## Rotating Fourier Series

### Chapter 7. How to calculate scn-the centers of gravity of the trajectories F(njω0t)?

**Chapter 7.1 Introduction
**In

**chapters 4, 5 and 6**we extracted from the rotating trajectory

**F(njω0t)**its “center of gravity”

**scn**as a

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

**nth**harmonic with pulsation

**ω=n *ω0**of the function

**f(t)**, because

**cn=2*scn**. At that time, we did not know the

**scn**formula and we accepted the result without proof. Now we will see it as well as the relationship of the

**scn**of the

**nth**trajectory with the

**nth**harmonic

**f**

**(t)**.

**Chapter 7.2 Center of gravity and weird formulas**

**
Fig. 7-1
**Weird formulas

Don’t go into details yet. Let me just say that the formulas are among the most important in mathematics, as is the

**mc2**(“mcsquare”) in physics. They are not intuitive, because what is not here? Complex functions, integrals, infinity,

**ω**pulsations . The

**equations**correspond to some

**trajectories**on the complex plane

**Z**. And the so-called The

**“center of gravity”**of the

**trajectory**will make it much easier to understand these strange and difficult formulas.

**Chapter 7.3 Centers of gravity of the trajectories F(njω0t), formulas and centrifuge rotating at nω0 with the function f (t)**

The center of gravity **scn** for the **n-th F (njωt)** trajectory is a **complex number**, from which we can easily read the harmonic of the function **f(t)** with pulsation **ω=n*ω0**. As if we threw the periodic function** f(t)** into the centrifuge rotating at the speed **ω=n*ω0**. And what’s falling out of it? Of course that the** harmonic** with pulsation **ω=n*ω0**! Until now, we have determined the **center of gravity** **scn** intuitively. In simple cases, when **scn=(0,0)**, it was exactly as expected. Such as, for example, the center of gravity of a circle. Now it’s time for the exact pattern on the **scn**. There will be no strict proof. But I will try to make it as obvious as riding a bicycle. Although is cycling obvious? The first cyclists in the **19th** century were treated like circus performers. How is it so, he is driving and isn’t coming down?

**Fig. 7-2
**Relationship of the center of gravity scn of the trajectory

**F(njω0t)**with complex Fourier coefficients

**c0, cn= a+jbn**and harmonics

**hn(t)**of the function

**f(t)**

**a.**Trajectory

**F(njω0t)**

A point moves on the real axis

**Re Z**according to the periodic function f(t). When the

**centrifuge**i.e.

**Z plane**with the function

**f(t)**rotates at the speed

**-nω0**(sign

**–**because it is

**clockwise**), the

**trajectory**is drawn according to the complex function

**F(njω0t)**. For

**n=0**the centrifuge is

**stopped**. The trajectory and what to do with it should already be known to you after

**chap. 4, 5 and 6**. If something is wrong, I recommend at least

**p.**

**4.3 of chap. 4.**

**b.**Formula for the

**center of gravity scn**of the trajectory

**F(njω0t)**.

The

**integer**function is the trajectory

**F(njω0t)**. The integral (divided

**by 2π**) can be treated as a point

**scn**, with

**average**distant of the trajectory

**F (njω0t)**from

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

**T=2π sec**. Then,

**for n=1**, the

**Z**plane will perform

**1**revolution, for

**n=2**->

**2**turns, etc …

Note:

Note:

This formula is known in a more general version, where

**T**is arbitrary and not just

**T=2π sec**. Why”? Because for any

**T**the result on

**scn**will be the same! In other words, the harmonic parameters

**a0, a1, a2 …**e.g. a

**square wave**with

**50%**duty cycle do not depend on the

**T**period of this wave. They are the same for e.g.

**T=1sec, 2sec, 3sec … 2π sec≈6.28 sec**, etc.

**c.**The formula for the coefficient

**c0**of the

**Fourier Series**for

**n=0**.

It is a

**constant component**of the function

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

**real number c0=a0**.

Substitute

**n=0**into the formula

**b**, the classical

**mean**of the

**periodic**function

**f(t)**will be obtained in the period

**T=2π**

**sec**.

**d.**The formula for the

**remaining nth**

**cn**coefficients of the

**Fourier Series**for

**ω=1ω0, ω=2ω0 … ω=nω0**. They are simply

**doubled**centers of gravity

**scn**.

**e.**A complex number

**scn**as a vector with components

**an**and

**jbn**

In the figure, the parameters

**a,b,ϕ**are

**positive**, but they can be

**negative**!

The “center of gravity”

**scn**as a

**complex number**can be represented in an

**algebraic**and

**exponential**version

**f.**

**scn**in the algebraic

**version**

**scn=an+jbn**

–

**cn**is the complex amplitude of the

**nth**harmonic

**nω0**

**-an**with the

**cosine**amplitude of the

**nth**harmonic

**nω0**

**-bn**with the

**sinusoidal**amplitude of the

**nth**harmonic

**nω0**

**g.**

**cn**in the

**exponential**version

**cn = | cn |*exp(+jϕn)**

The module

**|cn|**and the phase

**ϕn**for pulsations

**ω=nω0**are clearly visible.

**h.**

**nth**harmonic

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

**cosine**and

**sinusoidal**components.

**i.**

**nth**harmonic

**hn(t)**as cosine with phase shift

**ϕ**.

Module

**|cn|**is a

**“pythagoras”**of

**an**and

**bn**, and

**tg(ϕ)=bn/an**.

And now the concrete. How to calculate harmonics of some periodic function with known graph and period, eg **T=10 sec**?

**First.** We already know that if the shape of the function in the period is known, then the parameters **an, bn**, i.e. **harmonics**, do not

depend on the period** T** of the function. Let it be e.g. normalized **T=2πsec**.

**Secondly.** From the formula **c** we calculate **c0=a0** as the mean value of the function over the period **T=2π sec**.

Let us assume, for example **c0=a0=0.5**

**Thirdly**. From the formula **b** we calculate the **centers of gravity** of the trajectory **scn** for pulsation **ω=1/sec, ω=2/sec …, ω=n/sec …
**Let’s assume

**sc1=4-2j**

sc2=2-1j

sc3=1+0.5j

sc2=2-1j

sc3=1+0.5j

**Note:**

The formula

**b**makes you dizzy. Integrals, complex numbers, infinities… Apage Satanas. But what do we have

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

**Chapter. 11**

**Fourthly.**From the formula

**cn2*scn**we will calculate the parameters

**cn**for pulsations

**ω=1/sec, ω=2/sec …, ω=n/ sec …**in the

**algebraic**version

**f**

c1 = 8-4j

c1 = 8-4j

**c2 = 4-2j**

**c3 = 2 + 1j**

or in the

**exponential**version

**g**

**c1≈8.94 * exp (-j26.6 °)**

**c2≈4.47 * exp (-j26.6 °)**

**c3≈2.23 * exp (+ j26.6 °)**

Let me remind you that e.g.

**8.94**is the module

**|8-4j|**that is, “Pythagoras with 8 and 4 ″ a

**tg(26.6 °)≈4/8.**

**Fifth**Now we can present the next harmonics

**h1 (t), h2 (t), h3 (t) …**

**acc**

**. h**

**h1(t)=8cos(1t)-4cos(1t)**

h2(t)=4cos(2t)-2cos(2t)

h3(t)=2cos(3t)+1cos(3t)

or acc

h2(t)=4cos(2t)-2cos(2t)

h3(t)=2cos(3t)+1cos(3t)

**. i**

h1(t)=≈8.94*cos(1t-26.6°)

h2(t)=≈4.47*cos(2t-26.6°)

h3(t)=≈2.23*exp(3t+26.6°)

h1(t)=≈8.94*cos(1t-26.6°)

h2(t)=≈4.47*cos(2t-26.6°)

h3(t)=≈2.23*exp(3t+26.6°)

**Chapter 7.4 “Centers of gravity” sc of the trajectory F (-njω0t) for various functions f (t)
Chapter 7.4.1 Introduction**

The formula

**Fig.7-2a**relates to the trajectory

**F(njω0t)**when the plane with the function

**f (t)**rotates at different speeds

**ω=n*ω0**. For

**n=0**, the

**Z**plane does not rotate. The trajectory should already be obvious to you, especially after the animations. We will consider the formula

**Fig.7-2b**for the

**center of gravity**

**scn**of the trajectory

**F(njω0t)**for different functions

**f(t)**and different

**n*ω0**. Then we will examine the relationship of the

**centers of gravity scn**with successive harmonics

**hn(t)**of the function

**f(t)**.

We will examine the next 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:

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

**1.**Formula

**Fig.7-2b**indicates the point

**scn**on the

**Z**plane, which is the

**mean**distance between

**z=(0,0)**and the trajectory

**F(njω0t)**when the

**Z**plane performs

**n**revolutions

**0 … 2π**. In other words, the point

**scn**is the

**center of gravity**of the trajectory

**F(njω0t)**.

**2.**From the

**center of gravity scn**of the trajectory, we can easily calculate the

**nth**harmonic of the function

**f(t)**for

**n*ω0**. In

**Chapter 11**we will check with

**WolframAlfa**free program the formulas from

**Fig.7-2**. Then you will find even more that the

**formulas**in

**Fig.7-2**are as obvious as riding a bicycle.

**Chapter 7.4.2 Centers of gravity 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. That is, the centers of gravity

**scn**of the trajectory for each pulsation

**ω**should be zero, ie

**scn=(0,0)**.

**2.**Constant component

**c0=1**.

We will check whether the formulas

**Fig. 7-2b**and

**7-2c**confirm this.

We will examine the trajectory

**F(njω0t)**for

**f(t)=1, n=1 and ω0=1/sec**according to from the

**Fig.7-2a**. Somebody can sniff at this

**constant**function

**f(t)=1**. Is it a periodic function?

**Yes**! because every period

**T**(even any

**T**!) the function repeats itself.

**Fig. 7-3
**Complex function

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

**scn**.

**a.**Complex function

**1*exp (-1j1t**) as a

**rotating vector**.

Click here. During the period

**T=2π/ω≈6.28sec**(ω=1/sec

**)**, the vector will perform one revolution. What if we added successive vectors during the rotation?

**b.**A circle as a trace of a rotating vector, i.e. a trajectory

**1exp(-1j1t)**. We know without any calculations that the

**center of gravity**is

**scn=(0,0)**. Will the formula

**c**confirm this? You can see successive positions of 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 like the

**“center of gravity”**of the trajectory of the complex function

**1exp(jωt**). More

**precisely**, to get the

**average**distance, one has to

**divide**the sum of the

**vectors**, that is,

**z=(0,0)**by

**12*30º**

**=360º**, otherwise by

**2π**.

**c.**More precise formula for the center of gravity

**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 speed

**ω0=1/sec**. The definition of the center of gravity in

**b**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 of

**Δα=d(ωt)**from

**α=0**to

**α=2π**. The

**summation**changes to an i

**ntegration**from

**0**to

**2π**. And where does the

**division**by

**2π**come from? It is necessary to calculate the

**average**distance of the trajectory with a rotation of

**0…2π**from

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

**mean**and the

**sum**of the vectors are equal to

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

**Conclusions**

The trajectory

**1exp (-1j1t**) results in

**sc1=(0,0)**for

**ω=-1/sec**. So the harmonic with

**ω=1/sec**on the basis of the formula

**c**is also

**zero**! What if the centrifuge was operated at different speeds

**ω**? Then also

**scn=(0,0)**. So the constant function

**f (t)=1**has no harmonic, only a constant component

**c0=1**. It is not very revealing, but thanks to this, for the simple function

**f(t)=1**, the formulas in

**Fig. 7-2**are obvious. And that’s it!

**Chapter 7.4.3 Centers of gravity 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
**Trajectories

**F(n1t)=0.5cos (4t)*exp(-n1t)**for

**n=0,1,2**….

**8**

**ω=0**(n = 0)

The

**stationary**complex plane

**Z**on which the function

**f(t)=0.5cos(4t)**performs “

**swinging**” harmonic

**motions**.

It swings around

**sc0=c0=(0,0)**and the constant

**f(t)**component is obviously

**zero**.

**ω=4/sec**(n=4)

Only at this speed

**ω**, a trajectory with a

**non-zero**center of gravity

**sc4=(0.25.0)**will be created. Our intuition tells us that it is mean-distant from

**z=(0,0)**and this result can be calculated by the formula

**Fig.7-2b**. Notice that you only see the movement of the trajectory at the beginning. Then it follows the same path and therefore the trajectory is apparently stationary.

**Remaining ω**(n = 1,2,3,5,6,7,8)

Centers of gravity

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

**Fig.7-2b**.

**Conclusions:**

**Constant**component

**c0**.

It clearly confirms the formula

**Fig.7-2c**->

**c0=(0,0)=0**.

**The**

**fourth**harmonic

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

**ω=4/sec**

according to

**Fig.7-2d**–>

**c4=2*sc4=2*(0.25.0)=(0.5.0)=0.5+j0**–>

**a4=0.5**and

**b4 = 0**

according to

**Fig. 7-2h**–>

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

**Harmonics**for the remaining

**n*ω0**

They don’t exist because the centers of gravity

**scn**of their trajectories are

**zero**.

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 of

**Fig. 7-2**.

**Chapter 7.4.4 Centers of gravity 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,… 8 and ω0=1/sec
**The function

**f (t)**has

**2**harmonics, ie

**0.7*cos (2t)**and

**0.5*cos (4t)**with a constant component

**c0=1.3**.

I wonder how they will be centrifuged?

We will examine the trajectories

**F(njω0t)=f t)*exp(-njω0t)**for:

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

**n=0,1,2 and 4**

**ω0=1/sec**

Therefore, these will be trajectories for

**ω=0, ω=1/sec, ω=2/sec and ω=3/sec.**The remaining trajectories, ie for

**n=3,5,6,7**and

**8**, are rotating, similarly as

**for n=1**around center of gravity

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

**Chapter 5**.

**Fig. 7-5
**Trajectories

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

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

**4**

**ω=0**

The stationary complex

**Z**plane where the function

**f (t)=1.3+0.7cos (2t)+0.5cos (4t)**performs “swinging” movements around

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

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

**f (t)**.

**ω=2/sec, ω=4/sec**

At these velocities

**ω**, trajectories with

**non-zero**centers of gravity

**sc2=(0.35.0)**and

**sc4=(0.25.0)**will be created. They are on

**average distant**between the trajectory points

**F(2j1t)**and

**F(4j1t)**and

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

**Fig.7-2b**.

**ω=1/sec**and the remaining

**ω**

The center of gravity

**sc1=(0,0)**and the others also follow the formula

**Fig. 7-2b**

**Conclusions:**

**Constant component c0**

acc. to

**Fig. 7-2c**–>

**c0=a0=1.3**It

is also the center of gravity

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

**ω=0**

**Second harmonic h2 (t)**, i.e.

**for ω=2/sec**

acc. to

**Fig.7-2d**–>

**c2 = 2*sc2=2(0.35.0)=(0.7.0)=0.7+j0 -> a2=0.7**and

**b2=0**

acc. to

**Fig.7-2h**–>

**h2 (t)=0.7cos (4t)**

**Fourth harmonic h4 (t)**, i.e. for

**ω=4/sec**

acc. to

**Fig.7-2d**–>

**c4 = 2sc4 = 2(0.25.0)=(0.5.0)=0.5+j0**–>

**a4=0.5**and

**b4=0**

acc. to

**Fig. 7-2h**–>

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

The

**other**harmonics do not exist because the their

**scn=0**.

**Chapter 7.4.5 Centers of gravity scn of the trajectory F(-njω0t) for the function
f(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,… 8 and ω0=1/sec
**The function

**f(t)**with the period

**T=2πsec**consists of

**three cosines**with different amplitudes

**A**,phases

**ϕ**and

**ω**

and the constant component

**c0**.

**Fig. 7-6
**

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

In

**Chapter 6**, we examined

**9**trajectories

**F(-njω0t)**of the 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.

Each harmonic can be broken down into

**cosine**and

**sinusoidal**components 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)**.

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)**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

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—>

c5=0.4-j0.2

**h5(t)=0.4*cos(5t)+0.2*sin(5t)≈0.45*cos(5t-26.6°)**

We put the function

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

**nω0=n*1/sec**. Let’s turn on the rotation at

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

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

**ω=0**(the centrifuge is standing!),

**ω=-1**

**/sec**,

**ω=-2/sec**,

**ω=-3/sec**and

**ω=-5/sec**. The trajectories

**F(nj1t)**of those

**n**which correspond to the centers of gravity

**scn**will arise.

**Rys. 7-7
**Trajectories

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

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

**5**

**ω=0**

The stationary complex

**Z**plane where the function

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

**“swinging”**movements

around

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

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

**f (t)**.

**ω=1/sec, ω=3/sec**and

**ω=5/sec**

At these velocities

**ω**, trajectories with

**non-zero**centers of gravity

**sc1=(0.45,-0.3), sc3=(0.3,+0.2)**and

**sc5=(0.2,-0.1)**will be created.

**ω=1/sec**and the remaining

**ω**

At these velocities

**ω**, trajectories have

**zero**centers of gravity like for example

**sc1=(0,0)**.

**Conclusions:**

**Constant**component

**c0.**

acc.

**Fig.**

**7-2c**–>

**c0=0.5**

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

**ω=1/sec**

acc.

**Fig. 7-2d**–>

**c1=2*sc1=2*(0.45,-0.3)**

**=0.9+j0.6–>a1=0.9**i

**b1=0.6**

acc.

**Fig. 7-2h**–>

**h1(t)=**

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

acc.

**Fig. 7-2i–>h1(t)≈**

**1.08*cos(1t-33.7°)**

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

**ω=3/sec**

acc.

**Fig. 7-2d**–>

**c3=2*sc3=2*(0.3,+0.2)**

**=0.6-j0.4–>a3=0.6**i

**b3=-0.4**

acc.

**Fig. 7-2h**–>

**h3(t)=**

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

acc.

**Fig. 7-2i**–>

**h3(t)≈0.72*cos(3t+33.7°)**

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

**ω=5/sec**

acc

**. Fig. 7-2d**–>

**c5=2*sc5=2*(0.2,-0.1)**

**=0.4+j0.2–>a5=0.4**i

**b4=0.2**

acc.

**Fig. 7-2h**–>

**h5(t)=**

**0.4*cos(5t)+0.4*sin(5t)**

acc.

**Fig. 7-2i**–>

**h5(t)≈**

**0.45*cos(5t-26.6°)**

**i.e.**

The remaining harmonics

The remaining harmonics

**for ω=2/sec, ω=4/sec, ω=6/sec, ω=7/sec, ω=8/sek…**

doesn’t exist, because their

**scn=0**.

**Chapter 7.5 Harmonics 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 center of gravity

**scn**, which is already

**almost**the nth harmonic

**f(t)**.

Why

**almost**? Because

**scn**is not yet the

**nth harmonic**of the function

**f(t)**, but only a

**coefficient**from which we can easily calculate the

**nth harmonic**in

**complex**or

**real**form.

**1**.

**cn=2*scn**

**=an-jbn**

2. cn=an-jbnis the

2. cn=an-jbn

**complex**amplitude of the

**nth**harmonic

**hn(nω0t)**

3.a

3.a

**nth**harmonic in the

**complex**version

So the vector

**hn(jn*ω0t)=(an-jbn)*exp(njω0t)**rotating at the speed

**nω0**

**3.b**

**nth**harmonic in the

**real**version

**hn(t)=ancos(nω0t)+bnsin(nω0t)**–>

**Fig.7.2h**

or

**–>**

hn(t)=|cn|*cos(nω0t+ϕ)

hn(t)=|cn|*cos(nω0t+ϕ)

**Fig.7.2i**

The centers of gravity

**scn**of different trajectories

**Fn(jnω0t)**for different functions

**f(t)**are found in

**Fig.7-4**,

**Fig.7-5**and

**Fig.7-7**.

These are the “almost”

**nth**harmonics of

**f(t)**.

We will now present these drawings in a

**simplified**version, i.e.

– there will only be

**doubled**centers of gravity

**scn**for nth trajectories as

**cn**coefficients

– the drawings will be presented successively every

**1**second when

**ω**changes from

**ω=0**to

**ω=8/sec**

In this way,

**harmonics detector**of

**periodic functions f (t)**will be created, namely

**for:**

Harmonic detetector

Harmonic detetector

–

–

**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°)**

**Chapter7.5.2 Harmonic detetector for f(t)=0.5cos(4t)
**

**Fig.7-8
**Harmonics detector for

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

You put the periodic function

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

**ω=0**to

**ω=8/sec**. In this way, for each pulsation

**ω**, the successive harmonic

**hn(t)**is calculated according to the formulas from

**Fig.7-2**to which the complex number

**cn**corresponds. 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**don’t exist!

**Chapter7.5.3 Harmonic detetector for f(t)=1.3+0.7cos(2t)+0.5cos(4t)**

**Fig.7-9
**

**Harmonic detetector**for

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

**ω=0**

Constant component

**c0=1.3**

**ω=2/sec**

**component corresponds to harmonic**

c2=0.7

c2=0.7

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

**ω=4/sec**

**c4=0.5**component corresponds to harmonic

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

The remaining

**harmonics hn(t)**doesn’t exist.

**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 detetector**for

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

**ω=0**

Constant component

**c0=0.5**

**ω=1/sec**

**component corresponds to harmonic**

c1=1.08*exp(-j33.7°)

c1=1.08*exp(-j33.7°)

**h1(t)=1.08*cos(1t-33.7°)**

**ω=3/sec**

**c3=c3=0.72*exp(+j33.7°)**component corresponds to harmonic

**0.72*cos(3t+33.7°)**

**ω=5/sec**

**c5=0.45*exp(-j33.7°)**component corresponds to harmonic

**h5(t)=0.45*cos(5t-26.6°)**

The remaining

**harmonics hn(t)**don’t exist.