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

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:
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
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
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. 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°)

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:
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 integration 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–>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 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°)
The remaining harmonics i.e. 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-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(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)–>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
Harmonic detetector 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°)

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  
c2=0.7 component corresponds to harmonic 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  
c1=1.08*exp(-j33.7°)  component corresponds to harmonic  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.