F(1jt),F(3jt),F(5jt)**

Trajectories with

**non-zero**centroids

**sc1=(0.45,-0.3), sc3=(0.3,+0.2) and sc5=(0.2,-0.1)**. These points

**scn**are the average distance between the trajectory points

**F(nj1t)**and

**z=(0,0)**.

**for n=2, i.e.**for

**ω=2/sec–>F(2jt)**

The centroid

**sc2=(0,0)**also falls under the formula

**Fig. 7-2c**. Trajectories

**F(4jt), F(5jt), F(6jt), F(7jt), F(8jt)**with also

**zero**centroids

**scn**are not shown.

**Constant component c0**.

acc. to formula

**Fig.7-2c–> c0=0.5**

**First harmonic h1(t)**i.e.

**for ω=1/sec**

acc. to

**Fig. 7-2e–>c1=2*sc1=2*(0.45,-0.3)=0.9+j0.6–>a1=0.9**and

**b1=0.6**

acc. to

**Fig. 7-2i–>h1(t)=–>0.9*cos(1t)+0.6*sin(1t)**

acc. to

**Fig.7-2j–>h1(t)≈–>1.08*cos(1t-33.7°)**

**Third harmonic h3(t)**i.e. for

**ω=3/sec**

acc. to

**Fig. 7-2e–>c3=2*sc3=2*(0.3,+0.2)=0.6-j0.4–>a3=0.6**and

**b3=-0.4**

acc. to

**Fig. 7-2i–>h3(t)=–>0.6*cos(3t)-0.4*sin(3t)**

acc. to

**Fig. 7-2j–>h3(t)≈–>0.72*cos(3t+33.7°)**

**Fifth harmonic h5(t)**i.e. for

**ω=5/sec**

acc. to

**Fig. 7-2e–>c5=2*sc5=2*(0.2,-0.1)=0.4+j0.2–>a5=0.4**and

**b4=0.2**

acc. to

**Fig. 7-2i–>h5(t)=–>0.4*cos(5t)+0.4*sin(5t)**

acc. to

**Fig.7-2j–>h5(t)≈–>0.45*cos(5t-26.6°)**

**Other**

**harmonics**, i.e. for e.g.

**ω=2/sec, ω=4/sec, ω=6/sec, ω=8/sec, ω=8/sec…**

They do not exist, because the centroids

**scn**of these trajectories are

**zero**.

**Chapter 7.5 Harmonic Detector****Chapter 7.5.1 Introduction**

We already know that the **nth** trajectory **F(njω0t)** of the periodic function **f(t)** with fundamental pulsation **ω0** corresponds to the centroid** scn,** which is almost the **nth** harmonic **f(t)**.

Why almost?

Because** scn** is not yet the nth harmonic of the **f(t)** function, but only a **complex** coefficient from which we can easily calculate the **nth** harmonic in **complex** and **real** form.**1. cn=2*scn=an-jbn****2. An=cn=an-jbn** is the **complex** amplitude of the **nth** harmonic **hn(nω0t)****3.a****nth** harmonic in the **complex version**

So the vector **hn(jn*ω0t)=(an-jbn)*exp(n*jω0t)** rotating at a speed of** n*ω0****3.b****nth** harmonic in the** real version****hn(t)=an*cos(nω0t)+bn*sin(nω0t)–>Fig.7.2i**

or**hn(t)=|cn|*cos(nω0t+ϕ)–>Fig.7.2j**

We found the centroid** scn** of various trajectories **Fn(jnω0t)** for

various functions **f(t)** in **Fig. 7-4, Fig. 7-5** and **Fig. 7-7**. These are “almost” the **nth** harmonics of **f(t)**.

We will now present these animations in a **simplified** version, i.e.**– There** will only be **doubled** centroids scn for the **nth** trajectories as coefficients **cn****– Animations** will be presented **successively** every **1** second when **ω** changes from **ω=0** to** ω=8/sec**

In this way, **harmonic detectors** of periodic functions **f(t)** will be created, namely:**Harmonic** detector for:**– f(t)=0.5cos(4t)****– f(t)=1.3+0.7cos(2t)+0.5cos(4t)****– f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)**

**Chapter 7.5.2 Harmonic detector for f(t)=0.5cos(4t)**

**Fig 7-8**

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

You put a periodic function

**f(t)=0.5cos(4t)**into the centrifuge, which rotates with successively increasing speed

**from ω=0**to

**ω=8/sec**. In this way, for each

**ω**pulsation, the next harmonic

**hn(t)**corresponding to the

**complex**number

**cn**is calculated according to the formulas in

**Fig.7-2**. There is only one

**non-zero**harmonic for

**h4(t)=0.5cos(4t)**as vector

**c4=(0.5,0)=0.5**. The remaining

**cn=0**, i.e. the corresponding

**harmonics**do not occur!

**Chapter 7.5.3 Harmonic detector for f(t)=1.3+0.7cos(2t)+0.5cos(4t)**

**Fig.7-9**Harmonic detector for

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

For

**ω=0**, the constant component

**c0=1.3**

For

**ω=2/sec c2=0.7**corresponding to the harmonic

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

For

**ω=4/sec c4=0.5**corresponding to the harmonic

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

Other harmonics do not occur.

**Chapter 7.5.4 Harmonic detector for f(t)=0.5+1.08*cos(1t-33.7°)+0.72*cos(3t+33.7°)+0.45*cos(5t-26.6°)**

**Fig.7-10**Harmonic detector for

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

For

**ω=0**, the constant component

**c0=0.5**

For

**ω=1/sec, c1=1.08*exp(-j33.7°)**corresponding to the harmonic

**h1(t)=1.08*cos(1t-33.7°)**

For

**ω=3/sec**,

**c3=0.72*exp(+j33.7°)**corresponding to the harmonic

**h3(t)=0.5cos(4t)0.72*cos(3t+33.7°)**

For

**ω=5/sec, c5=0.45*exp(-j33.7°)**corresponding to the harmonic

**h5(t)=0.45*cos(5t-26.6°)**

Other harmonics do not occur.

**Note:**

Harmonic detectors in

**chapter 7.5**they study harmonics in the range

**ω=0…+∞**. In

**Series**and

**Fourier Transform**, however, it is more often assumed that

**ω=-∞…0…+∞**. That is, for negative and positive

**ω**pulsations. A

**detector**for them is included in the article

**“Fourier Transform” Fig. 5-5**.

**Chapter 7.6 Why does the “extension and compression” of the periodic function f(t) not affect the parameters an and bn of the Fourier series?**

Look at the formulas **Fig. 7-2b** and **Fig. 7-2c**. It turns out that if **f(t)** is only “stretched/squeezed” in time (but not vertically!), then the formulas do not depend on the period **T**.

Let our periodic function be e.g. °).

So it is **f(t)** from **Fig.7-6** “squeezed” in time by **1.75**.

**Fig. 7-11**

The **periodic** function from **Fig.7-6**, only “squeezed” in time, and its 2 trajectories.**Fig. 7-11a**

The function** f(t)** in **Fig.7-6** had a pulsation **ω0=1/sec** (**T=2π sec**), now **ω0=1.3/sec (T≈3.59sec)**. It is “squeezed” in time by **1.75**.**Fig. 7-11b**

Trajectory **F(1jω0t)**, i.e. for rotations **1*ω0=1*1.3/sec**.

It is exactly the same as in **Fig. 7-7 ω=1/sec**. Only the trajectory speed is **1.3** times faster.**Note:**

The centroid **sc1** of the trajectory** F(1jω0t)** and** c1=0.9-j0.6** will also be the same.**Fig. 7-11c**

Trajectory **F(2jω0t)**, i.e. for rotations **2*ω0=2*1.3sec**.

It is exactly the same as in **Fig. 7-7 ω=2/sec**. Only the **speed** of the trajectory is **1.3** times higher.**Conclusion**:

The centroid **sc2=0** of the trajectory **F(1jω0t)** and **c2=(0,0)** will also be the same. The periodic function **f(t)** has no harmonic with pulsation **ω=2*1.3/sec**.

This way we could check all other trajectories for** ω=3*1.3/sec, ω=4*1.3/sec…ω=8*1.3/sec.**

The shapes and centroids scn will of course be the same as in** Chapter. 6**.

That is, to calculate the centroid **scn** of the trajectory, the “limited” formula **Fig. 7-2c** when **T=2π** is enough.

So the coefficients **co, c1, c2…cn** of the **Fourier Series** do not depend on the period **T** of the function** f(t)**.

**Chapter 7.7 Why the centroid scn of the trajectory F(njω0t) is either:****1-scn=(0,0)-zero?****2-scn=(an,bn)-half the amplitude for the nth harmonic or cn/2?**

Each function** f(t)** with period **T**, i.e. **ω0** pulsations, consists of a constant component **c0=a0** and harmonics with** nω0** pulsations for** n=1,2,…∞**.

-When the trajectories **F(n*jω0t)=f(t)*exp(-n*jω0t)** rotate with a speed **ω≠nω0**, i.e. different from** ω** of each harmonic, a situation similar to the animation occurs **Fig. 7-4 ω≠- 4**. Then the trajectories of all the harmonics **rotate** individually around **scn=(0,0)**. So **F(n*jω0t)** as their sum will also rotate around **scn=(0,0)**. So the **case 1-scn=(0,0)-zero**

-When the trajectories **F(n*jω0t)=f(t)*exp(-j*jω0t)** rotate with the** speed ω=nω0**, i.e. equal to the nth harmonic, a situation similar to the animation occurs **Fig. 7-4 ω=- 4**. Then the trajectories** ω≠nω0** of the remaining harmonics rotate each around its own **scn=(0,0)**. And only one harmonic **ω=nω0** rotates around **scn**. This will shift the **centroid** of all harmonics to just this **scn**. So the **case 2-scn=(an,bn)**-half the amplitude for the **nth** harmonic, which is **cn/2**.

One more thing. In the animation **Fig.7-4**, the trajectories rotated with velocities **ω=nω0**. So with … not any! However, in the considerations we assumed that only **ω≠nω0** are arbitrary.** i.e. ω=-0.384*1/sec**. But I guess it’s obvious that when **f(t)** has no harmonic for **ω=-0.384*1/sec**, the trajectories** F(n*jω0t)=f(t)*exp(-n*j0.384t)** will rotate too around** scn=(0,0)**.